Financial mathematics: educational manual 9786010405806

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

N. Sh. Alzhanova

FINANCIAL MATHEMATICS Educational manual

Almaty «Qazaq university» 2014

UDC 519.8(075.8) LBC 22.19я73 A 45 Recommended for publication by the Scientific Council of the Higher School of Economics and Business, Editorial and Publishing Council of the National University of Kazakhstan named after Al-Farabi and teaching and methodology section of Republican educational and methodological Council by specialty group "Social sciences, economics and business" of higher and postgraduate education at the MES at Kazakh University of Economics named after T.Ryskulov (Protocol №1 24 January 2013)

Reviewers: Doctor of economics, professor R.U. Rakhmetova Doctor of economics, professor B.M. Mukhamediyev Doctor (PhD) A.M. Tleppayev

Alzhanova N.Sh. Financial mathematics: educational manual – Almaty: Qazaq A 45 university, 2014. – 96 p. ISBN 978-601-04-0580-6 In the manual theoretical, methodical and practical bases of application of the most widespread methods of financial calculations are visually stated. Work is constructed according to the university training program on financial mathematics. Basic theoretical provisions are illustrated by examples of the solution of standard tasks. The special attention is paid to opportunities of independent work, for this purpose examples of the solution of tasks are given, there is a section of practical tasks for the independent decision with answers. It is recommended as the manual for students and teachers of economic higher education institutions and faculties. The grant is also intended for the actuarial, insurance, auditor companies and bank workers.

UDC 519.8(075.8) LBC 22.19я73 ISBN 978-601-04-0580-6

© Alzhanova N.Sh., 2014 © KazNU al-Farabi, 2014

FINANCIAL MATEMATICS

I

3

ntroduction

The financial calculations which subject is the quantitative analysis of results of activity of the investment, credit, insurance, currency companies, represent as a one of the most widespread and popular branches of economic sciences today. The offered manual is written on the basis of a course of lectures which was read by the author for a number of years for students of economic specialties of the Kazakh national university named after al-Farabi, in it is stated bases of bases, the alphabetical principles of carrying out financial and actuarial calculations, possession with which in modern conditions is not only useful, but also it is obligatory as a starting condition of competent, operational and enterprising work of any perspective expert. The author sought to available and whenever possible to the evident description of ideas and quantitative approaches for search and decision-making in problems of financial mathematics. It is presented rather full statement of the theory according to the university program on financial mathematics in grant. Creation of the manual has to serve the best assimilation of the stated material: each chapter comes to the end with the list of questions, a set of tests and tasks for fixing of the received knowledge. Tests and tasks can be also used when carrying out a practical training. The offered grant is not only educational, but also the practical grant, that is the practical work, which development

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demands the preliminary qualitative analysis of an essence of the studied financial phenomena, clear understanding of the purposes and sense of estimated indicators, competent, expeditious and their enterprising use for acceptance, realization of protection of own financial interests and decisions. The edition of this grant, in our opinion, will help students of economic specialties and the directions to seize methods and approaches for a choice of effective investment decisions, and practical orientation of a material will be reliable help in investment questions. The grant is intended also for beginners of the actuarial, insurance, auditor companies and bank workers.

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CHAPTER

1

Models of operations development under the scheme of simple interest

1.1. When calculating simple interest Building on a simple interest rate (i). Given an initial value of money P and implemented its accretion, or growth, i.e. process of increasing the value of money by charging interest (Figure 1.1).

P

S=?

Time

Fig. 1.1. Expanded in the course of time

Accretion (future) the amount of money in a certain period is denoted S; number of interest periods, i.e. interest accruals period – n; the interest rate for the period с i. Then simple dekursivnye (common) interest is calculated as follows: P  i  I1 ,

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where P  n  i  I n – the amount of money in interest, accrued for all interest periods n; I 1 – the amount of money in interest accrued per unit time. Building process of money by charging simple interest looks like an arithmetic progression: P; P  P  i ; P  2P  i; P  3 P  i , and etc. with the first term P and the difference P  i , and analytically for n periods can be expressed by the following formula:

S  P  P  i    P  i  P  P  n  i  P (1  n  i )

(1.1.1)

where P  n  i  I n . Formula (1.1.1) expresses the essence of practical calculations related to the calculus: • The sum of the loan provided under simple interest; • Size of a time deposit with interest. In this situation, when n – the number of interest periods, i – rate for the period is as follows: • n – term financial transaction; • i – rate per period, interest is charged at the end life of the period. Example. A loan of 350 thousand KZT granted for two years at 10% per annum. Define the refundable amount, if simple interest is charged every year, and the debt is extinguished lumpsum payment. Solution. Using formula (1.1.1), we get S  P (1  n  i )  350 (1  2  0 ,1)  420 thous . KZT .

But this kind of calculation is rare. For these calculations, more likely to use the formula (1.1.2), where the analytical expressed the calculation principle for the case while given annual rate of i , and the period of operation is expressed in days, at least – in months. Set

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7

deadlines for operations through t. To translate the term of a financial transaction in a share of the year using equalizing denominator Y, indicating the duration of the year, expressed in the same units as t . The ratio t/Y substitute instead of n in (4.1.1) and to obtain a formula that is most often used, and is a form of (1.1.1):

t t   S  P 1   i   P  P   i  P  I n Y Y  

(1.1.2)

Formula (1.1.2) is used: • In determining the absolute value of interest and accrued interest on the whole amount for maintenance of demand deposits; • When servicing current accounts; • When calculating the amount of the debt, with interest at the term of operation less and lump sum repayment of debt; • When replacing and consolidating short-term payments; • In determining the amount of interest payments in drawing amortization plans (repayment) of debt. 1.2. Ordinary and accurate simple interest Note that t and Y in the case of measuring them in days can be expressed exactly or approximately (table 1.2.1). Table 1.2.1 Indices t and Y Measurement The Exact Approximate

t Actual days in the month (January – 31 February – 28 (29), March – 31, etc.) The number of days in all months assumed to be 30

Y In fact, 365 days a year or 366 A year is 360 days

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Depending on the values of t and Y, measured in different ways, in practice there are the following methods of payment: ACT ) – this means to 1. t and Y are measured accurately ( 365 charge the exact percentages (365) with the actual period of operation ( ACT ). To determine t here is used a special table of days serial number in the year: from the ending number of days operation subtract the started day (if the date of issue and date of the loan is counted as one.) In Kazakhstan, on this principle all banking operations are conducted; ACT 2. t measured accurately, and Y approximately ( ). This 360 method is used to calculate the ordinary (commercial) interest with the actual timing of the operation. Since the calculation of t/Y denominator is smaller than the calculations in the case of 1, i.e., 360 less than 365, so the amount of accrued interest on other things being equal will be more than 1.3889%. 3. t and Y are measured approximately. This method is used to calculate the ordinary (commercial) interest to the approximate timing of the operations for certain types of calculations with the population. Keep in mind those different methods of counting days and annual bases leads to different results. Example. Deposit of 750 thousand KZT placed at 10% from 01.01 to 01.04 in current year. Charge the exact percentages to the actual period of ACT operation ( ) and commercial interest on the actual timing of 365 ACT the operation ( ). 360 Decision. According to the condition t = 31+ 28 +31 = 90 days (1st April we don’t count, because this is the day of the loan) or at the table: t = 91 (the ordinal number of 1st April) – 1 (the ordinal number of 1st January) = 90 days. Then

F I N A N C I A L MC A HAP T E M ATER T I C S1

I(

9

ACT t 90 ) P  i  750   0,1  18.493 thous. KZT. 365 365 365

I(

ACT t 90  i  750   0,1  18.75 thous. KZT . ) P 360 360 360

Different results are obtained when using different methods to repay short-term debt through a series of interim payments: the actuarial method and rules of the merchant. When servicing current accounts banks are faced with a continuous chain of receipts and expenditures, as well as the need to interest on the ever-changing amount. In banking practice in this situation the rule – the total charge for the duration of the amount of interest equal to the amount of interest accrued for each of the constants in an interval of time amounts is used.

1.3. Variable rates of simple interest Inflation often forces changing the simple rate of simple interest. Consider the period of the contract (t 0 , t 0  T ) . Suppose that during this period the annual interest rate changes (m – 1) once in moments (t1  t 2  ...  t m  2  t m 1 ) . The period of the contract is divided into m subintervals with constant annual rate, so, that the subinterval rate (t 0 , t1 ) is j0, for subinterval (t1 , t 2 ) – j1 , …., and the last subinterval (t m 1 , t m ) is equal to j m -1. If the initial amount is P, then the coefficient or factor compounding the whole interval:

A(t 0 , t m ) 

m 1 S (t m )  1   (t s 1  t s ) j s S (t 0 ) s 0

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Note. Here using conventional in mathematics refer to several terms. For example, 2

2

s0

k 0

a0  a1  a2  :  as   ak ,

4

b1  b2  b3  b4  :  bk , k 1

besides the summation index can be defined either by a letter. If all rates % will be the same j1  j2  ... jm 1 , so this formula becomes:

S  P1  (t m  t 0 ) j 0 

Example. The contract involves the following scheme for interest calculation. For 1 year 60%. In the following six months the rate rises to 10%. Determine the coefficient of compounding for 2.5 years. Solution. Using the formula

A(t 0 , t m ) 

m 1 S (t m )  1   (t s 1  t s ) js , S (t 0 ) s 0

we get А(2,5) = 1+1· 0,6 + 0,5 · 0,7 + 0,5 · 0,8 + 0,5 · 0,9 = 2,8 1.4. Discounting and accounting for simple interest rate

In financial practice often faces the problem of the inverse definition augmented amounts: a known amount of S, which must be paid in some time n, to determine the amount of the loan received by P. This kind of situation can occur, for example, in the development of the contract, as well as in cases when the interest on the amount S is held directly while issuing loans. In these cases its said that the sum S is discounted, the process of charging and holding interest is called accounting, and the difference S  P discounted. In

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FINANCIAL MA E M ATER T I C S1 CTHAP

practice these calculations do not occur often, for example, to determine the amount of capital that should be invested by a certain percentage to get the required amount of money, to determine the interest charged on the loan, as well as to determine the present value of the short-term interest of the security. With reference to the mathematical discounting true definition: the current value – an amount that is invested under the existing interest rate to a specific payment date would have a value equal to the largest amount of the payment promised at that point in the future /Uotshem T.J, Parromou K. Quantitative Methods in Finance: Tutorial: Translation from English. / Ed. MP Efimova – Moscow: Finance, units, 1999/. Bringing in the actual loans and deposits operation is usually with a predetermined rate of interest, but to estimate the present value of expected incomes rates are chosen, including the calculation. Moreover the selection rates are guided by the current yield of the investing alternative in similar financial instruments given the level of risk. As the guide can act zero coupon bonds yield, short-term bank deposits and treasury bills. Mathematical discounting for the simple rate of interest:

Р where

S , 1  ni

(1.4.1)

1 – discount factor, 1  ni

n  t /Y . If in the formula (4.1.1) instead of P substitute difference

between S  P  I lefts:

the

I SРS or

current

and

future

S , so the 1  ni

value

(income)

S S  S ni  S S ni ,   1  ni 1 ni 1 ni

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t i Y I t 1 i Y S

(1.4.2)

Example. After 90 days of signing the contract, the debtor will pay one million KZT. Loan is 20% per annum (interest ordinary). What is the initial amount and the discount? Decision. Applying (1.4.1.) and (1.4.2), we obtain

Р

S  1000000 /(1  0.2  90 / 360 )  952380 .95 KZT , 1  ni I=S-Р=1000000-952380,95=47619,05 KZT.

1.5. Bank or a commercial accounting

This type of accounting is used for purchases (accounting) of bills and other short-term liabilities. The essence of this operation is that the bank before the maturity date or other obligatory payment buys it from the owner, who is a creditor for less than the amount that must be paid to him at the end of the period, e.g. gets (account) it with discount For the calculation of interest, after taking into account the discount rate of applied bills, which will be expressed by d symbol. By definition, a simple annual rate of discount is as d=S-Р/Sn, the size of the discount or account held by the Bank, is D=Snd, from where Р=S-D=S-Snd=S(1-nd)

(1.5.1)

Example. Entrepreneur wants to get a loan in the amount of 50,000 KZT for six months the Bank provides charging simple interest at the discount rate 24% per year. How much money the entrepreneur will have to pay for bank?

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13

Solution. Using formula (1.5.1), we obtain

S

P 50000  KZT . 1  nd 1  0,5  0,24

Controlling questions and problems In questions of 1.1. – 1.8. from the list of the responses choose the correct one. 1.1. What is meant by interest (interest money) in financial calculation: a) hundredth of debt amount; b) ratio of the sum paid for the use of the loan to the value of the debt; c) absolute value of the income from the provision of money in debt; d) relationship of the debt to the amount paid for the use of credit? 1.2. What is meant by interest rate: a) amount charged per year for every 100 KZT of primary loan; b) ratio of the money percentage paid for a fixed period of time to the value of the loan; c) absolute value of the income from the provision of money in debt; d) the relative value of income from the provision of money in debt? 1.3. What is meant by the interest period: a) one year; b) the time from the moment of receiving loan until full repayment of the debt; c) time period, in to which the interest rate is related; d) the time from the moment of receiving loan to the partial repayment of the debt? 1.4. What is meant by the augmented sum: a) the original amount of the debt, together with assessed on the accrued interest to the end of the term; b) amount charged for the used credit; c) income received by the creditor for the year; d) the original amount of the debt, discounted by the end of the term?

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1.5. What is meant by simple interest: a) calculating option, when the rate of interest applied to the same initial amount for the duration of the loan; b) calculating option, when the rate of interest applied to the amount of any accrued interest in the previous period; c) calculating option, when the interest rate changes discretely in time; d) calculating option, when interest rates changes continuously over time? 1.6. Specify compounding formula for simple interest rates: a) S  P (1  n  i ) ; b) S  P (1  n  d ) ;

P  S (1  n  i) 1 ; 1 d) P  S (1  n  d ) .

c)

1.7. Specify the mathematical formula of discounting in the case of simple interest rate: 1

a) P  S (1  n  i) ; b) P  S (1  n  i ) ; c) S  P (1  d  n ) ; d) P  S (1  d  n ) . 1.8. Specify the formula for a simple bank account discount rate: 1

a) P  S (1  n  i) ; b) S  P (1  n  i ) ; c) S  P (1  d  n ) ; d) P  S (1  d  n ) . Solve problems 1.9. – 1.31. 1.9.

A client placed in a deposit account 1 million KZT for 3.5 years at the rate of simple interest equal to 17%. Determine the accrued interest at the end of the term. 1.10. Enterprise prepares a loan agreement with a bank in the amount of 3

F I N A N C I A L MCAHAP T E M TER A T I C1 S

1.11.

1.12.

1.13.

1.14. 1.15. 1.16. 1.17.

1.18.

15

million KZT for the period from 5 January 2004 to 20 March 2004, with the rate of simple interest of 15% per annum. Calculate interest on credit in the calculation of common interest with the number of days the loan. Entrepreneur February 7 apply to the bank for a loan up to 14 May of the same year under the simple interest rate of 18% per annum. Bank held at the time of the loan interest for its entire term, outstanding entrepreneurs 50 thousand KZT. How much money should be returned to the bank, if the calculation of the accrued interest used common interest with the number of days and a leap year? Company filed 1st March in the bank for a loan of 150 thousand KZT undertaking to repay, with interest at the end of the year. What a way to charge simple interest is beneficial for the company and what – to the bank, if you are using the interest rate of 26% and nonleap year? Bank granted a loan of 1 million KZT. The agreement adopted simple interest rate for the first 0.5 years, equal to 20% per annum, and each subsequent 0.5 years, the rate will increase by 3% compared to the previous one. Term of the contract is 2 years. Determine the accrued amount for the entire term of the contract. How long contributed five thousand KZT increase to 6 thousand KZT in the calculation of interest on a simple interest rate of 32% per annum? How long to put the amount of money available under a simple interest rate of 20%, so that it increases to 2.5 times? After 180 days of signing the contract, the debtor will pay 2.5 million KZT. Loan is 10% per annum (interest ordinary), determine the discount (in the mathematical discounting). Entrepreneur after a while will need an amount of 25 thousand KZT, yet he has only 22 thousand KZT. In order to accumulate the required amount the entrepreneur is going to put in the bank 22 thousand KZT. The proposed bank interest rate is 30% per annum. The number of days necessary to accumulate the required amount if the bank charges a simple interest calculation using the exact percentages, and non-leap year? Bank for using within four months 960 thousand KZT must pay 70 thousand KZT. Determine the cost of funds in the form of a simple annual interest rate, in charge of common interest.

