A Level Further Mathematics for AQA: Discrete Student Book (AS/A Level) [Cambridge Elevate ed.] 9781108445207, 9781108445214

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Brighter Thinking

A Level Further Mathematics for AQA Discrete Student Book (AS/A Level) Jan Dangerfield

Contents Introduction How to use this resource 1 Graphs 1: The language of graphs 2: Properties of graphs 3: Planarity and isomorphism Mixed practice 1 2 Networks 1: Network optimisation problems using spanning trees 2: Route inspection problems 3: Travelling salesperson problem Mixed practice 2 3 Network flows 1: Flows and cuts 2: Variations on flow problems Mixed practice 3 4 Linear programming 1: Formulating constrained optimisation problems 2: Graphical solutions 3: The simplex algorithm Mixed practice 4 Cross-topic review exercise 1 5 Critical path analysis 1: Analysing precedence networks 2: Scheduling Mixed practice 5 6 Game theory for zero-sum games 1: Games with stable solutions 2: Mixed strategies Mixed practice 6 7 Binary operations 1: Properties of binary operations 2: Using Cayley tables Mixed practice 7 8 Groups 1: The group axioms 2: Generators of a group Mixed practice 8 Cross-topic review exercise 2 Practice paper Set notation Answers Acknowledgements Copyright

Introduction You have probably been told that mathematics is very useful, yet it can often seem like a lot of techniques that just have to be learnt to answer examination questions. You are now getting to the point where you will start to see where some of these techniques can be applied in solving real problems. However, we hope that, as well as seeing how maths can be useful, anyone working through this resource will realise that it can also be incredibly frustrating, surprising and ultimately beautiful. The book is woven around three key themes from the new curriculum.

Proof Maths is valued because it trains you to think logically and communicate precisely. At a high level, maths is far less concerned about answers and more about the clear communication of ideas. It is not about being neat – although that might help! It is about creating a coherent argument that other people can easily follow but find difficult to refute. Have you ever tried looking at your own work? If you cannot follow it yourself it is unlikely anybody else will be able to understand it. In maths we communicate using a variety of means – feel free to use combinations of diagrams, words and algebra to aid your argument. And once you have attempted a proof, try presenting it to your peers. Look critically (but positively) at some other people’s attempts. It is only through having your own attempts evaluated and trying to find flaws in other proofs that you will develop sophisticated mathematical thinking. This is why we have included lots of common errors in our Work it out boxes – just in case your friends don’t make any mistakes!

Problem solving Maths is valued because it trains you to look at situations in unusual, creative ways, to persevere and to evaluate solutions along the way. We have been heavily influenced by a great mathematician and maths educator George Polya, who believed that students were not just born with problem-solving skills – they were developed by seeing problems being solved and reflecting on their solutions before trying similar problems. You may not realise it but good mathematicians spend most of their time being stuck. You need to spend some time on problems you can’t do, trying out different possibilities. If after a while you have not cracked it then look at the solution and try a similar problem. Don’t be disheartened if you cannot get it immediately – in fact, the longer you spend puzzling over a problem the more you will learn from the solution. You may never need to integrate a rational function in the future, but we firmly believe that the problem-solving skills you will develop by trying it can be applied to many other situations.

Modelling Maths is valued because it helps us solve real-world problems. However, maths describes ideal situations and the real world is messy! Modelling is about deciding on the important features needed to describe the essence of a situation and turning that into a mathematical form, then using it to make predictions, compare to reality and possibly improve the model. In many situations the technical maths is actually the easy part – especially with modern technology. Deciding which features of reality to include or ignore and anticipating the consequences of these decisions is the hard part. Yet it is amazing how some fairly drastic assumptions – such as pretending a car is a single point or that people’s votes are independent – can result in models that are surprisingly accurate. More than anything else this book is about making links – links between the different chapters, the topics covered and the themes above, links to other subjects and links to the real world. We hope that you will grow to see maths as one great complex but beautiful web of interlinking ideas. Maths is about so much more than examinations, but we hope that if you absorb these ideas (and do plenty of practice!) you will find maths examinations a much more approachable and possibly even enjoyable experience. However, always remember that the results of what you write down in a few hours by yourself in silence under exam conditions are not the only measure you should consider when judging

your mathematical ability – it is only one variable in a much more complicated mathematical model!

How to use this resource Throughout this resource you will notice particular features that are designed to aid your learning. This section provides a brief overview of these features.

In this chapter you will learn how to: use the language of graphs recognise and use some special types of graph use graphs to model problems. If you are following the A Level course you will also learn how to: understand and use the term planarity understand and use the term isomorphism.

Learning objectives A short summary of the content that you will learn in each chapter.

Before you start... GCSE and A Level Mathematics Student Book 1

You should be familiar with the terms integer, rational number, real number and the notation for these sets.

1 Which of these numbers belong to the set , the set of positive integers?

Further Mathematics Student Book 1, Chapter 7

You should know how to carry out matrix addition and matrix multiplication.

2 Calculate: a b

Before you start Points you should know from your previous learning and questions to check that you’re ready to start the chapter.

WORKED EXAMPLE

The left-hand side shows you how to set out your working. The right-hand side explains the more difficult steps and helps you understand why a particular method was chosen.

WORK IT OUT Identify the correct solution and find the mistakes in the two incorrect solutions.

Key point A summary of the most important methods, facts and formulae.

Common error Specific mistakes that are often made. These typically appear next to the point in the Worked example where the error could occur.

Tip Useful guidance, including ways of calculating or checking answers and using technology. Each chapter ends with a Checklist of learning and understanding and a Mixed practice exercise, which includes past paper questions marked with the icon . Between some chapters, you will find extra sections that bring together topics in a more synoptic way. CROSS-TOPIC REVIEW EXERCISE Questions covering topics from across the preceding chapters, testing your ability to apply what you have learned. You will find practice paper questions towards the end of the resource, as well as a glossary of key terms (picked out in colour within the chapters) and answers. Maths is all about making links, which is why throughout this resource you will find signposts emphasising connections between different topics, applications and suggestions for further research.

Rewind Reminders of where to find useful information from earlier in your study.

Fast forward Links to topics that you may cover in greater detail later in your study.

Did you know? Interesting or historical information and links with other subjects to improve your awareness about how mathematics contributes to society. Colour-coding of exercises The questions in the exercises are designed to provide careful progression, ranging from basic fluency to practice questions. They are uniquely colour-coded, as shown here.

1

Show that, for the graph in question 1, the sum of the degrees of the vertices is twice the number of edges.

2

A graph has vertices and edges. What is the sum of the degrees of the vertices?

3

A graph is drawn with vertices labelled , , , , and . An edge is drawn between two vertices if the larger is a multiple of the smaller. a Draw this graph. b List a cycle in this graph. c Write down a trail that starts at and travels through every vertex in the graph once.

4

Explain why it is impossible to draw a graph with exactly five vertices that have degrees , , , and .

5

A simple graph has six vertices. The degrees of the vertices are , , , , and .

6

Explain why there is no simple graph with exactly four vertices with degrees , , and .

7

There are different trees with vertices. Draw an example of each and give its degrees.

8

A connected graph has vertices and edges. Explain why the graph must be simple.

9

A connected graph is semi-Eulerian if exactly two of its vertices are of odd degree. a A graph is drawn with vertices and edges. What is the sum of the degrees of the vertices? b Draw a simple semi-Eulerian graph with exactly vertices and edges, in which exactly one of the vertices has degree . c Draw a simple semi-Eulerian graph with exactly vertices that is also a tree. d A simple graph has vertices. The graph has two vertices of degree . Explain why the graph can have no vertex of degree .

10 A graph has vertices. The vertex degrees are

.

a How many edges does the graph have? b Explain how the vertex degrees show that the graph is not simple-connected. c How many different graphs fit this description?

Black – practice questions that come in several parts, each with subparts i and ii. You only need attempt subpart i at first; subpart ii is essentially the same question, which you can use for further practice if you got part i wrong, for homework, or when you revisit the exercise during revision. Green – practice questions at a basic level. Blue – practice questions at an intermediate level. Red – practice questions at an advanced level. Purple – challenging questions that apply the concept of the current chapter across other areas of maths. Yellow – designed to encourage reflection and discussion.

– indicates content that is for A Level students only. – indicates a question that requires a calculator.



1 Graphs

In this chapter you will learn how to: ​• use the language of graphs ​ recognise and use some special types of graph • ​ use graphs to model problems. • If you are following the A Level course you will also learn how to: ​• understand and use the term planarity ​ understand and use the term isomorphism. •

Before you start… GCSE Mathematics

You should be able to use a mapping to represent relationships.

1 What relationship is shown in this mapping?

Section 1: The language of graphs Key point 1.1 A graph is a set of vertices connected by edges. ​• A vertex is shown as a point on a graph. Vertices are sometimes labelled, and sometimes not. ​• An edge is a line or curve with a vertex at each end.

WORKED EXAMPLE 1.1

a List the vertices in this graph. b List the edges in the graph. a The vertices are , , , and . b

.





and

Not all graphs will have labelled vertices. You can identify an edge by the vertices at its ends. Edge same as edge for example.

is the

Key point 1.2 ​• A walk is a set of edges joined end to end, so the end vertex of one edge is the start vertex of the next. ​• A trail is a walk in which no edges are repeated. Vertices can be repeated in a trail, although often they are not. ​• A cycle is a trail that starts and finishes at the same vertex. Other than the start being the same as the finish, vertices are not repeated in a cycle.

Fast forward The terms trail and cycle will be used in route inspection problems and the travelling salesperson problem in Chapter 2.

WORKED EXAMPLE 1.2 Using the graph in Worked example 1.1: a give an example of a walk that passes through every vertex b give two examples of trails, one with a repeated vertex and one with no repeated vertices c explain why

is not a trail

d give an example of a cycle e explain why

is not a cycle.

a e.g.

This is not a trail because the edge

b An example of a trail with a repeated vertex is .

This trail repeats the vertex .

is repeated.

An example of a trail with no repeated vertices is c There is no edge

This trail has no repeated vertices.

. .

d For example: e Vertex is travelled through twice.

A cycle has no repeated vertices, apart from starting and finishing at the same vertex.

Key point 1.3 ​• Two vertices are directly connected, or adjacent, if there is an edge with these vertices at its ends. An indirect connection between two vertices passes through other vertices and involves more than one edge. ​• A graph is connected if it is possible to get from any vertex to any other, directly or indirectly. The position of the vertices and the shapes of the edges in a graph (including whether they cross each other or not) are irrelevant. All that matters is which vertices are adjacent (directly joined) to each other.

Key point 1.4 ​• An edge that directly connects a vertex to itself is called a loop. ​• A graph has a multiple edge if there are two or more edges that directly connect the same pair of vertices. ​• A graph with no loops and no multiple edges is called a simple graph.

WORKED EXAMPLE 1.3 Which of these graphs have: a a loop or loops b a multiple edge or multiple edges?

A

D B

C

E

F Graph 1

Graph 2

G

J H

I

K

L Graph 3

M

Graph 4 N

P

O Graph 5

a Graph 1 and graph 3

Graph 1 has a loop at Graph 3 has loops at and at

b Graph 1 and graph 4

Graph 1 has a multiple edge Graph 4 has a multiple edge

.

Key point 1.5 ​• The degree of a vertex is the number of edges that end at that vertex.

WORKED EXAMPLE 1.4 a Write down the degree of each vertex in the graphs used in Worked example 1.3. b Work out the sum of the vertex degrees for each graph. c Why is the sum of the vertex degrees equal to twice the number of edges? a Vertex

Degree Vertex

Degree

The degree of each vertex is the number of edges that end at that vertex. Each end of the loop contributes to the degree at and has degree .

b Graph Graph Graph Graph Graph c Each edge has two ends, each of which contributes towards a vertex degree.

Graphs and have edges each, Graphs and have edges each, Graph has edges,

The sum of the vertex degrees is the total number of edge ends, which is the same as twice the number of edges.

Key point 1.6 For any graph, the sum of the vertex degrees is twice the number of edges, which means that the sum of the vertex degrees is always even. An immediate consequence of this is that a graph cannot have an odd number of vertices with odd degrees.

WORKED EXAMPLE 1.5 A graph has vertices and edges. a What is the sum of the degrees of the vertices? b Explain why the graph must have loops or multiple edges. c Draw three different graphs that fit this description. a

edges; sum is twice the number of edges.

b If there are no loops or multiple edges, then the maximum degree at each vertex is , giving degree sum .

With no loops or multiple edges, the maximum degree sum is .

The degree sum is greater than , so there must be loops or multiple edges. c For example:

The first graph shown has multiple edges; the second has multiple edges and a loop; the third is made up of two disconnected graphs and has multiple edges and loops.

Key point 1.7 ​• A subgraph of a graph is formed by using some or all of the vertices of a graph together with some or all of the edges that connect these vertices. A subgraph is a graph contained within another graph. This could result in an unconnected vertex. However, subgraphs are usually connected.

Key point 1.8 ​• Subdivision means inserting a vertex of degree into an edge. Subdivision increases the number of vertices by and the number of edges by . WORKED EXAMPLE 1.6

a Show that graph is a subgraph of graph . b Explain how graph is obtained from graph by subdivision. a Remove the edge joining the two vertices of degree

It is irrelevant that the edges cross. Graph 2 is the same as a square.

b Insert a vertex into one of the edges joining a vertex of degree to a vertex of degree . This turns that edge into two edges joined by a vertex of degree .

Key point 1.9 ​• A simple graph, on a given number of vertices, with the maximum possible number of edges is called a complete graph. Each vertex is connected by a single edge to each of the other vertices. Recall from Key point 1.4 that a simple graph has no loops or multiple edges. ​• The complete graph with vertices is denoted by

and has

edges.

Common error Be careful not to confuse complete graphs and connected graphs (see Key point 1.3).

WORKED EXAMPLE 1.7 Draw the complete graph

. You could also draw example:

For example:

without edges crossing, for

WORKED EXAMPLE 1.8 Explain why

has

edges.

has vertices, each of which is connected to the other vertices. So each vertex has degree The degree sum is

Alternatively, there are edges from the first vertex, another edges from the second vertex, …

.

.

This gives a total of

The degree sum is twice the number of edges. So the number of edges is

.

.

Key point 1.10 ​• A bipartite graph is a simple graph that can be partitioned into two sets so that every edge joins a vertex from one of these sets to a vertex in the other set. No edge connects two vertices in the same set. ​• A simple bipartite graph, on a given number of vertices in each set, with the maximum possible number of edges is called a complete bipartite graph. Each vertex in the first set is connected by a single edge to each vertex in the second set. ​• The complete bipartite graph with vertices in the first set and vertices in the second set is denoted by and has edges.

Did you know? The classic ‘utilities problem’ where three houses need to each be connected to three utilities – gas, water and electricity – without any of the connections crossing uses the complete bipartite graph .

WORKED EXAMPLE 1.9 Draw the graph

.

For example: A

B

D

C

E

F

The two sets are usually presented either as two horizontal rows or two vertical rows, so that the two sets can be identified easily. The vertices need not be labelled.

WORKED EXAMPLE 1.10 Explain why

has

edges.

has vertices in the first set, each of which is connected to the vertices in the second set. So the first set consists of vertices each with degree , and the second set consists of vertices each with degree . There are endings in the first set and .

in the second set, so the number of edges is

Alternatively, each of the vertices in the first set has degree , giving degree sum , and each of the vertices in the second set has degree , giving degree sum . The total degree sum is . The degree sum is twice the number of edges. So the number of edges is

.

It is useful to have a way of describing graphs, particularly large ones.

Key point 1.11 ​• An adjacency matrix shows the number of edges that directly connect each pair of vertices. An adjacency matrix can also be used when there are no vertex labels given.

WORKED EXAMPLE 1.11 a Draw the graph represented by this adjacency matrix.

b What is the significance of the row sums and the column sums of the adjacency matrix? c If a loop is drawn at vertex , how would this be represented in the adjacency matrix? a A

D

C B

E

b The row sums are the same as the corresponding column sums. These give the vertex degrees.

c A loop at vertex contributes to the degree of vertex , so it is represented by an entry in the adjacency matrix.

Tip An adjacency matrix is symmetrical about the lead diagonal (top left to bottom right).

WORKED EXAMPLE 1.12 Construct the adjacency matrix for the graph with vertices , , , with edges between and , and , and , two edges between and , and a loop from to itself.

Each edge is shown by a in the cell corresponding to the vertices at its

ends, the repeated edge between and is shown by a and the loop from to itself is shown by a (as in Worked example 1.11). Cells that correspond to vertices that are not directly connected are labelled with .

Key point 1.12 ​• The complement of a simple graph is the set of edges that, when added to the graph, makes a complete graph. ​• Every pair of vertices that are not directly connected in the original graph are joined by an edge in the complement; every pair of vertices that are joined by an edge in the original graph are not directly connected in the complement. ​• The adjacency matrix for a simple graph consists of s and s. The adjacency matrix for the complement has a where the matrix for the graph has a , and a where the matrix for the graph has a , apart from the lead diagonal which consists of s in both adjacency matrices.

WORKED EXAMPLE 1.13 The graph represents six people. Edges connect people who have met one another.

Draw the complement of this graph and describe what the edges in the complement represent in this case. Original graph

The edges in the complement show people who have not met one another.

EXERCISE 1A

Complement

1

Is this graph simple? Explain your answer.

2

Show that, for the graph in question 1, the sum of the degrees of the vertices is twice the number of edges.

3

A graph has vertices and edges. What is the sum of the degrees of the vertices?

4

How many edges does the complete graph on five vertices have?

5

A graph is drawn with vertices labelled , , , , and . An edge is drawn between two vertices if the larger number is a multiple of the smaller. a Draw this graph. b List a cycle in this graph. c Write down a trail that starts at and travels through every vertex in the graph once.

6

Explain why it is impossible to draw a graph with exactly five vertices that have degrees , , , and .

7

A simple graph has six vertices. The degrees of the vertices are , , , , and . a Explain why must be odd. b What is the value of if the graph has edges? c What is the maximum possible number of edges that the graph could have?

8

Explain why there is no simple graph with exactly four vertices with degrees , , and .

9

Write down a possible adjacency matrix for a connected graph with four vertices and three edges for which there is: a a vertex with degree b no vertex with degree .

10 A graph has adjacency matrix:

a How many edges does this graph have? b Show that the graph is a bipartite graph. An extra vertex, , is created by subdivision of the edge

.

c Show that the resulting graph is still bipartite. 11

a How many edges are there in the complement of the complete bipartite graph

?

b Describe what the edges in the complement represent. A bipartite graph is a subgraph of subgraph has fewer edges than

. The subgraph is a connected graph with vertices. The .

c How many edges does this subgraph have? 12 A connected graph has vertices and edges. Explain why the graph must be simple. 13 A medieval river crossing puzzle involves a farmer, a wolf, a goat and a sack of cabbages. Initially, the farmer, wolf, goat and sack of cabbages are together on the north bank of the river. The farmer can use a small boat to cross the river. The boat is only big enough to carry the farmer and one of the other three items. The goat cannot be left with the cabbages or with the wolf, unless the farmer is also present. The wolf can be left alone with the cabbages. The problem is to find a way to get everything across to the other side of the river using as few crossings as possible. The problem can be modelled using a graph in which the vertices are labelled to show what is on the north bank of the river at the beginning of each crossing. Initially, the farmer and all of the items are all on the north bank; this is the vertex . The first crossing must involve the farmer taking the goat across the river, so the second vertex is . The final crossing must involve the farmer taking the goat from the north bank to the south bank; this is the vertex

.

a List the nine possible vertices, remembering that the goat cannot be left with the cabbages or with the wolf, unless the farmer is also present. b Draw a bipartite graph to represent the possible river crossings. c Find a solution to the problem.

Section 2: Properties of graphs Key point 1.13 ​• A graph that is both simple and connected is called a simple-connected graph. ​• A simple-connected graph with the minimum possible number of edges is called a tree. A tree on vertices has edges.

WORKED EXAMPLE 1.14 There are two different trees with vertices. a Draw the two different trees with vertices. b List the set of vertex degrees for each tree. a

Any other tree on vertices can be rearranged by moving the vertices around to make one of these graphs.

b

and

The trees have different sets of vertex degrees so they must represent different graphs. These are the graphs from question 9 in Exercise 1A.

The vertices at the ends of the branches in a tree have degree . A tree on vertices must have a minimum of vertices of degree and a maximum of vertices of degree .

Key point 1.14 ​• A traversable graph is one that can be drawn as a trail without going over the same edge twice.

WORKED EXAMPLE 1.15 Which of these graphs are traversable? A

B

C

D

, and . There is an easy way to check if a graph is traversable by using the degrees of the vertices.

Key point 1.15 ​• An Eulerian graph is a connected graph that has no vertices of odd degree. Eulerian graphs are traversable, with the trail starting and finishing at the same vertex. ​• A semi-Eulerian graph is a connected graph that has exactly vertices of odd degree. SemiEulerian graphs are traversable, but the trail starts at one of the odd vertices and finishes at the other odd vertex. It is easy to show that each time a trail enters and exits a vertex it uses up edges, so for a graph to be

traversable the degrees must all be even, apart from the start and finish which will have odd degree (or even if the trail finishes where it started). However, it is much more difficult to prove that every connected graph with or exactly vertices of odd degree is traversable. WORKED EXAMPLE 1.16 A simple-connected graph has vertices and edges. The vertex degrees are

and .

a Show that the graph must be semi-Eulerian. b Explain why the missing degrees must be and . Drawing an example of a graph that fits the description is not enough, because there might be more than one possibility. a

so

The sum of the degrees is twice the number of edges.

is odd

The sum of the degrees is even.

One of the missing degrees is odd and the other is even.

Odd even odd Odd odd or even even even Exactly odd degrees semi-Eulerian

The graph has exactly odd degrees, so it must be semiEulerian. b Simple-connected:

Connected: so

or

. Simple:

.

.

Key point 1.16 ​• A Hamiltonian graph contains a cycle that passes through every vertex exactly once, apart from starting and finishing at the same vertex. The cycle is called a Hamiltonian cycle. There will usually be edges in the graph that are not used in the cycle.

Common error Be careful not to confuse Hamiltonian graphs (which contain a closed trail that uses every vertex once) and Eulerian graphs (which contain a closed trail that uses every edge once).

WORKED EXAMPLE 1.17 In chess, a knight moves two squares horizontally followed by one square vertically or two squares vertically followed by one square horizontally. A classic chessboard puzzle, known as the knight’s tour, asks whether a knight can visit every square of an chessboard and whether this can be done with the knight finishing back where it started.

The centre of each square on the chessboard is a vertex. The solution requires a Hamiltonian cycle made up of edges that are knight’s moves. There are several possible solutions.

WORKED EXAMPLE 1.18 Draw a Hamiltonian cycle on the edges of a cube.

Each corner of the cube is a vertex. The solution requires a Hamiltonian cycle made up of edges of the cube. A possible solution is shown.

Did you know? Hamiltonian graphs are named after the Irish mathematician William Rowan Hamilton (1805– 1865) who invented the icosian game, which is now known as Hamilton’s puzzle. This involves finding a Hamiltonian cycle, using the edge graph of a dodecahedron. Try drawing a -dimensional skeleton graph of a dodecahedron and investigate this problem. Show that all Platonic solids, when considered as graphs, are Hamiltonian graphs.

EXERCISE 1B 1

Which of these graphs are trees?

2

Which of the graphs in question 1 are semi-Eulerian?

3

Explain why a tree can never be Eulerian.

4

Is this graph Eulerian, semi-Eulerian or neither? Explain your reasoning.

5

a What is the sum of the vertex degrees for a tree on five vertices? b What is the minimum possible number of vertices of degree for a tree on five vertices? c What is the maximum possible vertex degree for a tree on vertices? d List the possible sets of vertex degrees for trees on vertices.

6

A simple, connected Eulerian graph has vertices and edges. Deduce the degrees of the vertices.

7

This question is about the seven graphs ( )

, ( )

, ( )

, ( )

, ( )

, ( )

and ( )

a How many edges does each of these graphs have? b Which of the six graphs are Eulerian? 8

a Give an example of a graph that is Hamiltonian but not Eulerian. b Give an example of a graph that is Eulerian but not Hamiltonian.

9 10

There are different trees with vertices. Draw an example of each and state its degrees. a Are complete graphs Hamiltonian? b Are complete bipartite graphs Hamiltonian?

11 How many different trees on: a

vertices

b vertices c

vertices can be made as subgraphs of

?

12 A graph has vertices. The vertex degrees are

.

a How many edges does the graph have? b Explain how the vertex degrees show that the graph is not simple-connected. c How many different graphs fit this description?

.

Section 3: Planarity and isomorphism Key point 1.17 A planar graph is any graph that can be drawn with no edges crossing. A planar graph can be drawn as one layer, without needing any ‘bridges’ where one edge jumps over another.

Common error A planar graph need not actually be drawn with no edges crossing; all that matters is that it can be manipulated (topologically) into a graph with no edges crossing.

WORKED EXAMPLE 1.19 Give two reasons why it is useful to be able to draw a graph without having any edges crossing. It avoids ambiguity about whether there is a vertex where edges cross or if there is a ‘bridge’. It avoids ‘contamination’ between edges, for example short-circuits in an electrical component. WORKED EXAMPLE 1.20 A simple-connected planar graph is drawn on vertices. The graph has the maximum possible number of edges. Draw a graph that fits this description. For example: The maximum possible number of edges is A complete graph with vertices has edges, but some of these cross. You should be able to convince yourself that edges is impossible for a planar graph.

A convex polyhedron can be represented as a planar graph. The edges of the polyhedron are represented by the edges in the graph and the faces of the polyhedron are represented by regions (one of the faces is represented by the region that is ‘outside’ the graph). For example, this graph is a representation of a cube.

WORKED EXAMPLE 1.21 A cube has faces,

edges and vertices.

a Count the number of faces, edges and vertices for: i a tetrahedron (triangular-based pyramid) ii

a square-based pyramid

iii an octahedron (eight triangular faces, like two square-based pyramids joined at their square bases) iv a hexagonal prism (a prism with hexagonal cross-section). b Use your results from part a to conjecture a relationship between the numbers of faces, edges and vertices of a convex polyhedron. a

Faces Edges Vertices Cube i ii iii iv

b

Or any equivalent expression.

Key point 1.18 Euler’s formula says that for any connected planar graph (or convex polyhedron) where is the number of vertices, is the number of edges and is the number of faces (or regions). The regions of a graph are called faces and include the ‘outside’ region, which is sometimes called the infinite face.

WORKED EXAMPLE 1.22 a The graphs shown are all planar graphs. Show that Euler’s formula holds for each of these graphs.

b The graphs shown are non-planar graphs. Does Euler’s formula hold for either of these graphs?

Graph 4

a Graph 1

Graph 2

Graph 5

Graph 1 has ‘faces’, the two triangular regions and the infinite face.

Graph 2 has ‘faces’, the region enclosed by the edges and the ‘outside’ region (infinite face).

Graph 3

Graph 3 has ‘faces’, the two enclosed by the edges and the infinite face. Graph 4 has triangular ‘faces’, faces enclosed by edges and the infinite face.

b Graph 4 No If Euler’s formula holds then , but there are more than regions.

There are ways of choosing vertices to form a triangular face plus the infinite face.

Graph 5 No

If Euler’s formula holds then , but there are more than regions. Sometimes it is easy to see how a graph can be manipulated so that no edges cross, but at other times it can be difficult to know whether a graph is planar or non-planar.

Key point 1.19 Kuratowski’s theorem says that a necessary and sufficient condition for a finite graph to be planar is that it does not contain a subgraph that is a subdivision of or .

Rewind Recall from Section 1 that subdivision means inserting a vertex of degree into an edge. Subdivision increases the number of vertices by and the number of edges by . The graphs and are minimum cases for non-planarity. If either of these is contained within a graph, possibly after simplifying the graph by eliminating any vertices of degree , then the graph is also non-planar. WORKED EXAMPLE 1.23 a Use Kuratowski’s theorem to show that this graph is planar.

b Describe how the graph in part a can be drawn with no edges crossing. An edge is added to the graph in part a connecting to . c Show that the resulting graph is non-planar. a Only

and have degree

, so

Vertex can be removed so that edges

is not a subgraph.

