LOGO MATH 2040 Introduction to Mathematical Finance Instructor: Dr. Ken Tsang 1 Chapter 2: Solution of problems in interest Introduction: the basic problem Equations of value Unknown time & interest rate Determining time periods Examples 2 Introduction • How to solve an interest problem? –Use basic principle – Develop a systematic approach 3 Obtaining numerical results • Use a calculator with exponential functions. • Use the table of compound interest functions. • Series expansion could also be used. • Use software like MatLab or R 4 A common problem • A common problem is to determine a(t), where t is not an integer. • Use compound interest for integral periods of time and simple interest for fractional periods. • This method is equivalent to the method of using the first two terms of the binomial expansion assuming that 0 < k < 1. 5 Linear interpolation • The use of simple interest for a fractional period is equivalent to performing a linear interpolation between (1 + i)n and (1 + i)n+1, where n is the integral part of the period. • (1 + i) n+k (1 k)(1 + i)n + k (1 + i) n+1 = (1 + i)n(1 + k i) • (1 – d) n+k (1 k)(1 d )n + k (1 – d) n+1 = (1 – d)n(1 – k d) 6 Example • Find the accumulated value of $5000 at the end of 30 years and 4 months at 6% p.a. converted semiannually: (1) assuming compound interest throughout, and (2) assuming simple interest during the final fractional period. • (1) By direct calculation, 5000(1.03)60.667 = 30044.27 • (2) By assuming simple interest in the final fractional period, we have 5000(1.03)60(1.02) = 30047.18 7 Exact simple interest • There are three commonly used methods to count the number of days for the period of investment. • The first one is called the exact simple interest method. • In this method, the exact number of days are counted, and a year contains 365 days. • This method is often denoted by “actual/actual”. 8 Ordinary simple interest • The second one is called the ordinary simple interest method. • In this method, it is assumed that there are 30 days in each month and 360 days in a year. • This method is often denoted by “30/360”. • For simplicity, a formula for calculating the days in the “30/360” method is: 360(Y2 – Y1) + 30(M2 – M1) + (D2 – D1) 9 The banker’s rule • The third one is called the banker’s rule. • This method is a hybrid of the previous two. • It used the exact number of days in each month and 360 days in a year. • This method is often denoted by “actual/360”. • Bankers use this rule to charge interest for loans to their clients, and the rule is more favorable to the bankers. 10 Remarks on Determining time periods • Leap years lead to complications. – February 29 counted as one day and 366 days for the year. – February 29 counted as one day and 365 days for the year. – February 29 is not counted and 365 days for the year. • • In counting the number of days, either the first day or the final day is counted, but not both. The above methods are also used for compound interest calculation. 11 Example • Find the amount of interest that $2000 deposited From June 17 to September 10 of the same year will earn, if the rate of interest is 8%, on the following: – – – • Exact simple interest (actual/actual) Ordinary simple interest (30/360) Banker’s rule (actual/360) Answers for the three methods are: – – – 85 days. Interest = 2000(0.08)(85/365) = 37.26 Number of days = 360(0) + 30(9 – 6) + (10 – 17) = 83. therefore interest = 2000(0.08)(83/360) = 36.89 85 days. Interest = 2000(0.08)(85/360) = 37.78. 12 The basic problem • A typical problem on interest involves four basic quantities: – – – – The original investment – principal The length of the investment period – the period The rate of interest The accumulated value at the end of the period • In principle, if three of the four quantities are known, the fourth can be determined. 13 A fundamental principle in the theory of interest • Value of an amount of money depends on the time when it is payable. • This time value of money reflects the effect of interest. • Two or more amounts of money payable at different time cannot be compared directly. 14 Equation of value • The time value of money at any given point in time, t, will be either a present value or a future value from the payment time. • To compare two or more amounts of money payable at different time, they have to be accumulated or discounted to a common date, the comparision date. • The equation which accumulates or discountes each payment to the comparision date is called the equation of value. 15 Time diagram • It helps to draw out a time line and plot the payments and withdrawals accordingly. 16 Example • A $600 payment due in 8 years is equivalent to receive $100 now, $200 in 5 years and $ X in 10 years. If i = 8% p.a., find $ X. • The line diagram is 17 Example – solution 1 • Compare the values at t = 0. 18 Example – solution 2 • Compare the values at t = 5. 19 Example – solution 3 • Compare the values at t = 10. • All three solutions leads to the same answer, because they all treat the values of the payments consistently at a given point of time t. 20 Unknown time • As discussed before, if any of the four basic quantities are known, then the fourth can be determined. • Consider the situation where the length of investment is not known. • The easiest approach is to use logarithm. 21 Example - logarithm • How long does it take money to double if the interest rate i = 6%?. Very simple evaluation using MATLAB 22 Example – interest table + interpolation • If logarithms are not available, then use interest tables and perform a linear interpolation. • Using interest tables, we have and 23 Doubling a payment – rule of 72 • For doubling a single payment we have the rule of 72 • This rule gives a good approximation for the period to double the principal over a wide range of interest rates. 24 Comparing rule of 72 with exact values Rate of interest Rule of 72 Exact value Error 4% 18 17.67 +2% 6 12 11.90 +1% 8 9 9.01 0.1% 10 7.2 7.27 1% 12 6 6.12 2% 18 4 4.19 5% 25 Tripling a payment – rule of 114 • For tripling a single payment, we have the approximate rule of 114. 