1. LOGO 1 MATH 2040 Introduction to Mathematical Finance Instructor: Dr. Ken Tsang

2. 2 Examples Chapter 2: Solution of problems in interest Introduction: the basic problem Unknown time & interest rate Determining time periods Equations of value

3. 3 Introduction How to solve an interest problem? – Use basic principle –Develop a systematic approach

4. 4 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

5. 5 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.

6. 6 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)

7. 7 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

8. 8 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”.

9. 9 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(Y 2 – Y 1 ) + 30(M 2 – M 1 ) + (D 2 – D 1 )

10. 10 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.

11. 11 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.

12. 12 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.

13. 13 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.

14. 14 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.

15. 15 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.

16. 16 Time diagram It helps to draw out a time line and plot the payments and withdrawals accordingly.

17. 17 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

18. 18 Example – solution 1 Compare the values at t = 0.

19. 19 Example – solution 2 Compare the values at t = 5.

20. 20 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.

21. 21 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.

22. 22 Example - logarithm How long does it take money to double if the interest rate i = 6%?. Very simple evaluation using MATLAB

23. 23 Example – interest table + interpolation If logarithms are not available, then use interest tables and perform a linear interpolation. Using interest tables, we have and

24. 24 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.

25. 25 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% 107.2 7.27  1% 12 6 6.12  2% 184 4.19  5%

26. 26 Tripling a payment – rule of 114 For tripling a single payment, we have the approximate rule of 114.

27. 27 Multiple payments Let S t 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 S t, 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.

28. 28 Multiple payments - solution To find the true value of t, we form the equation: Taking logarithm on both sides, we have

29. 29 Multiple payments – approximate solution Suppose we use, 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 then the present value using the method of equated time will be less than the present value using exact t.

30. 30 Weighted average of time > exact time 1 The present value of the payment at time t k is. So the arithmetic weighted mean of present values is And the geometric weighted mean of present values is

31. 31 Weighted average of time > exact time 2 Since the geometric means are less than arithmetic mean  Method of equated time > exact t

32. 32 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

33. 33 Example – cont’d Using logarithm, we have n log e 1.03 = log e 1.5. Because log e 1.03 = 0.405465 and log e 1.5 = 0.029559, we can determine n to be 13.717. So the number of years is 6.859 Using MATLAB

34. 34 Example – cont’d From the interest table, we have (1.03) 13 = 1.46853 and (1.03) 14 = 1.51259 Performing a linear interpolation, So the number of years is 6.857. Note that the two answers are very close.

35. 35 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 v t = 100 v 2 + 200 v 3 + 500 v 8, which can be solved for t = 5.832. By method of equated time, we have As expected, the true value t is less.

36. 36 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

37. 37 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 v 5 + 300 v 10 = 360. If we put y = v 5, 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.

38. 38 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 3 rd year from now accumulate to $1000 at the end of the 10 th year from now?

39. 39 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.

40. 40 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%

41. 41 Worked example 2 At what effective rate of interest will sum of the present value of $2000 at the end of the 2 nd year from now and $3000 at the end the 4 th year from now be equal to $4000? The equation of value is 2000v 2 + 3000v 4 = 4000. That is simplified and re-written as 2v 2 + 3v 4  4 = 0. Solving it as a quadratic equation in v 2, we get the meaningful root 9 v 2 = 0.868517, which gives the answer i = 7.30%.

42. 42 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.

43. 43 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 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.

44. 44 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

45. 45 A simple plot by MATLAB

46. 46 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')

47. 47 A fancier MATLAB plot From this plot we can determine j approximately.