,

become a single

If is a subgraph, there must be five vertices with degree or more. , from

is formed by

edge

.

subdivision.

This leaves a graph with six vertices: , and . If is a subgraph, these vertices can be split into two sets of three, where there is an edge joining each vertex in the first set to each vertex in the second set. There is no edge or so would need to be taken together as a set, with as the other set. But there is no edge so is not a subgraph. Hence the graph is planar.

The graph does not contain a subgraph that is a subdivision of or .

b Move edge so that it passes outside of (and does not cross other edges) and the edge so that it passes outside of (and does not cross other edges). c Edge is added and vertex is removed so that edges , become a single edge . Vertices and each have degree . is now a subgraph, with the two sets as and .

There are edges from each of and to each of and .

So the graph is non-planar.

Two graphs might look very different from one another but have the same structure, in the sense that they can be transformed into one another by relabelling and moving the vertices and edges around without cutting any of the edges.

Key point 1.20 Two graphs are isomorphic if they have the same structure.

Tip From the ancient Greek: isos = equal, morphe = form or shape.

Did you know? The study of these properties is part of the branch of Mathematics known as topology. Isomorphic graphs are topologically equivalent. You can show isomorphism by using a reasoned argument or by setting up a correspondence between the sets of vertices. WORKED EXAMPLE 1.24 Show that these graphs are isomorphic.

Straighten out edge

and relabel the vertices:

Set up a correspondence between the sets of vertices.

WORKED EXAMPLE 1.25 Show that these graphs are isomorphic.

Relabel and

Compare the vertex degrees to suggest a possible correspondence.

Common error A necessary condition for two graphs to be isomorphic is that they have the same vertex degrees. However, this is not a sufficient condition, as two graphs can have the same degree but not be isomorphic.

WORKED EXAMPLE 1.26 Draw two non-isomorphic graphs that both have degrees

.

These trees each have degrees , but in the first the vertex of degree is connected to two of degree and two of degree , while in the second the vertex of degree is connected to one of degree and three of degree .

EXERCISE 1C 1

Which of these graphs are planar?

2

Which of these graphs are isomorphic? A

B

C

3

Use Euler’s formula for the planar graph

4

Show that this graph is planar by drawing it after it has been manipulated so that no edges cross.

5

Show that these graphs are not isomorphic.

6

Use Euler’s formula to find the number of regions in the planar graph described by this adjacency

to find the number of regions.

matrix.

7

A standard dice is modelled as a planar graph with vertices labelled and . The faces of the dice are represented by the vertices of the graph and edges connect vertices that correspond to adjacent faces, but not vertices that correspond to opposite faces. Use Euler’s formula to find the number of regions in the graph.

8

This graph connects vertices.

a Find three different subgraphs, each with vertices and edges. b Use the degrees of the vertices to show that the subgraphs are all different. 9

Give a relabelling of the vertices that shows that these graphs are isomorphic.

A

C

10

B

D

a Write the vertex degrees for the graph with this adjacency matrix.

b Show that the graph is Hamiltonian. c Use Kuratowski’s theorem to show that the graph is non-planar. 11 The complete graph

is drawn on the vertices

but with the edge

missing.

a Use Kuratowski’s theorem to show that the graph is planar. b Use Euler’s formula to find the number of regions in the graph. Regions are bounded by three edges that between them share three vertices. c How many ways are there to choose vertices from , excluding any triples that include both and ? d Explain why the answers to parts b and c are different.

Checklist of learning and understanding

​• ​ • ​ •

A vertex is shown as a point on a graph. Vertices are sometimes labelled, and sometimes not.

​•

A trail is a walk in which no edges are repeated. Vertices can be repeated in a trail, although

An edge is a line or curve with a vertex at each end. A walk is a set of edges joined end to end, so the end vertex of one edge is the start vertex of the next. often they are not.

​•

A cycle is a trail that starts and finishes at the same vertex. Other than the start being the same as the finish, vertices are not repeated in a cycle.

​•

Two vertices are directly connected, or adjacent, if there is an edge with these vertices at its

ends. An indirect connection between two vertices passes through other vertices and involves more than one edge.

​• ​ • ​ •

A graph is connected if it is possible to get from any vertex to any other, directly or indirectly. An edge that directly connects a vertex to itself is called a loop. A graph has a multiple edge if there is more than one edge that directly connects the same pair of vertices.

​• ​ •

A graph with no loops and no multiple edges is called a simple graph. The degree of a vertex is the number of edges that end at that vertex.

​• ​ •

For any graph, the sum of the degrees of the vertices is twice the number of edges. An immediate consequence of this is that a graph cannot have an odd number of vertices with odd degrees.

​•

A subgraph of a graph is formed by using some or all of the vertices of a graph together with some or all the edges that connect these vertices. A subgraph is a graph contained within another graph. This could result in an unconnected vertex. However, subgraphs are usually connected.

​•

Subdivision means inserting a vertex of degree into an edge. Subdivision increases the number of vertices by and the number of edges by 1.

​•

A simple graph, on a given number of vertices, with the maximum possible number of edges is called a complete graph. Each vertex is connected by a single edge to each of the other vertices.

​• ​•

The complete graph with vertices is denoted by

and has

edges.

A bipartite graph is a simple graph that can be partitioned into two sets so that every edge joins a vertex from one of these sets to a vertex in the other set. No edge connects two vertices in the same set.

​•

A simple bipartite graph, on a given number of vertices in each set, with the maximum possible number of edges is called a complete bipartite graph. Each vertex in the first set is connected by a single edge to each vertex in the second set.

​•

The complete bipartite graph with m vertices in the first set and n vertices in the second set is denoted by

​•

and has

edges.

An adjacency matrix shows the number of edges that directly connect each pair of vertices. An adjacency matrix can also be used when there are no vertex labels given.

​•

The complement of a simple graph is the set of edges that, when added to the graph, make a complete graph.

​•

Every pair of vertices that were not directly connected in the original graph are joined by an edge in the complement; every pair of vertices that were joined by an edge in the original graph are not directly connected in the complement.

​• ​• ​ •

A graph that is both simple and connected is called a simple-connected graph. A simple-connected graph with the minimum possible number of edges is called a tree.

​• ​•

The adjacency matrix for the graph consists of 0s and 1s. The complement has a where the graph had a and a where the graph had a 1, except for the leading diagonal which is all 0s in both the graph and its complement.

A tree on n vertices has

edges.

A traversable graph is one that can be drawn as a trail, without going over the same edge twice.

​•

An Eulerian graph is a connected graph that has no vertices of odd degree.

​• ​•

Eulerian graphs are traversable, with the trail starting and finishing at the same vertex.

A semi-Eulerian graph is a connected graph that has exactly vertices of odd degree.

​•

Semi-Eulerian graphs are traversable, but the trail starts at one of the odd vertices and finishes at the other odd vertex.

​•

A Hamiltonian graph contains a cycle that passes through every vertex (exactly once, apart

from starting and finishing at the same vertex).

​• ​ • ​•

The cycle is called a Hamiltonian cycle. There will usually be edges in the graph that are not used in the cycle.

A planar graph is any graph that can be drawn with no edges crossing.

​•

A planar graph can be drawn as one layer, without needing any ‘bridges’ where one edge jumps over another.

​•

Euler’s formula says that for any connected planar graph (or convex polyhedron) where is the number of vertices, is the number of edges and is the number of faces (or regions), including the infinite face.

​•

Kuratowski’s theorem says that a necessary and sufficient condition for a finite graph to be planar is that it does not contain a subgraph that is a subdivision of

​•

Two graphs are isomorphic if they have the same structure.

or

.

Mixed practice 1 1

A connected graph has five vertices with vertex degrees graph have? Choose from these options.

. How many edges does the

A B C D 2

A simple-connected semi-Eulerian graph has six vertices. The degrees of five of the vertices are and . What could be the degree of the sixth vertex? Choose from these options. A B C D

3

A graph is drawn with vertices labelled of their labels is a multiple of .

and . Edges connect two vertices if the sum

a Draw this graph. b List all the cycles in this graph. 4

By considering the number of edges, show that all simple-connected graphs that connect four vertices using five edges are isomorphic.

5

Explain why a graph with four vertices, of which one vertex has degree and the other three vertices have degree , is not simply connected.

6

A graph is described by the adjacency matrix

a Find a subset of the vertices that, together with the edges that directly connect them, form a subgraph that is . b What can you conclude from this about the original graph? 7

Two of these graphs are isomorphic.

a Which two graphs are isomorphic? b Show that the two graphs from your answer to part a are isomorphic. c Show that each of the other two graphs is not isomorphic to any of the other graphs. 8

A simple-connected Eulerian graph is drawn that has exactly nine vertices and

edges.

a Write down the minimum possible vertex degree. b What is the maximum number of vertices that can have this minimum degree? c What is the maximum number of vertices of degree that the graph could have? 9

A three-dimensional noughts and crosses game is played using a cube of cells. Label the cells , where represent the coordinates of the cell. A simple-connected graph is constructed in which the vertices represent the cells and edges join two cells if those cells are not in the same straight line of cells. a Which cells are directly connected to

?

b Show that the graph is Eulerian. 10 A graph has exactly four vertices. The degrees of the vertices are

and .

a Explain why, if the graph is connected, it must be Eulerian. b Show that the graph need not be connected. c How many edges does the graph have? d What is the greatest value of for which it is possible to draw a connected graph with the degree sequence

, with no loops.

11 A graph has adjacency matrix:

a Write a Hamiltonian cycle for this graph. b Use Kuratowski’s theorem to show that the graph is non-planar. 12 A connected graph is semi-Eulerian if exactly two of its vertices are of odd degree. a A graph is drawn with vertices and edges. What is the sum of the degrees of the vertices? b Draw a simple semi-Eulerian graph with exactly vertices and edges, in which exactly one of the vertices has degree . c Draw a simple semi-Eulerian graph with exactly vertices that is also a tree. d A simple graph has vertices. The graph has two vertices of degree . Explain why the graph can have no vertex of degree . [© AQA 2016]

2 Networks

In this chapter you will learn how to: ​• use networks to model problems ​• use spanning trees to solve network optimisation problems ​ solve route inspection problems • ​ find and interpret upper and lower bounds for a travelling salesperson problem. •

Before you start… Chapter 1

You should be able to understand the terms cycle and tree.

1

a Find a cycle through every node in this graph. b Find a tree that connects every node in this graph.

Section 1: Network optimisation problems using spanning trees Key point 2.1 A network is a weighted graph. This means that each edge of the graph is assigned a numerical value (a weight) that is attached to it. The weights could, for example, represent distances, costs or travel times between points.

Key point 2.2 ​• The vertices in a graph are called nodes in a network. ​• The edges in a graph are called arcs in a network.

Rewind You learned about vertices and edges of graphs in Chapter 1.

Key point 2.3 ​• The arcs in a network can be undirected or they can be directed (like one-way streets). ​• A simple network (with no multiple arcs or loops) can be represented as a weighted matrix, showing the weight of each arc and a blank, or when no arc exists. The weighted matrix can be called a distance matrix, cost matrix, time matrix, according to what the weights represent.

Fast forward You will use directed networks in Chapter 3 when you look at network flows. The weight of an arc will represent its maximum capacity.

WORKED EXAMPLE 2.1 A simple network is shown. The nodes in the network represent towns and the weights on the arcs represent travel costs in £ units.

Represent the network as a weighted matrix (cost matrix). In this network the arcs are undirected.

With a directed network the usual convention is that the arcs are from the nodes on the rows to the nodes on the columns.

Often you will present a weighted matrix as a table and omit the brackets around the matrix.

Tip The weights will usually be non-negative but need not be integers, unless this is given in the question.

Key point 2.4 ​• A spanning tree is a tree that connects all the nodes in a network. ​• A least weight tree connecting all the nodes of a network is a minimum spanning tree or minimum connector. ​• The total weight of a tree is the sum of the arc weights used in the tree. If, for example, the arc weights are the costs of laying cables between the points, the minimum spanning tree represents the cheapest way to join the places by cables, so that all of the places are connected, directly or indirectly.

Common error There are various strategies for finding minimum spanning trees. You should always show your working. It is important that you draw the tree, or that you list the arcs in the tree, and also that you give the total weight. One way to find a minimum spanning tree is to use Kruskal’s algorithm.

Key point 2.5 Kruskal’s algorithm: ​• Start to list the arcs in increasing order of weight. ​• Step 1: Add an arc of minimum weight in such a way that no cycles are created. ​• Step 2: If a spanning tree is obtained stop; otherwise return to Step 1.

Tip Kruskal’s algorithm starts with the least weight arc and works through the arcs in increasing order of weight. It builds a collection of little trees that are eventually combined to make a single spanning tree.

WORKED EXAMPLE 2.2 Use Kruskal’s algorithm to find a minimum spanning tree for this network.

Start by listing the arcs in order of increasing weight. Where there are arcs with the same weight they can appear in any order. It is not usually necessary to write out the whole list, but every vertex must be in the list.

…

✔ ✔ ✔ ✔ ✔

Use Step 1 repeatedly to add arcs to the network (working down the list). Each time check that no cycles are created. Add . Add . Add Add Add

. . .

…

✔ ✔ ✔ ✔

At this point the next arc in the list is create a cycle .

but this would

Add . Although every node is now included the two parts are not connected so a spanning tree is not yet formed.

✔

✔

Continue until a spanning tree is formed. Delete all remaining arcs and add the weights of the arcs that have been used to find the total weight of the spanning tree.

Total weight

WORKED EXAMPLE 2.3

You should show all the working on one diagram. The ticks and crossing out on the list are evidence that you have used the algorithm correctly.

Find a different minimum spanning tree for the network in Worked example 2.2.

Use

instead of

.

Total weight A second way to find a minimum spanning tree is to use Prim’s algorithm.

Key point 2.6 Prim’s algorithm: ​• Start with any node (this will usually be given in the question). ​• Step 1: Add an arc of minimum weight joining a node already included to a node not already included. ​• Step 2: If a spanning tree is obtained stop; otherwise return to Step 1.

Tip Prim’s algorithm builds a tree by growing out from the starting node until every node is included.

WORKED EXAMPLE 2.4 Use Prim’s algorithm, starting at , to find a minimum spanning tree for this network.

Choose Choose Choose to

, the least weight arc from to . , the least weight arc from to . (or ), a least weight arc from .

Note that the arc can start at any of the nodes already included.

Continue in this way until a spanning tree is obtained. Add the weights of the arcs that have been used to find the total weight of the spanning tree. The working should all be shown on one diagram. The list is evidence that you have used the algorithm correctly.

Total weight

Tip There might be more than one minimum spanning tree. The choice of the starting node and the order in which arcs of equal weight are chosen might give different trees. If a computer is being used to solve a problem, the network is likely to have been given as a distance table, rather than drawn as a diagram. Prim’s algorithm is easily converted into a form that is suitable for use of a distance network. This is called Prim’s algorithm in tabular form.

Key point 2.7 Prim’s algorithm in tabular form: ​• Start with the table (or matrix) of weights for a connected graph. ​• Step 1: Cross through the entries in an arbitrary row and mark the corresponding column. ​• Step 2: Choose a minimum entry from the uncircled entries in the marked column(s). ​• Step 3: If no such entry exists Stop, otherwise go to Step 4. ​• Step 4: Circle the weight

found in Step 2, mark column and cross through row .

​• Step 5: Return to Step 2.

Tip Step 1: Columns that are marked correspond to nodes that are in the tree. Entries in rows that are crossed out cannot be chosen, so an arc cannot connect two nodes within the tree. Step 4: Circled entries correspond to arcs that are in the tree. column .

is the entry on row and

WORKED EXAMPLE 2.5 Use Prim’s algorithm, starting at , to find a minimum spanning tree for the network described by this distance matrix.

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Step 2: Choose , the least weight arc, not crossed out or circled, from columns . Step 4: Record this by circling the entry (and recording in a list at the side). This entry is . Mark column and cross through (remainder of) row . Step 2: Choose

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arcs

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Mark column 2 and cross through (remainder of) row .

↓

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Step 4: Record this by circling the entry (and recording in a list at the side). This entry is . arcs

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Step 1: Cross through row and mark column . Step 2: Choose , the least weight arc, not crossed out, from column .

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Continue in this way until there are no entries that are not circled or crossed out in any of the marked columns. Add the circled values to find the total weight of the spanning tree. arcs

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Total weight Minimum spanning trees are useful in solving a variety of network optimisation problems. Later in this chapter, you will use minimum spanning trees to find bounds on the solution of the travelling salesperson problem. WORKED EXAMPLE 2.6 The distance matrix shown represents the lengths, in km, of gas pipelines that could be used to connect villages , and . Currently, the main gas supply only reaches as far as village and it is required to connect the other villages to the main supply.

a What total length of pipeline is needed if each of b What total length of pipeline is needed if

and needs to be directly connected to ?

and can be connected to via other villages

(indirectly connected to )? a b

Use arcs in minimum spanning tree.

WORK IT OUT 2.1 Find a minimum spanning tree for the network described by the distance matrix shown.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

The three shortest arcs are

and

.

Solution 2

Solution 3

EXERCISE 2A 1

How many arcs are there in a minimum spanning tree for a network with nodes? Choose from these options. A B C D

2

Find the weight of a minimum spanning tree on this network.

10 9

11

8

14 3

How many different spanning trees are there on the network in question 2?

4

Find the total weight of a minimum spanning tree for the network described by this weighted matrix.

5

For the network in question 4, suppose that a direct connection is made between and . What is the greatest weight that this arc can have if it can be used in a minimum spanning tree?

6

A minimum spanning tree on a certain network uses the arc with the greatest weight. What can be deduced about the network?

7

Five buildings on a university campus are connected by footpaths. It has been suggested that shelters should be installed along some of the footpaths. Only footpaths with shelters can be used when it is raining. In this table, the nodes represent the buildings and the arcs represent the footpaths. The weight of each arc is the cost (in units of £ ) of installing the shelters on the footpath represented by that arc.

The university want to minimise the money spent on installing shelters. a Which footpaths should have shelters installed for it to be possible to travel between any of the buildings when it is raining? b If an additional £

can be spent on the shelters, which footpath should it be spent on? Explain

your choice. 8

Give an example of a network on

and in which

minimum spanning tree that does not use 9

has the minimum weight but there is a

.

Following a heavy snowstorm a snowplough is used to clear the snow from some of the roads between five houses. In a weighted network the houses are represented as nodes and the roads are represented as arcs. The weight of each arc is the time taken (in minutes) to clear the snow from that road.

a Find the minimum time needed to clear a route to each house, given that the time needed to drive the snowplough along any road that has already been cleared of snow can be ignored. b Find the minimum time needed to clear a route to each house, given that instead, the time needed to drive the snowplough along a road that has already been cleared of snow is exactly half the time needed to clear the snow from that road. 10 Why is it not possible to have directed arcs in a minimum spanning tree?

Section 2: Route inspection problems Key point 2.8 A route is like a cycle or closed trail except that edges can be repeated.

Rewind You found cycles and trails in Chapter 1.

Key point 2.9 ​• A route inspection problem involves finding a least weight route that uses every arc of a network. ​• The problem was originally studied by the Chinese mathematician Kwan Mei-Ko in 1960, and is also known as the Chinese postman problem.

Key point 2.10 ​• Any network formed by weighting an Eulerian graph can be traversed without having to repeat any arcs. ​• Otherwise, arcs need to be doubled up, in the way that uses the least weight possible, so that the graph that was weighted to form the network becomes Eulerian.

WORKED EXAMPLE 2.7 Solve the Chinese postman (route inspection) problem for this network.

The odd-degree nodes are and . The least weight path connecting to is

The solution needs all nodes of even degree. In this case there is just one pair of odd nodes so this is the minimum total weight to be added to make all nodes of even degree. Total weight of original graph question.

The minimum weight is . The route uses each arc once and uses and a second time, giving degree degree and degree .

. This would often be given in the

For example: There is no need to write out a route unless it is asked for.

Common error Shortest routes between odd nodes are usually found by inspection. Do not assume that a direct route is the shortest route – an indirect route could be shorter. WORKED EXAMPLE 2.8

a Find the weight of the least weight tour that uses every arc in the network shown. The sum of the weights in the network is .

b Given that the tour starts and ends at , find how many times it passes through i ii iii a The odd-degree nodes are and . Four odd nodes so means that there are three pairs of least weight paths (each containing all four odd nodes). Note that the least weight path need not be the direct arc between two nodes. Least weight pairing is b i

2

.

has degree so it is passed through twice (in twice and out twice).

ii 2

has degree but degree .

iii 2

Similarly for .

is repeated, effectively making the

Common error In Worked example 2.8, you should show that you have considered all the ways of joining the odd nodes in pairs (for example, and ) and not just write down the shortest such pairing.

Key point 2.11 The standard Chinese postman problem starts and finishes at the same node. Sometimes a question will indicate that the route need not start and end at the same node. In this case, a semi-Eulerian network is needed, where the route will start at one odd-degree node and finish at another.

WORKED EXAMPLE 2.9 a Solve the problem in Worked example 2.8, but allowing the route to start and end at different nodes. b Find how many times the route passes through i ii iii

.

a The odd-degree nodes are and . One pair can remain odd, so you just need to find the shortest pairing of the other pair. Route starts and ends at and , repeating b i

.

The route either starts or ends at and also passes through once.

ii iii

EXERCISE 2B 1

2

Classify each graph as Eulerian, semi-Eulerian or neither.

A

B

C

D

The distance matrix for a network is shown. Which are the odd-degree nodes?

3

What is the least weight path connecting the odd nodes in the network in question 2?

4

Explain why the solution to the Chinese postman problem for the network in question 2 has total weight .

5

A groundsworker wants to use a line-painting machine to mark out a design on a pitch. The machine is faulty so once it has been switched off it cannot be switched on again. This means that some lines will be painted twice. The design is shown.

The small circle has radius metre. It passes through the centre of the larger circle and touches the larger circle. The straight lines are diameters of the larger circle. Each metre uses

litres of paint.

a What is the minimum amount of paint that the groundsman needs? b Why, when the groundsman is pushing the machine, would he need to walk further than the distance travelled by the machine? c What else do you need to know to be able to calculate the minimum distance that the groundsman travels when marking out the lines? 6

The least weight route that uses every arc at least once is required for this network.

When the Chinese postman algorithm is used, one group of odd nodes is

.

a How many groups of odd nodes are there altogether? b Explain why the total weight of the shortest paths for any group of odd nodes that includes must be greater than . c List the weights and total weight for each group of odd nodes that includes d Explain why

.

must be included in the group with least total weight.

A closed route is constructed that uses every edge once and repeats the shortest routes between and and and . e How many times does this route pass through ? 7

For the network in question , which arcs would be repeated in a route that covers every arc but starts and ends at different nodes?

8

The total length of the arcs in a network is

metres.

The degrees of the nodes are: Node Degree The lengths, in metres, of the shortest paths between the odd nodes are:

The shortest route that starts at and uses every route is required. a Where does the route end? b Use the Chinese postman algorithm to find the length of the route. 9

The Chinese postman algorithm is to be used on a network with eight odd nodes. a Show that, unless any additional information is given,

groups of odd nodes would need to be

considered. b Show that the lengths of shortest paths would need to be calculated. If it takes seconds to calculate each shortest path and seconds to find the total weight of each group of odd nodes, the time taken for a network with eight odd nodes will be over minutes.

c Calculate the corresponding time for a network with ten odd-degree nodes.

Did you know? Find out about the Königsberg Bridges problem. This problem was solved by the Swiss mathematician Leonhard Euler (1707–1783). The town of Königsberg, which was in Prussia but is now in Russia and is called Kaliningrad, had seven bridges connecting the two sides of the River Pregel to two islands in the river. Euler proved that it was impossible to walk a route that used every bridge once without repeating any bridges. Use your knowledge of the Chinese postman (route inspection) problem to understand Euler’s method.

Section 3: Travelling salesperson problem Key point 2.12 The travelling salesperson problem (TSP) involves finding a least weight cycle through all the nodes of an undirected graph (the least weight Hamiltonian cycle).

Rewind You learnt how to identify Hamiltonian graphs in Chapter 1.

Did you know? The travelling salesperson problem is one of the most widely studied problems in computational mathematics but no efficient strategy is known for the general case. Determining whether or not an efficient strategy exists (one that is better than an exhaustive check of every possibility) is equivalent to the ‘P versus NP’ problem, for which the Clay Mathematics Institute has offered a one million dollar prize. Although there is no general strategy to solve the travelling salesperson problem, it is easy to find upper and lower bounds for the minimum weight. Any Hamiltonian cycle will provide an upper bound for the problem. A problem might suggest a suitable cycle. If not, you can try the nearest neighbour algorithm.

Key point 2.13 The nearest neighbour algorithm involves starting at some node and travelling the least weight arc from this node to another node. From each node that is reached the least weight arc from this node to a node that has not yet been visited is chosen. This process is repeated until every node has been visited. If there is an arc that directly connects the final node to the start node, then this arc can be used to close the cycle.

WORKED EXAMPLE 2.10 Apply the nearest neighbour algorithm, starting at , to the network represented by this distance matrix.

Starting from the nearest adjacent vertex is . You do not want to revisit a vertex, until all vertices have been used.

From move to . From to and from to . Then return from to .

– –

Each row has one circled entry and each column has one circled entry.

– – –

Common error The nearest neighbour algorithm is similar to the construction of a minimum spanning tree in that both require you to find the least weight arc from a node that has been included to a node that has not yet been included. The difference is that, for a minimum spanning tree, every node that has been visited is considered whereas, for nearest neighbour, only the most recent node is considered. If the network is not based on a complete graph, the nearest neighbour algorithm might ‘stall’. This means that a node is reached for which all the adjacent nodes have already been visited. This means that the path stops before a Hamiltonian cycle has been achieved. WORKED EXAMPLE 2.11 Find the weight of the cycle ACFHGEDBA for this network.

This is a cycle through every node, so the least weight Hamiltonian cycle has weight at most .

WORKED EXAMPLE 2.12 Apply the nearest neighbour algorithm to the network in Worked example 2.11, starting at

and

in turn. When using the nearest neighbour algorithm, the next arc is chosen from the current node to a node that has not been visited. At the least weight arc is , but has already been visited.

Starting from :

Starting from :

The route stalls at because and have already been visited, but nodes and have not yet been visited.

Starting from :

If you had used the nearest neighbour from in Worked example 2.12, it would have given the same cycle as from , but starting . Starting from

or would have led to a stall.

Both cycles found by using the nearest neighbour algorithm in Worked example 2.12 are Hamiltonian cycles so the least weight cycle has total weight better upper bound.

and also

. Since

is smaller than

this is the

Key point 2.14 The best upper bound is the smallest of the upper bounds that have been found (the least upper bound).

Tip Sometimes you can use shortcuts or switches to improve an upper bound. For example, you could improve the solution from Worked example 2.11 by changing to and then travelling

instead of

.

The context might permit nodes to be repeated (the salesman travels through a town that he has already visited). In this case, it is better to travel instead of , even if has already been visited.

Key point 2.15 A tour is a closed route through every node that might pass through nodes more than once.

It is useful to try to give a lower bound for the travelling salesperson problem as well as an upper bound. If the lower bound and the upper bound are equal, then this is the weight of the solution to the travelling salesperson problem and the route that gave the upper bound is an optimal tour. If the lower bound and the upper bound are close to one another, then the route that gave the upper bound is either an optimal tour or is a very good tour.

Key point 2.16 For a network based on a complete graph, a lower bound can be found by removing a node and all the arcs from that node and then constructing a minimum spanning tree for the remaining nodes. The weight of the minimum connector is then added, that is, the two least weight arcs that will reconnect the node to the minimum spanning tree. An optimal tour uses two arcs to any given node, the total weight of these is less than or equal to the sum of the weights of the two least weight arcs from the given node. The remainder of the optimal tour is a ‘string’ connecting the other vertices. This must be at least as long as the minimum spanning tree for these vertices. If the graph is not based on a complete network, then the method could fail to give a lower bound. WORKED EXAMPLE 2.13 Show what happens when the method from Key point 2.16 is applied to the network shown when the node that is removed is: a b

.

a Reduced network with deleted.

Delete and arcs that are directly connected to (arcs that are incident on ). Minimum spanning tree: Total weight

.