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1.19. Businessman received a bank loan for 150 days at an interest rate of 30% per annum, while the bank had retained a commission of 1.5% of the loan. Find the yield of such financial transactions for the bank in the form of a simple annual interest rate, the bank charges a simple interest on the original amount of the loan, assuming a year of 360 days. Will the value of return change for a loan in 90 days? 1.20. After 90 companies must obtain the bill one million KZT. The company sells the promissory note to the bank. The latter take into account this bill for a simple discount rate of 20% per annum (year has 360 days). Determine the discount. 1.21. When issuing loans at an interest rate of 42% per annum was charged fee of 2.5% of the loan. Simple interest is charged on the exact amount of the original loan, a leap year. What term loan has been issued, if the yield of the transaction to the lender in the form of a simple annual interest rate of 64%? 1.22. When issuing a bank loan for 80 days at an interest rate of 38% per annum straight commission fee. Simple common interest is charged on the original amount of the loan, non-leap year. Determine what percentage of the value of loans made commissions, if the yield of such financial transactions for the bank in the form of a simple annual interest rate was 40%. 1.23. Bank issued one businessman 30 thousand KZT for 80 days, then received money from him awarded the second employer for 60 days, and finally received the second entrepreneur amount paid to third employer for 160 days. All loans were granted under a simple interest rate of 30% per annum, and the interest accrued ordinary. How much money should return the bank third entrepreneur? Determine profitability for all financial operations of the bank in the form of an annual simple interest rate. 1.24. Banks to lend his four clients A, B, C and D – in the following way: A customer – for 45 days at 28% per annum, and all money received from a client, immediately issued to clients in 120 days at 33% per annum; the entire amount received from customers to customers for 100 days with a 32% interest and received the money from the client C, D issued to the client for 40 days at 30% per annum. Client D at the end of the bank returned 37,632 KZT. How much money was a client, if in all cases the ordinary settlement is simple interest?

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1.25. Promise intends to present to any bank account bill of $ 50 thousand KZT. 45 days before the date of maturity. One bank offers discount bill at the accounting rate of 30% per annum. Another bank has to consider the bill on a simple interest rate of 30% per annum. Whose conditions are more favorable for the drawer? 1.26. During the bill left out a year and a half before the time for the simple discount rate of 12%, paid 4.5 thousand KZT. Determine the nominal value of the notes. 1.27. Entrepreneur wants to get a loan of 50 thousand KZT for six months. Has agreed to provide a loan on the terms of simple interest accrued on the account at 24% per annum. How much money the entrepreneur will have to pay for bank? 1.28. Client received 10th February loan from a bank for a simple discount rate of 30% per annum, and must return the entire debt of 27 May of the same year. What will be the return on the transaction for the bank in the form of a simple annual interest rate if it is a leap year, and: a) the time base for the account and the interest rate is the same and is equal to the number of days in a year, and b) the interest rate for the time base is 360 days, and for the interest rate – 366 days? 1.29. How long must place the amount of money available under a simple interest rate of 34% per year, it increased to 1.5? What will be the answer if accretion is the simple discount rate of 34% per annum? 1.30. Certificate of deposit type discount period 45 days, sold at a price determined by a simple discount rate of 32% per annum and the estimated number of days in a year, equal to 360. Determine the equivalent value of a simple annual interest rate, which determines the cost of funds of the bank, with an estimated number of days in the year, to 365. 1.31. Bank accounts bill for 180 days prior to the Interest Rate 34% per annum, using the time base of 360 days. Determine the profitability of such an operation in a simple annual interest rate on the time base equal to 365. 1.32. How long to put the amount of money available under a simple interest rate of 30% per annum to the accrued interest were 1.8 times greater than the original amount?

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FINANCIAL MATEMATICS

1.33. Outstanding loans at an interest rate of 40% per annum, with a commission of 2% of the loan. Simple interest is charged on the exact amount of the original loan, in a leap year. For what period should be granted a loan to yield such a deal for the lender in the form of a simple annual interest rate of 100%?

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FINANCIAL MATEMATICS

CHAPTER

2

Models of operations development under the scheme of compound interest

2.1. Interest upon interest

In financial practice a significant part of calculation is carried out using compound interests. The principal difference from the simple interest is in that the base for the calculation of the interest payment (discount) varies over the life of a financial transaction by the periodic addition (removal) of previously accrued income (discount), while the base using simple interest remains unchanged. A calculation by the rule of compound interest is often called the interest on the interest, and the procedures for addition of accrued interests are their reinvestment or capitalization. Due to the constant growth of the base in consequence of reinvestment percents rise in the initial amount of money are with the acceleration (Figure 5.1.). As a general rule, compound interest is applied in the medium and long-term financial transactions. But in any case, if the accrued interests (for example, the deposit) are capitalized, the calculations of the final augmented amounts should be based on the formulas of compound interest, as well as:

20

FINANCIAL MATEMATICS 

Terms of increased on the outstanding interest amount, if interests are calculated and added to the principal amount of the debt;  Repeatedly accounting of registered securities (accounting and recounting under the same conditions);  Determining the rent for leasing services;  Assessment of zero-coupon bonds;  Determining the changes in the value of money under the influence of inflation;  Discounted amount of cash over a number of periods in the project analysis. Calculation of effectiveness of short-term operations should be carried out as the formula of compound interest on the basis of the possibility of re-investment income in the same conditions.

S Interest сompound

Simple interest P 0

1st period

Time

Fig. 2.1. Accretion the course of time

2.2. Accretion of compound interest rate (i)

If the calculation is carried out at a decursive interest rate i, then a formula for determining augmented amount through n periods can be derived by tracing the path of compounding with the accounted

F I N A N C I A L MCAHAP T E M TER A T I C2 S

21

capitalization of interests at the end of each of the n periods:

P  P  i  P (1  i ) – to the end of the 1st period,

P (1  i )  ( P (1  i ))i  P (1  i ) 2 – to the end of the 2nd period, and, finally, by the end of n – year period of accrued sum has the form: (2.2.1) P (1  i ) n  S , where i – rate of interests for the period; n – financial transactions period and the numbers of interests periods, as interests are calculated at the end of each segment of the term. According to the general theory of statistics, if you know the continued growth of interest, then, to get a base, it is necessary to multiply all the available continued growth. The interest rate for the period – a chain growth rate, 1+i- chain growth rate. Since we consider a constant rate over the period, i.e. growth rates are constant, the overall base rate of growth for the entire period has the form of: f n ,i  (1  i )(1  i )  (1  i )  (1  i ) n . Expression (1  i ) n is called the coefficient (multiplier) building. Its value rates for the period i and the number of interest periods n tabulated. We have identified a multiplier compounding

f n,i  (1  i) n , therefore (2.2. 1 ) can be written as: S  S  f n ,i . Consequently, the factor compounding shows how many times

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FINANCIAL MATEMATICS

the initial amount of money under specified conditions is increased (n , i). Example. Loan 5 million KZT is issued under the compound interest for 4 years. Interest (20% per annum) is calculated annually and added to the principal amount of the debt. Determine the amount of the debt to maturity. Solution. Using equation (2. 2.1) , we obtain S  P (1  i ) n  5 (1  0 ,2 ) 4  10368 m ln KZT

Let us return to the analysis of accretion process on compound interest and assume that is more often than once a year, and capitalization. 2.3. Effective and nominal interest rates

If interest is charged, and not added after a year, and more often (m times a year), then there is intra-capitalization of interest. Accretion runs faster than a one-time capitalization. In such kind of situation in a financial transaction it is negotiated no rate for the period, and the annual interest rate (denoted by j), which is calculated based on the rate for a particular interest period (j /m). In this case, the annual base rate (j) is called the nominal in contrast to the effective rate ( i), which is characterized by a complete real effect (income) operations with accounting the intra-capitalization. The effective rate value provides the same result when calculating the interest once a year on it, and that m – fold expanded in a year at the rate of (j /m) (based on j). So, (1  i ) n  (1  j / m) mn ; for the year: (1  i )  (1  j / m ) m ; hence (2.3.1) i  (1  j / m) m  1;

HAP F I N A N C I A L M A TCE M A T ITER CS 2

j /m  m 1 i 1.

23

(2.3.2)

Example 1. Bank charges on deposit 8% of the nominal. What is the real rate of return of contributions (effective rate) in the calculation of interest: a) quarterly; b) semi annual

Solution. a) The real rate of return on a quarterly determined by the formula (2.3.1) 0,08 4  iкв  (1  j / m) m  1  (1  )  1  0,082432;  i кв  8,25 %, 4 similarly define the real profitability on half-year: b)  iпг  (1  j / m ) m  1  (1  0,08 ) 2  1  0,0816 ;  iпг  8,16 % 2

In foreign literature, the calculated rate on the deposit b) indicates APR. According to the formula (2.3.1) has calculated effective profitability any financial operation, not even related with compound interest. Thus for j takes the given or perivousl accrued profitability of the formula (2.3.1), and then m  Y / t – a possible number of times repeating this operation for a year. Accrued amount for intra capitalization m time is determined by the formula (2.3.3) S  P  (1  j / m ) mn where j/m – rate for the period, calculated on the basis of the base (nominal rate j ) and the number of times per year interest charge (m). mn – the number of interest periods, calculated on the basis of the number of times accrued percent per year (m) and the period of the financial operation (n) in years.

24

FINANCIAL MATEMATICS

If the period of the financial operation is not determined in years, the formula has another form:

S  P  (1  j / m)

m t

Y

(2.3.4)

Example 2. Debit card is accrued and added interest quarterly based on 9%. We determine how much money will have the cardholder after 7 months, if she has provided 500 U.S. dollars. Solution. Using the formula (2.3.4), we obtain S  P  (1  j / m)

m t

Y

 500 (1  0.09 : 4 )

4 7

12

 500 (1  0.09 : 4 )

2

1 3



 522 U .S . dollars

In our example, interest is charged only for the full period, so 1/3 of the degree should be rejected (i.e., seven months have passed 2 full interest period – 2 quarters). There are cases happened, when a fractional part of the interest period as interest is charged one way or another. When using the same interest rate compounding would be more than often occurred capitalization . We take for example the interest rate of 12% per annum, a sum of money in 1 CU and on the year m  1, 2, 4 (quarterly), 12 (monthly), then S, according to (2.3.3) will be respectively: Capitalization (number of times Accrued for the year cost of per year interest charge) sum in 1 CU 1 1.1200 2 1.1236 4 1.1255 12 1.1268 2.4. Continuous charged percent

In developing the investment decisions in the project analysis sometimes take, that m   is continuous charged percent at the end of any number of small time intervals. Rate for as much as a small

F I N A N C I A L M AC T EHAP M A TTER ICS 2

25

period of time is called the power of growth, and calculate the value of accrued as follows:

S  P  e j n ,

(2.4.1)

where e – mathematical (exponential) constant, and e j n - the value, mn j  that is sought by  1   at m   . m  For comparing interest rates with different frequency of interest charged each other they often lead to the corresponding rates with continuous charged percent. Example. Rate it best: get 20000 KZT after three years 68,000 KZT in 7.5 years, if you can put money on a continuous rate of 28% per year? Solution. Use the formula of accretion for continuous charged interests S  P  e j n . We express this formula P:

Р1 

S 1 20000  3.0.28  8634, KZT . e n е

In the same way we find Р2 

S2 68000  7 . 5 . 0 . 28  8321 , 037 KZT n e е

Obviously, the first option, namely, 20,000 KZT after 3 years is more profitable. 2.5. Discounting of compound interest rate

Determination of S and P is called a direct account. Accordingly, the reverse calculation, called accounting "100" gives the meaning of the present value of money. Formula of mathematical accounting for the compound interest rate has the form of:

26

FINANCIAL MATEMATICS

P

S (1  i ) n

(2.5.1)

Admeasurement (1  i)  n  pn; i are called discounting or accounting factor. Its values are tabulated according to the discounting period ( n ), i.e. the number of interest periods and the discount rate (i). Example 1. The redemption price of a 3-year bond with no coupon payments (the so-called zero-coupon) is 1000 CU Period of up to 2 years polgasheniya bonds. Calculate its current value if the financial market has developed a two-year rate of return – 7%.

Solution. Using formula of mathematical accounting (2.5.1), we obtain

P2 

S 1000   873,44 CU . n (1  i) (1  0,07 ) 2

These calculations have of great practical significance in the analysis of the project to bring the money, evaluated as at different dates (usually it is the future value of money), at one moment of time (such as the present). Example 2. We determine the current value of money, the future value of which is estimated in 10 periods in 2000 CU Discounting rate – 3% for the period.

Solution. According to the formula (2.5.1), we obtain S 1  S  p 10 ; 0 , 03  2 000   2 000  0 ,74409  (1  i ) n (1  0 ,03 ) 10  1488 ,18 CU P

The repeated accounting discount securities (accounting and reaccounting) under the same conditions on the compound discounting rate d is:

F I N A N C I A L M ACTHAP E M A TER TICS 2

P  S (1  d ) n

27

(2.5.2)

where d – the discount rate; n – until the end of the financial transaction, which is equal to the number of times records. Discounting rate on the compound accounting is widely used not only in banking deals, but also in the activities of other financial institutions, such as insurance. Example 3. Let customers in the 50-year old signed a contract of insurance on survival for 5 years in the amount of 1000 KZT. We believe that all such contracts are concluded lived to 50 years. Let us assume that the normal rate of return in the period of insurance contract is 5% (0.05). Discount factor at a normal rate of return of 5% for the appropriate use of funds, the insurance is calculated as follows: V

1

V

2

1  0 . 952 ...; 1  0 . 05 1   0 . 907 ...; ( 1  0 . 05 ) 2





V

5



1  0 . 783 ...; ( 1  0 . 05 ) 5

n



1 . ( 1  0 . 05 ) n



V

As we see in Table 2.5.1, the number of persons aged 50 years (L 50) is equal to 87,064, the number of individuals surviving to age 55 (L 50 +5 ) is 82 827. Since the insurance amount ( S ) for each contract is 1,000 KZT, the insurance fund for payments at the end of the insurance period must be:

l 50  5  S  82 827  1000  82 827 000 KZT

28

FINANCIAL MATEMATICS

Since the fund will be in the back of investment, calculate the present value, future payments using the discounting factor: l 50  5  S  V 5  82 827 000  0 ,7835262  64 897 124 KZT Table 2.5.1 One-time net rate on survival by population groups Populations group The entire population Men Women Urban (Total) Men Women Farming (Total) Men Women

The number of The number of One-time net persons aged 50 surviving to age 55 rate on survival l 50 l 55 5 ε 50 87064 82827 745.396 81546 75503 725.4624 92837 90397 762.9330 88000 83879 746.8340 82820 93269 84091

76897 90884 79592

727.4911 763.4904 741.6063

77697 91440

71427 88861

720.2971 761.4273

Since the payment will be made only dozhivshim to 55 years, a one-time net rate on survival will be:

64 897 124  745, 39561 KZT 87 064 In general, this calculation can be done by the formula

n x 

l xn  V n  S, lx

where nεx – a one-time net rate on survival of the insured aged x years on n years; lx+n – the number of people surviving to the end of the insurance (in the example 82 827 people); V n – discount factor corresponding to the normal rate of

F I N A N C I A L M A TCEHAP M A TTER ICS 2

29

return and the term of the contract (in the example 0.7835262); lx – the number of people at the beginning of insurance (in the example 87 064 persons); S – the insured sum. One-time net rate on survival will fluctuate for the populations groups in the tables used to calculate mortality rates. Provided that: the insurance period – 5 years the rate of return – 5% on 1000 KZT insurance amount, the lump net rate on survival will have the meanings given in groups. 4th table. 2.5.1. Net rate for customers in the urban and rural population, as well as by gender of the insured are different. 2.6. The Calculating term of payment and interests rates

FER methods allow to determine the required period at the end or at the beginning of the financial operation and its extinction or size of required a particular interest rate, if given the parameters ( P , S, I ( j ), D , m , or P , S , n , m ). To do this, the above identities are solved relatively n or i ( j ), d . For example, the formula of returns rate in a compound annual rate of interest will have another form:

in

S  1. P

(2.6.1)

Example 1. A financial instrument is placed for a period of 2 years at a price of 1 million KZT. And redeemed at a price of 1.2 million KZT. Compute the rate of return in the form of compound interest. i2

1 200 000  1  0,0954 (9,54% 1 000 000

per annum.

