Slide Set 3 Multiple compounding periods, continuous compounding, changing interest rates

1. Slide Set 3 – Multiple compounding periods, continuous compounding, changing interest rates IEM 3503/IEM3513 Dr. Baski Balasundaram Industrial Engineering & Management

2. 2 Review: Sections 2.6, 2.7, 3.2, 3.4,3.6 Nominal interest rate Multiple compounding periods per interest period Effective interest rate Interest rates that vary over time Principal and interest in loan payments Special and limiting cases of equivalence factors

3. 3 Interest period In our analysis until now, we have assumed that “compounding” takes place yearly; the interest period is one year (eg. 12% per year); and hence, we use a yearly time scale for the cash flow diagrams (i.e. count periods, n, in years) If “year” is replaced with “month” or any other time period everywhere, our analysis used to derive equivalence factors is still valid…

4. 4 Nominal Interest rate In practice however, say your credit card for instance, quotes what is called a nominal interest rate (18% per year/ 18% APR), but compounding period (how often the interest you owe is actually calculated) is more often than just one year, typically, daily Keywords: compounding period; interest period; nominal interest rate Does this matter: To our analysis? To our decision making process?

5. Class Pb. 1 Suppose you have a credit card with $1000 balance and you stop using it for the next year. The card charges 18% APR, compounded daily. How much interest would you owe at the end of the year? 5

6. Class Pb. 1: Solution 18% per year is the same as 18/365 % per day = 0.0493% daily (nominal rate can be expressed over any period, 18% APR is the same as 1.5% per month) nominal rate when expressed over the compounding period, has a physical meaning F = 1000(1+ 0.000493)365 = 1000*1.1971 = $1197.10 Interest owed is $197.10 which is more than $180 18% APR with daily compounding is effectively 19.71% APY 6

7. 7 Nominal Interest Rates Interest may be computed (compounded): Annually – Once a year (at the end) Every 6 months – 2 times a year (semi-annual) Every quarter – 4 times a year (quarterly) Every Month – 12 times a year (monthly) Every Day – 365 times a year (daily) … Continuous – infinite number of compounding periods in a year.

8. 8 Terminology Compounding period - Length of time between compounding operations; cash flows are not recognized except as end of period cash flows if they occur inside a compounding period. Interest period - Interest rates are stated as % per time period. This time period is called the interest period. Compounding frequency - the number of times that compounding occurs within the interest period Nominal interest rate – an expression of interest rate which in itself carries no physical meaning, can be scaled up or down by changing the interest period

9. 9 Nominal Interest Rate For example, if the nominal interest rate were 1.5% per month, the nominal rate would also be: 0.05% per day, (30 days/mo.) 4.5% per quarter, 9% semiannually 18% per year Depending on the compounding period, one of these would carry a physical meaning: the rate at which money would grow in that account

10. Class Pb. 2 Find the present worth of the yearly cash flow series above if the interest rate is 12% per year compounded monthly P = 900(P/F, 1% ,12) + 900(P/F, 1% ,24) + 900(P/F, 1% ,36) + 900(P/F, 1% ,48) This just rendered the (P/A) factor useless! 10

11. Class Pb. 2: Alternate solution P = 900(P/F, 1% ,12) + 900(P/F, 1% ,24) + 900(P/F, 1% ,36) + 900(P/F, 1% ,48) P = 900[ 1/(1+i)12 + 1/(1+i)24 + 1/(1+i)36 +1/(1+i)48 ] with i = 0.01 Substitute (1+i)12 = 1+ieff P = 900[ 1/(1+ieff) + 1/(1+ieff)2 + 1/(1+ieff)3 +1/(1+ieff)4 ] with ieff = (1.12)12 – 1 P = 900(P/A, ieff , 4) 11

12. 12 Suppose the nominal rate’s interest period is used for the time periods on a cash flow diagram; but actual compounding happens more often, i.e. compounding period < interest period Adjusting the time scale to count in terms of compounding periods can destroy any pattern that might exist in the cash flow Then we can compute an effective interest rate, whose period equals nominal rate’s period (which in turn equals cash flow’s time scale) This larger rate accounts by its magnitude, the fact that compounding is done more often, and permits us to retain the original time scale and cash flow pattern Effective Interest Rate

13. 13 P(1+r/m) (1+r/m) (1+r/m)…(1+r/m)=P(1+ieff)=F P(1+r/m)m = P(1+ieff) = F ? ieff = (1+r/m)m - 1 r/m is the nominal rate per CP; i.e. the rate at which P grows per CP; ieff is the rate at which P grows per interest period T; Note that r and ieff are interest rates expressed over period T, but r does not have the “rate of growth” interpretation Effective Interest Rate