Two least weight arcs from : Note that two different arcs are used. Lower bound

.

b Reduced network with deleted.

From the original graph, delete and arcs that are directly connected to .

Minimum spanning tree: Total weight

.

Two least weight arcs from : or

or

Lower bound

.

Key point 2.17 The largest of the lower bounds that have been found (greatest lower bound) is the best of the lower bounds. For the problem in Worked examples 2.11, 2.12 and 2.13, So the tour optimal tour.

of weight

weight of

.

is a good solution. It might or might not be the

EXERCISE 2C 1

and are connected by weighted arcs

.

Use the nearest neighbour algorithm starting from . Give the weight of the resulting tour. 2

When the nearest neighbour algorithm is used on the network in question , the arc is included, whichever node is used as the start. What is the length of a tour that does not use the arc ?

3

A network has four nodes, with each node connected to each of the others. Ignoring the direction of travel, how many different cycles are there?

4

a Find the lower bound for the travelling salesperson problem for the network described by the shown distance matrix by deleting each of the nodes in turn. b Give the total weight of the best of these lower bounds.

5

a Find the upper bound for the travelling salesperson problem for the network in question 4 by using the nearest neighbour algorithm, starting from each of the nodes in turn. b Give the total weight of the best of these upper bounds.

6

a What is the total weight of the least weight Hamiltonian cycle for this network?

b What is the total weight of the least weight closed tour if nodes can be repeated? An additional arc of weight is added, connecting to . c What is the total weight of the least weight Hamiltonian cycle? d What is the total weight of the least weight closed tour if nodes can be repeated? 7

Five buildings on a university campus are connected by footpaths. It has been suggested that street lights should be installed along some of the footpaths. The cost (in £ ) of installing the street lights on each footpath is shown in the table. The nodes represent the buildings and the arcs the footpaths.

A security guard wants to be able to walk a circuit using only paths with street lights, starting and ending at and passing through each of

and once.

The university want to minimise the money spent on installing street lights. a Find a minimum spanning tree for the reduced network formed when all arcs directly joined to are removed. Hence find a lower bound for the cost for the university. b Use the nearest neighbour algorithm to find an upper bound for the cost to the university. c Give a reason why the cost might be greater than the upper bound. 8

A reduced network is formed by removing a certain node from a graph. The resulting lower bound for the travelling salesperson problem for the original graph has the same total weight as the best upper bound. What can you deduce about the minimum spanning tree for the reduced network?

9

A graph consists of a weighted tree. a Explain why it is not possible to construct a cycle that passes through every node exactly once. A closed route is constructed that passes through every node, sometimes more than once. b What is the weight of the closed route?

Checklist of learning and understanding

​•

A network is a weighted graph. This means that each edge of the graph is assigned a numerical value (a weight) attached to it.

​• ​• ​ • ​ • ​ •

The weights could, for example, represent distances, costs or travel times between points.

The vertices in a graph are called nodes in a network. The edges in a graph are called arcs in a network. The arcs in a network can be undirected or can be directed (like one-way streets). A simple network (with no multiple arcs or loops) can be represented using a weighted matrix, showing the weight of each arc and a blank, or

when no arc exists.

​• ​ •

A spanning tree is a tree that connects all the nodes in a network.

​• ​ • ​ • ​ •

The total weight of a tree is the sum of the arc weights used in the tree.

A least weight tree connecting all the nodes is a minimum spanning tree or minimum connector. A minimum spanning tree can be found by using Kruskal’s algorithm or using Prim’s algorithm. A route is like a cycle or closed trail except that edges can be repeated. A Chinese postman problem is the problem of finding a least weight route that uses every arc of a network. It is also called the route inspection problem.

​•

Any network formed by weighting an Eulerian graph can be traversed without having to repeat any arcs.

​•

Otherwise, arcs need to be doubled up, in the way that uses the least weight possible, so that the graph that was weighted to form the network becomes Eulerian.

​•

Sometimes a question will indicate that the route need not start and end at the same node. In this case, a semi-Eulerian network is needed, where the route will start at one odd-degree node and finish at another.

​•

The travelling salesperson problem (TSP) involves finding a least weight cycle through all the nodes of an undirected graph (the least weight Hamiltonian cycle).

​•

For a network based on a complete graph, an upper bound can be found by using the nearest neighbour algorithm.

​•

The best upper bound for the travelling salesperson problem is the smallest of the upper bounds that have been found (the least upper bound).

​•

For a network based on a complete graph, a lower bound can be found by removing a node and all the arcs from that node and then constructing a minimum spanning tree for the remaining nodes. The weight of the minimum connector is then added, that is, the two least weight arcs that reconnect the node to the minimum spanning tree.

​•

The best lower bound for the travelling salesperson problem is the largest of the lower bounds that have been found (the greatest lower bound).

​•

A tour is a closed route through every node that can pass through nodes more than once.

Mixed practice 2 1

A network is described by the distance network shown.

What is the total weight of the minimum spanning tree? Choose from these options. A B C D 2

A network is described by the distance network shown.

What is the total weight of the network? Choose from these options. A B C D 3

In the network shown in the diagram, the weights represent distances between nodes. The total length of the arcs is .

a List the degrees of the nodes in the network. b What is the length of the shortest route that starts and finishes at and uses every arc at least once? 4

a Find a minimum spanning tree for the network described by the distance matrix shown.

b Give the total weight of the tree.

5

a Apply the nearest neighbour algorithm to the network in question , starting at node . b Calculate a lower bound for the travelling salesperson problem by using the reduced network formed by removing .

6

A cake decorator wants to mark out a design in icing. To avoid making a mess when starting and stopping icing, he wants to draw the design as a continuous route. He will achieve this by duplicating some arcs, but wants to minimise the number of nodes that are travelled through more than twice. The design is shown.

A

B

C

D E

F H

G J

I

K

The route needs to start at and finish at . Show that to achieve this every node will be visited twice. 7

a Find a minimum connector for the network in this diagram.

A

7

3 6

D

B

5 4 C 7

8

7

E

b Show that this minimum connector is not unique. 8

Sarah is a mobile hairdresser based at . Her day’s appointments are at five places: and . She can arrange the appointments in any order. She intends to travel from one place to the next until she has visited all of the places, starting and finishing at . The table shows the times, in minutes, that it takes to travel between the six places.

a Sarah decides to visit the places in the order tour.

. Find the travelling time of this

b Explain why this answer can be considered as being an upper bound for the minimum travelling time of Sarah’s tour. c Use the nearest neighbour algorithm, starting from , to find another upper bound for the minimum travelling time of Sarah’s tour. d By deleting , find a lower bound for the minimum travelling time of Sarah’s tour. e Sarah thinks that she can reduce her travelling time to wrong.

minutes. Explain why she is

[© AQA 2013] 9

The network below shows towns . The weight on each arc shows the length of the road, in miles, between the towns. During the winter, the council treats some of the roads with salt so that each town can be safely reached on treated roads from any other town. It costs £ to treat a mile of road. A

D

5 4

7

8 8

C

7 9

6 8

B

11 F

5

10 E

G

7

9 4

H

a i Use Prim’s algorithm starting from , showing the order in which you select the arcs, to find a minimum spanning tree for the network. ii Draw your minimum spanning tree. iii Calculate the minimum cost to the council of making it possible for each town to be safely reached on treated roads from any other town. b On one occasion, the road from to is impassable because of flooding. Find the minimum cost of treating sufficient roads for safe travel in this case. [© AQA 2015] 10 A network connects four nodes. The arc weights are shown in the table. The arcs are all undirected apart from .

From

The total weight of the arcs that enable travel from any node to any other node is

.

What is the value of ? 11 A network connects six nodes. The arc weights are shown in the table. The arcs are all undirected apart from and .

From

The total weight of the undirected arcs is

and the total weight of all the arcs is

.

Find the weight of the least weight route that uses every arc that: a starts and finishes at b starts at and finishes at . 12 The cost (in pence) of travelling by train between eight stations is given in the table. The cost for any cell marked with

is

and for any cell marked with is

.

Bodmin Camborne Falmouth Looe Newquay Penzance Saltash Truro a Use the nearest neighbour algorithm, starting from Truro, to find an upper bound for the minimum cost of travelling between the eight stations, starting and finishing at Truro. The upper bounds found by using other stations as the start for nearest neighbour are: £

£

£

£

£

£

.

b Which of these is the best upper bound? c By considering the reduced network formed by removing Truro, find a lower bound for the minimum cost of travelling between the eight stations, starting and finishing at Truro. The lower bounds found by reducing the network by removing other stations are: £

£

£

.

d Which of these is the best lower bound? e What can you deduce about the minimum cost of travelling between the eight stations, starting and finishing at Truro?

3 Network flows

In this chapter you will learn how to: ​• model a flow problem by a network of directed arcs ​• find the value of a cut and understand its meaning ​ use and interpret the maximum flow–minimum cut theorem • ​ understand and use the terms supersource and supersink. • If you are following the A Level course, you will also learn how to: ​• use flow augmentation to find the maximum flow in a network ​ refine flow problems to deal with restrictions, such as nodes of restricted capacity. •

Before you start… Chapter 2

You should be able to understand the difference between a network of directed arcs and an undirected network.

A

1

2

5

B 3

2 C

4

D

a Find the longest route from to in the directed network shown. b Find the longest route from to , without repeating any arcs, if the arcs were not

directed.

Section 1: Flows and cuts Key point 3.1 A network flow problem involves finding the maximum possible continuous flow travelling from a start point to a finish point through a network of routes, along each of which the flow rate is restricted. Examples of a network flow problem include maximising the flow of water through a river system from its source to the sea, maximising the flow of oil through an engine or maximising the flow of blood through a heart.

Key point 3.2 ​• The arcs show the routes that the flow can take (for example, the pipelines). ​• The arc weights represent the capacities. The flow along any arc cannot exceed the capacity of the arc.

Rewind You learned about networks in Chapter 2, so you are familiar with arcs and arc weights.

Key point 3.3 ​• All flows start from the source node. ​• There are no flows into the source. ​• The source is sometimes denoted by ​• All flows end at the sink node. ​• There are no flows from the sink. ​• The sink is sometimes denoted by (for terminus). ​• At all other nodes, flow passes through the node. ​• The flow that enters the node must equal the flow that leaves the node.

Tip In some cases, the flow in an arc will be directed; this must be the case for the arcs from the source and the arcs into the sink. There might be other directed arcs if there are valves or other reasons why the flow can only be in one direction. Often every arc is directed.

WORKED EXAMPLE 3.1 This network shows the capacities of the arcs in a network with source and sink .

A 13

10

C 13

5

S

T 5

12 B

12 5

D

a Which arcs must be directed?

b Explain why the flow along

cannot exceed

c Show that the flow a

,

,

. ,

is feasible.

,

There are no flows into the source. There are no flows from the sink.

,

and , so b The capacities of the arcs leaving are at most units can leave . At most units can enter , so the flow in cannot exceed . c

Arc

Flow

The flow that enters node equals the flow that leaves node .

Capacity

means that there is a flow of units along the route . For example: Flow along along

and flow , so flow along .

For each arc, flow capacity. Node

Flow in

Flow out

For example: At , flow in is along flow out is along .

; and

Flow in = flow out at each node (apart from the source and sink) WORKED EXAMPLE 3.2 For the network in Worked example 3.1, find a flow of For example: ,

,

Flow

Key point 3.4

,

units.

By sending only along and .

, an extra can be sent along both

This might or might not be the maximum flow.

​• A saturated arc is one that is full to capacity (flow capacity). ​• One way to find a maximum flow is to look for a ‘bottleneck’ formed by saturated arcs. ​• A cut is a partition of the nodes into two sets, one of which contains the source and the other of which contains the sink. ​• Any arc that directly connects a node in the ‘source set’ to a node in the ‘sink set’ is a cut arc. ​• The value of a cut is the sum of the capacities of the cut arcs. ​• This is the maximum possible flow across the cut from the source side to the sink side, without looking at the rest of the network. If the flow in a cut arc is from the sink side to the source side, then it contributes to the value of the cut. ​• If an arc is cut multiple times, then a forward flow and a backward flow can be cancelled out (for example, if an arc is cut twice, then it cancels out and adds to the value; if it is cut three times, then two of them cancel out). ​• The value of a cut is sometimes referred to as the capacity of the cut.

Tip If the network is drawn on a planar graph, then the cut can be shown as a curve.

Common error The value of a cut is not the sum of the flows in the cut arcs, which would be the same for every cut.

WORKED EXAMPLE 3.3 There are

possible cuts for this network.

A 17

C

13

13

18

S

T

6 14

16 B List these Nodes on source side Nodes on sink side Cut arcs

19 16

D

cuts and find the value of each. If you think of the network as being drawn on a sheet of paper, a cut is a curve that starts at the top of the sheet of paper and ends at the bottom of the sheet of paper,

with and on different sides of the cut. If necessary, lightly shade one side of the cut so that you can see whether the cut arc passes across the cut from to or from to . It is enough to list the nodes on the two sides of the cut, as any two cuts with the same vertex sets will have the same value.

Value of cut Nodes on source side Nodes on sink side Cut arcs

Value of cut

Listing all the cuts is not practical even for a small problem.

Key point 3.5 ​• A cut with the minimum value is called a minimum cut. A minimum cut forms a bottleneck that restricts the flow. In Worked example 3.3 the minimum cut has value maximum flow is at most .

. No more than

can flow across this cut, so the

Key point 3.6 ​• The maximum flow–minimum cut theorem states that: the maximum flow through a network equals the minimum cut value. ​• When the maximum flow happens, all cut arcs in each minimum cut are saturated.

Tip Checking every possible cut is not a practical way to find a maximum flow. However, if a flow and

a cut can be found that have the same value, then this must be a maximum flow and a minimum cut.

WORKED EXAMPLE 3.4 This network shows a system of pipes and the capacities of those pipes, in litres per second.

a i ii

Explain why the maximum flow along Find the maximum flow along

is litres per second.

.

b Find the maximum flow using the routes: i and ii

and

iii

,

and

.

c What can you deduce from the results of part b about the maximum flow? d Work out the value of: i the cut through , ii

,

and

the cut

, i.e. the cut that splits the nodes into these two sets.

e What can you deduce from the results of part d about the maximum flow? f

Find a maximum flow and prove that it is the maximum by giving a cut of the same value.

a i ii

can only carry litres per second.

The minimum capacity along the route restricts the flow along the route.

litres per second.

is the minimum capacity along the route.

b i

litres per second

ii

litres per second

iii

litres per second

and and and share and so although can flow along either of these, only can flow in total.

c Maximum flow found in litres per second, so maximum flow through network ⩾ litres per second. d i ii

litres per second

,

litres per second

e Minimum cut found so maximum flow through network ⩽ For example, flow and litres per second.

,

,

,

, litres

per second. f

The flows found are less than or equal to maximum flow.

The cuts found are greater than or equal to the minimum cut and maximum flow = maximum cut.

,

to give a flow of

There are several ways to flow to there are several ways to flow to

and .

Flow = cut from d ii Maximum flow = maximum cut So

litres per second is the maximum flow.

is also the minimum cut value.

Key point 3.7 ​• In some situations, there might be multiple sources (for example, there might be several rivers supplying flow to a lake). Then a single supersource is used to supply the sources. ​• The capacity of the arc from the supersource to each source must be greater than or equal to the sum of the capacities of the arcs from that source. ​• Similarly, a flow might have multiple sinks (for example, a river might flow into several lakes). Then a single supersink is used that all the sinks drain into. ​• The capacity of the arc from each sink to the supersink must be greater than or equal to the sum of the capacities of the arcs into that sink. EXERCISE 3A 1

This network shows the capacities of the arcs in a network.

A 12

C

8

8

6

S

T

5 10

9 B

14 6

D

Which could be the flow along

? Choose from these options.

A B C D E F 2

For the network in question , work out the value of the cut that separates the nodes , and from the nodes , and .

3

For the network in question , what is the maximum possible flow in arc

4

For the network in question , work out the value of the cut

5

This network shows the capacities of the arcs in a network.

?

A 8

C

12

7

T1

9

T2

5

10

S 15

10

18

B

The sinks

8

and

D

are connected to a supersink, .

a What are the capacities of the arcs b Why can the arcs 6

and

and

?

never both be saturated?

This network shows the capacities of the arcs in a network. Three cuts are marked on the network. A 12

C

9

13

10

S 13

20

D 11

B

11 8

12

a

b

T

7 E

c

Calculate the value of each of the cuts marked. 7

This network shows the capacities of the arcs in a network. A 12

C

9

13

10

S 13

20

D B

11 12

11 8

T

7 E

a Find a maximum flow through this network. b Use the maximum flow−minimum cut theorem to prove that your flow from part a is the maximum. 8

For the network in question , if the arc

has a flow of , what is the maximum flow in:

a arc b arc c arc 9

?

For the network in question , what would be the effect on the maximum flow of each of these

changes? a The capacity of arc

is increased to

b The direction of flow in c The capacity of arc

.

is reversed.

is reduced to .

10 This network shows a system of pipes.

A

B

12 7

6

15

C 21

16 E 3 G

D 14

21

10 F 4

20

H

a Which node is the source? b Which node is the sink? c What is the maximum flow through vertex ? d Show that the maximum flow is

.

Section 2: Variations on flow problems A flow can be augmented to construct a maximum flow.

Key point 3.8 ​• The potential increase on an arc is the amount by which the flow in that arc can be increased without exceeding the capacity, and the potential decrease on an arc is the amount by which the flow in that arc can be decreased.

WORKED EXAMPLE 3.5 This network shows a system of pipes and their capacities. The numbers in circles represent an initial flow. A 18 8

8 8

C

10 8

E

12

11

12 8

S

T 15

20 7 B

10 10 7

D

13 7 7 7

F

a Give the value of the initial flow. b Mark the potential increase and the potential decrease on each arc. a

and

b For example, arc has a flow of and a capacity of , so it has a potential increase of and a potential decrease of .

Key point 3.9 ​• A potential (in either direction) of means that no more can flow in that direction. Otherwise the flow can be augmented. When no further augmenting paths are possible the maximum flow has been achieved. ​• This method is called the labelling procedure.

WORKED EXAMPLE 3.6 a Augment the flow from Worked example 3.5, using the flow augmenting route,

.

b Use further flow augmentations to find the value of the maximum flow. a Flow can be augmented by .

Increase flow along route by , so potentials in this direction are decreased by and potentials in the other direction are

increased by . Note: previously there was a flow of from to but now it is .

b For example: Augmenting path

Flow

There are several possible flow augmenting paths. The potentials would be updated as well at each augmentation. No further augmentations (cut through arcs , is saturated).

Key point 3.10 ​• It might be necessary to put a lower limit on the capacities, for example, if an engine needs to be lubricated. In this situation, both a lower capacity and an upper capacity are shown on the arcs.

Common error When calculating the value of a cut, arcs that flow from the source to the sink are at their upper capacities and arcs that flow from the sink to the source are at their lower capacities (which need not be zero). The value of the cut is the sum of the upper capacities for the arcs in which the flow is forwards across the cut minus the sum of the lower capacities for the arcs in which the flow is backwards across the cut.

Tip The potential decrease is flow minus lower capacity.

WORKED EXAMPLE 3.7 This network shows the lower and upper capacities of a system of pipes.

A

D

1,6

2,4

0,6

3,3

C

S 2,4

1,8

T

6,8 3,5 5,10

B

E

a What can you deduce about the flow in

?

b By first considering the flow through , what can you deduce about the flow in c What is the value of the cut through d Suppose that units flows along decrease in arc

,

and

?

. What are the potential increase and the potential

?

e Find a flow with value a Flow

,

?

.

.

Lower capacity upper capacity

b Flow through flow in . Hence flow in

so

Maximum flow into so flow through .

.

c

Flow in

and minimum flow out of

,

is backwards across the cut.

It is already known that the flow in must be , but this is not taken into account when calculating the value of a cut. This just means that the cut arcs cannot all be saturated so this cannot be a minimum cut. d Potential increase Potential decrease e

Path

. .

Flow

There are other ways to describe the flow. This is the only feasible flow through the network.

Key point 3.11 ​• A node might have a restricted capacity. This means that there is a limit to the amount that can flow through the node. ​• The node, , is split into two separate nodes,

and

, say, with

connected to all

the arcs that have flows to the node and connected to all the arcs that have flows from the node. An arc connects to with capacity equal to the node restriction. WORKED EXAMPLE 3.8 a Redraw this network to show the case when at most

can flow through .

A 17

C

13

13

18

S

T

6 14

16

19 16

B

D

b Find a maximum flow and a cut of the same value. a

Split into an ‘in’ node and an ‘out’ node joined by an arc of capacity .

b Flow Flow Flow

along along

The saturated arcs are and . This cut crosses twice so they cancel out. The flow in arcs , and is from the sink side to the source side. So the value of the cut is .

EXERCISE 3B 1

What is the maximum flow along arc

in this network? Choose from these options.

A B C D

A 4,7

3,8 1,3

3,4

S 2,4

4,7 B

D

C

1,6 T

2,9 3,5

2,5

E

2

Find the minimum flow through C in the network in question .

3

Calculate the value of the cut through

4

Explain why there is no feasible flow through this network.

A 2,5

and

in the network in question .

0,6

S

C

2,4

T

6,7 3,5

4,6

B 5

,

D

1,6 3,4

1,4

,

E

This network shows potential increases and potential decreases for a flow through a network.

6

A 7

8 4

7

12

12

3

5 4

S

7

6

B

16

has lower capacity

T

9 0

10 13

16

Arc

C

D .

By considering the potential increase and potential decrease in arc about:

, what can you deduce

a the flow b the upper capacity of this arc? 6

For the network in question , the upper capacity of ?

7

The flow in question is augmented, using the route

is

. What is the flow in the arcs

.

a What is the maximum additional flow along this route? b What are the new potential increases and decreases on the arcs 8

,

and

?

The network shows a system of pipes with their lower and upper capacities. The numbers in circles represent an initial flow.

a What is the maximum possible flow through ? The flow through is restricted to a maximum of b Explain why this makes no difference.

.

and

The flow through is restricted to a minimum of and a maximum of . c Redraw the network to show this restriction. d Find the maximum flow in 9

in terms of .

The flow in question is augmented, using a route that includes

.

a Give this flow augmenting route. b Give the maximum amount by which the flow can be augmented, in terms of . c Give the new potential increase and potential decrease for the arc

.

d By using further flow augmenting routes, find the maximum flow, in terms of if necessary. e Prove that this flow is maximum.

Checklist of learning and understanding

​•

A network flow problem involves finding the maximum possible continuous flow travelling from a start point to a finish point through a network of routes in each of which the flow rate is restricted.

​• ​ • ​ • ​ • ​• ​ •

The arcs show the routes that the flow can take. The arc weights represent the capacities. All flows start from a source and end at a sink. At all other nodes, the flow that enters the node equals the flow that leaves the node.

A saturated arc is one that is full to capacity (flow = capacity). A cut is a partition of the nodes into two sets, one of which contains the source and the other of which contains the sink.

​• ​ •

The value of a cut is the sum of the capacities of the cut arcs. A minimum cut is a cut with the minimum value. A minimum cut forms a ‘bottleneck’ that restricts the flow.

​•

The maximum flow–minimum cut theorem states that: the maximum flow through a network equals the minimum cut value.

​•

A supersource can be used to supply multiple sources. The capacity of the arc from the supersource to each source must be greater than or equal to the sum of the capacities of the arcs from that source.

​•

A supersink can be supplied from multiple sinks. The capacity of the arc from each sink to the supersink must be greater than or equal to the sum of the capacities of the arcs into that sink.

​•

The potential increase on an arc is the amount by which the flow in that arc can be increased without exceeding the capacity, and the potential decrease is the amount by which the flow in that arc can be decreased.

​•

A potential (in either direction) of means that no more can flow in that direction. Otherwise the flow can be augmented. When no further augmenting paths are possible the maximum flow has been achieved. This method is called the labelling procedure.

​•

A node might have a restricted capacity. The node, , is split into two separate nodes, , say, with

connected to all the arcs that have flows to the node and

the arcs that have flows from the node. An arc connects node restriction.

to

and

connected to all

with capacity equal to the

Mixed practice 3 1

A system of pipes is shown.

What is the value of the cut through

,

,

? Choose from these options.

A B C D 2

A system of pipes is shown.

A

5

C

9

7

S

T

5 4

7 B

6

D

The maximum flow through the system is

.

Which of the arcs must be saturated in a maximum flow? Choose from these options. A B C D 3

The diagram shows a network of pipes, with source and sink , and the capacities of those pipes.

A

D

12

15

8

16

12

S

T

C 16

11 17

B

20 E

a What is the maximum possible flow along

?

b What is the maximum possible flow along the route 4

?

The diagram shows the network from question with two cuts

A

D

12

15

8

C2

T

C 16

11 C1

17

B

.

16

12

S

and

20 E

Find the value of each cut. 5

a Find a cut of value b Find a flow of value

for the network in question 4. for the network.

c What can you deduce from your results in parts and a and b? 6

The network shows a set of pipes with source and sink . The weights on the arcs give the capacities.

A 15

18

11

18 B

20

C

23

S

D

14

T 24

17

20 E

a What is the maximum possible flow along

?

b Draw a diagram to show the potential increase and the potential decrease on each arc when the flow in part a is used. c Use further flow augmenting paths to find the maximum flow from to . Show the potential increase and the potential decrease on each arc when the maximum flow is achieved. d State the value of the maximum flow. e Find a cut that proves that the flow is a maximum.

7

This network represents a flow system from a source . The system has two sinks, A 21

10

12

C 5

8

.

T1

20

S

and

6 3

B

T2

a Add a supersink and weight the arcs that you have added. b i Explain why the arc

can never be full to capacity.

ii What is the maximum possible flow in the arc

?

c Find, by inspection, a maximum flow through the network. State the value of this flow. d Find a maximum flow in which the flow in arc 8

is .

The diagram shows a network of pipes with source and sink . The capacity of each pipe is given by the number on each arc. B

5

10 10 60

A

30

5 C

20

10

10 20

a Find the values of the cuts b Find, by inspection, a flow of respectively. c

30 10

G

10 F

H

20

10 10

D 10 10

C1

E

J 30

C2

I

and

.

units, with flows of

i On a certain day, the section this day.

,

and

along

,

and

,

is blocked. Find, by inspection, the maximum flow on

ii Show that the flow obtained in part c i is maximal. [© AQA 2014, adapted] 9

The weights in this network show the lower and upper arc capacities.

a Find the value of the cut that separates

b i By first considering node find the minimum flow through . ii Write down the maximum possible flow through . c Find a flow of value in which exactly flows through . d How much can the flow be increased using the augmenting path The maximum flow is

.

e Find a cut of value

.

?

10 a The diagram shows a network of pipes. The capacity of each pipe is given by the number not circled on each arc. The numbers in circles represent an initial flow.

a

Find the value of the initial flow.

b i

ii

Use the initial flow and the labelling procedure to find the maximum flow through the network. List your flow augmenting paths and adjust the potential increases and potential decreases on the arcs. State the value of the maximum flow. List a possible flow along each arc corresponding to this maximum flow.

c

Confirm that you have a maximum flow by finding a cut of the same value. List the cut arcs.

d

On a particular day, there is a restriction at vertex which allows a maximum flow through of . Find, by inspection, the maximum flow through the network on this day. [© AQA 2015, adapted]

11 The network shown represents a system of pipes through which fluid flows from a source to a sink. The circled numbers show the flows through some of the arcs.

A

3

5

D

F 2

2 B

G 3

8

3 C

E

H

a Identify the sink. b Explain why the direction of the flow in the arc between and must be from to . c If the arc from to has no flow in it, find the flow in each of the other arcs. d If the flow in the arc from to is , find the flow in each of the other arcs.

4 Linear programming

In this chapter you will learn how to: ​• formulate constrained optimisation problems ​ use graphical methods to solve constrained optimisation problems. • If you are following the A Level course you will also learn how to: ​• use slack variables ​ use the simplex algorithm to optimise an objective • ​ interpret a simplex tableau. •

Before you start… GCSE

You should be able to solve linear inequalities in two variables, representing the solution set on a graph.

1 Shade on a graph the set of points that satisfy the inequalities

simultaneously. GCSE

You should be able to solve simultaneous equations.