26 Multiple payments • Let St represent a sequence of payments made at time t for t = 0, 1, …, n. • How to replace the multiple payments with a single payment S, equal to the sum of all St, such that the present value of S at time t is equal to the present value of the multiple payments? • Determine the value of t. 27 Multiple payments - solution • To find the true value of t, we form the equation: • Taking logarithm on both sides, we have 28 Multiple payments – approximate solution • Suppose we use t , the weighted average of time for the multiple payments, to approximate the value of t. • This method is called the method of equated time. • If we can prove that t t , then the present value using the method of equated time will be less than the present value using exact t. 29 Weighted average of time > exact time 1 • The present value of the payment at time tk is . • So the arithmetic weighted mean of present values is • And the geometric weighted mean of present values is 30 Weighted average of time > exact time 2 • Since the geometric means are less than arithmetic mean • Method of equated time > exact t 31 Example • Find the length of time necessary for $1000 to accumulate to $1500 if invested at 6% per annum compounded semiannually: (1) by use of logarithm, and (2) by interpolating in the interest table. • Let n be the number of half-years required. The equation of time is 1000(1.03)n = 1500. (1.03)n = 1.5 32 Example – cont’d • Using logarithm, we have n loge 1.03 = loge 1.5. • Because loge 1.03 = 0.405465 and loge 1.5 = 0.029559, we can determine n to be 13.717. • So the number of years is 6.859 Using MATLAB 33 Example – cont’d • From the interest table, we have (1.03)13 = 1.46853 and (1.03)14 = 1.51259 • Performing a linear interpolation, 1.50000 1.46853 n 13 13.714 1.51259 1.46853 • So the number of years is 6.857. Note that the two answers are very close. 34 Another example • Payments of $100, $200 and $500 are due at the ends of years 2, 3 and 8 respectively. Assuming an effective rate of interest of 5% per annum, find the point in time at which a payment of $800 would be equivalent: (1) by an exact method, and (2) by the method of equated time. • The exact time equation is: 800 vt = 100 v2 + 200 v3 + 500 v8, which can be solved for t = 5.832. • By method of equated time, we have 100 2 200 3 500 8 t 6. 100 200 500 • As expected, the true value t is less. 35 Unknown rate of interest – single payment • It is quite common to have a financial transaction where the rate of return needs to be determined. • For example, suppose $1000 investment triples in 10 years at nominal rate of interest convertible quarterly. Find i(4). Using MATLAB 36 Unknown rate of interest – multiple payments • Interest can also be determined if there are only a small number of payments and the equation of value can be reduced to a polynomial that is not too difficult to solve. • For example, at what effective interest rate will the present value of $130 at the end of 5 years and $300 at the end of 10 years be equal to $360? • The equation of value is 130 v5 + 300 v10 = 360. • If we put y = v5, the equation becomes a quadratic equation in y, which can be solved, and consequently i can be determined, giving y = 0.9 and i = (0.9)0.2 1. 37 Linear interpolation • Unfortunately, most of the times, a quadratic equation or low degree polynomial equation is not available. • In this case we can use linear interpolation. • For example, at what effective interest rate will an investment of $100 immediately and $500 at the end of the 3rd year from now accumulate to $1000 at the end of the 10th year from now? 38 Linear interpolation – cont’d • The equation to solve is 100(1 + i)10 + 500(1 + i)7 = 1000. • Put f (i) = (1 + i)10 + 5(1 + i)7. The problem is solved if we can determine what value of i will lead to f (i) = 10. • By trial and error, we find that f (9%) = 9.68 and that f (10%) = 10.39. So by linear interpolation, we have • The actual answer is 9.46%. • The linear interpolation can be repeated until the desired level of accuracy is attained. 39 Worked example 1 • At what interest rate convertible quarterly would $1000 accumulate to $1600 in six years? • We need to determine x = i(4)/4. So the equation of value is 1000(1 + x)24 = 1600. • Solving the equation, we get x = 0.019776. • The answer is i(4) = 4 x = 7.91% 40 Worked example 2 • At what effective rate of interest will sum of the present value of $2000 at the end of the 2nd year from now and $3000 at the end the 4th year from now be equal to $4000? • The equation of value is 2000v2 + 3000v4 = 4000. • That is simplified and re-written as 2v2 + 3v4 4 = 0. • Solving it as a quadratic equation in v2, we get the meaningful root 9 v2 = 0.868517, which gives the answer i = 7.30%. 41 Worked example 3 • Suppose an investment of $1000 now plus another investment of $2000 at the end of 3 years from now will accumulate to $5000 at the end of 10 years from now. What is the nominal interest rate convertible semiannually? • We need to determine j = i(2) /2. • So the equation of value is: 1000(1 + j)20 + 2000(1 + j)14 = 5000 • We cannot solve this equation easily, so we use linear interpolation. • We may also use the graphic capability of MATLAB to help. 42 Worked example 4 – cont’d • Put f (j) = 1000(1 + j)20 + 2000(1 + j)14 5000 • Then we need to find j so that f (j) = 0. • By trial and error, we get f (0.030) = 168.71 and f (0.035) = 227.17 • Performing one linear interpolation, we get 168 .71 j 0.0300 0.0050 0.0321 . 227 .17 168 .71 • This implies that i (2) = 2 (0.321) = 0.0642, or 6.42%. • A higher level of accuracy can be achieved if the linear interpolation is repeated until the desired accuracy is attained. 43 A simple MATLAB plot routine • Use the following command lines in MATLAB – – – – x = 1:0.005:1.2 y = x.^20 + 2*x.^14 – 5 Z=0 plot(x,y,x,z) • and we get the plot 44 A simple plot by MATLAB 45 A fancier MATLAB plot routine x=1:0.005:1.2; y=x.^20 + 2*x.^14-5; z=0 plot(x,y,x,z); title('y=(1+j)^2^0+2(1+j)^1^4-5'); xlabel('1+j'); ylabel('y') 46 A fancier MATLAB plot From this plot we can determine j approximately. 47