30

FINANCIAL MATEMATICS

Also calculated figure, called "real accumulated income of investments" through the purchase of shares as well as bonds. In this case, the formula (2.6.1) in the numerator is the sum of all accrued investment and reinvested earnings from them, and the denominator – the initial investment. Their relationship shows the relative growth for the entire period (" n ") of investments, and the extraction of the root level " n " and the subtraction of "1" makes annual interest. Example 2. The initial size of capital expenditure for 4 years was 1,000,000 KZT accrued amount of dividends received annually (assuming their investment within 4 years) 100 000 KZT, while the value of investments by the end of the period is 2.5 million KZT. We determine the real accumulated return. i4

2 500 000  100 000  1  0,2698 ( about 27% per annum ). 1 000 000

Now to find out which forms of payments (percent) more favorable for the depositor (creditor) and the borrower. For this calculation we make the future value of money by various methods discussed above, compare them and draw conclusions. Let the loan or the amount of money equal to the revenue of 100 CU, interest dekursivnaya rate – 0.03, anticipative (discount) rate – 0.03, during the build-up – 5 years. P  100 СU i  0,03 (3% per annum ) d  0,03 (3% per annum ) n  5 years

1. Determining the future value at the rate of dekursivnoy i: a) for simple interest

F I N A N C I A L M ACT EHAP M A TTER ICS 2

31

S  P(1  ni)  100  (1  5  0,03)  115 CU b) for compound interest

S  P (1  i ) n  100  (1  0,03) 5  115,93 CU 2. We find the income paid at the time of the loan or advance on your staff: a) simple interest P 100   117 ,65 CU; 1  nd 1  5  0,03 I  S  P  117 ,65  100  17 ,65 CU;

S

b) for compound interest P 100   116 ,45 CU; n (1  d ) (1  0,03) 5 I  S  P  116 ,45  100  16 ,45 CU.

S

The Ratio of simple dekursivnyh rate and advance percent imply that if these rates are equal, ie, d  i the value of the annual income does not change. For example, the annual income of 15 CU can be obtained from the original amount of 100 CU , given for 5 years at a dekursivnoy rate i  0,03 and income in 15 CU can be obtained, providing 85 CU with a discount rate d  0,03 . But as the interest income on advances obtained at the time of the loan, giving the amount of the debt, then, consequently, interest on advance payment to the lender is more profitable than the debtor. This advantage is increased if payment is not for one year, and for several years. In our example, the income of 16.45 CU creditor receives the amount giving a loan of 100 CU at the time of debt with compound interest at the advance years n=5 and d  0,03 revenue of 15.93 CUFor years n  5 and when i  0,03 decursive compound interest rate.

32

FINANCIAL MATEMATICS

Advantage of various forms of interest associated with the period of compounding money. If compare the fund accretion on ordinary decursive interest rate "to 100" [[ S  P (1  ni ) ] ] and compound decursive interest rate "with 100" [ S  P (1  i ) n ], then the money compounding periods longer than 1 year ( n > 1) compound interest is more "stronger" than the simple ones. Thus, when i  0,03 , n  5 and by simple interest income for the five years from 100 CU is equal to: I  P  n  i  100  5  0,03)  100  0,15  15 СU for compound interest: I  P (1  i ) n  P  100 (1  0 ,03) 5  100  15 ,93 CU

However, the build-up period of less than one year simple dekursivnye interest is "stronger" than compound. Prove it. Let i  0,03 , P  100 CU, t  1 : 2 year . We determine S: Y a) simple dekursivnym interest:

t  1  0,03    S  P 1  i   100 1  i   101,5 CU Y  2    b) interest on dekursivnym coumpound: S  P1  i 

t /Y

 1001  0,03

1/ 2

 101,49 CU

Consequently, when compounding periods less than one year to the lender simple dekursivnye interest is preferred. Comparing compound dekursivnye and coumpound advanced interest, it can be concluded that the accretion of money according to the rule "100" on the coumpound advanced interest is "stronger" than

F I N A N C I A L M ACT HAP E M A TTER ICS 2

33

the accretion on compound dekursivnym on the rule "to 100", because with equal dekursivnoy and interest rates ( i  d ) value

S

P greater value than the S  P (1  i ) n n (1  d )

"100"

"to 100"

If we compare the advance interest, we can see that the compound advanced interest on the rule of "100" at a period longer than one year ( n > 1) "weaker" than the simple advanced according to the rule of "100." In our example, d  0,03 , n  5 and P = 100 CU

S

P  117 ,65 СU 1  nd ("100" simple)

and

S

P  116,5 CU (1  d ) n ("100" complex)

However, when compounding periods less than one year coumpound advanced is "stronger" than the simple interest upfront. For example, P = 100 CU , year. d  0,03 t / Y  1 / 2 year Then P 100 S   101,52 СU t 1 1  d 1   0,03 Y 2 ("100" simple)

34

FINANCIAL MATEMATICS

S

P 100   101,53 CU t /Y (1  d ) (1  0,03) 1 / 2 ("100" complex)

These features and relationships are taken into account in the implementation of loan, deposit and other operations. Questions and problems

In questions of 2.1. – 2.7. from the list of the responses choose the correct one. 2.1. What is meant by compound interest: a) variant of calculating interest, in which the base year of measuring the time taken, conventionally consisting from 365 or 366 days, and the number of days in each month of the loan shall be 30; b) calculation variant, in which the accrued interest is attached with the amount of loan, and the amount received is the basis for the next calculation of interest; c) variant of calculating interest, in which capitalized interest is produced; d) calculation version, in which the accrued interest is attached with the amount of loan, and the amount received is not a base for the next calculation of interest? 2.2. Specify formula Building on compound interest rates: a) S  Pn (1  i ) ;

b) S  P n (1  i ) ; c) S  P (1  i ) n ; d) S  P (1  n i ) n .

F I N A N C I A L M AC T EHAP M A TTER ICS 2

35

2.3. Specify the mathematical formula for the discounting rate on the compound interest rate: a) P  S (1  i )  n ; b) P  S (1  n d ) ;

c) P  S (1  n i ) 1 ; d) P  S (1  d ) n ? 2.4. Specify the formula for bank accounting on compound discount rate: a) P  S (1  iсл )  n ; b) P  S (1  n d сл ) ;

c) P  S (1  n d сл ) 1 ; d) P  S (1  d сл ) n . 2.5. What happens to the augmented amount, if the frequency of compounding interest increases on the interest rate? a) value augmented amount does not change; b) value augmented amount increases; c) value augmented amount is reduced; d) value augmented amount varies little? ' 2.6. Effective annual discount rate provides the same result as the a) discounting once a year at a nominal discount rate; b) discounting on the compound discount rate several times a year; c) discounting a few times a year at a nominal discount rate, divided by the number of periods of discounting; d) accretion on the compound discount rate. 2.7. Difference between anticipative and dekursivnym accrual method disappears when a) calculation of simple interest;

36

FINANCIAL MATEMATICS

b) calculation of compound interest; c) when mixed scheme calculation; d) with continuous compounding percent. Solve problems 2.8 – 2.37. 2.8.

2.9.

2.10. 2.11.

2.12. 2.13.

2.14.

What is more profitable: get 4.6 thousand KZT 4 years or 5.2 thousand KZT after 5 years, if you can put money on deposit under coumpound interest rate of 16%? Assess the situation from the perspective of the future, and from the perspective of the present. Determine under which compound interest rate money can be put on deposit, if 10 thousand KZT. Now would be equivalent to 37.129 thousand KZT after 5 years. What will be the answer if bank pays compound interest quarterly? Determine the nominal annual interest rate if the effective rate is 30% and compound interest is calculated: a) quarterly b) monthly. Accrued by the end of the seventh year, the amount of 240 thousand KZT. Find its current value if the compounding interest is calculated: a) the half-yearly on the interest rate of 30% per annum, and b) on a quarterly basis at an interest rate of 40% per annum .. How much money is preferable for compound interest rate of 29% per year: 100 thousand KZT today, or 700 thousand KZT after 8 years? Loan agreement signed for 4 years. The initial amount of the debt is 2000000 KZT. The agreement provides for a variable rate of compound interest, which is defined as 10% per annum in the first year, every year thereafter the rate will increase to 10%. Determine the amount accrued at the end of the contract term. A client placed in bank 100 thousand KZT for 5 years at an interest rate of 36% per annum. Determine accrued during that time the amount in the calculation of compound interest: a) annually b) by half year, and c) a quarterly basis, and d) a monthly e) weekly e) daily. Believe a year of 360 days.

F I N A N C I A L M ACTHAP E M A TER TICS 2

37

2.15. Determine the time for which there is a doubling of the original amount in the calculation of simple and compound interest, if the interest rate is equal to: a) 5% b) 10% c) 15% d) 25% e) 50% e) 75% f) 100%. 2.16. How long does the initial amount of 20 thousand KZT increase to 60 thousand KZT, if compound interest at an interest rate of 28% per annum calculated: a) annually b) quarterly c) each month? 2.17. Depositor would like 4 years to double the amount that is placed in the bank for deposit. What kind of annual nominal interest rate should the the bank request in calculating compound interest on a quarterly basis? 2.18. You have the ability to get a loan or on conditions of 32% per annum with monthly compound interest, or under the 33% per annum with quarterly compound interest. Which option is preferable if the interest payment will be made at a time along with the repayment of the loan? 2.19. Calculate the effective interest rate if the bank pays interest monthly and capitalizes percent on the basis of a nominal rate of 40% per annum. 2.20. Firms need to save $ 2 million, so that after 10 years to purchase the building as an office. The safest way is to purchase a risk-free accumulation of government securities, generating annual revenue at the rate of 8% with semi-annual interest calculations. What should be the initial contribution of the company? 2.21. Promissory note at the amount of 1 million KZT, maturity of which comes after 3 years, counted by the bank on the compound accounting rate of 10% per annum. Determine the discount. 2.22. Construction company sells an apartment worth 450 thousand KZT in loan under coumpound interest rate of 25% per annum. The same firm established bank, accumulating funds for the construction of apartments and is payable on the money placed in a compound interest at an interest rate of 25% per annum. Mr. N made in the bank some money for 3 years before buying an apartment, the same amount – at the time of buying an apartment, another

38

2.23.

2.24.

2.25.

2.26.

2.27. 2.28.

FINANCIAL MATEMATICS

70 thousand KZT – after 2 years and 120 thousand KZT – after 3 years from the date of purchase of an apartment, paying off his debt thereby completely. Determine how much money Mr. N brought to the bank till and at the time of purchase of an apartment. Construction company sells an apartment worth 520 thousand KZT on credit under compound interest rate 20% per annum. The same firm established bank, accumulating funds for the construction of apartments and is payable on the money placed in a compound interest at an interest rate of 20% per annum. Mr. N introduced in the bank 100 thousand KZT a year before getting an apartment and another 150 thousand KZT. – after 2 years receiving the apartment. A year later, after he had made some money, and a year to pay off debts, making 300 thousand KZT. Determine how much money Mr. N made the bank one year after receipt of the apartment. Bill was taken into account for the 21 months till the maturity, and the owner of the promissory note was 0.8 times the amount written on the bill. On what kind of annual compound discount rate was taken into account this bill? Which discount conditions when discounting the compound discount rate are more beneficial to the bank: a) 32% of annual, semi-annual discounting, and b) 33% per annum, quarterly discounting? In 4 years till the maturity of debt obligation, and its owner has received 5 thousand KZT. Determine the amount written in liabilities if the accounting is performed on a compound discount rate and discounting makes: a) half-year lending rate by 40% per annum, and b) on a monthly basis at the discount rate 30% per annum. How much money will be put into a bank deposit to 5 years to get 80 thousand KZT, if there is a continuous interest charges at a rate 22%? Bank provides loan for 9 years under a compound interest rate 32% per annum with calculating interest every quarter. What continuous rate should the bank establish to 9 years to get the

F I N A N C I A L M AC T EHAP M A TTER ICS 2

2.29.

2.30.

2.31.

2.32.

2.33. 2.34.

2.35.

2.36.

39

same income? Will it change the result, if the term of the loan is 3 years? Bank provided a loan for 6 years at a continuous rate of 27% for the year. Determine the profitability of such a financial transaction for the bank in the form of: a) simple annual interest rate, and b) the annual effective interest rate. Providing a loan for 5 years at a continuous rate. Determine the value of this rate, if the return of the transaction to the lender in the form of an annual effective interest rate was 38%. Does the value of the continous rate depends on term loan? Entrepreneur can get a loan whether on the terms of the quarterly compounding interest rate 36% per annum, or in conditions of continuous accrual of interest with intensity of 34% for the year. Which option is better for an entrepreneur? Depositor would like to increase the amount by 2.5 times in b years that is placed in the bank for deposit. What should be the strength of growth, if the bank charges a continuous interest? What should be the strength of growth in order to ensure an increase in the amount if exposed to 4 times? Rate it best: get 20 thousand KZT after 3 years or 68 thousand KZT after 7.5 years, if you can put money on deposit at a continuous rate of 28% per year? Under what kind of continuous rate can put the money on deposit, if 10 thousand KZT of current is equivalent to 30 thousand KZT after 4 years? What kind of a compound interest rate with calculating interest in half year solves this problem? Put some capital in the bank at a continuous rate of 30%. After 2 years and 3 months old account was closed and received the sum of 189.755 thousand KZT. Determine the value of augmented amount that would have been obtained and a half years. Investor put it in the bank 8 thousand KZT under the accrual of interest on the continuous growth power of 26%. After a year and a half investor removed from the account 5 thousand

40

FINANCIAL MATEMATICS

KZT, and 2 years after that, he put seven thousand KZT. After 2 years and 6 months investor put a sum that in his account in a year was 60 thousand KZT. Determine how much money the investor put last time. 2.37. After 10 years, the liability will be paid at a sum of 10 million KZT. Determine the present value of liability, provided that the compound interest rate is applied 10%.

FINANCIAL MATEMATICS

CHAPTER

41

3

Accrued amount of payment flows

3.1. Classification and derivation augmented the amount of rent

Flow of payments, all of the members that are positive, and the time intervals are constant are called the financial rent or annuity. Financial annuity has the following parameters: a member of the rent – the value of each individual payment, Rent period – the time interval between two consecutive installments, Rent time period- is measured from the inception of the rent until the end of his last period, Interest rate – the rate used in the compounding and discounting charges forming the rent. Depending on the duration of rent period is divided by the annual and p-term, where p – the number of payments per year. According to the number of charges interest distinguishes rent with charge once a year, m times continuously. Moments of interests charges cannot coincide with moments of rent payments. In probability paying members distinguishes faithful and contingent rent. Faithful rent is unconditional payment, such as repayment of the loan. Contingent rent payment is made conditioned from some random event. Therefore, the number of its members is

42

FINANCIAL MATEMATICS

not known beforehand. For example, the number of pension payments depends on the life expectancy of a pensioner. According to the number members distinguish the rent with a finite number of members, or limited, and endless or eternal. As an eternal act of rent, for example, payments on bonds with unlimited or non-fixed terms. Depending on the availability of starting shift of rent relative to the beginning of the contract or at any other time of the rent is divided into immediate and delayed or deferred .Immediate rent term begins immediately of pending overdue. Rents are distinguished by the moment of charges payment. If payments are made at the end of each period, such rents are called usual, or postnumerando rents. If payments are made at the beginning of each period so rents are called prenumarando rents. Sometimes a payment is provided in the middle of the year. Largest members of rent distinguish constant (with equal members) and variable rent. If the amounts of payments vary on any mathematical law, they often appear to bring the standard formula, which greatly simplifies the calculation. Accrued amount of cash flow – is the sum of all its elements with the interest accrued thereon. Consider the formation process of augmenting the amount of rent postnumerandno. Imagine a stream of payments shown in the figure. Payments: R1 R2 R3 ….. Rn Years: 0

1

2

3 ….. n

Rn

R3(1+i)n3

R2(1+i)n2

Fig. 3.1

R1(1+i)n1

F I N A N C I A L MC A HAP T E M ATER T I C S3

43

As it can be seen from fig. 3.1, payments completed at the end of the first year, the end of the first period rent will increase by (1 + i) n -1 times and becomes a sum of R 1 (1 + i) n -1; payment will increase in the second year (1 + i ) n -2 times and becomes the sum of R 2 (1 + i ) n -2 , etc. The general formula for the value of rent element, the year t at the end of the rent period is expressed as R t (1 + i ) n – t . The amount of rent may be determined by direct counting formula n

S   Rt (1  i ) nt ,

(3.1.1)

t 1

where n – number of the rent members; R t – value of the rent element in the year t; i – used in the calculation of the rate of compound interest. Since, by the definition of financial rent R1  R 2  R 3  ...  R n , then let the value of each element in R , we get: n

S  R (1  i) n t ,

(3.1.2)

t 1

Suppose that we create a fund, which contributes an annual fee of size R. These contributions accrue interest at a rate per annum complex in the amount of i. After years sum of n in the fund will be S. Using the formula (3.1.2) in practical calculations with a significant period of rent is very time-consuming. In order to simplify the calculations, we use a different approach. As already noted, by the end of the first payment of rent period will turn into R 1 (1 + i ) n -1 , the second in the R 2 (1 + i ) n -2 , etc. If you place these values in reverse chronology, we obtain the following sequence: R , R (1 + i) , R (1 + i ) 2 , ..., R (1 + i ) n -1 . We see that this sequence is a geometric progression with first member R and the denominator (1 + i ).