2 Solve the simultaneous equations.

Section 1: Formulating constrained optimisation problems Key point 4.1 ​• A standard linear programming problem involves maximising or minimising an objective function that is a linear function of non-negative variables which are constrained by linear inequalities. ​• Formulating a linear programming problem involves: ​• identifying relevant variables, including units when appropriate ​ formulating constraints in these variables • ​ writing down an objective function and stating whether it is to be maximised or minimised. • WORKED EXAMPLE 4.1 A car ferry can carry standard cars and large cars. A standard car has length up to metres and mass up to tonnes when fully loaded. A large car has length between metres and metres and mass between and tonnes when fully loaded. Allowing for some space between cars, the ferry company allows metres for each standard car and metres for each large car. The mass of a standard car is assumed to be tonnes and the mass of a large car to be tonnes. The ferry has enough space for metres of cars and can carry up to tonnes of cars. The profit for each standard car transported is £ and for each large car is £ . The ferry company wants to know the maximum profit that it could make on a single ferry crossing. a Identify appropriate variables for the ferry company’s problem. b Write an objective function in terms of the variables. c Identify the constraints on the values of the variables and represent these algebraically. a

number of standard cars number of large cars

b Maximise

The variables are the number of cars of each type. Defining the variables involves stating what each letter represents and stating appropriate units (in this case ‘number of’). Maximise the profit (in £).

c Constraints on the values of and : – total length – total mass tonnes – non-negativity (‘trivial’ constraints) – restriction to avoid decimal bits of cars. Representing the constraints algebraically:

and

are integers.

You can solve a linear programming problem in two variables graphically.

Key point 4.2 ​• The region where all the constraints of a linear programming problem are satisfied is called the feasible region. ​• It is useful to shade out the unwanted regions of the graph to leave the feasible region as the unshaded (clean) region. WORKED EXAMPLE 4.2

Sketch a graph of the feasible region for the problem from Worked example 4.1.

passes through and [or use , for example]. Check a point that is not on the line, for example is on the side of the line where the inequality is satisfied, so shade out the other side. Similarly for the other constraints. The feasible region is the unshaded region.

Common error Be careful about the gradients when plotting the lines. For example, in Worked example 4.2, the line

passes through

and

or it is the line

with gradient

.

Tip The constraints might be given as text or in a table. Sometimes the constraints involve ratios rather than specific values. It could be useful to substitute specific values to help plot the boundary line for such a constraint.

WORKED EXAMPLE 4.3 A small café sells two different takeaway lunch deals. Type costs £ and consists of a filled roll, a piece of fruit and a bottle of water. Type costs £ and consists of a sandwich and two pieces of fruit. The café has pieces of fruit available and bottles of water. There is no restriction on the number of rolls or amount of bread, but there is only enough filling for rolls or for sandwiches, or for some mixture of rolls and sandwiches. The café wants to find the maximum amount of money that they can make from selling lunch deals. Set up the problem as a linear programming formulation. number of type lunch deals sold. number of type lunch deals sold Maximise

Define the variables. Objective £

subject to: Pieces of fruit. Bottles of water. Filling. There is enough filling for , or for , . If the constraint is not obvious, then

one way to deal with this is to suppose that each roll uses, say, units of filling; then there are units of filling and each sandwich uses units of filling. This gives , which is equivalent to the given constraint. Non-negativity. Additional restriction.

are integers. WORKED EXAMPLE 4.4

A gardener has of garden. She wants to plant vegetables on at least of the garden and she wants the area used for flowers to be at least twice as big as the area used for vegetables. The remainder of the garden (that is not used for flowers or vegetables) will be covered in turf. Preparing the ground for vegetables and planting vegetables costs £ per , preparing the ground for flowers and planting flowers costs £ per and laying turf costs £ per . The gardener wants to know what to buy to keep the cost to a minimum. Set up the problem as a linear programming formulation. area planted as vegetables  area planted as flowers  Minimise

The total area is .

, so the area that is covered in turf is

Cost in £ . Equivalently, maximise



.

subject to: At least

of vegetables.

Area used for flowers is at least twice the area used for vegetables (for example, if , then ). Non-negativity for area that is turfed. Non-negativity for and .

WORKED EXAMPLE 4.5 A person consumes about Between

and

calories each day.

of these calories should come from carbohydrates, between

and

from

protein and the rest from fats and oils. Carbohydrates come from fruit, vegetables and starchy foods. Some vegetables are also protein rich. A portion of protein rich vegetables contains on average calories of carbohydrates and calories of protein. A portion of standard (not protein rich) vegetables contains on average calories. It is recommended that at least of the calories from carbohydrates should be from vegetables. A nutritionist wants to know the minimum number of portions of vegetables (including protein rich vegetables) to recommend. Set up the problem as a linear programming formulation. number of portions of protein rich vegetables number of portions of standard vegetables

Minimise

Total number of portions of vegetables.

subject to:

So

Protein from vegetables.

Carbohydrates from vegetables. So Minimise subject to:

, are integers. WORKED EXAMPLE 4.6 Ben is taking part in a TV quiz show. He has hours remaining before the show is filmed. Ben wants to spend some time studying for his specialist round, some time studying for the general knowledge round and some time sleeping. Ben believes that each hour spent studying for the specialist round will increase his overall score by points and each hour spent studying for the general knowledge round will increase his overall score by point. Of the time that Ben spends studying, at least must be spent on the general knowledge round. Ben needs to spend at least hours sleeping. Each hour spent sleeping will increase his overall score by points. Ben wants to know how long to spend studying for each round and how long to spend sleeping to maximise the increase in his overall score. a Set up the problem as a linear programming formulation. b Reformulate your linear programming formulation in terms of the time spent studying for each round if all the time when Ben is not studying is spent sleeping. a

hours spent studying for specialist round hours spent studying for general knowledge round hours spent sleeping Maximise

Total increase in score.

subject to: At most At least

hours available. of study time on general knowledge.

So At least hours sleeping. Maximise subject to:

b

Total time used is equal to

So

hours.

so Maximise subject to:

EXERCISE 4A 1

A linear programming problem involves two variables: amount of coffee sold (litres) amount of tea sold (litres) One of the constraints is that the amount of coffee sold must be more than three times the amount of tea sold. Which of the

points is feasible? Choose from these options.

A B C D 2

The feasible region of a linear programming problem satisfies Which of the

,

,

and

.

points is not in the feasible region? Choose from these options.

A B C D 3

Graph the feasible region for the constraints:

. 4

A linear programming problem is described. A manufacturing process involves two processes, mixing and finishing. Only one item can be mixed at any one time and only one item can be finished at any one time.

Each type item produced takes process.

minutes in the mixing process and

minutes in the finishing

Each type item produced takes process.

minutes in the mixing process and

minutes in the finishing

The profit on each type item produced is £

and the profit on each type item produced is £

.

The manufacturer wants to maximise the daily profit. a Define suitable variables for the problem. b Set up the problem as a standard linear programming formulation. 5

Set up this problem as a standard linear programming formulation. Jamie makes two types of energy drink using water and dry ingredients that he buys in bulk. The table shows the amount of each dry ingredient (in grams) needed for 1 litre of each energy drink and the total amount of each ingredient that is available. Jamie wants to maximise the total amount of energy drink that he makes. Ingredients: Drink Xtra Yepp Amount available

6

Work out the exact coordinates of the vertices of the feasible region for the problem in question 5.

7

a Which of the constraints are redundant (do not contribute to the feasible region) in the formulation for question 5? b Use algebra to explain why the constraints named in part a really are redundant.

8

The constraints of a linear programming problem are:

where is a positive constant. a For what values of is the second constraint redundant? b Find the vertices of the feasible region, in terms of .

Section 2: Graphical solutions Key point 4.3 ​• The objective function for a linear programming problem is of the form , where is fixed. This is a straight line of fixed gradient, with different values of giving different values of the objective. This gives a set of straight lines of fixed gradient (a set of parallel lines), also called objective lines. As the value of the objective increases, the objective lines slide across the feasible region, while keeping their gradient constant. The higher the value of , the greater the value of the objective. ​• The optimum vertex is the point where the sliding objective line last touches the feasible region. This will occur at a vertex of the feasible region (or along an edge if both ends of the edge are optimal). ​• To find the optimal value for a linear programming problem, you can either: ​• slide an objective line across the feasible region, in the direction of increasing value of the objective, to determine the optimal vertex (or edge) and then calculate the value of the objective at that vertex (or edge) or ​• calculate the value of the objective at every vertex of the feasible region and deduce the optimum.

Common error You can find the optimal vertex approximately from the graph, but the values might not be accurate enough. For an accurate answer, you should solve the simultaneous equations. You can do this by using a solver on a calculator.

WORKED EXAMPLE 4.7 a Use a sliding objective line to find the maximum value of the constraints and . b The objective is changed to maximise

, where

on the feasible region defined by . At what value of does the

optimal vertex change? c The objective is now changed to maximise , and the constraint . What is the optimal value of in this situation?

is changed to

y

a 8

You graphed the feasible region in Before you start ... question 1 at the start of this chapter. Note that in this example the feasible region is the shaded region rather than the unshaded region. Choose a point in the feasible region, for example . At this point, the value of is . Graph the line (i.e. ) passing through and . Do the same for another value of (for example passing through and ). The objective lines will be parallel. Slide the objective lines in the direction of increasing until they reach a vertex of the feasible region.

7 6 5 4 3 2 1 O

1

2

3

4

Optimum vertex is .

x

, where

b The gradient of the objective line is

.

The objective line is

.

As increases, the gradient of the objective line will decrease. The optimal objective line will still pass through until it lies along the edge defined by the boundary

The line that defines the boundary has gradient

.

. .

When

, all points on the

edge from

to

become optimal solutions. For

At

, the vertex

becomes optimal.

, the optimal

vertex is at

.

y

c

8 7

The objective line now has gradient

6

optimal vertex is

5

, so

, and the

.

Changing to will bring that edge of the feasible region down to pass through and .

4 3 2 1 O

1

2

3

x

4

Vertices of new feasible region are , and . The optimal vertex is now at , where

.

Tip The vertex checking method is usually easier but the sliding objective line method is better when changes are made to the coefficients in the objective or constraints. It can also be useful when the variables are restricted to taking integer values.

WORKED EXAMPLE 4.8 Solve the problem from Worked example 4.1. Maximise subject to:

and

are integers.

You graphed the feasible region in Worked example 4.2. You can find the vertices of the feasible region by solving the equations that define adjacent boundaries simultaneously, using the equation solver on a calculator. For accuracy, you should use exact values (fractions), unless the decimals are exact. The vertices of the feasible region are at

The optimum is .

at

However, the original problem also had the restriction that and are both integers.

Informally, it is usually good enough to check integervalued feasible points near the optimal vertex.

gives

Common error The optimal integer-valued point is usually near the optimal vertex. However, if the gradient of the objective line is similar to an edge of the feasible region at the optimum vertex, then the optimal integer-valued point is not necessarily near the optimal vertex. When the variables are restricted to being integers, it might be possible to use the feasible region and superimpose a grid of integer points. WORKED EXAMPLE 4.9 By using a grid of integer points, or by considering the maximum feasible value of for each feasible value of , show that the optimal vertex for the problem in Worked example 4.8 is at

.

The feasible region was graphed in Worked example 4.2.

The maximum integer value of for each integer value of , and the corresponding value of are shown.

For to , the upper boundary of the feasible region is the line .

For to , the upper boundary of the feasible region is the line .

The maximum value of .

is

at the point

EXERCISE 4B 1

The diagram shows the feasible region for a linear programming problem with two objective lines marked.

Which is the optimal vertex when the objective is to minimise ? Choose from these options. A B C D 2

The optimal vertex for a linear programming problem is The objective is to maximise

.

.

Find the maximum value of on the feasible region. 3

The vertices of the feasible region for a linear programming problem are . The objective is to maximise

.

Find the maximum value of on the feasible region.

and

4

The constraints of a linear programming problem are:

Which are feasible points? Choose from these options. A B C D 5

The borders of the feasible region for a linear programming problem are .

and

Find the coordinates of the vertices of the feasible region. 6

The objective for the linear programming problem in question 5 is to maximise a positive constant. There is more than one optimal vertex. Find the maximum value of .

7

Solve the linear programming problem: maximise subject to

and 8

.

Solve the linear programming problem: maximise subject to

and 9

.

Solve the linear programming problem: maximise subject to

and

are integers.

10 Solve the linear programming problem: maximise

where is

subject to

and

are integers.

Section 3: The simplex algorithm Key point 4.4 ​• Slack variables are non-negative variables that are added to the constraints of a linear programming problem to make inequality constraints into equations.

WORKED EXAMPLE 4.10 Use slack variables to rewrite the constraints from Worked example 4.1 as equations. becomes becomes

Slack variables are usually denoted by , , etc. or by , , etc., where these are .

and also The constraints are now:

together with four non-negativity constraints

Key point 4.5 ​• The simplex algorithm is an algebraic method that can be used to solve a linear programming problem in the form: Maximise a linear function of the variables subject to a set of constraints, each of which is of the form: a linear function of the variables ⩽ a non-negative constant and each variable is greater than or equal to . ​• Unlike the graphical method, the number of variables is not restricted to two. ​• The simplex algorithm uses a matrix formulation of the problem in which the objective and the constraints are written as equations in non-negative variables (by adding slack variables) and the coefficients are presented in tabular form as a simplex tableau. ​• To apply the simplex algorithm, the objective must be written in the form: Maximise where

a linear function of the variables a constant

and the constraints must be written in the form: linear function of the variables and a slack variable a non-negative constant. ​• The standard format for a simplex tableau has a row representing the objective followed by rows for the constraints (the non-trivial constraints). The columns represent the objective, followed by the variables and then the slack variables; the final column represents the right-hand side of the equations.

WORKED EXAMPLE 4.11 Represent this linear programming problem as an initial simplex tableau. Maximise subject to:

. Rewrite the objective in the form:

Maximise

linear function of variables constant Note that the coefficients of not .

and are now

and

subject to: Write each of the constraints as a linear function of variables ⩽ non-negative constant. ,

,

,

.

,

,

Then add slack variables to form equations.

Write the coefficients as an initial simplex tabulation. The final column represents the right-hand side of the equations. The tableau represents the matrix equation

The simplex algorithm works by carrying out iterations, as follows, until the objective row has no negative entries. ​• Choose a column with a negative entry in the objective row. The pivot row is the one for which the non-negative value of the entry in the final column divided by the positive value in the pivot column is minimised. The pivot element is the element in the pivot row of the chosen column. ​• Divide all entries in the pivot row by the value of the pivot element. ​ Add to, or subtract from, all other old rows a multiple of the new pivot row, so that the pivot column • ends up consisting of and a single .

Tip You will sometimes be told which column to choose, but if there is a choice it is usual to choose the column with the most negative entry in the objective row (or, if there is more than one such column, to choose from amongst these the one for which the corresponding column appears earliest in the tableau). Similarly, if there is more than one row giving the minimum ratio, choose any one of them.

WORKED EXAMPLE 4.12 Identify a suitable pivot for the tableau from Worked example 4.11. Pivot on the column because it has the most negative value in the objective row. Compare ratios:



so pivot on the in the third row of the column.

You do not need to consider the ratio for the last row because the entry in the -column is not positive.

Tip Always show how you worked out the ratios and state the pivot.

WORKED EXAMPLE 4.13 Carry out one iteration of the simplex tableau from Worked example 4.11.



Divide each element in the pivot row by to form the new pivot row with a in the column (pivot column). Then: replace row by row replace row by row replace row by row so that each of these rows has a in the column (pivot column).

Common error If you choose the wrong element for the pivot, then the value in the final column in the row that should have been the pivot row will be negative.

Tip To avoid rounding errors, use fractions if the values obtained are not exact decimals.

Key point 4.6 ​• A variable that corresponds to a column in the tableau that consists entirely of and a single is called a basic variable. The other variables are called non-basic variables. An iteration of the simplex tableau consists of making a non-basic variable into a basic variable and making a basic variable into a non-basic variable.

Key point 4.7 ​• To interpret the output of a simplex tableau: ​• the value of each basic variable is found by reading down the column to the and

across to the value in the final column ​• the value of each non-basic variable is .

Common error Note that the values of the variables are read from the final column and not from the objective row.

WORKED EXAMPLE 4.14 Interpret the output of the simplex tableau from Worked example 4.13.



The constraint is at its limit ( so no slack).

The basic variables are and . The values of these are found by reading down to the and then across to the final column. and . and are because the value in the final column is . and are because they are non-basic (their columns do not consist of and a single ).

Key point 4.8 ​• If the objective row has no negative entries, then an optimum solution has been achieved.

WORKED EXAMPLE 4.15 Carry out another iteration of the simplex algorithm on the tableau from Worked example 4.14 and interpret the output. Pivot on the -column. Compare ratios:

so pivot on the in the fourth row of the -column.

The value of the objective has increased to The constraints

and

.

are at their limits.

There is still a negative value in the objective row, so this is not the optimum.

Divide row by (row is unchang Then: replace row by row replace row by row row is unchanged so that each of these rows has a -column (pivot column).

WORKED EXAMPLE 4.16 a Use the simplex algorithm to find the optimum solution to the linear programming problem from Worked examples 4.11 to 4.15. Maximise subject to:

. b Investigate the effect of choosing the first pivot from a different column. a Continuing from the tableau found in Worked example 4.15, the next pivot is column row as this is the only row with a positive entry in the pivot column. The resulting tableau is:

The optimum is when All three constraints are at their limit.

and

.

and b If the first pivot is chosen from the -column, the result of the first iteration is:

Choosing the next pivot from the -column results in the same tableau as in Worked example 4.15, which then leads to the same solution as in part a. Or, choosing the next pivot from the y-column (remembering that the pivot element must be positive) gives:

The next iteration gives the same final tableau as in part a when the first pivot was from the -column. For this problem, this is the only solution and it is achieved after a total of three iterations whichever column is chosen for each of the pivot choices.

Fast forward You will use the simplex algorithm again in Chapter 6 Game theory for zero-sum games to solve problems in which two players have more than two strategies to choose from and wish to maximise their winnings. You can also use the simplex algorithm to minimise an objective, provided the variables are nonnegative and the origin is a vertex of the feasible region.

Key point 4.9 ​• To use the simplex algorithm to minimise an objective, change the objective from minimise to maximise where

.

WORKED EXAMPLE 4.17 Use the simplex algorithm to solve the linear programming problem: minimise subject to:

. Initial tableau:

and So



Pivot on column row :

and Minimum non-negative ratio is in row . Divide entries in row by to get Then:

.

replace row by row replace row by row

.

Pivot on column row :

There is still a negative entry in the objective row. Perform a second iteration. After this iteration there are no negative entries in the objective row so the optimum has been achieved.

Maximum

, so minimum

when and are both , so there is no slack in either constraint.

Tip You can use a spreadsheet to check the pivot operations.

WORK IT OUT 4.1 Set up an initial simplex tableau to represent the linear programming problem: minimise subject to:

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

EXERCISE 4C

EXERCISE 4C 1

Represent this linear programming problem as an initial simplex tableau. Maximise subject to:

2

Which element is the pivot choice for the next iteration of this simplex tableau for a maximisation problem?  

3

Does this tableau represent an optimal solution for a maximisation problem? Explain how you know.  

4

Read off the values of the variables in this tableau.  

5

Which element is the pivot choice for the next iteration of this simplex tableau for a maximisation problem?  

6

Carry out one iteration of the simplex tableau, using the 5 in column row 3 as the pivot.  

7

Use the simplex algorithm to solve the linear programming problem: maximise subject to:

. 8

a Apply the simplex method to the initial tableau shown, choosing the initial pivot from the column.  

b Apply the simplex method to the initial tableau shown, choosing the initial pivot from the column.  

c By considering the original linear programming problem, interpret the output from parts a and b. 9

Use the simplex algorithm to solve the linear programming problem: minimise subject to

. 10 Use the simplex algorithm to solve this problem: Emma is taking part in a TV quiz show. She has hours remaining before the show is filmed. Emma wants to spend some time studying for her specialist round, some time studying for the general knowledge round and some time sleeping. Emma believes that each hour spent studying for the specialist round will increase her overall score by points and each hour spent studying for the general knowledge round will increase her overall score by points. Of the time that Emma spends studying, at least must be spent on the general knowledge round. Emma needs to spend at least hours sleeping. Each hour spent sleeping will increase her

overall score by

points.

Emma wants to know how long to spend studying for each round and how long to spend sleeping to maximise the increase in her overall score. Use the simplex algorithm to solve Emma’s problem.

Checklist of learning and understanding

​•

A standard linear programming problem involves maximising or minimising an objective function that is a linear function of non-negative variables which are constrained by linear inequalities.

​•

Formulating a linear programming problem involves:

​• ​ • ​ • ​•

identifying relevant variables, including units when appropriate formulating constraints in these variables writing down an objective function and stating whether it is to be maximised or minimised.

The region where all the constraints of a linear programming problem are satisfied is called the feasible region.

​• ​•

It is useful to shade out the unwanted regions of the graph to leave the feasible region as the unshaded region.

The objective function for a linear programming problem gives a set of straight lines of fixed gradient, also called objective lines.

​•

As the value of the objective function increases, the objective line slides across the feasible region, while keeping its gradient constant.

​•

The optimum vertex is the point where the sliding objective line last touches the feasible region. This will occur at a vertex of the feasible region (or along an edge if both ends of the edge are optimal).

​•

Slack variables are non-negative variables that are added to the constraints of a linear programming problem to make inequality constraints into equations.

​•

A simplex tableau gives the coefficients of the objective and constraints of a linear programming problem when written in the form: maximise , where

a linear function of

the variables equals a constant (usually ), with each constraint written in the form: linear function of the variables plus its slack variable equals a non-negative constant, where each variable and slack variable is non-negative.

​•

The simplex algorithm involves carrying out iterations, as follows, until the objective row has no negative entries.

​•

Choose a column with a negative entry in the objective row. The pivot row is the one for which the non-negative value of the entry in the final column divided by the positive value in the pivot column is minimised. The pivot element is the element in the pivot row of the chosen column.

​• ​ •

Divide all entries in the pivot row by the value of the pivot element. Add to, or subtract from, all other old rows a multiple of the new pivot row, so that the pivot column ends up consisting of and a single .

​•

A variable that corresponds to a column in the tableau that consists entirely of and a single is called a basic variable. The other variables are called non-basic variables.

​• ​•

An iteration of the simplex tableau consists of making a non-basic variable into a basic variable and making a basic variable into a non-basic variable.

To interpret the output of a simplex tableau:

​•

the value of each basic variable is found by reading down the column to the and across to the value in the final column

​• ​•

the value of each non-basic variable is .

To use the simplex algorithm to minimise an objective, change the objective from minimise to maximise where .

Mixed practice 4 1

The feasible region for a linear programming problem has vertices The objective is to maximise

.

.

What is the optimal vertex? Choose from these options. A B C D 2

A problem is represented in this initial simplex tableau.

What is the value in the cell that is the initial pivot? Choose from these options. A B C D 3

A linear programming problem involves two variables and . The values of and must be non-negative, the total of and must be at most and must be no more than three times . Write the constraints on and as inequalities.

4

The feasible region for a linear programming problem has vertices . Solve the linear programming problem when the objective is: a maximise b minimise

5

.

Use non-negative slack variables, and , to rewrite the constraints

as equations. 6

Without doing any further work, what can you deduce from this tableau about the maximum value of ?



7

The feasible region for a linear programming problem has vertices . The objective is to maximise

.

For what values of is the vertex 8

and

optimal?

a Write out the linear programming formulation that is represented by this initial simplex tableau when the objective is to maximise .

b Which entry would be the pivot for the first iteration of the simplex method? c Carry out one iteration of the simplex method. d Interpret the tableau from part c. 9

A baker has enough flour to make loaves or buns. He has plenty of all the other ingredients. The only other restriction is the baker’s time; he has hours available to prepare the loaves and buns. A loaf takes Each loaf gives a profit of

minutes to prepare and a tray of buns takes

pence and each bun gives a profit of

The baker makes loaves and

pence.

buns, where and are non-negative integers.

The baker wants to maximise his profit. a Formulate the baker’s problem as a linear programming problem. b Find how many loaves and how many trays of buns the baker should make. c How much profit could the baker make? d Why might the profit be less than this? 10 Maximise subject to:

     and

.

minutes.

a Display the linear programming problem in a simplex tableau. b i The first pivot to be chosen is from the -column. Identify the pivot and explain why this particular value is chosen. ii Perform one iteration of the simplex method. iii Perform one further iteration. c Interpret your final tableau and state the values of your slack variables. [© AQA 2016] 11 A company producing chicken food makes three products, Basic, Premium and Supreme, from wheat, maize and barley. A tonne

of Basic uses

of wheat,

of maize and

of barley.

A tonne of Premium uses

of wheat,

of maize and

of barley.

A tonne of Supreme uses

of wheat,

of maize and

of barley.

The company has

tonnes of wheat,

The company must make at least The company makes £ tonne of Supreme.

tonnes of maize and

tonnes of barley available.

tonnes of Supreme.

profit per tonne of Basic, £

per tonne of Premium and £

per

They plan to make tonnes of Basic, tonnes of Premium and tonnes of Supreme. a Write down four inequalities representing the constraints (in addition to

).

b The company want exactly half the production to be Supreme. Show that the constraints in part a become:

c Illustrate the constraints graphically and label the feasible region. Your graph should have an -axis from to

and a -axis from to

.

d Write an expression for , the profit for the whole production, in terms of and only. e i By drawing an objective line on your graph, or otherwise, find the values of and which give the maximum profit. ii State the maximum profit and the amount of each product that must be made. [© AQA 2015] 12 This tableau has been produced after one iteration of the simplex algorithm.

The value of is positive. The tableau is not yet optimal. a What was the value in the cell that was used for the pivot in the first iteration? b Find the optimal solution, in terms of . The optimal value of is . c Find the value of . 13 Solve the linear programming problem: maximise subject to:

14 Use the simplex method to solve the linear programming problem: minimise subject to:

CROSS-TOPIC REVIEW EXERCISE 1 1

A connected graph has vertex degrees , , , . Which descriptions are true? Choose from these options.

2

A

is Eulerian.

B

is semi-Eulerian.

C

is simple connected.

D

has

edges.

A network is described by this distance matrix.

– – – – What is a least weight cycle through all the vertices? Choose from these options. A B C D 3

The network shows the capacities of a system of pipelines. a Which node is the source and which is the sink? b Find the value of the marked cuts and . c Using the values of the cuts and , what can you deduce about the maximum flow through the network?

4

A paintballing battle takes place in a forest. The tracks through the forest are represented by the arcs in the network shown.

The nodes correspond to ‘safe zones’ and the arc weights represent travel times between safe zones, in minutes.

a Find the quickest route from to . At the end of the battle the organisers need to check along every track for damage. b Find the minimum time needed to travel a route that starts and ends at and uses every arc at least once. 5

A graph has four vertices. The vertex degrees are , , and . a Explain why the graph must be semi-Eulerian. b What is the minimum number of arcs that the graph can have? c Explain why the graph cannot be simple-connected.

6

By graphing the feasible region, find the minimum value of , and .

subject to the constraints

7

a Carry out one iteration of this simplex tableau when the objective is to maximise .

b Interpret the resulting tableau, including whether the constraints are limiting or not. 8

Coloured paint is mixed by using different ratios of red, yellow, blue, white and black paint. The total amount of paint used must not exceed one litre. The amount of black paint must be at most

of the total.

The amount of blue paint must be more than twice the amount of yellow paint. The desirability value of a mixed colour is given by the amount of blue plus the amount of black minus the amount of red. The desirability is to be as large as possible. Formulate a linear programming problem to represent this information. 9

A document, which is currently written in English, is to be translated into six other European Union languages. The cost of translating the document varies as it is harder to find translators for some languages. The costs, in euros, are shown in the table below.

Danish

English

French

German

Hungarian

Italian

Spanish

Danish

English

French

German

Hungarian

Italian Spanish

a i On a copy of the table, showing the order in which you select the arcs, use Prim’s algorithm, starting from , to find a minimum spanning tree for the graph connecting , , , , , and . ii Find the total cost of using the arcs in your minimum spanning tree. iii Draw your minimum spanning tree. b It is given that the graph has a unique minimum spanning tree. State the final two edges that would be added to complete the minimum spanning tree in the case where: i Prim’s algorithm, starting from , is used ii Kruskal’s algorithm is used. [© AQA 2014] 10 Six towns are linked by train routes. The costs £ of using a direct train service between pairs of towns are shown in the cost matrix. A dash means that there is no direct service, so, for example, to travel from to would involve travelling to one of the other towns first.

a i Construct a minimum spanning tree for the network. ii Give the total weight of the minimum spanning tree. iii Give the cost of the cheapest route from to using only arcs from the minimum spanning tree. b i Find the cheapest route from to and its cost. ii Find the cheapest route from to .

The costs of the cheapest (direct or indirect) routes are shown.

c

i Apply the nearest neighbour algorithm, starting at , to find the cost of a closed route through all the vertices. ii By initially reducing the network by removing vertex , find a lower bound for the least cost closed route through the six towns.

d Using only your answers to part c, what can you deduce about the cheapest cost route that travels through each town at least once, starting and ending at ? 11 The network shows a system of pipes from to and their capacities. a Explain why: i the flow in arc

cannot be

ii the flow in arc

cannot be

b i Describe a flow of

.

from to .

ii Show that this is a maximum flow by finding a cut of value

.

12 The network shows a system of pipes with lower and upper capacities for each pipe, in litres per second.

a i Find the value of the cut . ii Hence state what can be deduced about the maximum flow from to . b The diagram shows a partially completed feasible flow of

litres per second from to .

What are the missing flows on the arcs c

,

and

?

i Using your feasible flow from part b as an initial flow, draw a diagram to indicate potential increases and decreases of the flow along each arc. ii Use flow augmentation to find the maximum flow from to . You should indicate any flow augmenting paths in a table and modify the potential increases and decreases of the flow on your network. iii State the value of the maximum flow and draw a diagram to show a maximum flow. [© AQA 2016]

13 A graph has adjacency matrix:

a Apply Euler’s formula to find the number of regions, given that the graph is planar. b i Use Kuratowski’s theorem to show that the graph is planar. ii Verify that the graph is planar by drawing it with no arcs crossing. c Describe an isomorphism between the graph and a triangular prism. 14 Two graphs are shown.

The adjacency matrices for two more graphs are given.

0

1

1

1

0

0

1

1

1

1

1

0

2

1

0

1

0

0

1

0

1

2

0

0

1

1

0

0

1

2

1

1

0

0

1

1

1

1

0

0

0

0

1

1

0

1

Graph 3

0

2

0

0

Graph 4

a Show that two of the graphs are isomorphic. b Show that all other pairs are non-isomorphic. 15 a Write out the linear programming problem that is represented by this initial simplex tableau when the objective is to maximise .

b Carry out one iteration of the simplex method. The final tableau is:

c Write down the optimal value of and the values of , and at the optimum vertex. 16 The network shows a system of pipes with their lower and upper capacities. The numbers in circles show an initial flow.

a What is the value of the initial flow? b Copy the graph and label the arcs to show the potential increases and potential decreases. c Augment the flow using the flow augmenting route

.

d Use further flow augmentations to find the maximum flow from to .

e Prove that this is the maximum flow. 17 When the route inspection algorithm is used on a network with two odd nodes, one shortest path needs to be found (for example,

).

When the route inspection algorithm is used on a network with four odd nodes, six shortest paths need to be found. These form three sets of two paths (for example, and ; and ;

and

).

a i How many shortest paths need to be found when the route inspection algorithm is used on a network with six odd nodes? ii How many sets of three paths are formed? b i How many shortest paths need to be found when the route inspection algorithm is used on a network with odd nodes. ii How many sets of paths are formed? 18 The constraints of a linear programming problem are

where , and are slack variables. a A student claims that the first constraint means that is at most twice as large as . What should the student have said? b If

, what values can take?

The objective is to maximise

, where is a positive constant.

c Find an expression for the maximum value of when

.

d Use the constraints to deduce the maximum possible value of . e If the variables , and must be integer values, find an expression for the maximum value of when .

5 Critical path analysis

In this chapter you will learn how to: ​• construct and interpret precedence (activity) networks, using activity-on-node ​ carry out critical path analysis • ​ calculate and interpret float. • If you are following the A Level course you will also learn how to: ​• construct and interpret Gantt (cascade) diagrams and resource histograms ​ carry out resource levelling and deal with resourcing restrictions. •

Before you start… Chapter 2

You should be able to use networks.

1 A network is shown. The longest route from to that does not repeat any vertices has weight Find this route.

Section 1: Analysing precedence networks A project is made up of activities. Some activities cannot start until others have finished. Critical path analysis is used to schedule the activities efficiently.

Key point 5.1 ​• A project is represented by an activity network, in which the activities are represented by the nodes and the arcs represent events (moments in time between activities).

WORKED EXAMPLE 5.1 Stan is making a cake for a competition. The activities involved are: Switch the oven on and let it warm up. Mix the ingredients for the cake. Prepare the cake tins. Put cake mix into tins and put into oven. Mix the filling. Prepare the icing. Make the decorations. Remove cake from oven and let it cool. Put layers together using filling. Ice cake. Decorate cake. Activity cannot happen until

and are all completed.

Activity must happen after . Activity must happen after and . Activity must happen after and . Activity must happen after and . Show this information as an activity network.

A B

D

Any of can happen first, or, if Stan has helpers, they could happen at the same time. and could also be done at the start because they have no predecessors. must happen after and . must happen after , which also means that it must happen after and , but there is no need to specify this. Only immediate predecessors need to be listed. Similarly for and .

H

C E

I

F

J

G

K

Key point 5.2 ​• The nodes can be represented using box notation to show the activity name and duration, together with space for working.

Tip For example, if the duration of activity is minutes, then you would show it as:

WORKED EXAMPLE 5.2 The table shows the activities involved in a project and their immediate predecessors. Draw an activity network for this precedence table.

Activity

Duration

Immediate

(days)

predecessors

and have no predecessors. and both follow and both follow

If there are enough workers available, the time needed to complete the project is limited by the longest (maximum weight) route(s).

Key point 5.3 ​• A forward pass involves moving through an activity network and finding the earliest start and finish time for each activity. ​• When an activity has two or more predecessors they must all be finished before the activity can start. ​• The earliest start times are recorded in the left-hand lower box. WORKED EXAMPLE 5.3 Carry out a forward pass on the network from Worked example 5.2 to find the earliest start times.

could start as early as day and finish after

days.

could start as early as day and finish after

days.

cannot start until has finished, so its earliest start time is day and its earliest finish is after days. cannot start until has finished, so its earliest start time is day and its earliest finish is after days. cannot start until both and have finished, so its earliest start time is day and its earliest finish is after days.

Key point 5.4 ​• The minimum project completion time is the largest of the earliest finish times.

Key point 5.5 ​• A backward pass involves moving through an activity network from the end(s) towards the start(s) and finding the latest finish time and latest start time for each activity, without delaying the entire project. ​• The latest finish times are recorded in the right-hand lower box.

Common error The latest finish times should not be confused with earliest finish times.

WORKED EXAMPLE 5.4 Carry out a backward pass on the network from Worked example 5.3 to find the latest finish times.

The minimum project completion time is

days.

The latest finish time for and , without delaying the entire project, is day The corresponding latest start times are and has to be finished before and can start so the latest finish for is the smaller of The latest finish time for is day . The latest start time for is day . Similarly, the latest finish time for

is day and the latest start time for

Tip The notation used will usually be defined in the question. The standard format is:

is day .

and

Other options include using six cells, showing the latest start time and earliest finish time as well as the information given here.

Key point 5.6 ​• Any activity for which the difference between the earliest start time and the latest finish time is the same as the activity duration is a critical activity. ​• The critical activities form a critical path through the network. There could be more than one critical path. ​• Any delay in a critical activity will delay the completion of the project. ​• Delays in non-critical activities might or might not delay the completion of the project. ​• The float on an activity is the amount by which the difference between the earliest start time and the latest finish time of the activity exceeds the activity duration. ​• Float = (latest finish time − earliest start time) − (activity duration). ​• The float represents how long the activity could be delayed without delaying the completion of the project. Delays on more than one activity might interfere with one another. ​• Critical activities have zero float. WORKED EXAMPLE 5.5 a Calculate the float for each activity in Worked example 5.4. b The start of activity is delayed by days. What effect does this have on the minimum project completion time? a

b If is delayed by days, the minimum project completion time will increase by days to days.

The float on is days, so it can be delayed by up to days with no effect on the minimum completion time. The critical activities are then and .

EXERCISE 5A 1

An activity network includes an activity with immediate predecessors The boxes show the earliest start and latest finish times for and .

and as well as the duration for

What is the earliest start time for ? Choose from these options. A

and (and no others).

B C D 2

A partial activity network is shown. There are no other predecessors for activity .

Calculate the duration of activity . 3

An activity has duration hours. It has earliest start time

hours and latest finish time

hours.

Calculate the float for this activity. 4

Explain why, for the project in question 1, activity cannot be critical.

5

The table shows the activities involved in a project, their durations and immediate predecessors. Activity

Duration

Immediate

(minutes)

predecessors

Find: a the minimum project completion time b the critical path(s). 6

The table shows the activities involved in a project, their durations and immediate predecessors. Activity

Duration

Immediate

(days)

predecessors

a Find the earliest start time for each activity. b Calculate the minimum project completion time. c List the critical path(s). 7

A project involves seven activities, and . The table shows the durations of these activities and their earliest start times and latest finish times (in hours). Activity

Duration

Earliest start time

Latest finish time

a Write down the minimum project completion time. b Find the critical path(s). c Calculate the float for each non-critical activity. d Which activities are immediate predecessors for ? 8

For the project in question , a delay means that the duration of activity increases from hours to hours and the duration of activity increases from hours to hours. a What is the new project completion time? b Which activities are now critical?

9

The activities involved in a project, their durations, in days, and immediate predecessors are given in the table. Activity

Duration

Immediate

(days)

predecessors



Activity is not a critical activity. a Which activities are critical? b What values can take?

Section 2: Scheduling Scheduling involves timetabling activities subject to restrictions.

Key point 5.7 ​• Activities with float need not be started immediately. ​• Some of the float can be shared between two or more activities. A delay in one of these activities has an effect on how much other activities can then be delayed.

WORKED EXAMPLE 5.6 The activity network represents a project. The times are in minutes.

a i ii

Write down the critical path. List the float on each non-critical activity.

The start of activity is delayed by minutes. This is equivalent to increasing the duration of to minutes. b i ii

What is the new minimum completion time? Find the float on each non-critical activity.

Instead, the duration of activity is changed to c i ii

minutes.

What effect does this have on the project completion? How long could the start of activity be delayed without the project completion being further affected?

a i

The critical activities are

ii b i

minutes

ii

c i

The earliest finish for becomes and the earliest finish for becomes No other activities are affected. becomes critical. The latest finish for changes to to

Project completion time minutes It increases by minutes.

ii Start time of could be delayed by up to minutes.

and the latest finish for and

can start at and finish at , had minutes of float so the duration could be increased to before it had an effect on the minimum project completion time. latest finish time for . minutes float on .

There are various graphs that are useful when visualising the float and activity start and finish times. These include Gantt charts and resource histograms.

Did you know? The first cascade chart was devised in the 1890s by a Polish engineer, Karol Adamiecki. Cascade charts are now known as Gantt charts after the American engineer who, in the early 1900s, devised the version of the chart that is widely used today. The charts were originally drawn by hand, but since the advent of computer technology it has become much easier to draw them and make amendments when necessary during a project. You can investigate the usefulness of Gantt charts, and some of the software that can be used to draw them, online.

Key point 5.8 ​• A Gantt chart has a horizontal time axis and a row for each activity. Each activity is shown as a block of time. ​• Sometimes you will be told that each activity starts at its earliest start time and sometimes that each activity starts at its latest start time. ​• Float is shown using dashed lines. ​ The critical activities are sometimes shown in a single row. •

Tip For clarity, it might be appropriate to use shading or use horizontal lines through the middle of the boxes, for the activity or the float.

Tip A Gantt chart does not show interference between activities, but you can see these from the precedences in the activity network.

WORKED EXAMPLE 5.7 Construct a Gantt chart for the project in Worked example 5.6 (before the delays in parts b and c took effect). Show each activity starting at its latest start time.

For reference:

When using a Gantt chart to construct a schedule, the non-critical activities can ‘move’ horizontally within their boxes, provided the activity precedences are taken into account.

Tip Unless a question says otherwise: once an activity has been started, it must continue without interruption until it has

finished if an activity needs more than one worker, they all start together and all work on the activity without interruption until the activity has finished each worker is able to complete any of the activities all equipment and resources required are available. WORKED EXAMPLE 5.8 The number of workers required for each activity in the project in Worked example 5.7 is given in this table.

Activity Number of workers a Explain why workers cannot complete the project in

minutes.

b Construct a schedule to show how the project can be completed in a For completion in

minutes by workers.

minutes, activities and must overlap by at least minutes (earliest

finish for workers.

and latest start for

), but and together need

For reference:

b For example: Time (mins)

Worker

Worker

Worker

Worker

You can show a schedule as a diagram or a timetable, for example. Here the activities have been scheduled at their earliest start times first and then moved to reduce the number of workers where necessary. Activity uses workers.

Key point 5.9 ​• A resource histogram has a horizontal time axis and shows the resource (usually the number of workers) on the vertical axis. ​• The critical activities are usually placed along the bottom of the histogram. ​• Usually each activity will start at its earliest start time, but sometimes you will be given other conditions (for example, to start each activity at its latest start time). ​• Float is not shown on a resource histogram. ​ The individual activities need not appear as rectangular blocks, although often they will. •

Common error There must be no ‘holes’ or ‘overhanging blocks’ in a resource histogram. A resource must look like a histogram in Statistics, with vertical blocks. It can have horizontal gaps but no vertical gaps between blocks. Resource levelling involves trying to share out the resources more evenly. Usually this means trying to minimise the number of workers needed. A resource histogram (together with a Gantt chart or activity network) is useful for deciding which activities can be delayed so that the number of workers is minimised. WORKED EXAMPLE 5.9 a Represent the problem from Worked example 5.8 as a resource histogram. Show each activity starting at its latest start time for completion in the minimum project completion time. The duration of activity is reduced to 11 minutes. b Show that 4 workers can complete the project in 46 minutes. a

In this example the critical activities are , and , but requires no workers and the diagram is made neater by putting below . b Start with and , once has finished do . By the time finishes is already complete, so (and ) can start, finishing at time and can be followed by . For reference:

Activity Workers

EXERCISE 5B 1

Which graph(s) could be the shape of a resource histogram? Choose from these options. A Workers 6 5 4 3 2 1 0

5

10

Time

A B

C

D

2

A Gantt chart is shown. The times are in days.

E D C B A 0

5

Time

a How do you know that the activities are shown starting at their earliest start times? Each activity needs one team of workers. b Give a schedule that enables two teams of workers to complete the project in days.

3

The table shows the activities in a project, the duration (in hours) and the number of workers needed for each. Activity Duration Workers Explain why workers cannot complete the project in

4

hours.

The table shows the activities involved in a project, their durations, in hours, and their immediate predecessors. Activity

Duration

Immediate predecessors

a Calculate the minimum time in which the project can be completed. b What would the minimum time be if the duration of is: i

reduced to

ii

increased to

iii

hours hours

hours and the duration of is

hours?

5

Draw a Gantt chart for the project in question 4 b iii. Show each activity starting at its earliest start time.

6

The Gantt chart for a project is shown.

T S R Q P 0

1

2

3

4

5

6

7

8

9

10

11

12

13 Days

Activities and each have two immediate predecessors. a Construct a table to show the duration and immediate predecessors for each activity. Each critical activity needs one worker and each non-critical activity needs two workers. b What is the minimum time needed to complete the project for: i

three workers

ii

two workers?

The duration of activity can be reduced to day if two workers are allocated to it. c What is the minimum time needed to complete the project for:

7

i

three workers

ii

two workers?

The table shows the activities involved in a project, their durations, in minutes, and immediate predecessors. Activity

Duration

Immediate predecessors

a What is the minimum completion time for the project? b The start of activity is delayed by minutes. How much does this delay the completion of the project? Instead of the start of activity being delayed, there is a delay in the start of one of the other activities. c What is the greatest delay that there can be in the start of: i

activity

ii

activity

without the minimum project completion time being affected? 8

The table shows the activities involved in a project, their durations, in hours, their immediate predecessors and the number of workers needed. Activity

Duration

Immediate

(hours)

predecessors

The minimum project completion time is

Workers

hours.

a By considering the number of worker-hours needed, show that the project cannot be completed by workers in hours. The critical activities are

and .

b Explain why at least workers are needed to complete the project in c Find the minimum time in which workers can complete the project.

hours.

d How many hours can be saved if workers are available?

Checklist of learning and understanding

​•

A project is represented by an activity network, in which the activities are represented by the nodes and the arcs represent events (moments in time between activities).

​•

A forward pass involves moving through an activity network and finding the earliest start and finish time for each activity. When an activity has two or more predecessors they must all be finished before the activity can start.

​• ​ •

The minimum project completion time is the largest of the earliest finish times. A backward pass involves moving through an activity network from the end(s) towards the start(s) and finding the latest finish time and latest start time for each activity, without delaying the entire project.

​•

Any activity for which the difference between the earliest start time and the latest finish time is the same as the activity duration is a critical activity. The critical activities form a critical path through the network. There might be more than one critical path.

​• ​ • ​•

Any delay in a critical activity will delay the completion of the project. Delays in non-critical activities might or might not delay the completion of the project.

The float on an activity is the amount by which the difference between the earliest start time and the latest finish time of the activity exceeds the activity duration.

​• ​ •

Float = (latest finish time − earliest start time) − (activity duration). The float represents how long the activity could be delayed without delaying the completion of the project. Delays on more than one activity might interfere with one another.

​• ​•

Critical activities have zero float.

A Gantt chart has a horizontal time axis and a row for each activity. Each activity is shown as a block of time.

​•



​•

Sometimes you will be told that each activity starts at its earliest start time and sometimes that each activity starts at its latest start time.

​• ​ •

Float is shown using dashed lines. The critical activities are sometimes shown in a single row.

A resource histogram has a horizontal time axis and shows the resource (usually the number of workers) on the vertical axis.

​• ​ •

The critical activities are usually placed along the bottom of the histogram.

​• ​ •

Float is not shown on a resource histogram.

Usually each activity will start at its earliest start time, but sometimes you will be given other conditions (for example, to start each activity at its latest start time). A resource histogram has no vertical gaps (no ‘holes’ and no ‘overhanging blocks’).

Mixed practice 5 1

An activity has earliest start time hours, latest finish time hours and hours of float. What is the duration of the activity? Choose from these options. A

hours

B

hours

C

hours

D 2

hours Activity

Immediate

Duration

predecessors

(days)

What is the minimum project completion time for this project? Choose from these options. A

days

B

days

C

days

D 3

days

The activity network for a project is shown.

C

A

D

B

F

G

E The activity durations, in hours, are: a Find the earliest start time for each activity. b Find the latest finish time for each activity. c Find the minimum project completion time. d List the critical path(s). 4

a If, in question 3, the duration of activity is hours instead of hours, what changes will there be in: i

the minimum project completion time

ii

the critical path(s)?

b When the duration of activity is hours, how much extra time could be spent on each non-critical activity before it affects the minimum project completion time? 5

The table lists the activities involved in a project, their durations, in minutes, and their immediate predecessors. Activity

Duration

Immediate

(minutes)

predecessors

Which activity could be delayed by minutes without there being any change to the minimum project completion time? 6

The table lists the activities in a project, their durations, in hours, and their immediate predecessors. Activity

Duration

Immediate

(hours)

predecessors

The duration of is unknown and is recorded as . Activity is not critical. a What can be deduced about the value of ? A delay in the delivery of some of the materials needed means that the start of activity is delayed. No other activity is affected. b What is the longest delay for the minimum project completion time to remain unchanged? 7

A major project has been divided into a number of tasks, as shown in the table. The minimum time required to complete each task is also shown. Activity

Immediate

Duration

predecessors

(hours)

a Construct an activity network for the project. b Find the earliest start time for each activity. c Find the latest finish time for each activity. d List the critical path(s). [© AQA 2014] 8

The table shows the activities in a project, their durations (in hours) and their immediate predecessors. Activity

Duration

Immediate

(hours)

predecessors

a Calculate the earliest start time and the latest finish time for each activity. b Construct a Gantt chart for the project, with each activity starting at its earliest start time. Each activity requires workers. c Construct a schedule to show how workers can complete the project in 9

hours.

The diagram shows an activity network for a project. Each activity requires one worker. The duration required for each activity is given in hours.

a Find the earliest start time and latest finish time for each activity. b i ii

Find the critical path. Find the float time of activity .

c Draw a resource histogram to illustrate how the project can be completed in the minimum time, assuming that each activity is to start as early as possible. d i

Given that there are two workers available for the project, find the minimum completion time for the project.

ii

Write down an allocation of tasks to the two workers that corresponds to your answer in part d i. [© AQA 2016]

10 The activities in a project, their durations (in hours) and immediate predecessors are shown in the table. Activity

Duration

Immediate

(hours)

predecessors

a Find the minimum project completion time. b List the critical path(s). Each activity requires one worker. c Draw a resource histogram showing each activity starting at its earliest start time. There are two workers available, Sam and Cal. Activities and must be done by Cal.

Activities and can be done by Sam or by Cal or done jointly by Sam and Cal. All other activities can be done either by Cal alone or jointly by Sam and Cal. Any activity that is done jointly has its duration reduced by hour: for example, if Sam and Cal share activity , then it takes them hours (each) instead of hours if Cal does without Sam’s help. d Show how Cal and Sam can complete the project together in no more than

hours.

6 Game theory for zero-sum games

In this chapter you will learn how to: ​• ​ • ​ • ​ • ​ •

understand, interpret and construct pay-off matrices find play-safe strategies and the value of the game prove the existence or non-existence of a stable solution identify and use dominated strategies find optimal mixed strategies for a game, including use of graphical methods.

If you are following the A Level course you will also learn how to: ​• convert higher-order games to linear programming problems.

Before you start… Chapter 4

You should be able to find graphical solutions to a linear programming problem.

1 Graph these constraints, with on the vertical axis and on the horizontal axis.

Hence find the maximum feasible value of .

Section 1: Games with stable solutions Game theory has applications in gaming, in warfare and in economics. This chapter gives you an introduction to some of the more important mathematical techniques used in some of these applications. The methods that people use to make their decisions can be of interest to psychologists.

Key point 6.1 The games considered are between two players or two teams who choose between a number of strategies and declare their choices simultaneously.

WORKED EXAMPLE 6.1 For each of these games, state whether or not it is a two-person, simultaneous play game. A: Chess B: Roulette C: Rock, paper, scissors D: Football E: Snap If you are not familiar with any of these games you could look them up. Some of these games have variations that are used in other cultures. A: Two-person but not simultaneous play. B: Simultaneous play but not two-person.

The players take turns so chess is not a simultaneous play game.

C: Two-person, simultaneous play.

The classic example of the kind of game that is analysed in game theory.

D: Two-person but not simultaneous play. E: Simultaneous play but not usually two-person.

Roulette usually involves several people, each playing against the ‘house’, not against each other.

Football involves two teams but they do not declare choices. Snap usually involves more than two players.

Key point 6.2 ​• With a two-person simultaneous play game, a table can be used to record the amount that each player wins for each combination of strategies. This is called a pay-off matrix. The two players do not need to have the same number of strategies to choose from. ​• A game is usually be played repeatedly and each player aims to maximise the total amount (points or score) that they win.

WORKED EXAMPLE 6.2 Two people play rock, paper, scissors. Each chooses one of rock, paper or scissors and they show their choices at the same time. Rock wins over scissors, scissors wins over paper and paper wins over rock. The winner gets points and the loser gets points. If both show the same choice, then they get point each. Construct the pay-off matrix. Second player

The rows represent the possible strategies (choices)

rock paper scissors rock First Player

paper

for one player and the columns represent the possible strategies (choices) for the other player. The entries in the cells are of the form , where is the number of points won by the player on rows and is the number of points won by the player on columns, when that combination of strategies is played.

scissors

Many of the games that are considered can be formulated as zero-sum games.

Key point 6.3 ​• A zero-sum game is one in which, for every combination of strategies, the amount that one player wins is equal to the amount that the other player loses. ​• A game can be converted to a zero-sum game if the total amount won by the two players is the same for every combination of strategies. ​• The pay-off matrix for a zero-sum game is usually written showing just the amounts won by the player on rows. The player on columns wins the negatives of the values in the table.

WORKED EXAMPLE 6.3 a Describe how the game of rock, paper, scissors from Worked example 6.2 can be converted to a zerosum game. b Write out the pay-off matrix for this zero-sum game. a For each combination of cells, the total amount won is points . If each player pays point to play, then the

Second player rock

total amount won and lost is .

paper

scissors

rock First Player

paper scissors

For example, if both play scissors, they each pay point and they each get point back, so the net gain is points each. b

Second player rock paper scissors rock First Player

The pay-off matrix shows the number of points won for the player on rows.

paper scissors

In the game of rock, paper, scissors, there is no obvious winning strategy and the play comes down to trying to double guess your opponent. In other games, some choices will be better than others. WORKED EXAMPLE 6.4

Is there a best strategy for player or for player for the zero-sum game with the pay-off matrix shown?

Player

Player

No strategy is always best for either player.

Remember that player wins the negative of the values in the table.

You might have an opinion about which strategy you would choose if you were player or player

For example, player could argue that gives the maximum possible score of (and also the maximum row total), but if often plays this would be a poor choice for .

, but what you choose depends on whether you are an optimist or a pessimist. A pessimist would usually consider the worst possible outcome and then select their strategy accordingly.

For example, player could argue that gives a worst possible score of , while with the score could be .

Key point 6.4 ​• The play-safe strategy for either player is the strategy for which the worst possible score is best (the least worst strategy). There could be more than one play-safe strategy for either player. ​• In a zero-sum game: ​• for the player on rows, the play-safe strategy is the row for which the row minimum is largest ​• for the player on columns, the play-safe strategy is the column for which the column maximum is smallest. WORKED EXAMPLE 6.5 a Find the play-safe strategies for the zero-sum game from Worked example 6.4, with this pay-off matrix. Player

Player

b What would happen if player knew that player would play safe? c What would happen if player knew that player would play safe? a

Row min.

Col. max. The play-safe strategies for are and ; the play-safe strategy for is . b If knew that would play safe, then would play .

c If knew that would play safe, then would have to guess whether was going to play or . If plays , then would be best playing (but then would change to ). If plays , then would be best playing .

The row minimum is greatest in rows and . If plays or , the worst possible outcome (for ) is to lose points. The column maximum is least in column . If plays , the worst possible outcome (for ) is to lose points. Remember that player wins the negative of the values in the table.

would play . would win points, which is better than losing (by playing ) or losing (by playing ). would play either or . Remember that player wins the negative of the values in the table.

Key point 6.5 ​• Sometimes the pay-offs mean that if either player chooses to use their play-safe strategy, then the best option for the other player is also to use their play-safe strategy. Such a game is called a stable game. ​• A zero-sum game is stable if the row maximin value is equal to the column minimax value. ​• If both players play safe, then neither of them will have any incentive to change.

Common error For a game that is not zero-sum, the play-safe strategy for the player on columns corresponds to the column maximin of the scores for the player on columns. With a zero-sum game, the scores for the player on columns are the negatives of the entries in the table so the column maximin of the table showing the scores for the player on columns is the same as the negative of the column minimax of the table showing the scores for the player on rows. For a zero-sum game, the row maximin and column minimax should be used; the game is stable if the row maximin equals the column minimax.

Key point 6.6 ​• When a game is stable, the value of the row maximin equals the value of the column minimax. This is called the value of the game for the player on rows. The value for the player on columns is the negative of the value for the player on rows. ​• The value of a stable game is the amount the player can expect to win per game, on average, even if the other player plays in the best way possible.

Common error The play-safe strategies are the labels of the chosen rows and columns and not the values of the row minima or column maxima.