44

FINANCIAL MATEMATICS

In general, the sum of a geometric progression with the number of elements n is determined by the formula

S n  b1 

qn  1 q 1

(3.1.3)

where b 1 – the first element; q – denominator geometric progression. In our case, like this b1  1, q  1  i , so S  R

(1  i ) n  1 (1  i ) n  1  R i (1  i )  1

(3.1.4)

This formula simplifies calculations of the rent amount augmenting. Denote the factor by which to multiply R after s n ; I ; index n; I indicates the duration of the rent and the value of interest rate. In the future, this factor is called compounding factor rents. This coefficient factor is itself the amount of accrued rent, the term of which is 1. n 1

S n; i   (1  i) t  t o

(1  i) n  1 i

(3.1.5)

Thus, S  R  s n; i

(3.1.6)

As you can see, the coefficient Building rent depends on the length (number of members of the rent), and interest rates. With the increase of each of these parameters, its value increases. When, i  0 S  R n at n  1 S  R . Easy to tabulate the values of the coefficient. Example. During the 4 years in the account at the end of each year comes 50 thous.KZT, on which once a year charges interest at

F I N A N C I A L M ACTHAP E M ATER T I C S3

45

compound annual rate of 10%. Required to determine the amount of the current account by the end of that period. Solution. Using (3.1.4), we obtain S  R

(1  i) n  1 (1  0,1) 4  1  50 000   50 000  4,641  232 050 KZT i 0,1

3.2. Annual rent with interest charged m times a year

Let's see how complicated formula, if rent payments are made once at the end of the year, and interest is charged m times a year. This means that each time the rate is applied j/m, where j – the nominal rate of interest. Then the terms of rent with charges until the end of interests have the form R (1  j / m) m ( n 1) , R (1  j / m) m ( n  2 ) ,..., R

If we look at this sequence from right to left, so we will see that this is a geometric progression again, the first of which is R , the denominator (1  j / m ) m , and the number of members equal to n . Sum of the terms of this progression will be augmented the amount of rent:

SR

(1  j / m) mn  1 (1  j / m) m  1

(3.2.1)

Example. To create a fund company invests annually to the Bank by 24 thousand KZT under the annual nominal interest rate of 32%. We define the amount that would be accumulated in the fund after 8 years if the payments are made at the end of the year, and compound interest charged by quarter.

Solution. Using (3.2.1), we obtain

46

FINANCIAL MATEMATICS

(1  j / m) mn  1 (1  0,32 / 2) 28  1  24000   (1  j / m) m  1 (1  0,32 / 2) 2  1 9,748004204  24000   676944,7364KZT 0,3456 SR

3.3. Rent p – urgent with interests charged once a year ( m  1 )

In this case, the payments are made several times during the year (by quarter, quarterly, monthly, etc.), but the interests on these payments charge once a year. If R – annual amount of payments, the payment is equal to the size of an individual R/p. During the period of rent made np payments. Then the sequence of payments with the charged interest till the end of the interest term is also itself a geometric progression, written in reverse order: by the end of the last period of rent payment will R R R be R/p, the penultimate  (1  i ) 1 / p . Then  (1  i) 1 / 2 p ,  (1  i ) 1 / 3 p p p p etc. So the amount of considered rent is a sum of terms of this geometric progression: 1

np

R (1  i ) p  1 (1  i ) n  1 S    R 1 1 p   (1  i ) p  1 p  (1  i ) p  1   

(3.3.1)

Example. Entrepreneur from investing in some projects will be for four years to get to the end of each semester 12 thousand KZT. We define the possible amount that will get entrepreneur after 4 years, if possible to put money in the bank at the complex interest rate 24% per annum with interest annually.

Solution . Using (3.3.1), we obtain

F I N A N C I A L M ACTHAP E M A TER TICS 3

47

(1  i ) n  1 ( 1  0 , 24 ) 4  1  24000   1 1     p 2 2  ( 1  0 , 24 )  1  p  (1  i )  1      4 ( 1, 24 )  1  12000   144 167 KZT 1   2 ( 1 , 24 )  1     S  R

3.4. Rent p -term, p  m

Contracts often charge interest and receipt of payment for the same time. Thus, the number of payments p in the year and the number of charges per cent m in the year are the same, i.e. p=m .Then, to obtain the sum can be augmented to use the analogy with the annual rent and a one-time interest charged at the end of the year, for which (1  i ) n  1 SR i The difference is only in the fact that all of the parameters presently characterize the rate and payment for the period, not per year. Thus, we obtain S

R (1  j / m) mn  1 (1  j / m) mn  1  R m j/m j

(3.4.1)

Example. Under the previous example, assume that an employer will put money in the bank [a complex interest rate 24% with interest charged every six months and we determine possible amount they receive through 4 years.

Solution. Using (3.4.1), we obtain

48

FINANCIAL MATEMATICS

S R

(1  j / m ) mn  1 (1  0,24 / 2 ) 8  1  24000   147 596 KZT j 0 ,24

3.5. Rent p -term, with p  m

This is the general case of p -term rent with interest charged m times a year. Member of the rent is R/p. The sum of the last term rent interest is not charged. Penultimate term rents will be in the fund of the year 1/p, payment prior to the penultimate, – 2/p part of the year, etc. Building coefficient on payment that is in the fund of one year will be equal (1  j / m) m . And then respectively, coefficient for the last term rent will be (1  j / m ) m / p

for the previous –

, then – (1  j / m ) We obtain a geometric (1  j / m ) progression with first term R/p and common term (1  j / m ) m / p . In result we get is an accrued amount. 2m / p

S

3m / p

R (1  j / m) ( m / p ) np  1 (1  j / m) mn  1   R m/ p p (1  j / m) 1 p[(1  j / m) m / p  1]

(3.5.1)

Example. Entrepreneurs with the goal to purchase equipment, do at the end of each quarter, equal contributions to the bank under the annual nominal interest rate of 28%, with compound interest calculated by quarter. How big should each contribute to a entrepreneurs could save 250 thousand KZT for three years.

Solution. Using (3.5.1), we can express the value of R / p :

3.6. Annual Rent with continuous calculated Interests

Denote a continuous rate q. We obtain a geometric progression with first term R, the denominator of e q. Sum annual rents with continuous calculated interests are determined by the formula:

FINANCIAL MC A THAP E M ATER T I C S3

S  R

e qn  1 eq 1

49

(3.6.1)

Rent k – urgent with continuous calculated interests. Equation (6.6.1) is transformed as follows:

S  R

e qn  1 k (e q / k  1)

(3.6.2)

Example. Available with an annual rent payment of 100 million KZT. Interests are calculated continuously at 20% per annum. Term rent – 5 years. Find the amount accrued for the cases: a) the annual rent; b) quarterly rent. Solution. Using (3.6.2), we obtain:

e 0 , 25  1  777 ,4 m ln . KZT e 0 , 2  1) e 0 , 2 5  1  837 ,8 m ln . KZT б) S  100  0 , 2 / 4 e  1) а) S  100 

Questions and Problems

In questions of 3.1. – 3.8. from the list of the responses choose the correct one. 3.1. Which of the following operations can be attributed to the notion of "stream of payments": a) number of consecutive payments and receipts; b) number of consecutive payments; c) number of consecutive revenue; d) number of consecutive accrual of interest on a loan agreement? 3.2. What is the "sum of the accrued rent": a) the sum of all terms of the payment sequences;

50

FINANCIAL MATEMATICS

b) the sum of all payments; c) the sum of all income; d) the sum of all terms of payments sequences with interest calculated on them at the end of the term rent? 3.3. What is called financial rent: a) stream of payments, all of terms that have positive values; b) stream of payments, which are constant time intervals; c) regular payments made by the debtor to repay the debt; d) stream of payments, all of whose positive values, and the time intervals are constant? 3.4. What is rent postnumerandno: a) rent payments formed after a specified moment of time; b) rent payments that come at the end of each period; c) rent payments that is adjusted for audited inflation; d) rent payments that is adjusted by the amount of the tax? 3.5. Specify a coefficient building the ordinary annual rent with a single calculated interest in the year: (1  i ) n  1 a) ; i 1  (1  i )  n b) ; i (1  i ) n  1 ; c) (1  i ) m / p

d)

1  (1  i )  n . (1  i ) m / p

3.6. If the expected rate of return is constant for 10 years, two years after the completion of the investment, this revenue stream is a) constant, postponed, limited rent; b) constant, immediate, limited rent; c) variable deferred, and limited rent; d) constant, postponed, eternal rent.

F I N A N C I A L M ACTHAP E M ATER T I C S3

51

3.7. What is rent prenumerandno: a) rent payments formed after a specified moment of time; b) rent payments that come at the beginning of the period; c) rent payments that is adjusted for audited inflation; d) rent payments is adjusted by the amount of the tax? 3.8. Specify the p -term rent when p = m . : (1  j / m) mn  1 a) S  R ; j

b) S  R

(1  i ) n  1 ; 1   p (1  i ) p  1  

(1  i ) n  1 ; i (1  j / m) mn  1 d) S  R  . p[(1  j / m) m / p  1]

c) S  R 

Solve the problems 3.9-3.30.

Annually postnumerandno to the fund paid on 10000 KZT for 20 years, on which compounding annual interest 10% is calculated on a quarterly basis. Determine the amount accrued at the end of the term. 3.10. Annually to the fund on 10000 KZT for 20 years. Payments are made in equal installments at the end of each quarter. Compound interest at the rate of 10% per annum calculated quarterly. Determine the amount accrued at the end of the term. 3.11. For 20 years, the account at the end of each year goes on 10000 KZT. Annual discounting is made on compounds rate of 10% per annum. Determine the present value of rent. 3.12. 1.2. You offer to lease land for five years, by selecting one of the two options for payment of rent, a) 15 thousand KZT. – at the end of each year; b) 130 thousand KZT. – at the end of 3.9.

52

3.13.

3.14.

3.15.

3.16.

3.17. 3.18.

FINANCIAL MATEMATICS

five years. Which option is preferable, if the bank offers 24% per annum for deposits? In which pay at the end of each year, both options are almost practically equivalent? Entrepreneur of the investment in some projects will be for four years to get to the end of each quarter 12 thousand KZT. Identify potential amounts that in four years will receive an entrepreneur, if you can put money in a bank at a complex interest rate 24% per annum with interest: a) annually b) every six months, and c) a quarterly basis? During 6 years every six months to the bank paid 10 thousand KZT scheme: a) postnumerando b) prenumerando. Bank pays compound interest semi-annually at the rate of 20% per annum. How much money is in the account at the end of the term? Analyzed two variants of the accumulation of funds for the annuity scheme postnumerando: a) to put a deposit of 30 thousand KZT every six months, provided that the bank pays 18% with semi-annual compound interest, and b) to make an annual contribution of $ 63 thousand KZT under the 19% per annum, with an annual compound interest. How much money is in the account after 10 years with the implementation of each plan? Which plan is more preferable? Will it be your choice, if the interest rate in the second plan will be reduced to 18.5%? To create a fund company invests annually to the Bank by 24 thousand KZT under the annual nominal rate of 32%. Determine the amount to be accumulated in the fund after 8 years, if: a) the contributions made at the end of the year, and compound interest charged by quarter, b) contributions are made in equal installments at the end of each month (i.e. two thousand KZT ) and compound interest accrued quarterly c) contributions are made in equal installments at the end of each quarter (i.e., 6 thousand KZT) and calculated interest is continuous. Should I buy for 5500 KZT valuable papers, generating an annual income of $ 1,000 KZT for twenty years, if the bank offers a complex interest rate of 18% per annum? Bank offers ordinary annuity (postnumerando) rent for 10 years with quarterly payment of 4 thousand KZT. The annual

F I N A N C I A L M A TCEHAP M A TTER ICS 3

3.19.

3.20.

3.21.

3.22.

3.23.

3.24.

3.25.

53

interest rate for the entire period is constant, and compound interest is calculated quarterly. At what price can buy this rent, if payments will be: a) immediately, b) 4 years c) after 5.5 years, and complex interest rate is 32% per annum? A client wants to save up for your account 80 thousand KZT., exercising at the end of each year equal contributions to the bank under a complex interest rate 30% per annum. How big should be the contribution of each client to be able to accumulate the required amount for: a) 5 years b) 10 years? Entrepreneur with the goal of purchasing equipments do at the end of each quarter, equal contributions to the bank under the annual nominal rate of 28%, with compound interest calculated by semester. How big should each contribute to a entrepreneur could save 250 thousand KZT for: a) 3 years b) 8 years with only the scheme of compound interest? In contributions to the Bank by 15 thousand KZT at the beginning of each quarter for 8 years accrued quarterly compound interest at 20% per annum. How much money is in the account at the end of the term? Annually at the beginning of the year the bank does the next installment in the amount of 14 thousand KZT. Bank sets the nominal interest rate of 36%. How much money is in the account at the end of six years, if the interest is compounded: a) annually b) quarterly c) each month? A quarterly contributions to the bank in the amount of 10 thousand KZT by ordinary annuity (prenumerando) scheme bank pays compound interest at a nominal interest rate of 24% per annum: a) once a year, and b) every six months. How much money is in the account after 3 years? Proposed to invest 300 thousand KZT for 5 years, subject to refund this amount units (annually 60 thousand KZT). After 5 years of paid additional compensation in the amount of 120 thousand KZT. Whether to accept this offer if you can deposit money in the bank at the rate of 20% per annum (complex)? A client put it in the bank 20 thousand KZT, intending to withdraw from the account at the end of each year, 5.5

54

3.26.

3.27.

3.28.

3.29.

6.30.

FINANCIAL MATEMATICS

thousand KZT. How long a client can withdraw money from the account, the bank charges a compound interest rate of 24% per year: a) annually b) quarterly c) continuously? How much money should be placed in a bank at a nominal interest rate of 32% per year to over it in 10 years and have the opportunity at the end of each year withdrawn from the account 7 thousand KZT exhausting the bill in full, if the bank pays compound interest: a) annually; b) the half-year, and c) continuous? How much money should be placed in a bank at a nominal interest rate of 24% per annum to 9 years to be able to receive annually 12 thousand KZT, taking the money in equal installments every 2 months, and at the end of the ninth year, due to exhaust completely if Bank calculates compounding interest: a) annually b) quarterly c) continuously? If the rational use of land for crops, it can generate an annual income (net of expenses) to 190 thousand KZT. However, in the area discovered oil field, the development of which will allow for three years, starting next year, to receive an income of respectively 300, 700 and 500 thousand KZT. For organization works for this year necessary investments at the amount of 150 thousand KZT. After pumping oil land would be assessed at 40 thousand KZT and unusable for agriculture. Applying a complex interest rate of 20% per annum, to conclude, as advantageous to use this land. How much money should be placed in a bank at a nominal interest rate of 32% per annum to an infinitely long time to be able to receive annually 80 thousand KZT, taking the money in equal installments every 3 months (i.e. 20 thousand KZT) if the bank pays compound interest: a) annually b) quarterly c) continuously? You can invest the same amount of money in one of two projects. The first project will provide a perpetual ordinary annuity (postnumerando) with annual installments of 15 thousand KZT. The second project for two years will bring 30 and 80 thousand KZT. Which of these projects better, if the interest rate is 24% per annum? Is it possible to change the interest rate so that the response was reversed?

FINANCIAL MATEMATICS

CHAPTER

55

4

Modern cost of the rent

4.1. Formula of the modern cost of a rent

The modern cost of a rent is the sum of modern costs of elements of a rent. The modern cost of an element of a rent is defined by discounting of its size on the beginning of the period of a rent. Let's say that we were given payments R1, R2, R3… Rn . Let's define how much is the sum of all elements (payments) of a rent in a zero point. For determination of Modern cost of a rent as it was already noted, it is necessary to make discounting of sizes of its elements or, otherwise, to find their modern costs. Let's lead the sums of rent payments to a uniform temporary point of the beginning of the period (see figure 4.1).

0 R1(1+i) -

R

R

R

1

2

3

R2(1+i) 2

R3(1+i) 3

Rn(1+i) n

Fig. 4.1

R n

56

FINANCIAL MATEMATICS

If on the basis of modern costs of elements of a rent to make sequence in direct chronology, we will receive R(1+i) -1, R(1+i) -2, R(1+i) -3, …, R(1+i) -n.

(4.1.1)

As it was already noted, the modern cost of a rent is the sum of the specified (discounted) sizes of elements of a rent for the beginning of period, i.e. n

A  R (1  i) t ,

(4.1.2)

t 1

It is a direct method of determination of modern cost of a rent. As a row (7.1.1) is a geometrical progression with the first element R(1+i) –1 and a denominator (1+i) –1 , we have:

A  R  (1  i) 1 

(1  i)  n  1 1  (1  i)  n   R , (1  i) 1  1 i

(4.1.3)

where n – number of years of a rent. For a case of an eternal rent ( n   ) we will consider a formula of 1  (1  i )  n 1  modern cost. From this that in a formula lim n  i i n expression (1  i) at n   , follows that the modern cost of an eternal rent will be defined so:

A

R i

(4.1.4)

Example 1. According to the credit agreement the total amount of a debt (with percent) is repaid by equal parts within 5 years equal payments of 1 million tenges. In calculations the difficult rate of 20% per annum is used. To find the main amount of debt.

Decision. On a condition of n =5, R=1 million tenges, i=0,2, payments on repayment of a debt represent a rent, which have a validity

57

F I N A N C I A L M AC T EHAP M A TTER ICS 4

period of 5 years. Required size of an initial (without percent) loan can be considered as the modern cost of a stream of payments on credit 5 repayment: A  1  1  (1  0,2) / 0,2  2,991 million tenges





Example 2. There is a termless bond worth 10 thousand tenges with the constant coupon rate equal to 20%. The average standard of profitability on securities market is equal 25%. To find the current cost of the bond. The current cost of the bond can be considered as the modern cost of the infinite rent presented by sequence of coupon payments.

Decision. We use a formula of modern cost of an eternal rent (7.1.4): as R  10 000  0, 2  2000 - annually gained coupon income, A

R 2000   8000 thousandte nge i 0,25

Example 3. Let's assume that on the basis of annual assignments in the sum of 100000 dollars. Within 10 years it is formed the fund for payment of awards in the field of financial mathematics. The rate of percent is 10%. To find fund size in 10 years.

Decision. Using a formula of the increased sum of a rent, we find





S  100 000  (1  0,2)10  1 / 0,1  1 590 000 dollars

Ratio between the increased sum and modern cost of a rent. Let's divide expression (6.1.4) on (7.1.3): S (1  i ) n  1 i (1  i ) n  1     (1  i ) n A i 1  (1  i )  n 1  1 (1  i ) n

Thus,

S  A(1  i)n ,

(4.1.5)

A  S (1  i) n .

(4.1.6)

58

FINANCIAL MATEMATICS

Example 4. The modern cost of rent postnumerandno with term of 5 years – 500 million tenges. The interest rate is accepted at the level of 15% per annum. To define the increased sum of this rent. Decision. Using a formula (4.1.5), we will receive

S  500  (1  0,15) 5  1005,7 mil.tenges 4.2. Annual rent with charge of percent of m of times in a year

If percent are charged by m times in a year, the modern cost of the first m member of a rent will make R(1  y / m) , second member –

R(1  y / m)2 m , third member – R(1  y / m)3m etc. We have a geometrical progression:

R(1  y / m)  m , R(1  y / m) 2 m , R(1  y / m) 3m , ..., R(1  y / m)  nm . It is easy to see that the first element of this geometrical progression m m is equal R(1  y / m) , a denominator (1  y / m) , and number of members of a progression of n. Let's designate through A the sum of modern costs of members of a rent, we will receive

A  R(1  y / m)m 

(1  y / m)mn  1 1  (1  y / m)mn  R  (1  y / m)m  1 (1  y / m)m  1

(4.2.1)

Example. It is created the fund in which within 5 years annual contributions of 100 million tg are made, with a quarterly charge on them difficult percent on a rate of 20% per annum. To find the modern cost of this cash flow.

Decision. Applying a formula (4.2.1), we will receive A  100 

1  (1  0,2 / 4)4 5  577,5 mil.tg. (1  0,2 / 4)4  1

F I N A N C I A L M AC T EHAP M A TTER ICS 4

59

4.3. The r-urgent rent with charge of percent once a year ( m  1 )

Given to the beginning of period the first member of a rent will be equal ( R / р)(1  i)1 / р , the second – ( R / р)(1  i)2 / р , the third –

( R / р)(1  i)3 / р etc. Thus, it is possible to make a geometrical 1 / р progression with the first element ( R / р)(1  i) and a denominator

(1  i)1 / р . Number of members of this geometrical progression of np. Therefore, the sum of modern costs of elements of a considered rent will make: A

R (1  i )( 1 / k ) nр  1 1  (1  i )  n (1  i ) 1 / р   R  k (1  i ) 1 / р  1 k[(1  i )1 / р  1]

(4.3.1)

Example. For conditions of the previous example we will assume that rent payments are carried out on half-year and percent are charged once in a year.

Decision. According to a formula (4.3.1), we will calculate A  100 

1  (1  0,2) 5  313,4 mil.tg 2[(1  0,2)1 / 2  1]

4.4. The r-urgent rent at m  p As well as in the previous options of a rent, it is necessary to lead (to discount) all elements to the beginning of the period of a rent. The modern cost of the first element of a rent as a result of such reduction 1 will be ( R / m)(1  y / m) , a denominator of a geometrical progression

(1  y / m)1 , number of elements of a geometrical progression of np. Sum of a geometrical progression:

60

FINANCIAL MATEMATICS

A

(1  y / m) mn  1 R 1  (1  y / m) mn (1  y / m)1    R (1  y / m)1  1 m y

(4.4.1)

Example. In the previous example we will assume that rent payments are carried out on half-year and on half-year percent are charged.

Decision. In this case using a formula (4.4.1), we will receive

A  100 

1  (1  0,2 / 2) 2  5  307 ,2 mil.tg 0 ,2

4.5. Dependence between the modern size and the increased sum of a rent

Let A – the modern size of an annual rent postnumerandno, and S – its increased sum by the term to the end n, p  1, m  1 . Let's show that the accretion of percent for A sum for n of years gives the sum, equal S: A(1  i ) n  R 

1  (1  i ) n (1  i ) n  1  (1  i ) n  R  S i i

(4.5.1)

From this it follows that discounting of S gives A: S n  A

(4.5.2)

and coefficients of discounting and a accretion of a rent are connected by ratios

a ni (1  i) n  s ni ;

s ni 

n

 a ni ;

(4.5.3) (4.5.4)

61

F I N A N C I A L M ACT HAP E M A TTER ICS 4

4.6. Determination of parameters of a financial rent

Sometimes when developing contracts there is a need to determine A other parameters of a rent by the set increased sum of a rent of S or its modern cost: R, n, i, p, m. Such parameters as p and m, are usually set in a consent of two signing parties. There are parameters R, n, i. Two of them are set, and the third pays off. Such calculations can be repeatedly repeated at various values of set parameters, the consent of the parties won't be reached yet. Determination of the size of an annual amount of payment of R. Depending on what generalizing characteristic of a constant rent is set – S or A, two following options of calculation are possible:

R  S / sni

(4.6.1)

or

R  A / ani .

(4.6.2)

Definition of term of a constant rent. Let's consider the solution of this task on the example of a usual annual rent with the continuous set payments. Resolving initial formulas for S and A concerning n term, we receive the corresponding expressions:

S R

AR

(1  i )  1 i n

1  (1  i ) i

and

n

and

 S ln  i  1  R ; n  ln 1  i 

(4.6.3)

A    ln  1  i  R   . n ln 1  i 

The last expression for n obviously makes sense only at R  Ai . Definition of a rate of percent. To find i rate, we will consider expressions for S and A from (7.6.3) as the nonlinear equations of rather

62

FINANCIAL MATEMATICS

unknown i (again we assume that it is a question of a constant annual rent postnumerandno), which are equivalent for two another: (1  i ) n  1 S   sni i R

or

1  (1  i )  n A   ani i R

(4.6.4)

In these equations the only unknown is the interest rate of i. The solution of the nonlinear equations can be found only approximately. Some methods of the solution of such equations are known: linear interpolation, Newton-Rafson and etc. We will consider only the first of them. First of all it is necessary to find bottom (iH) and top (iB) of an assessment of a rate by means of approximate calculations. It is carried out by substitution in one of formulas (7.6.4) of various numerical values i and comparison of result with the right part of expression. Further correction of the bottom value of a rate is made on the following interpolation formula:

i  iН 

s  sН (iВ  iН ) , sВ  sН

(4.6.5)

in which sH and sB – values of coefficient of a accretion (or reduction coefficient) rents for interest rates of iH and iB respectively. The received value of a rate is checked, substituting it in the left part of the initial equation and comparing result to the right part. If the reached accuracy is insufficient, repeatedly apply a formula (7.6.4), having replaced in it value of one of approximate estimates of a rate on more exact, found on the previous iteration, and value of coefficient of a accretion corresponding to it (or reductions). Example. On the basis of annual contributions of 10 thousand tg. it is supposed to create fund of 120 thousand tg. within 5 years. What has to be a rate of percent considered in calculations? Decision. Calculations we will carry out on a formula (4.6.4),

F I N A N C I A L M ACTHAP E M A TTER ICS 4

63

s  120 / 10  12 . Let's assume that iН  35%, iВ  50%. . Let's find values of coefficients of a accretion of a rent for iH and iB rates: accretion coefficient at the rate of percent being on the lower bound: sН  [(1  0,35)5  1] / 0,35  9,95 ; accretion coefficient at the rate of 5 percent being on the upper bound: sВ  [(1  0,5)  1] / 0,5  13,19 . Required rate of percent:

i  0,35 

12  9,95  (0,5  0,35)  0,448 or 44,8% 13,19  9,95

Check:

(1  0,448)5  1 s  11,98 0,448 Value s  11,98 is close to the actual value s  12 , so the found rate of percent is calculated rather precisely. Control questions and tasks

In questions 4.1. – 4.8 . to choose the correct from the list of offered answers. 4.1. What understand as the term "modern size of a rent": a) sum of all members of a rent; b) the sum of all members of sequence of payments with the percent added on them by the end of term of a rent; c) the sum of all members of the rent discounted at the time of the beginning of a stream of payments or previous it; d) the sum of all members of the rent discounted for the end of a stream of payments or previous it. 4.2. To specify coefficient of reduction of modern cost of a constant rent at single charge of percent in a year:

64

FINANCIAL MATEMATICS

(1  i ) n  1 ; i 1  (1  i ) n b) ; i (1  i ) n  1 ; c) (1  i ) m / p а)

d)

1  (1  i)  n . (1  i) m / p

4.3. The coefficient of reduction of modern cost of an eternal rent at single charge of percent in a year is equal: (1  i ) n ; а) i 1  (1  i ) n b) ; i 1 c) ; i 1 d) . (1  i ) m / p 4.4. Coefficients of discounting and building of a rent are connected among themselves by the following ratio: а) ani (1  i) n  sni ;

b) ani (1  i) n  sni ; c) sni (1  i) n  ani ; d) ani (1  i) n  sni . 4.5. At increase in an interest rate value of coefficient of reduction of a rent a) doesn't change; b) decreases;

F I N A N C I A L M ACT EHAP M A TTER ICS 4

65

c) increases; d) it isn't defined. 4.6. As the current cost of a stream of payments is called a) the sum of all his members of a stream with the percent added on them by the end of the year; b) the sum of all his members discounted for the beginning of term of a stream of payments; c) total amount of the saved-up debt by the term end; d) total volume of investment. 4.7. The investor intends to put the sum to the bank that the son during five-year term of training could remove 10000 tenges at the end of every year and spend to the end of study all contribution. At an annual rate of difficult percent of 12% the sum of a contribution will make:

а) b) c) d)

54125 tenges 58 tiyn; 58459 tenges 76 tiyn; 36047 tenges 76 tiyn; 39127 tenges 68 tiyn.

4.8. To specify expression for calculation of the given sum for a usual r-urgent rent at m-fold charge of percent in a year in general case: 1  (1  i)  n ; а) A  R  р[(1  i)1 / р  1]

1  (1  y / m)  mn ; (1  y / m) m  1 1  (1  i )  n ; c) A  R  i 1  (1  y / m)  mn d) A  R  ; y b) A  R 

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FINANCIAL MATEMATICS

To solve tasks 4.9. – 4.31. 4.9.

4.10.

4.11.

4.12.

4.13.

4.14.

It is brought on 10000 tg. within 20 years in fund annually postnumerandno on which difficult percent of 10% per annum quarterly are charged. To define the increased sum for the end of term. In fund it is annually brought on 10000 tg. within 20 years. Payments are made by equal shares at the end of every quarter. Difficult percent on a rate of 10% per annum are charged quarterly. To define the increased sum for the term end. Within 20 years on the settlement account at the end of every year arrives on 10000 tg. Annual discounting is made on a difficult rate of 10% per annum. To determine the modern cost of a rent. Two options of accumulation of means according to the annuity scheme pre-numerals are analyzed: a) to put on the deposit the sum of 15 thousand tg. every quarter provided that the bank charges 20% per annum with quarterly charge of difficult percent; b) to make an annual contribution of 52 thousand tg. on the terms of 22% per annum at annual charge of difficult percent. What sum will be on the account in 8 years at implementation of each plan? What plan is more preferable? Will it change your choice if the interest rate in the second plan is increased to 23? The insurance company signed the contract with the enterprise for three years, having established an annual insurance premium in 6 thousand tg. Insurance premiums are located in bank under a difficult interest rate of 25% per annum. Define the sum which will be received by insurance company under this contract if contributions arrive: a) at the end of every year; b) equal shares at the end of every half-year of 3 thousand tg. ; c) equal shares at the end of every quarter of 1,5 thousand tg. To consider possibility of use and only difficult percent, and the mixed scheme. Insurance company, having signed for 4 years the contract with some firm, receives from it insurance premiums about 15

F I N A N C I A L M AC T EHAP M A TTER ICS 4

4.15.

4.16.

4.17.

4.18.

67

thousand tg. at the end of every quarter. These contributions the company places in bank under an annual nominal interest rate 36% per annum. Find the specified cost of the sum which will be received by insurance company under this contract if difficult percent are charged: a) quarterly; b) monthly; c) continuously. The enterprise intends to create in 5 years development fund of 300 thousand tg. What sum the enterprise has to assign annually on this purpose on condition of the placement of money to bank at the end of every year under an interest rate of 24% per annum with charge of difficult percent: a) annually; b) monthly? The client within 6 years makes an annual contribution to bank 12 thousand tg. under a difficult interest rate of 24% per annum. Determine the size saved up by the end of term of the sum if the scheme of difficult percent is applied only and: a) contribution is made at the beginning of every year; b) contribution is made by equal shares at the beginning of every quarter (i.e. on a quarter of an annual contribution); c) contribution is made by equal shares at the beginning of every month (i.e. by the one twelfth part of an annual contribution). For creation in 5 years of fund of 200 thousand tg. the firm makes annual equal contributions to bank under an annual nominal interest rate of 36%. Define, what size of annually contribution the firm has to do, if: a) contributions become in the end of the year, and difficult percent are charged monthly; b) contributions become equal shares at the end of every half-year (i.e. on a half of an annual contribution), and difficult percent are charged quarterly; c) contributions become equal shares at the end of every quarter (i.e. on a quarter of an annual contribution) and continuous percent are charged. The businessman obtained for 6 years the loan of 500 thousand tg. and annually it has to pay to the creditor percent on a rate of 15%. At the same time with obtaining the loan the businessman (for her repayment) creates insurance fund in which at the end of every year will do identical contributions that by the time of

68

4.19.