WORKED EXAMPLE 6.6

Which of these games are stable? a

b

c

d

a

Row min.

Col. max. Row maximin Game is stable.

column minimax

b

Row min.

Col. max.

Minimum value in row is and in row is so row maximin ; play-safe strategy for player on rows is . Maximum value in column is , in column is and in column is . and , so column minimax ; play-safe strategy for player on columns is .

Row maximin , column minimax Game is unstable.

c

Minimum value in row is ; minimum value in row is . so row maximin  ; playsafe strategy for player on rows is . This is shown by . Maximum value in column is ; maximum value in column is . so column minimax ; play-safe strategy for player on columns is . This is also shown by . The value of the game is .

Row min.

Minimum value in row is ; minimum value in row is . so row maximin ; playsafe strategy for player on rows is .

Maximum value in column is ; maximum value in column is . so column minimax ; play-safe strategy for player on columns is .

Col. max. Row maximin , column minimax Game is unstable. d

Row min.

Col. max. Row maximin

, column minimax

Minimum value in row is , in row is and in row is . Row maximin . , and are all play-safe strategies for player on rows. Maximum value in column is , in column is and in column is . Column minimax . , and are all play-safe strategies for player on columns.

Game is unstable.

EXERCISE 6A 1

A zero-sum game has this pay-off matrix. Tim

Sue

How many points does Tim win when Sue plays strategy and Tim plays strategy ? Choose from these options. A B C D 2

A zero-sum game has this pay-off matrix. Bel

Andy

Which is the play-safe strategy for Andy? 3

For the zero-sum game in question 2, if both players use their play-safe strategy how many points does Bel win?

4

The matrix shows the pay-offs for a game. The entries are of the form

, where is the score for

the player on rows and is the score for the player on columns. Player

Player

Explain why this game cannot be converted to a zero-sum game. 5

Prove that this zero-sum game is stable.

6

Alice and Bob play a card game. Each has three cards. Alice’s cards are labelled , and . Bob’s cards are labelled , and . Alice and Bob each select a card and reveal their choices simultaneously. The player whose card has the larger value scores the value on their card, the other player scores the difference between the values on the two cards. a Write the pay-off matrix to show the scores for Alice and Bob for each combination of strategies. Put Alice’s options on rows and Bob’s on columns. b Show that the sum of the play-offs is not the same for all combinations of strategies.

7

Find the play-safe strategy for each player for this game. Sandi

Rupi

8

The pay-off matrix for a game is shown. Fionn

Gina

The game can be converted to a zero-sum game. a Write down the pay-off matrix for the zero-sum game, showing the values for Gina only.

b Determine whether this game is stable or unstable. 9

The pay-off matrix for a zero-sum game is shown.

Determine any values of for which the game is stable.

Section 2: Mixed strategies Sometimes you can remove an entire row or column because a player could always do better by making a different choice. WORKED EXAMPLE 6.7 The pay-off matrix for a zero-sum game is shown.

Player

Player

a Player never wins points by playing . Explain why this could still be a winning strategy. b Identify a strategy that player should never choose. Explain your reasoning. c Is there a strategy that player should never choose? Explain your reasoning. a If plays , then ’s best choice is , because every other strategy would result in losing points. b

should never play

is likely to choose as then will either win points or score , no matter what chooses.

because gives a better

score whatever B chooses. : : : c No, for example, is better than when plays but worse when plays is better than when plays but worse when plays is better than when plays but worse when plays .

If chooses : If chooses : If chooses :

for but for but for but

. . .

Remember that player wins the negative of the values in the table.

Key point 6.7 ​• If every entry in a row is greater than the corresponding entry in another row, the first of these rows is said to dominate (or strictly dominate) the second of these rows. The player on rows would never choose a row that is dominated by another. ​• If every entry in a row is greater than or equal to the corresponding entry in another row, and the two rows are not the same, the first of these rows is said to weakly dominate the second of these rows. There is no advantage for the player on rows in choosing a row that is weakly dominated by another. ​• Similarly, one column dominates another if the pay-off for the player on columns is better in every row (remembering that in a zero-sum game the player on columns wins the negative of the value in the table), or weakly dominates if they are better or equal (and not all equal).

Common error A row is only dominated by another if the pay-off in each column is less than or equal to the corresponding value in the dominating row. A row is not dominated if the comparison in one column is with one row and in another column is with a different row. You can reduce a pay-off matrix by removing a dominated row or column. WORKED EXAMPLE 6.8 Use dominance to reduce the pay-off matrix from Worked example 6.7. In Worked example 6.7, row dominates row . ,



A dominated row (or column) can be removed and the payoff matrix reduced.

Key point 6.8 ​• For an unstable game, the best tactic, over a long run of games, is to use a mixed strategy. ​• A mixed strategy means that the players use random numbers to choose between the strategies, playing them with probabilities that are calculated to maximise the minimum expected pay-off.

WORKED EXAMPLE 6.9 An unstable zero-sum game has the pay-off matrix shown.

Cheryl

Darren

Darren decides to flip a fair coin to choose whether to play strategy or strategy . a How many points can Darren expect to win when Cheryl plays: i strategy ii

strategy ?

b Which strategy should Cheryl choose? a i

Equally likely to lose or gain .

ii

Equally likely to gain or lose .

b

Cheryl expects to lose she chooses .

points if she chooses and lose point if

WORKED EXAMPLE 6.10 For the game in Worked example 6.9, suppose that, instead of flipping a coin, Darren uses random numbers to choose whether to play strategy or strategy and chooses with probability (where ) and with probability

.

a How many points can Darren expect to win when Cheryl plays: i strategy ii

strategy ?

b For what values of does Darren expect to win more points when Cheryl plays strategy rather than strategy ? c What value of should Darren use and how many points does Darren expect to win in this case? a i



Lose with probability and gain with probability .

ii



Gain with probability and lose with probability .

b

when

Darren’s expected gain when Cheryl plays expected gain when Cheryl plays .

so c

when

When

Darren should use random numbers to choose between and , so that the probability of choosing is

.

Darren’s expected gain is

, whether he

chooses or . If

Darren’s expected gain is less than

when

Cheryl chooses .

Darren expects to win

If

points.

Cheryl chooses .

, Darren’s expected gain is less than

when

Key point 6.9 ​• To find a mixed strategy for the player on rows choosing between two strategies: ​• let the probability of playing the strategies be and ​• ​ • ​ • ​ •

calculate the expected pay-off, in terms of , when the other player uses each of their options sketch a graph showing the expected pay-off against for the feasible region is the area below all the expected pay-off lines for find the maximum feasible value of the minimum expected pay-off by sliding a horizontal profit line up across the feasible region until it reaches the highest point.

Tip Imagine scanning across the graph from to , and for each value of using the minimum expected pay-off. This lower boundary on the graph gives the row minimum for each value of , and the highest point on the lower boundary corresponds to the row maximin. The value of where the lower boundary is highest is the optimal choice for the mixed strategy. This value could occur at one of the extremes,

or

, but more often it will be at a value between these

extremes and you can use simultaneous equations to calculate the optimum value of .

Tip The highest point on the boundary might be at one end ( point where the lines cross. WORKED EXAMPLE 6.11

or

). It is not necessarily at the

Find the optimal mixed strategy for the player on rows for this zero-sum game.

There is no need to check that the game is unstable.

Choose randomly between the rows so that row is played with probability and row with probability .

Row min.

Expected pay-off for rows when columns plays is

.

Expected pay-off for rows when columns plays is . Col. max. Row maximin , column minimax Unstable game.

This is the graph from the answer to the question in the Before you start ... section at the start of this chapter. A sketch graph is sufficient. Maximum value occurs at highest point: . For example, use single-digit random numbers: let to mean ‘play row ’ let to mean ‘play row ’.

when Choose row with probability with probability

and row

WORKED EXAMPLE 6.12 Find the optimal mixed strategy for the player on rows for this zero-sum game.

Choose randomly between the rows so that row is played with probability and row with probability

.

Expected pay-off for rows when columns plays is

There is no need to check that the game is unstable. Row min.

. Expected pay-off for rows when columns plays is .

Col. max.

Expected pay-off for rows when columns plays is

Row maximin Unstable game.

.

column minimax

Expected pay-off for rows when columns plays is .

The maximin occurs at the highest point of the lower boundary. This is at the intersection of the expected pay-offs for and .

If each player plays optimally, the player on rows expects to win points per game. If the player on columns does not play optimally, then the player on rows expects to win more than points per game. when

.

Choose row with probability and row with probability

.

Tip If a game is stable, the optimal value of is (if the first row is the play-safe) or (if the second row is the play-safe). The mixed strategy is to always use the play-safe.

Tip If a column is dominated by another, then, on the graph, the line for the dominated column is entirely above the line for the other column. The line for the dominated column does not form part of the lower boundary. However, a line can be entirely above the lower boundary without the corresponding column necessarily being dominated by another. To find an optimal mixed strategy for the player on columns, you could reformulate the problem with that player on rows, but if you already know the optimal expected value of the game for the player on rows, then you can use another method.

Key point 6.10 ​• For a zero-sum game, the optimal expected value of the game for the player on columns is the negative of the optimal expected value of the game for the player on rows.

WORKED EXAMPLE 6.13

The optimal expected value for Darren of the game in Worked example 6.9 is with probability

, by playing strategy

.

Cheryl

Darren

Use this information to find the optimal mixed strategy for Cheryl. The optimal mixed strategy for Cheryl will give Cheryl an expected value of

.

Probability(Cheryl chooses strategy )

.

Expected pay-off for Cheryl if Darren plays strategy is

Remember that Cheryl wins the negative of the values in the table. Cheryl chooses randomly between strategies and , choosing strategy with probability and strategy with probability . Or calculate the expected pay-off for Cheryl if Darren plays strategy .

. , when

.

Then

,when

.

Cheryl should choose with probability

and with probability

.

WORKED EXAMPLE 6.14 Find the optimal mixed strategy for the player on columns for the zero-sum game from Worked example 6.12.

When the player on columns plays with their optimal mixed strategy, the expected value will be per game. The optimum solution for the player on rows was at the intersection of the expected payoffs for strategies and . The optimal mixed strategy for the player on columns will be a mixture of strategy and strategy .

The optimal mixed strategy for the player on rows is to choose with probability and win an expected value of per game.

Player on columns plays with probability and with probability .

and are played with probability .

Expected pay-off for columns if rows plays is

.

Expected pay-off for columns if rows plays is .

Remember that the player on columns wins the negative of the values in the table.

, when Player on columns should choose with probability

and with probability

You can solve higher-order games, where both players have more than two strategies to choose between, using linear programming.

Key point 6.11 ​• To formulate the problem of finding a mixed strategy as a linear programming problem: ​• add a constant throughout the pay-off matrix to make all the pay-offs non-negative ​• let the probability of playing the options be ​• calculate the expected pay-off, in terms of of their options ​• set up a linear programming problem: maximise subject to

(or

, when the other player uses each

constant that was previously added)

expression for expected pay-off for each option

and .

Tip You could then use the simplex algorithm to find the values of The reason for adding a constant is to ensure that the minimum expected pay-off, , is non-negative. The reason for maximising variable in the constraints.

, rather than maximising , is that appears as a

The reason for expected pay-off is so that the maximum is the value on the upper boundary for each set of probabilities. The reason for writing the total probability as in the simplex tableau with right-hand side (

(instead of ) is so that there is a row ) non-zero. Without this the value of the

objective would never increase. At the optimum, there will be no slack in this constraint because, otherwise, can still be increased. In practice, the solution would usually be found by using a simplex package on a computer.

Rewind You used the simplex algorithm in Chapter 4.

WORKED EXAMPLE 6.15 Set up the initial simplex tableau to find the optimal mixed strategy for the player on rows for this zero-sum game.



Add throughout the table to make all entries nonnegative.

Maximise subject to: Choose randomly between the rows so that row is played with probability , row with probability and row with probability .

Add slack variables and set up the initial tableau. A simplex solver gives the solution as , to give an expected value of

and .

WORKED EXAMPLE 6.16 Set up the initial simplex tableau to find the optimal mixed strategy for the player on columns for the zero-sum game in Worked example 6.14.

The player on columns wins the negative of the values in the pay-off matrix.

Add throughout to make all entries non-negative.

Maximise subject to:

Choose randomly between the columns so that column is played with probability , column with probability , column with probability and column with probability .

Add slack variables and set up the initial tableau. A simplex solver gives the solution as and , to give an expected value of .

WORKED EXAMPLE 6.17 Set up the initial simplex tableau to find the optimal mixed strategy for the player on rows for this zero-sum game.



Add throughout to make all entries non-

negative.

Maximise subject to: Choose randomly between the rows so that row is played with probability , row with probability and row with probability .

Add slack variables and set up the initial tableau. A simplex solver gives the solution as and expected value of

, to give an

.

EXERCISE 6B 1

One row of this pay-off matrix dominates over another.

Which row dominates which other row? 2

Sunny and Tom play a zero-sum game. The expected pay-offs for Sunny when Tom plays each of his strategies are and . Find the optimal value of .

3

A zero-sum game has the pay-off matrix shown.

A mixed strategy for the player on rows is to be found by formulating the problem as a linear programming problem. Before the constraints can be formulated, the pay-off matrix must be adjusted. Write out a suitable adjusted pay-off matrix. You are not required to formulate the linear programming problem or set up a simplex tableau. 4

The pay-off matrix for a zero-sum game is shown. Player

Player

Player chooses randomly between the rows so that row is played with probability and row with probability . Find and simplify an expression for the expected pay-off for player when player chooses strategy . 5

Show that the zero-sum game shown can be reduced using dominance.

6

The pay-off matrix for a zero-sum game is shown. Player

Player

Player chooses randomly between the columns so that column is played with probability and column with probability . Find and simplify an expression for the expected pay-off for player when player chooses each of strategies and . 7

The pay-off matrix for a game is shown. Fionn

Gina

Explaining your reasoning carefully, deduce whether column is dominant over any other column.

8

Ben and Cindy repeatedly play a zero-sum game. The pay-off matrix for the game is shown. Cindy

Ben

Ben and Cindy both play using their optimal mixed strategy. a Find the optimal mixed strategy for Ben. b How much should Ben expect to win per game, in the long run? c How much should Cindy expect to win per game, in the long run? 9

a Set up the initial simplex tableau to find the optimal mixed strategy for the player on rows for this zero-sum game. b Find the optimal mixed strategy.

10 a Set up the initial simplex tableau to find the optimal mixed strategy for the player on columns for this zero-sum game. b Find the optimal mixed strategy.

11 Rosie and Jim play a game using cards numbered which they choose a card to play. If either player plays the card with value they win

. They are each dealt cards from points, otherwise the player whose card has

the smaller value wins points, where is the larger value minus the smaller value. The game is zero-sum. Suppose that Rosie has been dealt the cards

(so Jim has

).

a Show that the game is not stable. b Set up the initial simplex tableau to find the optimal mixed strategy for Rosie. The final tableau is

where

(play card )

c What is the optimal mixed strategy for Rosie?

Checklist of learning and understanding

​•

​•

A pay-off matrix can be used to show the amount that each player wins for each combination of strategies in a two-person simultaneous play game. The two players need not have the same number of strategies to choose from.

​•

A game is usually played repeatedly and each player aims to maximise the total amount (points or score) that they win.

A zero-sum game is one in which, for every combination of strategies, the amount that one player wins is equal to the amount that the other player loses. A game can be converted to a zero-sum game if the total amount won by the two players is the same for every combination of strategies.

​• ​•

The play-safe strategy for either player is the strategy for which the worst possible score is best (the least worst strategy). There might be more than one play-safe strategy for either player.

​• ​•

The pay-off matrix for a zero-sum game is usually written showing just the amounts won by the player on rows. The player on columns wins the negatives of the values in the table.

In a zero-sum game, the play-safe strategy for the player on rows is the row for which the row minimum is largest and the play-safe strategy for the player on columns is the column for which the column maximum is smallest.

A stable game is one in which the best strategy for each player is to use their play-safe strategy. If both players play safe, then neither of them will have any incentive to change.

​•

A zero-sum game is stable if the row maximin value equals the column minimax value. This is called the value of the game for the player on rows. The value for the player on columns is the negative of the value for the player on rows.

​•

The value of a stable game is the amount the player can expect to win per game, on average, even if the other player plays in the best way possible.

​•

A row dominates another if the pay-off for the player on rows is better in every column. The player on rows would never choose a row that is dominated by another and the pay-off matrix can be reduced.

​•

A column dominates another if the pay-off for the player on columns is better in every row (remembering that in a zero-sum game the player on columns wins the negative of the value in the table). The player on columns would never choose a column that is dominated by another and the pay-off matrix can be reduced.

​•

For an unstable game, the best tactic, over a long run of games, is to use a mixed strategy. This means using random numbers to choose between the strategies, playing them with probabilities that are calculated to maximise the minimum expected pay-off.

​•

To find a mixed strategy for the player on rows who can choose between two strategies:

​• ​ • ​ • ​ •

let the probability of playing the strategies be and calculate the expected pay-off, in terms of , when the other player uses each of their options sketch a graph showing the expected pay-off against for the feasible region is the area below all the expected pay-off lines for

​•

find the maximum feasible value of the minimum expected pay-off by sliding a horizontal profit line up across the feasible region until it reaches the highest point.

​•

For a zero-sum game, the optimal expected value of the game for the player on columns is the negative of the optimal expected value of the game for the player on rows.

​•

To formulate the problem of finding a mixed strategy as a linear programming problem:

​• ​ • ​ •

add a constant throughout the pay-off matrix to make all the pay-offs non-negative

​•

set up a linear programming problem:

let the probability of playing the options be calculate the expected pay-off, in terms of their options maximise subject to:

(or

when the other player uses each of

constant that was previously added)

expression for expected pay-off for each option and

Mixed practice 6 1

The pay-off matrix for a zero-sum game is shown.

Which is the play-safe strategy for the player on rows? Choose from these options. A B C D No play-safe strategy. 2

The pay-off matrix for a zero-sum game is shown.

The player on rows plays with probability and with probability . What is their expected pay-off when the player on columns plays ? Choose from these options. A B C D 3

Pav and Mina play a zero-sum game. Mina

Pav

a Prove that the game is not stable. b Find the play-safe strategy for each player. 4

Rose and Jim repeatedly play a zero-sum game. The pay-off matrix for the game is shown. Jim

Rose

a Explain why Rose should never play strategy . b Hence reduce the pay-off matrix. 5

The pay-off matrix for a zero-sum game is shown. Player

Player

Player chooses randomly between the columns so that column is played with probability and column with probability

.

a Find and simplify an expression for the expected pay-off for player when player chooses each of strategies and . b Find the optimal mixed strategy for player . 6

Here is the pay-off matrix for the game in question 4. Jim

Rose

a Find the optimal mixed strategy for Rose. b Find the optimal mixed strategy for Jim. 7

Set up the initial simplex tableau to find the optimal mixed strategy for the player on rows for this zero-sum game.

8

a Two people, Adam and Bill, play a zero-sum game. The game is represented by the following pay-off matrix for Adam. Bill

Strategy

Adam

i

Show that this game has a stable solution.

ii

Find the play-safe strategy for each player.

iii State the value of the game for Bill. b Roza plays a different zero-sum game against a computer. The game is represented by the following pay-off matrix for Roza. Computer Strategy Roza

i

State which strategy the computer should never play, giving a reason for your answer.

ii

Roza chooses strategy

with probability . Find expressions for the expected gain for

Roza when the computer chooses each of its two remaining strategies. iii Hence find the value of for which Roza will maximise her expected gain. iv Find the value of the game for Roza. [© AQA 2012] 9

Two people, Roz and Colum, play a zero-sum game. The game is represented by the following pay-off matrix for Roz. Colum Strategy

Roz

a Explain what is meant by the term ‘zero-sum game’. b Determine the play-safe strategy for Colum, giving a reason for your answer. c i ii

Show that the matrix can be reduced to a by matrix, giving the reason for deleting one of the rows. Hence find the optimal mixed strategy for Roz. [© AQA 2012]

10 Rob and Chas play a zero-sum game. The pay-off matrix for the game is shown. Chas

Rob

a Formulate the problem of finding the value of the game as a linear programming problem. b Find the value of the game. 11 Dan and Jenny play a zero-sum game. The pay-off matrix for the game is shown. Jenny

Dan

Dan chooses randomly between his strategies so that

,

and

.

a Write down, in terms of , and , the expected pay-off for Dan when Jenny plays each of her strategies. b Use the fact that

to eliminate from each of the expressions from part a.

In the optimal mixed strategy for Dan,

,

and

.

c Find the value of the game. d Find the optimal mixed strategy for Jenny. 12 Kaz and Trev play a zero-sum game. The pay-off matrix for the game is shown. Trev

Kaz

Kaz chooses randomly between her strategies, choosing strategy with probability , strategy with probability , strategy with probability and strategy with probability . a Formulate the problem of finding the value of the game as a linear programming problem. In the optimal mixed strategy and are both . b Find the values of and . c Find the optimal mixed strategy for Trev.

13 In a zero-sum game, the player on rows chooses between different strategies and the player on columns chooses between different strategies. The problem of finding an optimal mixed strategy for the player on rows is set up as an initial simplex tableau. How many rows and how many columns does the simplex tableau have? 14 A zero-sum game has the pay-off matrix shown. Strategy

a i ii

Find the row maximin and column minimax, in terms of where necessary. Prove that the game is never stable.

iii Explain why can never be a play-safe strategy. b i ii

Find the range of values of for which row is dominated by another row. Find the optimal mixed strategy for the player on rows when the condition in part b i holds.

iii Find the optimal mixed strategy for the player on columns in this case.

7 Binary operations

In this chapter you will learn how to: ​• understand and use binary operations, including examples using modular arithmetic and matrix multiplication ​• ​ • ​ • ​ • ​ •

use and prove the commutativity of a binary operation use and prove the associativity of a binary operation construct a Cayley table for a given set under a given binary operation prove the existence of an identity element for a given set under a given binary operation find the inverse of an element belonging to a given set under a given binary operation.

Before you start… GCSE and A Level Mathematics Student Book 1

You should be familiar with the terms integer, rational number, real number and the notation for these sets.

1 Which of these numbers belong to the set , the set of positive integers?

Further Mathematics Student Book 1, Chapter 7

You should know how to carry out matrix addition and matrix multiplication.

2 Calculate a b

Section 1: Properties of binary operations Key point 7.1 ​• A binary operation is any mathematical procedure that has two inputs and one output. For a binary operation •:

In ordinary arithmetic, addition, subtraction, multiplication and division (by a number that is not zero) are all examples of binary operations. WORKED EXAMPLE 7.1 Which of these are binary operations for positive integers and ? a b c d e f All except d and f are binary operations.

d has only one input value. f has three input values.

Some binary operations have standard symbols (such as addition: or some other such symbol to denote the operation.

). In other cases, you can use

Common error The symbols used for binary operations could have different meanings in different questions, but will have a fixed meaning within a question.

Key point 7.2 ​• The inputs for a binary operation will usually both come from the same set. When the output is also from this set (for any valid inputs), the binary operation is said to be closed on this set. ​• It is not necessary to be able to achieve every element of the set as an output, but every output needs to belong to the set.

WORKED EXAMPLE 7.2 Which of the binary operations addition, subtraction, multiplication and division are closed on the set of positive integers? Addition is closed on

.

Adding two positive integers gives a positive integer.

or

Subtraction is not closed on

.

For example, and

Multiplication is closed on Division is not closed on

. .

are both positive integers but

is not.

Multiplying two positive integers gives a positive integer. For example,

.

and are both positive integers but

is not.

Although binary operations are often applied to numerical sets, they can also be applied to other types of set. WORKED EXAMPLE 7.3 A set consists of polygons with edge length unit. All such polygons with edges are denoted by . An operation combines two of the polygons by joining them along an edge (but not more than one edge) to form another polygon. a What is

?

b What is

?

c Is a closed binary operation? a

Joining a quadrilateral to a triangle will create a polygon with sides (because one edge of the triangle is joined to one edge of the quadrilateral internally). For example,

b

Generalising the argument from part a.

c Yes, it is always possible to combine two such polygons along an edge to form another of the polygons.

There is always a way to avoid having to join along more than one edge.

WORKED EXAMPLE 7.4 A set

consists of matrices of the form

, for real numbers .

a Is the operation of matrix addition closed on the set

?

b Is the operation of matrix multiplication closed on the set

?

a

Not closed, the entry in the cell in the second row, second column becomes , so the sum does not have the required form.

b

Closed, for all real set .

the product is another matrix in the

The positioning of the s means that, in this case, the product is particularly simple.

Key point 7.3 ​• For an integer , the residue modulo N (where is an integer

) is the remainder when

is written as an integer multiple of plus the remainder , where ​ For example, • ​ Even numbers have residue modulo and odd numbers have residue modulo . • ​• Modular arithmetic (sometimes called clock arithmetic) is arithmetic carried out on the set of residues modulo , for some given positive integer .

Common error Addition modulo is the same as addition followed by reducing modulo . For example, because . The multiples of modulo are , … . Similarly for multiplication modulo ; for example, of modulo are

because

. The powers

, … .

WORKED EXAMPLE 7.5 a Show that addition modulo , for any given positive integer

, is a closed binary operation.

b Show that multiplication modulo is also a closed binary operation. a A positive integer

means that

for some positive integer . Adding any number with residue to any number with residue gives a number with residue . b

Multiplying any number with residue by any number with residue gives a number with the same residue as .

Did you know? Modular arithmetic forms part of a branch of Mathematics called number theory. Number theory includes results such as Fermat’s last theorem as well as results about remainders and residues.

Key point 7.4 ​• A binary operation o is commutative if

for all

.

This means that swapping the order of the inputs does not change the output. ​• For example, addition of real numbers is commutative but subtraction is not. but, for example,

.

Tip You can use the commutativity of addition and multiplication of real numbers without proof.

WORKED EXAMPLE 7.6 Prove that multiplication of matrices of the form

, for real numbers , is commutative.

Find the product of two general matrices of the required form and then find the product with the inputs swapped. and are real numbers so

.

Multiplication of real numbers is commutative.

Hence,

Multiplication of matrices of this form is commutative.

In general, matrix multiplication is not commutative, but it is commutative for matrices of this form.

Key point 7.5 ​• A binary operation o is associative if

for all

, .

Inserting or removing brackets does not change the output. ​• For example, addition of real numbers is commutative but subtraction is not. , so writing is unambiguous but, for example, while

.

Tip You can use the associativity of addition and multiplication of real numbers without proof. You can use the associativity of matrix multiplication without proof.

WORKED EXAMPLE 7.7 Prove that the operation , used in Worked example 7.3, is associative.

Use the associativity of addition of real numbers to write as

The answer is the same as for . The operation is associative on this set.

EXERCISE 7A

EXERCISE 7A 1

A binary operation is defined on the set

as follows.

​• When both input letters are the same, the output is the same as the input letter. ​ When the input letters are different, the output is the third letter. • For example, What is

and

.

?

2

Is the operation in question 1 commutative?

3

By considering

4

Show that subtraction modulo is not commutative.

5

A set

, show that the operation in question 1 is not associative.

consists of matrices of the form

, for real numbers .

a Show that the operation of matrix addition is closed on the set

.

b Show that the operation of matrix addition is commutative on the set 6

A binary operation is defined on a Calculate

and

by

.

.

.

b Prove algebraically that

for all

.

c Show that the operation is not associative. 7

A binary operation is defined for integers

by

, if , if

.

a Prove that the operation is commutative. b Prove that the operation is not associative. 8

A binary operation is defined on . a Find the possible values of b Given that

by

for some constant

.

, for all , what can you deduce about the value of ?

c Given that the operation is commutative, show that it is also associative.

Section 2: Using Cayley tables Key point 7.6 ​• When a binary operation is closed on a (small) finite set, the operation can be listed using a Cayley table. The rows represent the first input, , the columns represent the second input, , and the entries in the table show

.

Common error The entry in row and column shows and the entry in row column shows This only matters when the operation is non-commutative.

.

Did you know? Arthur Cayley (1821–1895) was a British mathematician who excelled in many subjects at school. He graduated top of his year in mathematics at Cambridge University and became a fellow of the University. However, he was not prepared to take holy orders (a condition of fellowship at that time) and instead trained to be a lawyer. Cayley published prolifically on a wide range of mathematical topics. He is mostly remembered today for his work on matrices.