4.20.

4.21.

4.22.

4.23.

FINANCIAL MATEMATICS

return of a debt to save up 500 thousand tg. Define total annual expenses of the businessman if on the money being in fund, quarterly difficult percent on a nominal interest rate of 20% per annum are charged. Some firm buys an oil-bearing site which, according to experts, within 15 years will bring in the income in 600 thousand tg. annually, then oil stocks most likely will be exhausted. Annually the firm wishes to draw interest for the enclosed sum on a rate of 24%. At the same time the firm creates insurance fund in which at the end of every year will do identical contributions that by the end of the 15th year to save up the sum paid for a site with stocks of oil. On the money enclosed in fund, difficult percent on a rate of 20% per annum are charged. For what sum the firm buys a site? Some firm wants to create fund of 400 thousand tg. For this purpose at the end of every year the firm assumes to bring 80 thousand tg. in bank under 32% per annum. Find the term necessary for creation of fund if the bank charges difficult percent: a) annually; b) quarterly; c) continuously. For the purpose of accumulation on the account of 180 thousand tg. Mister N at the beginning of every quarter will bring 6 thousand tg. in bank under a nominal interest rate of 26% per annum. Define the necessary for this purpose term if the bank charges continuous percent. The worker signs the contract according to which in case of her permanent job at the enterprise to a retirement (in 60 years) the enterprise undertakes to transfer at the end of every year within 15 years into the account of the worker in bank the identical sums which will provide it after a retirement at the end of every year additional payments in 5000 tg. with the enterprise within 10 years. What sum the enterprise annually has to transfer if the worker is 45 years old and it is supposed, what the bank guarantees an annual interest rate of 22%? The businessman wants to open the bank account, having put such sum that his son being the first-year student, could remove from this account at the end of every year on 3600 tg.

F I N A N C I A L M ACTHAP E M A TER TICS4

4.24.

4.25.

4.26.

4.27.

4.28.

4.29.

69

having settled all contribution by the end of five-year term of training. What size there has to be a sum if the bank charges difficult percent on a rate of 30% per annum? The site is leased for 20 years. The sum of annual payment makes 30 thousand tg. and each five years there is an indexation of size of payment for 10%. Calculate the current price of the contract of the moment of its conclusion if the difficult bank interest rate is equal 25% per annum. The young man of 24 years had an opportunity to end a year course worth 12 thousand tg. and to hold higher position. There has to be a salary in a new position that the young man considered training expedient if now its annual salary makes 21,6 thousand tg how above and he considers return accepted for norm on investments of 16% per annum? In a new position the young man is going to work to a retirement, i.e. 40 years. How the answer will change if such possibility of training is considered by the man of 54 years? Before a retirement mister N wants to provide himself the additional annual income in the sum of 6 thousand tg. beyond all bounds long. What sum he has to place in the bank charging difficult percent on a rate of 28% per annum? There is a termless annuity postnumerandno with annual payments 1 thousand tg. It is required to determine the specified cost of this annuity at an annual interest rate: a) 5%; b) 10%; c) 100%. How will it change the calculated values for annuity pre-numerals with the same payments? Determine the current (specified) cost of termless annuity postnumerandno with annual receipt of 4,2 thousand tg. if the percent offered by the state bank on fixed deposits is equal 24% per annum, and difficult percent are charged on half-year. The company guarantees payment of dividends of 4 thousand tg. on an action at the end of every year during vaguely long time. Whether it makes sense to take shares of this company at the price of 18 thousand tg. if it is possible to place money for the deposit under 21% per annum with charge of difficult percent: a) annually; b) quarterly?

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FINANCIAL MATEMATICS

4.30. Farmer is suggested to sell the ground which was in its possession on which he grows up on the average 600 tons of potatoes a year. The price of one kilogram of potatoes (in dollars) same also is equal from year to year of 0, 3 dollars. The bank percent on currency deposits steadily keeps at the level of 15% per annum. Below what price fermer doesn't make sense to sell the land if costs of cultivation, collecting and realization of potatoes are estimated at 60 thousand dollars a year? 4.31. The firm is going to found fund for annual (in the end of the year) payments of grants to the workers. Define the sum which the firm has to place on the deposit in bank to ensure beyond all bounds long at the end of every year 15 thousand tg. if the bank charges: a) annually difficult percent on a rate of 30%; b) on half-year difficult percent on a rate of 30%; c) continuous percent with a force of growth of 30%.

FINANCIAL MATEMATICS

CHAPTER

71

5

Planning of repayment of debt

5.1. Debt repayment by a lump sum Let's consider in detail the practical application of the methods of financial and economic calculations given above for planning of repayment of debt. The loan can come back a lump sum or by installments. This or that way makes a reservation in advance at the conclusion of the transaction. Also specify: D – initial sum of debt; g – rate of percent for the credit; mq – number of times of charge of percent on the credit within a year; n – loan term. Examples on calculation of the money satisfying a debt if the debt is liquidated by a lump sum without creation of extinctive fund, we considered above and solved on a formula

FV  D(1  q / mq )

mq n

.

More often for debt repayment by a lump sum in a demanded

72

FINANCIAL MATEMATICS

time point (through n term) the debtor creates sinking (extinctive) fund which represents the target deposit where means, sufficient for single repayment of a debt gradually collect. If contributions carry out periodically, we deal with a financial rent of the room. Let's enter the additional parameters connected with formation of extinctive fund: p – number of contributions to extinctive fund in a year; N – number of years of establishment of extinctive fund (term); Let the debt be repaid by a lump sum in the caused time point. In such situation the debtor often resorts to creation of extinctive (depreciation) fund. Need of creation of extinctive fund sometimes makes a reservation in the contract of delivery of a loan. On periodic contributions to this fund percent on i rate are charged. If contributions to fund are constant, there are urgent payments appear the same: the urgent payment (percent are periodically paid to the creditor)   Dg  

(5.1.1)

  D : sN ;i

(5.1.2)

where

Here Dg – periodically paid percent; α – the annual sum of contributions extinctive fund; N – period of creation of extinctive fund. Values of coefficient of a accretion of a rent are determined by a formula N 1

s N ; i   (1  i ) t  t 0

(1  i ) N  1 i

The formula (5.1.2) assumes contributions to extinctive fund in the end of the year. If contributions are carried out р time in a year, instead of S N ; i the coefficient of a accretion of a r-urgent rent S N( p; i) which is calculated on a formula undertakes

F I N A N C I A L M ACTHAP E M A TTER ICS 5

sN( p; i) 

73

(1  i) N  1 . p[(1  i)1 / p  1]

The more contributions are made, the less their annual. If percent on an amount of debt aren't paid to the creditor, and join the main amount of debt, urgent payment γ consists of one element: D (1  g ) n (5.1.3)  sN ;i Debt repayment by single payment favourably to the debtor under a condition when means of extinctive fund take place on the interest rate exceeding a rate on which the debt is taken, i.e. i  g . If term of creation of fund is equal to loan term ( N  n ), the formula (5.1.3) gives the size of urgent payment which is less, than the expenses determined by a formula (8.1.1), at g  i . In cases when g  i , the specified methods lead to identical results. Example. For debt repayment by a lump sum in two years the debtor in credit institution creates extinctive fund in which means gradually collect. Let's determine the size of equal contributions at the end of the half-year for creation in two years of extinctive fund of 500 million tenges. The fund is created in credit institution which charges percent, quarterly proceeding from an annual rate of 80%. Let's enter additional conditions. Let at the same time with formation of extinctive fund the borrower annually during all term of the loan equal to two years, pay percent proceeding from 20% per annum. It is necessary to calculate the annual size of expenses on debt service. Using a formula (5.5.1), we will express the size of R / p : (1  j / m) m / p  1 500  (1  0,8 / 4) 2 / 4  1   500 : 7,499584  (1  j / m) mn  1 (1  0,8 / 4) 24  1  66,670365mil.tenge R/ p  S 

74

FINANCIAL MATEMATICS

The found size is the amount of semi-annual payments therefore R=133,340730 of one million tenges, the annual sum of percent on a debt will make 500  0,2  100 mil .tenge . Annual expenses on service of a debt will make:   100  R  100  133,340730  233,340730 mil.tenge According to calculations annual calculations for single repayment of the main amount of debt in 2 years and annual payment of percent on the credit make 233,340730 million tenge. Formation of a sinking fund expediently, if q  i . Otherwise it is more favorable for debtor to carry out debt repayment by installments. At equal urgent payments (γ) the payment plan of debt is formed as follows. 5.2. Debt repayment by equal urgent payments

This type of a loan is most widespread in practice of domestic foreign economic relations. Let expenses on a loan is constant, then the payment plan of a loan can be developed provided that the maturity date of a loan or total size of expenses on a loan as a whole is set. The distinctive feature of such plan – the sum of percentage payments decreases, and extinctive payments grow in time. Let's consider only the first case, namely when loan term is set. The first development stage of the plan – calculation of urgent payment. Further there are percentage payments, the sum of repayment of a debt and the debt rest: urgent payment (annual payments) 

D1  const an ; g

(5.2.1)

where an; g - coefficient of reduction of a constant annual rent with g rate; amount of repayment of a debt

FINANCIAL M T E MTER A T I C5S C AHAP

dt    Dt g  dt 1 (1  g ), t  1, ..., n; d1    D1 g  D1 : sn; g ;

75

(5.2.2) (5.2.3)

the debt rest for the beginning of year

Dt 1  Dt  dt  Dt (1  g )  

(5.2.4)

the rest of the satisfied debt for the beginning of year

Wt  d1 st 1; g ;

(5.2.5)

where st 1; g – coefficient of building of a constant annual rent for

t  1 years. Formula (8.8) is applied when the detailed payment plan isn't developed. Example. The credit to the amount of 80 000 tenges under 24% per annum charged on the scheme of difficult percent on an unpaid balance is given out. The debt needs to be repaid monthly within a year equal urgent payments. Make the debt payment plan if payments happen: 1) at the end of every month; 2) at the beginning of every month Decision. Let's consider a situation comprehensively. Total amount, by an assessment at the moment, will periodically decrease in future assessment by identical size, won't be settled yet absolutely. Obviously, 80000 tenges should be treated as the sum of the modern sizes equal to urgent payments γ.

Therefore:  

D1  7476.200728 an ; g

amount of repayment of a debt d t    Dt g  dt 1 (1  g ), t  1, ..., n; d1    D1 g  D1 : sn; g  5876.2 Let's issue results in table 5.2.1.

76

FINANCIAL MATEMATICS Table 5.2.1 Debt payment plan of equal urgent payments

Month 1 2 3 4 5 6 7 8 9 10 11 12

The rest for the beginning of year 80000 74123,8 68130,076 62016,478 55780,608 49420,02 42932,22 36314,66 29564,75 22679,85 15657,25 7306,32

Debt repayment 5876,2 5993,724 6113,59 6235,87 6360,59 6487,79 6617,56 6749,91 6884,90 7022,60 7163,06 7306,32

Percent 1600 1482,476 1362,60 1240,33 1115,61 988,40 858,64 726,29 591,295 453,59 313,145 169,88

The urgent payment 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2 7476,2

If payments are made at the beginning of every month, the size of payment is defined as 

D1 80000 : (1  i )1 / p   7343.37679 an ; g 10.89417065

Results of calculations are presented in the form of table 5.2.2. Table 5.2.2 Month 1 2 3 4 5 6 7 8 9 10 11 12

The rest for the beginning of year 80000 72656,65 66627,49 60489,27 54240,03 47877,75 41400,39 34805,88 28092,09 21256,86 14298,00 7213,27

Debt repayment 7343,37 6029,15 6138,21 6249,23 6362,27 6477,35 6594,5 6713,78 6835,22 6958,85 7084,72 7212,87

Percent 0 1314,18 1205,13 1094,104 981,07 865,992 748,83 629,55 508,11 384,48 158,61 130,47

The urgent payment 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37 7343,37

77

F I N A N C I A L M ACTHAP E M ATER T I C S5

In case of credit repayment by installments the equal sums the plan is formed a little differently. 5.3. Debt payment plan of equal sums

Percentage payments and urgent payment in this case always decrease in time. Let payments be made once in the end of the year, then urgent payment

 t  Dt g  D1 : n, t  1,..., n

(5.3.1)

where D1 : n – the sum going on repayment of a principal debt; D1 – initial amount of debt. The debt rest for the beginning of year

 n 1 Dt 1  Dt  , t  1,..., n  n 

(5.3.2)

If the debt is satisfied р time in a year and percent is similarly paid, urgent payment

 t  Dt g / p  D1 : pn

(5.3.3)

where t – number of payment period, t  1,..., pn . The debt rest for the beginning of period

 np  1   Dt 1  Dt   np 

(5.3.4)

Example. The credit of 250 000 tenges is given out for 5 years under 10% per annum. The debt is satisfied by equal parts, percent are charged on the remained amount of debt once a year and paid together with payment of a principal debt. To make the loan payment plan.

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FINANCIAL MATEMATICS

On an example condition: 1) the annual sum of repayment of a debt is equal: 250 000:5=50 000; 2) urgent (annual) expenses on service of a debt change year by year (γt):

 t  Dt  q  D / n , where Dt – t – D – n –

the debt rest for the beginning of year; serial number of the period (year); initial sum of the credit; credit term;

3) expenses in the first year: D  q  250000  0,1  25000 д.е.

 1  25 000  50 000  75 000 и т.д. The solution of an example we will issue in the form of the table. Debt payment plan of equal sums Month 1-й 2-й 3-й 4-й 5-й In total

The rest for the beginning of year 250 000 200 000 150 000 100 000 50 000 -

Debt repayment

Percent

The urgent payment

50 000 50 000 50 000 50 000 50 000 250

25 000 20 000 15 000 10 000 5 000 -

75 000 70 000 65 000 60 000 55 000 -

The loan can provide a grace period with payment of percent or with the corresponding building of the main amount of debt. In the first case urgent payments throughout a grace period consist of one percentage payments. In the second case the initial amount of debt is L increased up to the size D1 (1  g ) , where L – duration of a grace period.

F I N A N C I A L M ACTHAP E M ATER TICS5

79

Control questions and tasks

To choose the correct from the list of offered answers in questions 5.1. – 5.8. 5.1. One of ways of repayment of a debt provides charge of percent on all amount of debt and their accessions to a principal debt at the time of credit opening, and debt repayment with percent happens: a) equal sizes during all term of the credit; b) the payments decreasing eventually; c) the payments increasing eventually; d) urgent payments. 5.2. At debt repayment by the equal sums percentage payments and urgent payment: a) increase in time; b) percentage payments increase, urgent payments decrease in time; c) decrease in time; d) percentage payments decrease, urgent payments increase in time. 5.3. At debt repayment by equal urgent payments: a) the sum of percentage payments decreases, and extinctive payments grow in time; b) the sum of percentage payments increases, and extinctive payments decrease in time; c) the sum of percentage payments and extinctive payments don't change in time; d) the sum of percentage payments and extinctive payments decrease in time. 5.4. Debt repayment by a lump sum favourably to the debtor under a condition when means of extinctive fund take place on an interest rate: a) not exceeding a rate on which the debt is taken, i.e. when; b) exceeding a rate on which the debt, i.e. when is taken;

80

FINANCIAL MATEMATICS

c) to equal rate on which the debt is taken, i.e. when; d) irrespective of a rate on which the debt is taken. 5.5. The debt of 100 million tenges is satisfied by a constant annual rent within 5 years. For the rest of a debt percent on a rate of 20% per annum are charged. The amount of annual extinctive payment thus will make: a) 38,268 million tenges; b) 33,438 million tenges; c) 58,128 million tenges; d) 63,354 million tenges. 5.6. If percent on an amount of debt aren't paid to the creditor, and join to the main amount of debt, urgent payment γ consists of one element and is equal:

а)  

D (1  g ) n ; sN ;i

b)  

D (1  i ) n ; sN ;i

c)  

D (1  g ) n ; s N ;i

d)  

D (1  g ) n . aN ;i

5.7. The payment size at creation of extinctive fund (percent are periodically paid to the creditor) is equal   Dg   , where   D : sN ;i Values of coefficient of building of a rent thus are

determined by a formula: N 1

а) s N ; i   (1  i ) t  t 0

(1  i ) N  1 ; i

F I N A N C I A L M ACT HAP E M A TTER ICS 5

b) s N( p; i) 

81

(1  i) N  1 ; p[(1  i)1 / p  1]

(1  i) N  1 ; p[(1  i)1 / p  1] N 1 (1  i ) N  1 .   (1  i ) t  i t 0

c) sN( p; i)  d) s N ; i

5.8. Repayment of the credit can provide a grace period with payment of percent or with the corresponding building of the main amount of debt. In the first case urgent payments throughout a grace period consist of: а) one percentage payments; L б) sizes D1 (1  g ) , where L – duration of a grace period;; в) sizes of urgent payments; D (1  g ) n г) sizes   . sN ;i To solve tasks 5.9. – 5.19. 5.9.

The loan is obtained for five years in the sum of 600 thousand tg in bank under 24% per annum, charged on the scheme of difficult percent on an unpaid balance. It is necessary to return the equal sums at the end of every year. It is required to determine the size of annual payment and to make the payment plan. 5.10. You occupied for seven years 36 thousand tg. under 30%, charged on the scheme of difficult percent on an unpaid balance. It is necessary to return the equal sums at the end of every year. Define, what part of the main sum of the credit will be extinguished for the first two years. 5.11. The creditor allowed a loan in the sum of 120 thousand tg. under 25% per annum charged on the scheme of difficult percent on an unpaid balance. According to the financial

82

5.12.

5.13.

5.14.

5.15.

5.16.

5.17.

FINANCIAL MATEMATICS

agreement the debtor will repay a debt the equal sums 20 thousand tg. at the end of every month. Make the payment plan. The farmer got in shop a tractor for 300 thousand tg. on credit. According to the contract the debt needs to be repaid within a year of equal sums at the end of every month, and on an unpaid balance difficult percent on an annual nominal interest rate of 24% are charged monthly. At what price the contract can be acquired by bank if he charges on the lent money difficult percent on an annual nominal interest rate of 36% monthly? The loan is obtained for seven years in the sum of 600 thousand tg. in bank on the terms of charge of difficult percent on an unpaid balance. It is necessary to return the equal sums at the end of every year, and at change of an interest rate the amount of annual payment also changes. To make the payment plan if the annual interest rate in the first three years makes 20%, in the next two years – 25% and in the last two years – 30%. The businessman occupied for six years 45 thousand tg. under 20%, charged on the scheme of difficult percent on an unpaid balance. It is necessary to return the equal sums at the end of every year. Determine the size of percent which will be paid by the businessman in the fourth year. You occupied for five years 80 thousand tg. under 24%, charged on the scheme of difficult percent on an unpaid balance. It is necessary to return the equal sums at the end of every year. Define the total amount of percent to payment. The enterprise got the building for 840 thousand tg. with the following provisions: a) 25% of cost are paid immediately; b) the rest is repaid by equal annual payments within 10 years with charge of 22% per annum on outstanding part of the credit according to the scheme of difficult percent. Determine the size of annual payment. The businessman obtained the loan in the sum of 400 thousand tg. under 25% per annum, charged on the scheme of difficult

F I N A N C I A L M ACT HAP E M A TTER ICS 5

83

percent on an unpaid balance. According to the financial agreement the businessman will repay a debt the equal sums 150 thousand tg. at the end of every year. Make the payment plan. 5.18. The businessman wants to acquire the equipment worth 240 thousand tg. The debt can be repaid within a year the equal sums at the end of every quarter, and on an unpaid balance are charged the difficult percent at the rate of 20% per annum. Determine the size of quarter payment and make the payment plan. 5.19. 3.1.48. The credit is given out to the amount of 80 thousand tg. under 24% per annum charged on the scheme of difficult percent on an unpaid balance. The debt needs to be repaid monthly within a year the equal sums. Determine the size of each payment if it happens: a) at the end of every month; b) at the beginning of every month.

84

FINANCIAL MATEMATICS Table 1 The current cost of single receiving or payment of the sum of $1.00 at the end of the period

Period of receiving (payment) 1 2 3 4 5

1% 0,990 0,980 0,971 0,961 0,951

2% 0,980 0,961 0,942 0,924 0,906

4% 0,962 0,907 0,863 0,823 0,784

5% 0,952 0,907 0,863 0,823 0,784

6% 0,943 0,890 0,840 0,792 0,747

8% 0,926 0,857 0,794 0,735 0,681

10% 0,909 0,826 0,751 0,683 0,621

12% 0,893 0,797 0,712 0,636 0,567

14% 0,877 0,769 0,675 0,592 0,519

15% 0,870 0,756 0,658 0,572 0,497

16% 0,862 0,743 0,641 0,552 0,476

6 7 8 9 10

0,942 0,933 0,923 0,914 0,905

0,888 0,871 0,853 0,837 0,820

0,746 0,711 0,677 0,645 0,614

0,746 0,711 0,677 0,645 0,614

0,705 0,665 0,627 0,592 0,558

0,630 0,583 0,540 0,500 0,463

0,564 0,513 0,467 0,424 0,386

0,507 0,452 0,404 0,361 0,322

0,456 0,400 0,351 0,308 0,270

0,432 0,376 0,327 0,284 0,247

0,410 0,354 0,305 0,263 0,227

11 12 13 14 15

0,896 0,887 0,879 0,870 0,861

0,804 0,788 0,773 0,758 0,743

0,585 0,557 0,530 0,505 0,481

0,585 0,557 0,530 0,505 0,481

0,527 0,497 0,469 0,442 0,417

0,429 0,397 0,368 0,340 0,315

0,350 0,319 0,290 0,263 0,239

0,287 0,257 0,229 0,205 0,183

0,237 0,208 0,182 0,160 0,140

0,215 0,187 0,163 0,141 0,123

0,195 0,168 0,145 0,125 0,108

16 17 18 19 20

0,853 0,844 0,836 0,828 0,820

0,728 0,714 0,700 0,686 0,673

0,458 0,436 0,416 0,396 0,377

0,458 0,436 0,416 0,396 0,377

0,394 0,371 0,350 0,331 0,312

0,292 0,270 0,250 0,232 0,215

0,218 0,198 0,180 0,164 0,149

0,163 0,146 0,130 0,116 0,104

0,123 0,108 0,095 0,083 0,073

0,107 0,093 0,081 0,070 0,061

0,093 0,080 0,069 0,060 0,051

21 22 23 24 25

0,811 0,803 0,795 0,788 0,780

0,660 0,647 0,634 0,622 0,610

0,359 0,342 0,326 0,310 0,295

0,359 0,342 0,326 0,310 0,295

0,294 0,278 0,262 0,247 0,233

0,199 0,184 0,170 0,158 0,146

0,135 0,123 0,112 0,102 0,092

0,093 0,083 0,074 0,066 0,059

0,064 0,056 0,049 0,043 0,038

0,053 0,046 0,040 0,035 0,030

0,044 0,038 0,033 0,028 0,024

26 27 28 29 30

0,772 0,764 0,757 0,749 0,742

0,598 0,586 0,574 0,563 0,552

0,281 0,268 0,255 0,243 0,231

0,281 0,268 0,255 0,243 0,231

0,220 0,207 0,196 0,185 0,174

0,135 0,125 0,116 0,107 0,099

0,084 0,076 0,069 0,063 0,057

0,053 0,047 0,042 0,037 0,033

0,033 0,029 0,026 0,022 0,020

0,026 0,023 0,020 0,017 0,015

0,021 0,018 0,016 0,014 0,12

35 40 45 50 60

0,706 0,672 0,639 0,608 0,550

0,500 0,453 0,410 0,372 0,305

0,181 0,142 0,111 0,087 0,054

0,181 0,142 0,111 0,087 0,054

0,130 0,097 0,073 0,054 0,030

0,066 0,046 0,031 0,021 0,010

0,036 0,022 0,014 0,009 0,002

0,019 0,011 0,006 0,003 0,001

0,010 0,005 0,003 0,001

0,008 0,004 0,002 0,001

0,006 0,003 0,001 0,001

Notes: 1. To find the current cost of future sum: Current cost = Factor * Sum 2. To find future size, equal to this or that current cost: The sum = Current cost / Factor

FINANCIAL MATEMATICS

18%

20%

22%

24%

25%

26%

28%

30%

35%

40%

45%

50%

0,847 0,718 0,609 0,516 0,437

0,833 0,694 0,579 0,482 0,402

0,820 0,672 0,551 0,451 0,370

0,806 0,650 0,524 0,423 0,341

0,800 0,640 0,512 0,410 0,328

0,794 0,630 0,500 0,397 0,315

0,781 0,610 0,477 0,373 0,291

0,769 0,592 0,455 0,350 0,269

0,741 0,549 0,406 0,301 0,223

0,714 0,510 0,364 0,260 0,186

0,690 0,476 0,328 0,226 0,156

0,667 0,444 0,296 0,198 0,132

0,370 0,314 0,266 0,225 0,191

0,335 0,279 0,233 0,194 0,162

0,303 0,249 0,204 0,167 0,137

0,275 0,222 0,179 0,144 0,116

0,262 0,210 0,168 0,134 0,107

0,250 0,198 0,157 0,125 0,099

0,227 0,178 0,139 0,108 0,085

0,207 0,159 0,123 0,094 0,073

0,165 0,122 0,091 0,067 0,050

0,133 0,095 0,068 0,048 0,035

0,108 0,074 0,051 0,035 0,024

0,088 0,059 0,039 0,026 0,017

0,162 0,137 0,116 0,099 0,084

0,135 0,112 0,093 0,078 0,065

0,112 0,092 0,075 0,062 0,051

0,094 0,076 0,061 0,049 0,040

0,086 0,069 0,055 0,044 0,035

0,079 0,062 0,050 0,039 0,031

0,066 0,052 0,040 0,032 0,025

0,056 0,043 0,033 0,025 0,020

0,037 0,027 0,020 0,015 0,011

0,025 0,018 0,013 0,009 0,006

0,017 0,012 0,008 0,006 0,004

0,012 0,008 0,005 0,003 0,002

0,071 0,060 0,051 0,043 0,037 0,031 0,026 0,022 0,019 0,016 0,014 0,011 0,010 0,008 0,007 0,003 0,001 0,001

0,054 0,045 0,038 0,031 0,026 0,022 0,018 0,015 0,013 0,010 0,009 0,007 0,006 0,005 0,004 0,002 0,001

0,042 0,034 0,028 0,023 0,019 0,015 0,013 0,010 0,008 0,007 0,006 0,005 0,004 0,003 0,003 0,001

0,032 0,026 0,021 0,017 0,014 0,011 0,009 0,007 0,006 0,005 0,004 0,003 0,002 0,002 0,002

0,028 0,023 0,018 0,014 0,012 0,009 0,007 0,006 0,005 0,004 0,003 0,002 0,002 0,002 0,001

0,025 0,020 0,016 0,012 0,010 0,008 0,006 0,005 0,004 0,003 0,002 0,002 0,002 0,001 0,001

0,019 0,015 0,012 0,009 0,007 0,006 0,004 0,003 0,003 0,002 0,002 0,001 0,001 0,001 0,001

0,015 0,012 0,009 0,007 0,005 0,004 0,003 0,002 0,002 0,001 0,001 0,001 0,001 0,001

0,008 0,006 0,005 0,003 0,002 0,002 0,001 0,001 0,001 0,001

0,005 0,003 0,002 0,002 0,001 0,001 0,001

0,003 0,002 0,001 0,001 0,001

0,002 0,001 0,001

3. To find the period at known future cost, the current cost and profitability: Factor = Current cost / the Sum; find in a column 4. To find profitability at known future cost, the current cost and the period: Factor = Current cost / the Sum; find in a line

85

86

FINANCIAL MATEMATICS Table 2 The current cost of $1.00, received or paid at the end of every period (Annuity)

Number of period 1%

2%

4%

5%

6%

8%

10%

12%

14%

15%

16%

1 2 3 4 5

0,990 1,970 2,941 3,902 4,853

0,980 1,942 2,884 3,808 4,713

0,962 1,886 2,775 3,630 4,452

0,952 1,859 2,722 3,545 4,329

0,943 1,833 2,673 3,465 4,212

0,926 1.783 2.577 3.312 3.993

0,909 1.736 2.487 3.170 3.791

0,893 1.690 2.402 3.037 3.605

0,877 1.647 2.322 2.914 3.433

0,870 1,626 2,283 2,855 3,352

0,862 1,605 2,246 2,798 3,274

6 7 8 9 10

5,795 6,728 7,652 8,566 9,471

5,601 6,472 7,325 8,162 8,983

5,242 6,002 6,733 7,435 8,111

5,075 5,786 6,463 7,108 7,722

4,917 5,582 6,210 6,802 7,360

4.623 5.206 5.747 6.247 6.710

4.355 4.868 5.335 5.759 6.145

4.112 4.564 4.968 5.328 5.650

3.889 4.288 4.639 4.946 5.216

3,784 4,160 4,487 4,772 5,019

3,685 4,039 4,344 4,607 4,833

11 12 13 14 15

10,368 11,255 12,134 13,044 13,865

9,787 10,575 11,343 12,106 12,849

8,760 9,385 9,986 10,563 11,118

8,307 8,863 9,393 9,898 10,379

7,887 8,384 8,853 9,295 9,712

7.139 7.536 7.904 8.244 8.559

6.495 6.814 7.103 7.367 7.606

5.937 6.194 6.424 6.628 6.811

5.453 5.660 5.842 6.002 6.142

5,234 5,421 5,583 5,724 5,847

5,029 5,197 5,342 5,468 5,575

16 17 18 19 20

14,718 15,562 16,398 17,226 18,046

13,578 14,292 14,992 15,678 16,351

11,652 12,116 12,659 13,134 13,590

10,838 11,274 11,390 12,086 12,463

10,106 10,477 10,828 11,158 11,470

8.851 9.122 9.372 9.604 9.818

7.824 8.022 8.201 8.365 8.514

6.974 7.120 7.250 7.366 7.469

6.265 6.373 6.467 6.550 6.623

5,954 6,047 6,128 6,198 6,259

5,669 5,749 5,818 5,877 5,929

21 22 23 24 25

18,857 19,660 20,456 21,243 22,023

17,011 17,658 18,292 18,914 19,523

14,029 14,451 14,857 15,247 15,622

12,821 13,163 13,489 13,799 14,094

11,764 12,042 12,303 12,550 12,783

10.017 10.201 10.371 10.529 10.675

8.649 8.772 8.883 8.985 9.077

7.562 7.645 7.718 7.784 7.843

6.687 6.743 6.792 6.835 6.873

6,312 6,359 6,399 6,434 6,464

5,973 6,011 6,044 6,073 6,097

26 27 28 29 30

22,795 23,560 24,316 25,066 25,808

20,121 20,707 21,281 21,844 22,396

15,983 16,330 16,663 16,984 17,292

14,375 14,643 14,898 15,141 15,372

13,003 13,211 13,406 13,591 13,765

10.810 10.935 11.051 11.158 11.258

9.161 9.237 9.307 9.370 9.427

7.896 7.943 7.984 8.022 8.055

6.906 6.935 6.961 6,983 7,003

6,491 6,514 6,534 6,551 6,566

6,118 6,136 6,152 6,166 6,177

35 40 45 50 60

29,408 32,835 36,094 39,196 44,955

24,999 27,355 29,490 31,424 34,761

18,665 19,793 20,720 21,482 22,623

16,374 17,159 17,774 18,256 18,929

7,070 7,105 7,123 7,133 7,140

6,617 6,642 6,654 6,661 6,665

14,498 15,046 15,456 15,762 16,161

11.654 11.925 12.108 12.234 12.376

9.664 9.779 9.863 9.915 9.967

8.176 8.244 8.282 8.304 8.324

Note: 1. To find the current cost of a series of identical receiving money or their payments: Current cost = Factor * Annuity 2. To find the annuity giving this or that current cost: Annuity = Current cost / Factor