Fast forward You will use Cayley tables again in Chapter 8 when you look at groups.

WORKED EXAMPLE 7.8 a Construct a Cayley table to show the result of multiplying modulo . b Explain how the Cayley table shows that multiplication modulo is commutative. a The residues modulo are and . since since since since . since

b The Cayley table is symmetrical about the leading diagonal.

For example, .

Key point 7.7 ​• An identity element for a binary operation is an element that leaves the other input unchanged. ​• If is an identity element, then for all . ​• The identity for addition of real numbers is ; the identity for multiplication of real numbers is . WORKED EXAMPLE 7.9

. . .

a Use the Cayley table from Worked example 7.8 to find the identity element. b Show that the binary operation in Worked example 7.7 has no identity. a The row for is a repeat of the column labels, so so

, for all . Similarly, for column , for all (this follows from

commutativity). The identity element is .

b

would be an identity, but a polygon must have at least edges, so there is no identity.

Key point 7.8 ​• The inverse of an element belonging to a given set under a given binary operation is the element that gives the identity as the output. ​• If is an identity element and , then is the inverse of . Under addition of real numbers, the inverse of is

because

.

Under multiplication of real numbers, the inverse of is , provided , because . The number has no inverse under multiplication.

Fast forward You will use identity elements and inverses again in Chapter 8 when you look at groups.

Key point 7.9 ​• If an element is its own inverse it is called self-inverse.

WORKED EXAMPLE 7.10 Use the Cayley table from Worked example 7.8 to find the inverse of each element. has no inverse. is self-inverse. The inverse of is . The inverse of is . is self-inverse.

The identity is .

EXERCISE 7B 1

A binary operation is defined on the set of real numbers by

.

What is the identity? Choose from these options. A B C D 2

What is the inverse of under the operation in question ?

3

Construct a Cayley table for the binary operation defined on .

4

a Construct a Cayley table for the operation of addition modulo .

by

b Deduce the identity for addition modulo . c Write down the inverse of each element under addition modulo . 5

a Construct a Cayley table for the operation defined on the set

using these rules:

​• When both input letters are the same, the output is the same as the input letter. ​ When the input letters are different, the output is the third letter. • b Explain how you can tell from the Cayley table that the operation is commutative. c Explain how you can tell from the Cayley table that the operation has no identity element. 6

A binary operation o has the Cayley table shown.

a Write down the identity element. b Write down the inverse of each element. c Find the value of

.

d Construct the Cayley table for addition modulo . e Explain how you can now deduce that the operation o is associative.

7

a Write out the Cayley table for matrix multiplication on the set ,

where

and

.

b Which element is the identity and what is the inverse of each element? c Is matrix multiplication commutative on the set? The matrix d Work out

. .

e If is included in the Cayley table, how many other matrices are needed so that the table is closed? 8

a Give an example of a binary operation that has no identity. b Give an example of a binary operation that has a right identity but no left identity (i.e. there exists an element such that for all but there is no element such that for all ). c Prove that if an element exists that is both a right identity and a left identity, then it is the only identity (i.e. if there exists an element such that for all then there is no element for which for all or for all ).

Checklist of learning and understanding

​• ​ •

A binary operation is any mathematical procedure that has two inputs and one output.

​•

For an integer , the residue modulo (where is an integer ) is the remainder when is written as an integer multiple of plus the remainder , where .

​•

Modular arithmetic (sometimes called clock arithmetic) is arithmetic carried out on the

The inputs for a binary operation will usually both come from some set. When the output is also from this set (for any valid inputs), the binary operation is said to be closed on this set.

set of residues modulo for some given positive integer

.

​• ​ • ​ •

A binary operation o is commutative if

​•

An identity element for a binary operation is an element that leaves the other input

A binary operation o is associative if

. for all

.

When a binary operation is closed on a (small) finite set, the operation can be listed by using a Cayley table. unchanged. If is an identity element, then

​•

for all

for all .

The inverse of an element belonging to a given set under a given binary operation is the element that gives the identity as the output. If is the identity and then is the inverse of . If an element is its own inverse it is called self-inverse.

,

Mixed practice 7 1

The binary operation is defined by What is the value of

? Choose from these options.

A B C D 2

The binary operation is defined on

What is the value of

by:

? Choose from these options.

A B C D 3

A binary operation is defined on the set

by the Cayley table shown.

a Which element is the inverse of ? b Show that 4

.

A binary operation ∇ is defined on

by:

for all other values. a Write out the Cayley table for . b Show that

.

c Show that

.

d Show that is not associative. 5

Explain why, for any binary operation, the identity is always self-inverse.

6

A set

consists of matrices of the form

, for real numbers and .

a Show that the operation of matrix multiplication is closed on the set

.

b Show that the operation of matrix multiplication is commutative on the set 7

A binary operation has the Cayley table shown.

a Is the operation commutative? Explain how you know. b Write down the inverse of each element. c Is the operation associative? Explain how you know. 8

A binary operation is defined by

.

a Show that is not commutative. b Prove, algebraically, that

can only take the values and .

c Prove whether or not is associative. 9

A binary operation is defined on

.

Construct the Cayley table of given that: ​• ​ •

is the identity

​• is not commutative ​ each row contains the numbers •

and .

10 Describe what happens in question 9 if the final bullet point is replaced by: ​• each column contains the numbers

and .

.

8 Groups

This chapter is for A Level students only. In this chapter you will learn how to: ​• understand and use the language of groups ​ recognise and use finite and infinite groups and subgroups • ​• identify and use a generator of a group ​ understand and use Lagrange’s theorem • ​ recognise and use isomorphism between groups of small finite order. •

Before you start… Chapter 7

You should be able to use a Cayley table and understand the meaning of the terms identity, inverse, associativity.

1 A binary operation has this Cayley table.

a Identify the identity. b Which element is the inverse of ? c Other than the identity, which element is selfinverse? d Prove, algebraically,

that for all Chapter 7

You should understand the meaning of the term commutative.

2 A binary operation has this Cayley table.

Is commutative?

Section 1: The group axioms Key point 8.1 ​• A set forms a group under a binary operation if, for all ​• ​• there exists an element ​ there exists an element • ​ • .

,

for which for which

​• The four properties: closure, existence of an identity, existence of inverses and associativity are called the group axioms. ​• The group formed using the set and the binary operation is written

.

Rewind You worked with binary operations, identity elements and inverses in Chapter 7.

WORKED EXAMPLE 8.1 Show that

is a group under multiplication modulo .

The Cayley table shows that the operation is closed. Every entry in the table is either or

is the identity. is self-inverse and is self-inverse. So each element has an inverse. Multiplication modulo is associative on since multiplication is associative on the real numbers.

and If an operation is associative on a set, then it is associative on any subset.

Key point 8.2 ​• A group for which the operation is also commutative is called an abelian group. ​• The order of a group

is the number of elements in the set .

​• The period of an element is the smallest number of repeated applications of the element to get the identity. The identity always has period .

WORKED EXAMPLE 8.2 The group

is defined on the set

by the Cayley table:

Write down: a the order of the group b the period of each element. a

There are elements in the set .

b The identity is .

Row and column for are

Period of

.

Period of

.

Period of

.

Period of

.

and .

The identity always has period .

Tip When the binary operation is multiplication, , and so on, the identity can be written as . The period of element is the smallest positive integer for which . Similarly, when the binary operation is addition, , and so on, the identity can be written as . The period of element is the smallest positive integer for which .

Common error A useful consequence of the group axioms is that each element appears once in every row and once in every column. This can be useful for showing up errors in calculating the Cayley table for a group.

Key point 8.3 ​• A subgroup of a group is the binary operation together with a subset of such that is closed under includes the identity element and if an element is in then the inverse of that element is also in . ​• Because is associative on , it is also associative on , a subset of WORKED EXAMPLE 8.3 Show that

is a subgroup of the group

from Worked example 8.2. is a subset of

The operation is closed on

The identity Inverse of

.

is the identity in the reduced Cayley table.

. and inverse of

.

.

The identity, , appears in each row and each column of the reduced Cayley table.

Associativity does not need to be checked because is associative under .

Key point 8.4 ​• A subgroup is a proper subgroup of a group subset of , i.e. the set is not the whole of .

when the set is a proper

​• The order of a proper subgroup is not since any subgroup includes the identity. ​ For a finite group , the order of a proper subgroup is less than the order of the group . • ​ The group is an improper subgroup of itself. • ​• A trivial subgroup is a subgroup consisting of an operation and the set containing just the identity. ​• A non-trivial subgroup is a subgroup that is not trivial. ​• A proper non-trivial subgroup is a subgroup that is not the whole group and does not consist of the operation and the set containing just the identity. WORKED EXAMPLE 8.4 a In Worked examples 8.2 and 8.3: i is a proper or improper subgroup of the group ii

is

a proper or improper subgroup of the group

b Write down the trivial subgroup of the group a i

is a proper subgroup of .

?

.

In Worked example 8.3, it was shown that is a subgroup of . has two elements, so the subgroup has order , whereas has order .

ii Neither. is not a subgroup of . b

The inverse of is , but this is not in

.

The trivial subgroup has order .

Key point 8.5 The symmetries of a regular polygon are the orientations of the polygon formed by rotating or reflecting the polygon onto itself. The set of all such symmetries, together with the operation ‘followed by’ form a group. This group is a symmetry group of a regular polygon. For a regular -sided polygon, the symmetry group will have rotations and reflections. This symmetry group has order .

WORKED EXAMPLE 8.5 Construct the symmetry group of a square. A square can be rotated onto itself by rotating about the centre through

or reflected onto itself by

reflecting in the four lines of symmetry,

and . Original orientation

Rotation through

Rotation through

Rotation

Reflection in axis

Reflection in axis

Reflection in axis

Reflection in axis

through

For example, o rotate through then rotate through rotate through

.

The table is closed (every entry is one of

For example, o rotate through then reflect in horizontal axis,

or ).

which has the same effect as

is the identity. Every element is self-inverse, apart from and , which are inverses of each other.

.

For example, o reflect in then rotate through which has the same effect as .

Associativity can be assumed.

WORKED EXAMPLE 8.6 Find the subgroups of order and order of the symmetry group of a square from Worked example 8.5. The trivial subgroup of order . A subgroup of order , consisting of the identity and rotation through .

Another four subgroups of order , each consisting of the identity and a reflection.

Key point 8.6 ​• A cyclic group of order consists of the identity and elements, where at least one of these has period and the others have period either or a factor of . ​• The subgroup of the symmetry group of a regular -sided polygon consisting of all the

rotations but no reflections is the cyclic group of order . ​• The group formed by addition modulo on the set order .

is also the cyclic group of

​• If a non-trivial group has no proper non-trivial subgroup, then it is a cyclic group of prime order.

WORKED EXAMPLE 8.7 a Write out the Cayley table for

from Worked example 8.6.

b Write out the Cayley table for addition modulo on the set

.

a

This can be extracted from the Cayley table in Worked example 8.5.

b

You found this in question 6 in Exercise 7B, for example.

EXERCISE 8A

EXERCISE 8A 1

What is the order of the symmetry group of a regular hexagon? Choose from these options. A B C D

2

Write out the Cayley table for addition modulo on the set

3

What is the period of each element in the group

4

What is the order of the group

5

For the symmetry group of a square in Worked example 8.5, find a proper subgroup that includes both and .

6

How many proper subgroups does the symmetry group of a square in Worked example 8.5 have?

7

The set that

8

consists of

. ?

?

matrices of the form

where

are non-zero real numbers. Show

forms a group under the operation of matrix multiplication.

a Construct the symmetry group of an equilateral triangle. b Is this group abelian?

9

a Show that any cyclic group of order has no proper non-trivial subgroup. b Find a proper non-trivial subgroup of the cyclic group



.

Section 2: Generators of a group Key point 8.7 ​• Lagrange’s theorem says that for any finite group , the order of every subgroup of divides the order of .

Tip The proof of Lagrange’s theorem uses the concept of cosets of in , but this is beyond the scope of this chapter.

Common error Note that Lagrange’s theorem does not say that a group of order will have subgroups with orders that are the factors of , only that it cannot have subgroups with orders that are not factors of . For example, a group of order might or might not have subgroups of orders and , but cannot have a subgroup of order .

WORKED EXAMPLE 8.8 Using Lagrange’s theorem, what can you deduce about the orders of the proper non-trivial subgroups of: a the symmetry group of a square (as in Worked example 8.5) b the cyclic group

?

a The order of a subgroup divides the order of the group. The group in Worked example 8.5 has order , so there could be proper non-trivial subgroups of order or but there cannot be proper non-trivial subgroups of order or .

In this case, there are subgroups of order and of order , which you found in question 6 in Exercise 8A. Lagrange’s theorem does not say that there are subgroups of order and , only that there cannot be subgroups of any other order.

b The order of the group is , which is prime, so Lagrange’s theorem says that there are no proper non-trivial subgroups.

Key point 8.8 ​• An element is called a generator of a group if repeated application of that element forms every element in the group. ​• The set of generators of an element, , under the operation , is denoted by

WORKED EXAMPLE 8.9 Show that for a cyclic group of prime order, each element generates the group, apart from the identity. The group has (finite) prime order . Let the operation be , and the identity be . If is any element of the group, and is not the identity, then

If as well, with a positive integer less than , then would have to be a factor of .

This is because, if , then .

but is not a multiple of

is odd so either is even or is even. If is even, let ; if is even, let . Then is a subgroup of order , but by Lagrange’s theorem there are no non-trivial proper subgroups. So are all different and hence generates the group.

If

, with

, then

.

An example of this result was in question 9 in Exercise 8A, where



with identity , so each of the elements is a generator of the group. WORKED EXAMPLE 8.10 Show that no element of the symmetry group of a square, from Worked example 8.5, generates the entire group. No element has period 8, so no element generates the entire group. and , so Similarly, for

and

.

, so Similarly, for

. and

.

Each element generates a subgroup.

Key point 8.9 ​• Two groups of finite order are isomorphic if they have the same structure. There is a correspondence between the elements of one group and the elements of the other. ​• You can write ‘ is isomorphic to ’ as ‘ ’. ​• If and are finite and then and have the same number of elements (have the same order). ​• Every group of order is isomorphic to . ​• Every group of order , where is prime, is isomorphic to the cyclic group of order . ​• If a generator of each group can be found, then the correspondence can be between successive powers of the generators.

Tip Every group of order is either isomorphic to the cyclic group of order or to the subgroup

of the symmetry group of a square first used in Worked example 8.5, then used in Worked example 8.10.

This is the smallest non-cyclic group.

WORKED EXAMPLE 8.11 a Show that the groups not isomorphic. b Show that the subgroups

from Worked example 8.1 and and

from Worked example 8.2 are

, from Worked example 8.5, are not

isomorphic. c Show that the groups isomorphic.

and

, from Worked example 8.7, are

a These groups have different orders so they cannot be isomorphic. b Both groups have order , but in self-inverse, whereas in inverse.

only and are all four elements are self-

has one subgroup of order whereas has three subgroups of order 2.

c

is the identity, with period . has period . and each have period or

has order and has order 4.

is the identity with period has period . and each have period . or

Set up the correspondence.

The Cayley tables are the same with this relabelling. and and

EXERCISE 8B

EXERCISE 8B 1

A group has order . Which could not be the order of a subgroup? Choose from these options. A B C D

2

Construct the group generated by under addition modulo .

3

Use Lagrange’s theorem to show that the group generated in question 2 is isomorphic to the cyclic group of order .

4

a Construct the symmetry group of a rectangle that is not a square.

b Show that this group is not isomorphic to a cyclic group. 5

a Construct the group

under multiplication modulo

.

b Show that this group is isomorphic to one of the groups from question 4. 6

Find all the proper non-trivial subgroups of the cyclic group of order .

7

a Construct the Cayley table generated by the element under the operation

8

b Verify that

is a group.

c Show that

is isomorphic to the cyclic group of order .

.

a Construct the Cayley table generated by the element under the operation , where . b Verify that the operation does not form a group.

Checklist of learning and understanding

​•

A set forms a group under a binary operation if, for all

​• ​ • ​ • ​ • ​ •

there exists an element

for which

there exists an element

for which

,

. The four properties: closure, existence of an identity, existence of inverses and associativity are called the group axioms.

​• ​ • ​ •

The group formed using the set and the binary operation is written

.

A group for which the operation is also commutative is called an abelian group. The order of a group

is the number of elements in the set .

The period of an element is the smallest number of repeated applications of the element to

​• ​ •

get the identity.

​• ​ •

Because is associative on , it is also associative on , a subset of .

A subgroup of a group is the binary operation together with a subset of such that is closed under includes the identity element and if an element is in then the inverse of that element is also in . A subgroup

is a proper subgroup of a group

when the set is a proper

subset of , i.e. the set is not the whole of .

​• ​ • ​ • ​ •

The order of a proper subgroup is not since any subgroup includes the identity. For a finite group the order of a proper subgroup is less than the order of the group . The group is an improper subgroup of itself. A trivial subgroup is a subgroup consisting of an operation and the set containing just the identity.

​• ​ •

A non-trivial subgroup is a subgroup that is not trivial.

​•

The symmetries of a regular polygon is the set of orientations of the polygon formed by rotating or reflecting it onto itself.

​•

The symmetries together with these rotations and reflections form a group called the symmetry group of the polygon.

​•

For a regular -sided polygon the symmetry group will have rotations and reflections. The order of the symmetry group is .

​•

A cyclic group of order consists of the identity and elements, where at least one of these has period and the others have period either or a factor of .

​•

The subgroup of the symmetry group of a regular -sided polygon consisting of all the

A proper non-trivial subgroup is a subgroup that is not the whole group and does not consist of the operation and the set containing just the identity.

rotations but no reflections is the cyclic group of order .

​•

The group formed by addition modulo on the set

is also the cyclic group of

order .

​•

If a non-trivial group has no proper non-trivial subgroup, then it is a cyclic group of prime order.

​•

Lagrange’s theorem says that for any finite group , the order of every subgroup of divides the order of .

​•

An element is called a generator of a group if repeated application of that element forms every element in the group.

​•

The set of generators of an element, , is denoted by

, using the

operation .

​•

Two groups are isomorphic if they have the same structure. There is a correspondence between the elements of one group and the elements of the other.

​• ​ •

You can write ‘ is isomorphic to ’ as ‘

​• ​ • ​ •

Every group of order is isomorphic to

If and are finite and same order).

then and have the same number of elements (have the .

Every group of order , where is prime, is isomorphic to the cyclic group of order If a generator of each group can be found, then the correspondence can be between successive powers of the generators.



’.

Mixed practice 8 1

A group is defined in this Cayley table.

Which element is the inverse of ? Choose from these options. A B C D 2

A group has order

. What could be the order of a subgroup? Choose from these options.

A B C D 3

Construct the Cayley table generated by the element under the operation , where .

4

A group has order , but is not isomorphic to the cyclic group of order . Deduce the period of each element of the group.

5

A regular five pointed star is shown.

a How many elements are there in the symmetry group of the star? b Describe the groups formed as generators of each of the elements of the symmetry group. 6

A group has subgroups of order and . a Explain why the order of the group must be a multiple of

.

b Explain why the group must have at least one subgroup of order . 7

A binary operation is defined on a Find the value of the identity element. b Find the inverse of .

by

.

c Find a self-inverse element, other than the identity. 8

The set

consists of

matrices of the form

, where

.

forms a group under the operation of matrix addition modulo . a What is the order of this group? b Write down the inverse of

.

c Describe three non-trivial proper subgroups. 9

A group has order

. The binary operation is multiplication. The identity is , the element

has period and the element is not a power of . a List the elements of the group. b Simplify the powers of the element

.

c List the sets that form each proper non-trivial subgroup. 10

is a group of order

.

is a subgroup of , where and different elements of .

.

are also subgroups of , where

are all

a Which element is the identity of ? Explain how you know. b Explain why must equal . c Simplify the elements

and

.

11 A binary operation is defined on the set

by

.

a Construct the Cayley table for this operation. b Explain why

is not a group.

c Find all the subsets, of , for which

is a group.

12 Some entries in the first two rows of a Cayley table for a group are shown.

a Use

to find the value of another entry in the table.

b Use

to find the value of another entry in the table.

The group is isomorphic to the symmetry group of an equilateral triangle, the identity is and is a subgroup. c Complete the rest of the Cayley table.

CROSS-TOPIC REVIEW EXERCISE 2 1

In a project an activity has duration minutes and has latest finish time

minutes. What is

the latest time that the activity can start? Choose from these options. A

minutes

B

minutes

C

minutes

D 2

minutes

The pay-off matrix for a zero-sum game is:

The game is stable. What is a possible value of ? Choose from these options. A B C D 3

The table shows the points won by player 1 in a zero-sum game.

a Find the play-safe strategy for each player. b Deduce whether the game is stable or unstable. c Advise player on which strategy to play. 4

The table lists the activities involved in a project, their durations, in days, and their immediate predecessors.

Activity

Duration

Immediate

(days)

predecessors -

,

,

a Construct an activity network for the project. i

Find the earliest start time for each activity.

ii

Find the latest finish time for each activity.

b What is the minimum time in which the project can be completed? c List the critical activities. The duration of activity is increased so that the minimum time in which the project can be completed increases by days. d What is the new duration of activity ? 5

A binary operation

is defined on

by

a Construct a Cayley table for the operation b Calculate

6

and

.

.

c Explain how you know whether or not the operation

is commutative.

d Explain how you know whether or not the operation

is associative.

The diagram shows an activity network for a project. The duration required for each activity is given in hours. The project is to be completed in the minimum time.

a Find the earliest start time and latest finish time for each activity. b Find the critical path. c Find the float time of activity . d Given that activities and will both overrun by completion time for the project.

hours, find the new minimum [© AQA 2013]

7

a Use dominance to reduce this zero-sum game to a

game.

b Find the probability with which each strategy is chosen in an optimal mixed strategy. 8

The table shows the activities in a project, their durations (in hours) and their immediate predecessors.

Activity

Duration

Immediate

(hours)

predecessors -

a Calculate the earliest start time and the latest finish time for each activity. b Calculate the float for activity . c Construct a Gantt chart showing each activity starting at its latest start time. Activities and need one worker, activity needs no workers, and activities and need two workers. d Draw a resource histogram to illustrate the number of workers needed each hour to complete the project with each activity starting at its earliest start time. There are two workers available to complete the project. e What is the shortest time in which two workers can complete the project? 9

Stan and Christine play a zero-sum game. The game is represented by the following pay-off matrix for Stan. Christine Strategy

Stan

a Find the play-safe strategy for each player. b Show that there is no stable solution. c Explain why a suitable pay-off matrix for Christine is given by

[© AQA 2015] 10 The pay-off matrix for the player on rows in a zero-sum game is shown.

Determine whether the game is ever stable. 11 a Write out the Cayley table for matrix multiplication on the set , where

,

,

and

.

b Which element is the identity? c Write down the inverse of each element. 12 This Cayley table shows the effect of a binary operation

Show that

on the set

.

is not a group.

13 A binary operation is defined on

by

.

a Show that is commutative. b Find the identity for . c Construct the set

under the operation .

d Show that

is not a group.

14 The Cayley table shows the effect of a binary operation

Show that

on the set

.

is associative.

15 A binary operation is defined on the set

by

a Construct the Cayley table for . A group

, where is the binary operation defined by

b Identify the inverse of each element. c Find all the proper non-trivial subgroups of . 16 a Set up the initial simplex tableau to find the optimal mixed strategy for the player on rows for this zero-sum game.

The optimal mixed strategy for the player on rows involves choosing randomly between the rows so that has probability

and has probability .

b Calculate the value of the game for the player on rows. The game is reduced by removing row . c Show whether the table can be further reduced by dominance.

PRACTICE PAPER Time allowed: 1 hour. The total number of marks is 50. The marks for each question are shown in brackets [ ]. 1

Which graphs are semi-Eulerian? Choose from these options.

[1 mark] 2

The set forms a group under the operation of multiplication modulo of the group? Choose from these options.

. Which is a generator

A B C D [1 mark] 3

The feasible region for a linear programming problem is shown.

The objective is to maximise

.

The optimal solution is at the point

to one decimal place.

a Use the graph to deduce the optimal solution when and must be integers. [1 mark] The objective is changed to maximise

.

b For what values of is the integer solution from part a still optimal? [2 marks] 4

A graph has five vertices with degrees

listed in increasing order.

a Explain why must equal . [1 mark] b Deduce the number of edges in the graph.

[1 mark] c Explain why there is no graph with all of these properties: it has vertices, each with degree at least it has the same number of edges as in part b it is made of two simple subgraphs that are not connected to each other. [2 marks] 5

This network shows the lower and upper capacities, in litres per second, of a system of pipes through which water flows from to at the rate of litres per second.

a Explain why arc

must be at its maximum capacity. [1 mark]

b Find the flow along arc

, in litres per second. [1 mark]

c By finding a suitable cut, show that the flow of litres per second is the maximum flow from to . [2 marks] 6

The diagram shows a simplified map of some towns and the motorway routes between them.

The motorway distances (km) between the towns are shown in the table.

The total length of the motorways is a i ii

Find a route from to of length

. .

What is the length of the shortest route from to ? [2 marks]

Harry is a motorway maintenance engineer. He needs to check all the motorways to make sure that there are no potholes. Harry works in town and lives in town . He wants to find a route that starts at , ends at and covers each motorway shown on the map. b Use an appropriate algorithm to find the minimum distance that Harry will need to travel. [3 marks] 7

A linear programming problem is: maximise subject to

Formulate the problem as a simplex tableau and carry out one iteration of the simplex algorithm. [6 marks] 8

a Construct the group

. [2 marks]

b Find all the proper subgroups of . [2 marks] c Show that is isomorphic to

. [1 mark]

9

The table lists the activities involved in a project, their durations (in hours) and their immediate predecessors. Activity

Duration (hours)

Immediate predecessors

a Find the minimum project completion time. [4 marks] b

i Which activities are critical? ii Find the float on each non-critical activity. [2 marks]

Two workers are needed for each of activities each.

. The remaining activities require one worker

c Find the minimum time in which i ii workers can complete the project. [4 marks]

10 Alex and Bel play a zero-sum game. The pay-off matrix is shown. Bel

Alex

a Show how the game can be reduced to a

game. [3 marks]

b Find the play-safe strategy for each player. [3 marks] c If it is known that Bel will use her play-safe strategy, which strategy should Alex use? [1 mark] d Deduce whether the game is stable or unstable. [1 mark] e Describe how Alex should choose which strategy to use to maximise his expected pay-off over a number of games. [3 marks]

Set notation Set notation

Meaning is an element of is not an element of is a subset of is a proper subset of the set with element the set of all such that... the number of elements in set the empty set the universal set the complement of the set the set of natural numbers the set of integers the set of positive integers the set of non-negative integers the set of real numbers the set of rational numbers union intersection the ordered pair the closed interval the interval the interval the open interval the set of complex numbers

Answers 1 Graphs Before you start 1 Each word maps to the number of letters in the word.

Exercise 1A 1 No, the graph is not simple, there are two edges that connect and . 2 The degrees of the vertices are

The sum of the vertex degrees is

.

The graph has edges and 3 . The sum of the degrees of the vertices is twice the number of edges. ;

4 5 a

b For example: c For example: 6 The sum of the degrees is twice the number of edges, so must be an even number. It is impossible to have an odd number of vertices with odd degree. 7 a The sum of the vertex degrees must be even, so there must be two odd degrees. is odd but and are all even. So must be odd as well. b . The sum of the vertex degrees must be so . c

.

. The maximum value of is , when this vertex is connected to each of the others. This gives degree sum . This is twice the number of edges, so edges.

8 For a simple graph with four vertices, the maximum vertex degree is . Here there is a vertex of degree , so this vertex is connected to another by multiple edges or is connected to itself by a loop.

9 a For example:

b For example:

10 a 4 b The edges connect and . c

is replaced by

to and

but there is no edge between and or between any of . The edges connect

to

, with no edges between

any of

and or between any of

and .

11 a b The edges between the vertices in each of the two sets that form the bipartite graph. c The subgraph has a minimum of edges and a maximum of edges. 12 The graph is connected so it is possible to get from any vertex to every other vertex. If there were a loop at a vertex, , say, then this would leave only four edges to connect the other five vertices to each other and to , but this would need another five edges. If there were a multiple edge between two vertices, and , say, then this would only leave another three edges to connect the other four vertices to each other and to or , but this would need another four edges. In either case, there are not enough edges to make a connected graph. 13 a b

c To solve the problem, a path is needed from

to

This path is

that does not involve any wasted journeys.

(or the same with

and swapped).

The farmer should take the goat across and come back with nothing, then take the wolf across and bring the goat back, then take the cabbages across and come back with nothing, and finally take the goat across again (or the same but with the wolf and the cabbages swapped).

Exercise 1B 1

and .

2

and .

3 A tree has at least two vertices of degree . An Eulerian graph has no odd vertex degrees. 4 Semi-Eulerian. The vertex degrees are

. It has exactly two odd degree vertices.

5 a b c d 6 Each vertex degree is either or . There are five such degrees and they sum to So three vertices have degree and two have degree . 7 a b 8 a For example, b For example,

.

9

10 a Yes, provided the graph has at least three vertices. The complete graph

(

) can be

represented as a regular polygon with vertices with lines (diagonals) joining every pair of vertices. The cycle around the outside of the polygon is a Hamiltonian cycle for . b No, unless the two sets have the same number of vertices. Any path will alternate between the two sets, so a Hamiltonian cycle is only possible for

when

.

11 a b c 12 a b A simple graph with vertices cannot have a vertex with degree . The maximum vertex degree in a simple graph is , where is the number of vertices. c

Exercise 1C 1

2 3

and are planar. is non-planar. For example,

and . so

4 For example:

5 Each graph has four vertices and seven edges. Each graph has vertex degrees

.

However, in the first graph the two vertices of degree are directly connected whereas in the second graph they are not. 6 Each edge contributes two 1s in the matrix, so

.

so so

7

8 a For example:

b The graphs have different numbers of vertices with degree 1 (three, two and four vertices with degree 1). 9 or 10 a b A possible Hamiltonian cycle is c

so the graph is Hamiltonian.

and are each directly connected to that the graph is non-planar.

and , so

is a subset of the graph. This means

11 a The graph has only 9 edges so is not a subgraph, and it has only 5 vertices so subgraph. This means that the graph is planar. b

so

c

.

; of these include

and one of

, so triples.

Or d For example: if the graph is drawn as shown then the triple correspond to a region.

Mixed practice 1 1 2 3 a

is not a

,

. forms a triangle but does not

b 4

has six edges, so every simple-connected graph that connects four vertices using five edges is the same as the graph

with one edge removed. All such graphs are isomorphic.

5 The vertex with degree can only connect to one of the other three vertices. This vertex could have degree 3 but the other two vertices must have degree at most for a simple graph. 6 a b It is non-planar (using Kuratowski’s theorem). 7 a Graphs and are isomorphic. b Use the vertex degrees to suggest a possible correspondence between the vertices of graph and those of graph :

This gives:

which has the same structure as graph . c For example: In graph

has degree . No other vertex in any of the graphs has degree .

In graph degree .

and each have degree . Each of the other graphs has only two vertices of

8 a b c 9 a

b Each cell has degree , which is even, so the graph is Eulerian. 10 a There cannot be an odd number of odd vertex degrees, so must be even. There are no odd vertex degrees so the graph is Eulerian. b For example: the graph could consist of four separate vertices, two with one loop each and the other two with two loops each. c d 11 a For example: b Remove vertex by combining edges

and

as

.

Then merge each of

and the edge .

So the graph contains 12 a

into a single vertex. Each of

is directly connected to

as a subgraph. By Kuratowski’s theorem, the graph is non-planar.

.

b For example:

c For example:

d The graph is simple so each vertex of degree is connected to each of the other five vertices. The remaining four vertices must each be connected to the two vertices of degree Therefore, no vertex has degree .

2 Networks Before you start 1 a For example:

b For example:

Work it out 2.1 Solution 3 is correct.

Exercise 2A 1 2 3 4 5 6 The arc with the greatest weight is the only arc leading to the node at one of its ends. 7 a b

because, without this, getting from to using shelters means travelling

8 For example,

is a minimum spanning tree that does not use 9 a b

.

, say. .

minutes minutes

10 In a spanning tree, it must be possible to get from any node to any other node. If there is a directed arc, then there must be a way to get from the end of the arc to the start, and that means that there is a set of arcs that form a cycle.

Exercise 2B 1

is semi-Eulerian, is Eulerian, is neither and is Eulerian.

2

and .

3 4

(each arc appears twice in the table)

5 a b Because of turning corners. c For example, the distance from the groundsman’s feet to the point where the paint leaves the

machine. 6 a b The two least weight arcs are

and

but these are not the least weight paths joining

with two paths. c

d If is not used, then the least weight path from has weight at least and the other two paths connecting odd nodes must have total weight greater than , so the total weight is more than . e Repeat has degree and

,

are repeated, so is passed through times.

7 8 a b

9 a b c Over

minutes or over hours.

Exercise 2C 1 2 3 4 a b 5 a b 6 a There is no cycle, because to travel in and out of the arc

must be travelled twice, so is

repeated. b

, for example:

c

, for example:

d

for example:

7 a MST for reduced network: Two least weight arcs from are Lower bound b

so £ so £

c For example, there might be a security issue that means that a certain path should be included. 8 It is a string (a tree with just two nodes of degree ). The two least weight arcs from the node that was removed connect it to the ends of the string. 9 a There are no cycles in a tree so to form a closed route some arcs (and hence some nodes) will need to be repeated. b Every arc will need to be travelled twice, the total weight will be twice the weight of the tree.

Mixed practice 2 1

2 3 a b 4 a

b 5 a b 6

is odd but cannot travel through again, so must repeat either or . However, if is used, then they cannot start at without travelling through either or at least three times. So repeat and make even by repeating . Similarly, repeat and .

7 a One of these two networks.

Total weight b Not unique; there are two spanning trees with total weight 8 a

minutes

b This is a tour, but it might be possible to improve it. c d

e Minimum tour 9 a i ii

iii £ b Replace 10 11 a

with

; new cost £

.

.

b 12 a £ b £ c £ d £ e £

£

3 Network flows Before you start 1 a b

Exercise 3A 1

and .

2 3 4 5 a b The flow out of is at most

, so the flow into is at most

6 a b c 7 a For example:

and

Flow b The cut through

and

has value

maximum flow minimum cut So

.

maximum flow.

8 a b c 9 a Maximum flow increases to

.

b No change. c Maximum flow reduces to

.

10 a b c d Flow for example, Cut through

and

.

Exercise 3B 1 2 3 4 Vertex is infeasible. Maximum flow in Maximum flow out 5 a 24 b 33 6 7 a b Arc

Potential increase

Potential decrease

.

8 a b The maximum that can flow into is

, along

, so the flow through can never be as big as

.

c

d If

, maximum flow

Otherwise, maximum flow

.

.

9 a b If

, flow can be augmented by

c If increase

, potential increase and potential decrease .

d For example,

and potential decrease

, otherwise potential

and

And if Maximum flow

, otherwise, flow can be augmented by .

. .

e For example, cut through arcs

has value

.

Maximum flow minimum cut, so this must be the minimum cut and the maximum flow.

Mixed practice 3 1 2

and .

3 a b 4

;

5 a Cut through b For example, c Maximum flow

and flow

. flow

minimum cut.

6 a b

c For example, Augmenting path

Flow

flow

.

d e For example, Cut through

and

.

7 a

b i Maximum flow into

, so maximum flow out

:

can be at most . ii c

flow

flow

flow

d

flow

flow

flow

flow ; maximum flow flow

flow .

8 a b For example, flow flow flow flow flow flow flow c i For example, As above but remove ii Cut through Flow above has value

flow and

has value so maximum flow

.

Maximum flow minimum cut, so maximum flow 9 a b i ii

flow

, so minimum cut

.

.

.

.

c Arc Flow Arc Flow d e Cut through 10 a b i Initial state Arc Potential Arc Potential Arc Potential Arc Potential Arc Potential Arc Potential For example: Augmenting path

Final state Arc Potential Arc Potential Arc Potential Arc Potential Arc Potential

Flow

Arc Potential ii Maximum flow For example: Arc Flow Arc Flow Arc Flow c Cut through

.

d Flow as above but reduce, for example,

by .

Cut through 11 a b

and all have arcs that flow in and arcs that flow out, so must be the source, and hence the flow must be from to .

c Arc Flow d Arc Flow

4 Linear programming Before you start 1

2

Exercise 4A 1

and .

2

and .

3

4 a

number of type items produced each day number of type items produced each day These are really average numbers because on any individual day the number could be a fraction, but over a number of days there will be whole numbers of items.

b Maximise subject to: mixing time or finishing time or and 5

.

amount of Xtra (litres)

amount of Yepp (litres) Maximise subject to: or or or

or and

.

6

and

7 a

and

b

(from ) and

so

(from ) and

, so

, hence

( is redundant).

, hence

and

, so

( is redundant). 8 a b When

, the vertices are

When

and

, the vertices are

When

, the vertices are

, ,

,

,

,

1 B 2 3 4 A and C. 5 6 when

8

when

9 when 10

and and and

when

and

, or when

and

Work it out 4.1 Solution 2 is correct. In solution 1, the objective row reads

, so

In solution 3, the objective row reads

, so

Exercise 4C 1









and and

Exercise 4B

7

.

.

.

, hence

2 in column row . 3 Yes, because there are no negative values in the objective row. 4 5 in column row (there are no positive entries in column ). 6



7 8 a

The solution is

.

b









The solution is

.

c The original problem was: maximise subject to:

Both solutions (

and

) satisfy the constraints and give

.

The objective lines are parallel to the boundary line for the constraint . The vertices that have been found are the end points of this edge of the feasible region. Every point on this edge will also give the optimum value . 9 Maximise

, so

Pivot on column row :









Minimum 10

when

.

hours spent studying for specialist round hours spent studying for general knowledge round hours spent sleeping Maximise subject to:

,

,

Maximum

when

,



.

Emma should spend hours studying for her specialist round, knowledge round and hours sleeping.

hours studying for the general

Mixed practice 4 1 2 3

and

4 a

at

b

at

.

5

6 Current solution has , but this is not optimal because there is still a negative value in the objective row, so the maximum value is at least . The ratio

, so the maximum value is greater than

7 8 a Maximise subject to:

b in column z row . c

.

d Optimum value has been achieved. Maximum value of 9 a Maximise

when

and

.

(pence)

subject to:

and

both integers.

b The closest three points to the optimal vertex at are and maximised at , so the baker should make loaves and trays of buns. c £ d He might not sell all the loaves and buns. 10 a















b i Minimum non-negative ratio is . Pivot on the in row . ii









iii 











c Max

11 a



.

is

. b Substitute

:

c

d e i Draw an objective line with gradient to find optimum vertex or check profit at each vertex.

So maximum at ii £

by making

. tonnes of Basic,

tonnes of Premium and

12 a in column row . b

,

,

,

,

,

,

c 13

when

.

14





















tonnes of Supreme.



















Min

when

.



Cross-topic review exercise 1 1 2 3 a Source is , sink is . b

,

c Maximum flow

.

4 a b Total weight of arcs

.

The only odd nodes are and , so repeat Minimum journey time

.

minutes.

5 a There cannot be an odd number of odd vertex degrees, so must be odd. There are exactly two odd vertex degrees so the graph is semi-Eulerian. b c In a simple-connected graph with vertices, the maximum vertex degree is when a vertex is connected to each of the others. This graph has vertices of degree and . 6 Feasible region has vertices Minimum value of

is

. at

.

7 a

b

,

,

,

Maximum value of is

,

,

,

when

,

,

.

The first constraint is at its limit; the others still have slack. 8

amount of red (litres) amount of yellow (litres) amount of white (litres) amount of black (litres) Maximise subject to:

. 9 a i

( )

-

ii

euros

iii

b i ii 10 a i

followed by

.

followed by ,

,

,

. ,

ii iii £ b i

; £

ii

£

c i

£

ii £ d £

£

11 a i Max flow from

(along

) so max flow in (along

and

)

. Flow in

is at most

ii Max flow into (along ) so max flow out (along and ) . Flow in is at most and flow in is at most , so max flow into is and max flow out (along ) is . b i For example, ii Cut through

,

flow

,

and

.

flow ,

12 a i ii Less than or equal to b

,

litres per second.

,

c i

ii For example, Flow

Value

flow

,

flow

.

With appropriate changes made to potentials. iii

litres per second. Diagram must have

13 a

,

,

,

( to ) (sum of values in matrix

)

, so b i Graph cannot contain since no vertex has degree . Graph cannot contain since, for example, form a cycle. The graph has vertices so there are no spare vertices and each vertex has degree so there are no spare arcs. ii For example,

iii

and form the two triangular faces of the prism, rectangular faces.

,

and

form the three

14 a Graph and Graph are isomorphic:

b The vertex degrees for the four graphs are:

Considering the connections for the vertex of degree : In Graph it is adjacent to the two vertices of degree whereas in the other three graphs it is adjacent to one vertex of degree and one of degree , so Graph is non-isomorphic to any of the others. In Graph the repeated arc connects two vertices of degree whereas in Graphs and they connect a vertex of degree and a vertex of degree , so Graph is non-isomorphic to Graphs and

. 15 a Maximise subject to:

. b

c

when

,

and

.

16 a 11 b

c

flow This changes the potentials as: from

to ,

from to

from

to ,

from to

from

to ,

from to

.

d For example, Augmenting path

e Cut

Flow

,

has value

.

Maximum flow = minimum cut so is the maximum flow and the minimum cut. 17 a i (for example, , , , , , , , …) ii

(for example,

,

and

;

,

and

;

,

and

;

,

and

; …)

b i . ii The first node can be paired with any of the remaining

nodes; the next node with any of the

remaining

nodes, and so on. This gives

sets of n paths.

18 a y is at least twice as large as . b c d

if

,

.

from the first constraint, and , so

e

if

.

from the second constraint, so

.

5 Critical path analysis Before you start 1

Exercise 5A 1 2 3

hours.

4 If is critical, then it is part of a critical path, but the only immediate predecessors of are so one or both of these would be critical. However, neither nor is critical since . 5 a

minutes

b 6 a b

days

c

and

7 a

hours

b c d

and .

8 a

hours

b 9 a

and

b

. .

Exercise 5B 1

and .

2 a The float is shown after the blocks. b For example: Time

Team

Team

3 Total number of worker hours needed is . But 4 a

hours (

b i

hours

minutes)

hours

ii

hours

iii

hours

5

6 a

Activity

Duration (days)

Immediate predecessors

is only

.

and , and

b i ii c i ii 7 a

days days days days minutes

b No delay. c i

minutes

ii minutes 8 a b

and need to be done consecutively. The other activities need more than workers are needed.

worker hours, which is more than

, so

c If there are only workers, then the only activity that can happen alongside another is , so the time needed is hours. d If there are workers, then can happen alongside , so the time needed is hours.

hours. This saves

Mixed practice 5 1 2 3 a, b Activity

c

Earliest start time

Latest finish time

hours

d 4 a i Unchanged. ii

still critical but now and are also critical, so there are two critical paths:

and

.

b

hours,

hour

5 6 a b

hours (

minutes)

7 a

b See left-hand boxes in the diagram in part a. c See right-hand boxes in the diagram in part a. d 8 a

Activity

Earliest start time

Latest finish time

b

E D C B A 0

15

30

45

Hours

c Two workers do the critical activities and then ; the other two do followed by and then . 9 a Activity Earliest start time Latest finish time

b i ii hours c

d i

hours

ii For example, and . 10 a

and

hours

c

Workers

b

0

H (3) G (2) A (2) B (3) 2

H (3) C (2) 4

F (3) E (5)

D (3) 6

8

J (1) I (4) 10

d For example, Share Share

then Cal does . then Cal does

and Sam does .

12 Hours

6 Game theory for zero-sum games Before you start 1

Exercise 6A 1 2 3 4 The total score is not the same for every cell, for example, cell total score . 5

has total score but

has

Row min.

Col. max. Row maximin

, column minimax

,

row minimax column maximin, so game is stable. 6 a

Bob

Alice

b Total score is not the same for each cell, for example, if Alice plays and Bob plays the total score is but if Alice plays and Bob plays the total score is . 7

Sandi Row min.

Rupi

Col. max.

Play-safe strategy for Rupi is . Play-safe strategy for Sandi is . 8 a

      Fionn

Gina

b row minima: col. maxima: row maximin

, col. minimax

so game is unstable. 9 Row minima: row maximin

if

if

.

Column maxima: column minimax

, if

if

if

.

Game is stable when row maximin column minimax. This happens when

and not otherwise.

Exercise 6B 1

dominates over . and

.

2 3 Add (or a constant

) throughout.

4 5 The entries in column are bigger than the corresponding entries in column , so column is always worse than column for the player on columns. For the player on columns, rows plays

and rows plays

.

Column is dominated by column so column can be removed. 6

7 No dominance. Col. is better for Fionn than column when Gina plays or , but worse when Gina plays . Col. is better for Fionn than column when Gina plays but worse when Gina plays or . Col. is better for Fionn than column when Gina plays or , but worse when Gina plays . Col. is better for Fionn than column when Gina plays or , but worse when Gina plays . 8 a Ben chooses randomly between the rows so that row is played with probability probability . b c 9 a

and row with

b Play row with probability

and row with probability

.

10 a

Add to make entries

:



b Play column with probability 11 a

Jim

Rosie

Row maximin max Column minimax = min so not stable. b Add to make entries

:

and column with probability

.

c Choose randomly between and so that each has probability

.

Mixed practice 6 1 2 3 a Row maximin max Column minimax min Row maximin column minimax so not stable. b Play-safe for Pav is . Play-safe for Mina is . 4 a Row is (weakly) dominated by row . Jim plays

,

Jim b Rose

5 a

,

b Choose randomly between the columns so that column is played with probability with probability

and column

.

6 a Rose chooses randomly between the rows so that row is played with probability probability .

and row with

b Jim chooses randomly between the columns so that column is played with probability column with probability . 7 –1

8 a i Row maximin Col. minimax Row maximin col. minimax so stable. ii Adam

, Bill

iii b i The computer should never play

because it is dominated by

ii

iii

iv 9 a For each pair of strategies: Roz gain Colum gain

.

.

and

For example,

would give

b Column maxima are Column minimax

dominates

to Colum,

.

. .

Play-safe strategy is c i

to Roz and

.

since

and

so delete

Colum Roz

ii Let Roz play

with probability .

expected gain expected gain expected gain

Solving:

Roz plays

with probability

10 a Maximise subject to:

.

b 11 a

and

with probability

.

.

b

c d Jenny should choose randomly between strategies and , but never choose strategy , so that and

.

12 a Maximise subject to:

b

and

c Choose with probability

, with probability

and with probability .

13 rows (objective row, four constraint rows corresponding to the four strategies for the player on columns and one row for the constraint on the total probability). columns ( , three probability variables corresponding to the three strategies for the player on rows, five slack variables and a column for the values on the right-hand side of the equations). 14 a i Row maximin

for all values of .

Column minimax

ii Row maximin

column minimax

iii Column max b i

cannot dominate since will dominate if

.

but column minimax

.

. and

, so when

ii Choose with probability and with probability . iii Choose with probability

and with probability

.

.

7 Binary operations Before you start 1 and

.

2 a b

Exercise 7A 1 2 Yes. 3

but

4 For example,

.

but

.

5 a

which is in

.

b Addition of real numbers is commutative so

hence

6 a

b

c For example,

,

but 7 a

and

, so

b For example,

,

but 8 a If If

. or

is even), then

or ( is odd), then

b

. .

and . If is odd, then Hence,

, so can have any value, but if is even, then

.

c From part b, the only value of for which When

is

.

, so

.

(since + is commutative) so * is commutative

.

If and are both even or both odd, then

, but if one of

.

even

even

even

even

even

odd

even

odd

even

even

odd

odd

odd

even

even

odd

even

odd

odd

odd

even

odd

odd

odd

Exercise 7B 1 2 3

4 a

b c

is self-inverse, the inverse of is and the inverse of is .

5 a

b The Cayley table is symmetrical about the leading diagonal. c No row, or column, reads 6 a b

is self-inverse. The inverse of is . is self-inverse. The inverse of is .

c d

(in that order).

is odd and the other even, then

e The two Cayley tables have the same structure with



(or swap and ).

Rearrange the order of the rows and columns.

Addition is associative, so addition modulo is associative and hence the operation o is associative. 7 a

b Identity is .

and

are each self-inverse; has no inverse.

c Yes, the Cayley table is symmetrical about the leading diagonal. d e more:

.

8 a For example, the operation from question 5. b For example, subtraction for real numbers:

for all , but

only when

.

c If an element, , say, is both a right identity and a left identity for some operation o, then for all . If is another right identity, then

for all and hence

Similarly, if is a left identity, then that .

but

1 2 3 a b ∇ 0 1

0

1

, as shown.

, as shown. In either case, you can conclude

Mixed practice 7

4 a

but

2

2 b

and

c

and

d For example,

5 If is the identity of o, then In particular,

for all .

.

So is the inverse of . 6 a

which has the required form to belong to

.

b

7 a Yes, the table is symmetrical about the leading diagonal. b The inverse of is and the inverse of is . The inverse of is and the inverse of is . is self-inverse. c No, for example,

, but

8 a For example,

b

c

But

.

Hence and so @ is associative. 9 The first two bullet points give:

.

If each row contains

and , then

is not commutative, so

10

.

, but row contains the numbers

and , so this gives

8 Groups Before you start 1 a b c d is the identity, so

and

2 No, for example,

, but

.

Exercise 8A 1 2

3 has period . and each have period . For example,

, , , .

4 This group has infinite order. 5 In rows

and :

…

For closure, the subgroup must also include :

…

, so

This is now closed. So

is a proper subgroup.

6 These are the groups formed using and the subsets: and 7 Closure:

, since

non-zero and real

. are non-zero real

numbers. Identity: the identity is

Inverses: the inverse of

.

is

, since

are non-zero and real

are non-zero

real numbers. Associativity: matrix multiplication is associative, so it is associative on the set 8 a

rotate about centre through . rotate about centre through rotate about centre through reflect in axis through vertex at top. reflect in axis through vertex at bottom left. reflect in axis through vertex at bottom right.

.

b Non-abelian, for example,

, but

9 a Let the group be Suppose that

.

, where is the identity and each of and

.

Any subgroup that includes the element would need to include subgroup. But

and so there is no such proper

, so there is no proper subgroup that includes .

Similarly, b For example,

has period .

, so there is no proper subgroup that includes , and , so there is no proper subgroup that includes . or

.

Exercise 8B 1

, and .

2

3 The group in question 2 has order . By Lagrange’s theorem a group of order has no proper non-trivial subgroups. So every group of order is isomorphic to the cyclic group of order . 4 a identity, rotate through , reflect in short axis of symmetry, symmetry.

reflect in long axis of

b Apart from the identity, each element of the symmetry group has period . Each element is selfinverse.

In the cyclic group, has period but and each have period . Only and are self-inverse.

5 a

b Two elements are self-inverse and two are not. So this group cannot be isomorphic to the symmetry group of a rectangle. The identity is . Set up a correspondence between elements of cyclic group .

and the elements of the



This has the same structure as the Cayley table in part a. 6

By Lagrange’s theorem, the order of a proper subgroup must be a factor of , so any proper non-trivial subgroup can only have order or . The identity element is , the only self-inverse element is , so the only subgroup of order is .

and

, so and each have period .

is a group of order

and each have period so they do not generate any proper non-trivial subgroups. 7 a

b The Cayley table is closed. The identity is . Element Inverse Multiplication of real numbers is associative, so c Set up a correspondence between

8 a

is associative.

and the cyclic group

:

b There is no identity element;

, but

and

.

Mixed practice 8 1 2 3

4 The identity has period and the other three elements must each have period . No element can have period since, otherwise, that element would generate the group and then the group would be isomorphic to the cyclic group. 5 a

, five rotations and five reflections.

b The identity generates the trivial subgroup. Any of the rotations generates the subgroup of order , consisting of all the rotations. Each reflection generates a subgroup of order consisting of the identity and the reflection. There are five such subgroups of order . 6 a By Lagrange’s theorem, the order of a subgroup is a factor of the order of the group, so the order of the group is a multiple of both and . and are coprime so the order of the group must be a multiple of

.

b A subgroup of order is isomorphic to either the cyclic group of order or the symmetry group of a rectangle from question in Exercise 8B. Each of these has at least one subgroup of order 2. 7 a b c 8 a b c A non-trivial proper subgroup must have order . Subgroup where

.

Subgroup where

.

Subgroup where

.

9 a For example,

.

b

. c

.

10 a The identity is ; this is the only element that is common to , and . b c 11 a

is a subgroup of order so is isomorphic to the cyclic group of order . .

b There is no identity element. c 12 a b c

Cross-topic review exercise 2 1 2 3 a Play-safe for player is . Play-safe for player is . b Stable since value of play-safe strategy

for both players.

c Player should always choose strategy . Player will end up losing points per game, but this is less than the expected loss if strategy is chosen when player plays safe. 4 a

b days c

, ,

d days 5 a #

b

c Not commutative, from part a the table is not symmetrical, for example, d Not associative, for example from part b, 6 a Activity

but

.

EST LFT Activity EST LFT b c d

hours hours

7 a Row dominates row since

,

and

but

.

.

Column

dominates column since

Column dominates column since

, ,

and and

. .

Reduced game:

b

Strategy Probability Strategy Probability

8 a Activity

Earliest start time

Latest finish time

(hour)

(hour)

b hour c

T S R Q P 0 1 2 3 4 5 6 7 8 Hours

d

e

hours

9 a Stan

and , Christine

b Row maximin

column minimax

c Column dominates column . Column dominates column . Transpose and multiply by 10 Row minima are: , Row maximin

to show Christine’s gains. .

for for for

Column minimax are: Column minimax

,

,

.

for for for

Row maximin is always never stable.

and column minimax is always

, so they are never equal. The game is

11 a

b The identity is . c Each element is self-inverse. 12 The operation is closed and the identity is . However, the elements and do not have inverses since does not appear in the rows (or columns) for and . 13 a

, multiplication and addition of real numbers are commutative so .

b c

, the identity is . ,

,

d

, which is the identity so

for all , so has no inverse. *

14

and so if any of

for , then

The remaining cases are those where If

, then

For all other cases at least one of

. . . and the others are or .

.

So

.

So the operation is associative. 15 a

b

element inverse c

,

16 a

b c No. For the player on columns:

if the player on rows chooses , then is better than and is better than rows chooses , then is better than and is better than .

but if the player on

Practice paper 1

and .

2 3 a b 4 a The sum of the degrees must be even (there is always an even number of odd vertex degrees). so must be . b Degree sum

edges.

c One of the two disconnected subgraphs must have vertices and the other vertices. If the subgraphs are simple, then the maximum number of edges is when they are this only gives edges and are required. 5 a Min flow from (along

and

b Flow across cut through

,

So flow along c Cut through

)

max flow into (along

,

.

. ,

,

,

,

,

Value of cut

.

Maximum flow = minimum cut 6 a i ii b Route inspection: and will need to be odd, need to pair , , , .

Travel every arc and repeat Minimum distance 7

Pivot on the in row of column .

. .

and

), so flow in

, but .

,

8 a

,

,

,

,

. b

,

,

.

c

There is a one-to-one correspondence between the groups so they are isomorphic. 9 a

b i

, ,

ii c i

has hour of float, has hours of float, has hours of float. , , consecutively take

hours. , consecutively (with alongside) take

hours.

hours ii

and (or and ) take hours. (with alongside) then (with after ) take hours

10 a Row is dominated by row : ,

,

Column is dominated by column : ,

,

The reduced game has pay-off matrix: Bel

Alex b

Bel Row min. Alex Col. max. Row maximin

, column minimax

.

hours.

Play-safe for Alex is , play-safe for Bel is . c

; this gives the better pay-off for Alex when Bel plays .

d Unstable, since

.

e Choose randomly between rows so that row is chosen with probability and row with probability . If Bel plays : Alex expects to win If Bel plays : Alex expects to win

. .

Row is chosen with probability and row with probability .

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