6,215 6,234 6,242 6,246 6,249

87

FINANCIAL MATEMATICS

18%

20%

22%

24%

25%

26%

28%

30%

35%

40%

45%

50%

0,847 1,566 2,174 2,690 3,127

0,833 1,528 2,106 2,589 2,991

0,820 1,492 2,042 2,494 2,864

0,806 1,457 1,981 2,404 2,745

0,800 1,440 1,952 2,362 2,689

0,794 1,424 1,923 2,320 2,635

0,781 1,392 1,868 2,241 2,532

0,769 1,361 1,816 2,166 2,436

0,741 1,289 1,696 1,997 2,220

0,714 1,224 1,589 1,849 2,035

0,690 1,165 1,493 1,720 1,876

0,667 1,111 1,407 1,605 1,737

3,498 3,812 4,078 4,303 4,494

3,326 3,605 3,837 4,031 4,192

3,167 3,416 3,619 3,786 3,923

3,020 3,242 3,421 3,566 3,682

2,951 3,161 3,329 3,463 3,571

2,885 3,083 3,241 3,366 3,465

2,759 2,937 3,076 3,184 3,269

2,643 2,802 2,925 3,019 3,092

2,385 2,508 2,598 2,665 2,715

2,168 2,263 2,331 2,379 2,414

1,983 2,057 2,108 2,144 2,168

1,824 1,883 1,922 1,948 1,965

4,656 4,793 4,910 5,008 5,092

4,327 4,439 4,533 4,611 4,675

4,035 4,127 4,203 4,265 4,315

3,776 3,851 3,912 3,962 4,001

3,656 3,725 3,780 3,824 3,859

3,544 3,606 3,656 3,695 3,726

3,335 3,387 3,427 3,459 3,483

3,147 3,190 3,223 3,249 3,268

2,752 2,779 2,799 2,814 2,825

2,438 2,456 2,468 2,477 2,484

2,185 2,196 2,204 2,210 2,214

1,977 1,985 1,990 1,993 1,995

5,162 5,222 5,273 5,316 5,353

4,730 4,775 4,812 4,844 4,870

4,357 4,391 4,419 4,442 4,460

4,033 4,059 4,080 4,097 4,110

3,887 3,910 3,928 3,942 3,954

3,751 3,771 3,786 3,799 3,808

3,503 3,518 3,529 3,539 3,546

3,283 3,295 3,304 3,311 3,316

2,834 2,840 2,844 2,848 2,850

2,489 2,492 2,494 2,496 2,497

2,216 2,218 2,219 2,220 2,221

1,997 1,998 1,999 1,999 1,999

5,384 5,410 5,432 5,451 5,467

4,891 4,909 4,925 4,937 4,948

4,476 4,488 4,499 4,507 4,514

4,121 4,130 4,137 4,143 4,147

3,963 3,970 3,976 3,981 3,985

3,816 3,822 3,827 3,831 3,834

3,551 3,556 3,559 3,562 3,564

3,320 3,323 3,325 3,327 3,329

2,852 2,853 2,854 2,855 2,856

2,498 2,498 2,499 2,499 2,499

2,221 2,222 2,222 2,222 2,222

2,000 2,000 2,000 2,000 2,000

5,480 5,492 5,502 5,510 5,517

4,956 4,964 4,970 4,975 4,979

4,520 4,524 4,528 4,531 4,534

4,151 4,154 4,157 4,159 4,160

3,988 3,990 3,992 3,994 3,995

3,837 3,839 3,840 3,841 3,842

3,566 3,567 3,568 3,569 3,569

3,330 3,331 3,331 3,332 3,332

2,856 2,856 2,857 2,857 2,857

2,500 2,500 2,500 2,500 2,500

2,222 2,222 2,222 2,222 2,222

2,000 2,000 2,000 2,000 2,000

5,539 5,548 5,552 5,554 5,555

4,992 4,997 4,998 4,999 5,000

4,541 4,544 4,545 4,545 4,545

4,164 4,166 4,166 4,167 4,167

3,998 3,999 4,000 4,000 4,000

3,845 3,846 3,846 3,846 3,846

3,571 3,571 3,571 3,571 3,571

3,333 3,333 3,333 3,333 3,333

2,857 2,857 2,857 2,857 2,857

2,500 2,500 2,500 2,500 2,500

2,222 2,222 2,222 2,222 2,222

2,000 2,000 2,000 2,000 2,000

3. To find the number of the periods necessary for a covering of an investment: Factor = Investment / Annuity; find in a column 4. To find profitability of annuity for an available investment: Factor = Investment / Annuity; find in a line

88

FINANCIAL MATEMATICS

Answers for questions and tasks Chapter 1 Answers for questions 1.1. c); 1.2. b); 1.3. c); 1.4. a); 1.5. а); 1.6. а); 1.7. а); 1.8. d); Answers for tasks 1.9. 595000 tg.; 1.10. 90000 tg.; 1.11. 52549 tg.; 1.12. (The way 360/360 is favorable to the enterprise – 182500 tenges; for bank – a way 365/360 – 182043 tenges); 1.13. 1490000 tg.; 1.14. 0,625 year or about 228 days; 1.15. 7,5 year.; 1.16. 119047 tg.; 1.17. About 126 days; 1.18. 21,88%; 1.19. 34,31;36,55%; 1.20. 50000 tg.; 1.21. 0,1225 year or about 45 days; 1.22. 0,3%; 1.23. 38,08 thousand tg.,; 32,32%; 1.24. 29112 tg.; 1.25. It is more favorable the accounting of the bill on an interest rate; 1.26. 5488 tg.; 1.27. 56818 tg.; 1.28. 32,89%; 33,49%; 1.29. 1,471 year or 1 year and 172 days; 0,980 year , or about 358 days; 1.30. 33,80%; 1.31. 41,53%; 1.32. 6 years; 1.33. 0,03445 year , or about 13 days. Chapter 2 Answers for questions 2.1. b or c); 2.2. c); 2.3. а); 2.4. d); 2.5. b); 2.6. c); 2.7. d).

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Answers for tasks 2.8. It is more favorable to receive 4,6 thousand tg. in 4 years; 2.9. 30%,; 27,12%; 2.10. а) 27,12% ; б) 26,53%; 2.11. а) 33919 tg., б) 16642 tg.; 2.12. 100 thousand. tg. today; 2.13. 4804800 tg.; 2.14. а) 465259 tg., b)523384 tg., c) 560441 tg., d) 589160 tg., e) 601224 tg., f) 604421 tg.; 2.15. In years: а) 20 и 14,207, b) 10 and 7,273, c) 6,667 and 4,959, d) 4 and 3,106, e) 2 and 1,710, f)1,333 and 1,239, g)1 and 1. 2.16. а) 4,450 year; b) 4,059 year; c) 3,969 year; 2.17. 17,71%; 2.18. The first option is more preferable; 2.19. 46,41%; 2.20. 912800 tg.; 2.21. 271000 tg.; 2.22. 116406 tg.; 2.23. 261187 tg.; 2.24. 11,97%; 2.25. The first conditions for bank are favorable; 2.26. а) 29802 tg., б)16856 tg.; 2.27. 26630 tg.; 2.28. 30,78%, doesn`t change; 2.29. а) 67,55%, b) 31%; 2.30. 32,21%, doesn`t consist; 2.31. The second; 2.32. 15,27%, 23,10%; 2.33. It is better to receive 20 thousand tg. in 3 years; 2.34. 24,47%, 29,45%; 2.35. 151522 tg.; 2.36. 10894 tg.; 2.37. 3855432 tg.; Chapter 3 Answers for questions 3.1. а); 3.2. d) 3.3 d) 3.4 b) 3.5. a) 3.6 а) 3.7 f) 3.8. а).

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Answers for tasks 3.9. 598149 tg. 98 tiyn; 3.10. 620956 tg. 78 tiyn; 3.11. 85135 tg. 64 tiyn; 3.12. The option b); 16,152 thousand. tg. 3.13. а) 144,166 thousand tg or 144,606 thousand tg; b) 147,596 thousand tg; c) 149,549 thousand tg. 3.14. а) 213,843 thousand tg; b) 235,227 thousand tg; 3.15. а) 1534,803 thousand tg; b) 1556,661 thousand tg and at decrease in a rate – 1518,772 thousand tg. 3.16. а) 676,944 thousand tg; b) 826,388 thousand tg; c) 859,856 thousand tg; 3.17. No. 3.18. а) 47,698 thousand tg; b) 13,923 thousand tg; c) 8,772 thousand tg; 3.19. а) 8,847 thousand tg; b) 1,877 thousand tg. 3.20. а) 14,165 thousand tg; b) 2,372 thousand tg. 3.21. 607,441 thousand tg. 3.22. а) 281,767 thousand tg; b) 331,843 thousand tg; c) 346,940 thousand tg; 3.23. а) 173,161 thousand tg; b) 176,773 thousand tg. 3.24. no. 3.25. а) 9,583 year; b) 13,254 year; c) 17,895 year; 3.26. а) 20,513 thousand tg; b) 19,214 thousand tg; c) 17,805 thousand tg; 3.27. а) 46,886 thousand tg; b) 44,294 thousand tg; c) 43,355 thousand tg; 3.28. Under agricultural cultures (agricultural – 950 thousand tg; oil – 723,852 thousand tg.). 3.29. а) 278,267 thousand tg; b) 250 thousand tg; c) 240,133 thousand tg; 3.30. The second project is better; "yes" at a rate less than 18,13% Chapter 4 Answers for questions 4.1. c); 4.2. b); 7.3. c); 4.4. а); 4.5. b); 4.6. b); 4.7. c).

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Answers for tasks 4.9. 598149 tg. 98 tiyn; 4.10. 620956 tg. 78 tiyn; 4.11. 85135 tg. 64 tiyn; 4.12. а) 1185,956 thousand tg., b)1126,840 thousand tg. and at increase in a rate 1178,783 thousand tg.; 4.13. а) 22,875 thousand tg., b) 24,225 thousand tg., 24,305 thousand tg., c) 24,920 thousand tg., 25,020 thousand tg.; 4.14. а) 124,689 thousand tg., b) 122,618 thousand tg., c) 121,542 thousand tg.; 4.15. а) 32,274 thousand tg., b) 35,279 thousand tg.; 4.16. а) 163,384 thousand tg., b) 150,994 thousand tg., c) 148,328 thousand tg.; 4.17. а) 17,408 thousand tg., b) 16,341 thousand tg., c) 14,920 thousand tg.; 4.18. 123,426 thousand tg.; 4.19. 2363,302 thousand tg. 4.20. а) 3,442 year, b) 3,347 year, c) 3,312 year; 4.21. 4,079 year; 4.22. 230,26 thousand tg. 4.23. 8768 thousand tg.; 4.24. 124,020 thousand tg.; 4.25. More, than on 5,39 thousand tg.in a year; more, than on 6,952 thousand tg.in a year 4.26. 21,429 thousand tg.; 4.27. Postnumerandno: а) 20 thousand tg., b) 10 thousand tg., c) 1 thousand tg.; Prenumerandno: а)21 thousand tg., b) 11 thousand tg., c) 2 thousand tg.; 4.28. 16,509 thousand tg.; 4.29. а) yes; b) no. 4.30. 800 thousand dollars. 4.31. а) 50 thousand tg., b) 46,512 thousand tg., c) 42,874 thousand tg. Chapter 5 Answers for questions 5.1. а); 5.2. c); 8.3. а); 5.4. b); 5.5. b); 5.6. а); 5.7. а). 8.8. а).

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FINANCIAL MATEMATICS Answers for tasks 5.9. 218547 tg.; the rest of a principal debt for the beginning of the third year – 433015 tg., repayment at the end of the third year of the added percent and a principal debt: 103924 tg. and 114623 tg. 5.10. 4710 tg. or about 13,08% of the main sum. 5.11. The debt is satisfied for 0,535 year (6,42 months); the rest of a principal debt for the beginning of the third month – 84171 tg., repayment at the end of the third month of the added percent and a principal debt: 1580 tg. and 18420 tg. ; the last payment – 8547 tg. 5.12. 282,375 thousand tg. 5.13. The rest of a principal debt for the beginning of the third year – 497801 tg., repayment at the end of the third year of the added percent and a principal debt: 99560 tg. and 66894 tg. the rest of a principal debt for the beginning of the fifth year – 356170 tg., repayment at the end of the fifth year of the added percent and a principal debt: 89043 tg. and 93421 tg. 5.14. 5701 tg. 5.15. 65699 tg.; 5.16. 160,583 thousand tg. 5.17. The debt is satisfied for 4,923; the rest of a principal debt for the beginning of the third year – 287500 tg., repayment at the end of the third year of the added percent and a principal debt: 71875 tg. and 78125 tg.; the last payment – 139649 tg. 5.18. 67157 tg.; the rest of a principal debt for the beginning of the second quarter – 184035 tg., repayment at the end of the second quarter of the added percent and a principal debt: 8582 tg. and 58575 tg. 5.19. а) 7476 tg., b) 7343 tg.

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ecommended literature

1. Arslanova P. Livshits V. The principles of an assessment of investment projects in different systems of managing // Investments in Russia, 1995, №. 1-2. 2. Berens V., Havranek P. M., Guide to an assessment of efficiency of investments: The lane with English – the 2nd prod. reslave. And additional – M.: Interekspert, INFRA-M, 1995. 3. Bocharov V. V. Methods of financing of investment activity of the enterprises. – M.: Finance and statistics, 1998. 4. Idrisov A.B. Planning and analysis of efficiency of investments. – M.: 1994. 5. Kochovich E. Financial mathematics: Theory and practice of financial and bank calculations. – M.: Finance and statistics, 1994. 6. Kuznetsova O. A. Livshits V. N. Capital structure. The Analysis of methods of its account at an assessment of investment projects // Economic-mathematical methods, 1995, T 31 releasing 4. 7. Lipsits I.V. Kosov V. V. Investment project: preparation and analysis methods: Educational handbook. – M.: Beck, 1996 8. Smirnov A.L. The organization of financing of investment projects. – M.: JSC Konsaltbankir, 1993. 9. Project management / V.D. Shapiro, L.M. Nemchin, S.E. Nikeshin. – Spb: Two – Three, 1996 10. Holt R.N., Barens S.B. Planning of investments. – M.: Business LTD, 1994. 11. Chetyrkin E.M. Metods of financial and commercial calculations. – M., 1995.

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Contents Introduction ........................................................................................................ 3 Chapter 1. Models of development of transactions of the scheme of simple percent 1.1. Calculations when charging simple percent .................................... 5 1.2. Ordinary and exact simple percent .................................................. 7 1.3. Variable rates of simple percent ...................................................... 9 1.4. Discounting and the account of simple interest rates ...................... 10 1.5. Banking or commercial account ..................................................... 12 Controlling questions and problems ............................................................................13 Chapter 2. Models of development of transactions of the scheme of difficult percent 2.1. Percent for percent ........................................................................... 19 2.2. Building on a difficult rate of percent (i) ......................................... 20 2.3. Effective and nominal rates of percent ............................................ 22 2.4. Continuous charge of percent ......................................................... 24 2.5. Discounting on a difficult rate of percent ........................................ 25 2.6. Calculation of term of payment and interest rates ........................... 29 Questions and problems...................................................................................... 34 Chapter 3. The increased sum of a stream of payments 3.1. Classification and conclusion of a formula of the increased sum of a rent .................................................................................... 41 3.2. Annual rent with charge of percent of m of times in a year ............ 45 3.3. r- urgent rent with charge of percent once a year ( m  1 ) ................ 46 3.4. r- urgent rent, p  m ......................................................................... 47 3.5. r-urgent rent, at p  m….................................................................. 48

FINANCIAL MATEMATICS 3.6. Annual rent with continuous charge of percent ............................... 48 Questions and Problems .............................................................................49 Chapter 4. Modern cost of a rent 4.1. Formula of modern cost of a rent .................................................... 55 4.2. Annual rent with charge of percent of m times in a year ................. 58 4.3. r-urgent rent with charge of percent once a year ( m  1 ) ................ 59 4.4. r-urgent rent at m p ....................................................................... 59 4.5. Dependence between the modern size and the increased sum of a rent .................................................................................... 60 4.6. Determination of parameters of a financial rent .............................. 61 Control questions and tasks ............................................................................... 63 Chapter 5. Planning of repayment of debt 5.1. Debt repayment by a lump sum ...................................................... 71 5.2. Debt repayment by equal urgent payments ..................................... 74 5.3. Debt payment plan equal sums ....................................................... 77 Control questions and tasks ............................................................................... 79 Answers for questions and tasks ......................................................................... 88 Recommended literature ................................................................................... 92

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Educational issue

Alzhanova Nurzhan Sharipovna FINANCIAL MATHEMATICS Educational manual Computer page makeup: K. Umirbekova Cover designer: R. Shangaraev _www.maths.york.ac.uk

IB No. 7551 Signed for publishing 18.06.14. Format 60x84 1/16. Off set paper. Digital printing. Volume 5,6 printer’s sheet. Edition: 50. Order No 1943. Publishing house “Qazaq university” Al-Farabi Kazakh National University KazNU, 71 Al-Farabi, 050040, Almaty Printed in the printing offi ce of the “Kazakh Universitety” publishing house