14. 14 “r % per time period T, compounded ‘m’ times in time period T. E.g. 18% per year, compounded monthly (r = 18%, T = 1 year, CP = 1 month, m = 12) ieff = (1 + r/m)m - 1 per time period T E.g. ieff = (1 + 0.18/12)12 – 1 = 0.19562 or, 19.56% per year Effective Interest Rate

15. 15 Varying Statements of Interest Rates Effective Rate is directly stated “Effective rate is 8.243% per year, compounded quarterly; No nominal rate given; Compounding frequency m = 4; No need to calculate the true effective rate! It is already given: 8.243% per year!

16. 16 Varying Statements of Interest Rates Only the interest rate is stated “8% per year”. No information on the frequency of compounding; Must assume it is for one year! Assume that “8% per year” is a true, effective annual rate!

17. 17 What is the future worth of $100 after 1 year if a bank pays 12% interest per year compounded semiannually? Solution: First find effective interest rate per year r = 12% per year (nominal); T = 1 year; CP = 6 months; m = 2; Effective rate per year = (1 + r/m)m –1 i per year = (1 + 0.12/2)2 – 1 = 0.1236 = 12.36% F = $100 (1+0.1236) = $112.36 Class Pb. 3

18. 18 A bank pays a 12% per year compounded semiannually. What is the effective interest rate for a) 6 mo.? b) 12 mo? c) 24 mo.? Solution: a) Compounding period CP is 6 mo. The nominal interest is 12% per year, so the nominal interest rate per 6 mo is 6%. The effective interest rate per 6 mo. is also 6% since the compounding period equals to the interest rate period (6 mo.). Class Pb. 4

19. 19 r = 12% / year CP = 6 months i/6 months = r/6months = 12% / 2 = 6% b) What is the effective interest rate for 12 mo.? r/12 months = 0.12 (given); m = 2 i/12 months = (1 + .12/2)2 – 1 = 12.4% c) What is the effective interest for 24 mo.? i = (1+.24/4)4 - 1 = 26.2 %

20. 20 Numerical Problems, with multiple compounding periods Method 1: Determine the nominal interest rate over the given compounding period, and treat each payment separately. i.e. change the time unit of the cash flow diagram to CP Method 2: Calculate an effective interest rate for the interest period (equal to cash flow periods) and then proceed “as usual” i.e. retain the time unit and cash flow pattern as imposed by the cash flows for the cash flow diagram

21. 21 An engineer deposits $1,000 in a savings account at the end of each year. If the bank pays interest at the rate of 6% per year, compounded quarterly, how much money will be accumulated in the account after 5 years? Class Pb. 5

22. 22 Solution Method 1:    m = 4 qtrs per year; CP = 1 qtr; r = 0.06 per year = 0.015 per qtr; F = 1,000(F/P,1.5%,16) + 1,000(F/P,1.5%,12) + 1,000(F/P,1.5%,8) + 1,000(F/P,1.5%,4) + 1,000 = $5,652.50.

23. 23 Method 2:    i = (1 + r/m)m – 1 = (1 + 0.06/4)4 –1 = 0.06136 (effective interest rate per year). F = $1,000(F/A, 6.136%, 5). Note that there are no tables for i = 6.136%. Using the formula, we get

24. 24 A deposit of $3000 is made after 3 years and $5000 after 5 years, what is the value of the account after 10 years if the interest rate is 12% per year compounded semi-annually? Class Pb. 6

25. 25 Method 1 CP = 6 mon i per CP = .12 per year/2 = .06 semi-annually F = 3000 (F/P,6%,14) + 5000 (F/P,6%,10) = 3000 (2.2609) + 5000 (1.7908) = 15,736.70 Method 2 i = (1 + (.12/2))2 – 1 = 12.36% per year F = 3000 (F/P,12.36%,7) + 5000 (F/P,12.36%,5) = 15,736.70

26. 26 $100 is deposited each quarter for 5 years. What is the future value of the account if interest is 12% per year compounded monthly. Class Pb. 7

27. 27 Solution: CP = 1 month; quarterly cash flow Nominal rate per month is .12/12 = .01 The effective interest per quarter is: i = (1+ .03/3)3 - 1 = .0303 ? F = $100 (F/A, .0303, 20) = $2695.76

28. Continuous compounding Suppose we have nominal r% per year and m compounding periods in one year FWn = P(1+ ieff )n = P(1+r/m)mn As m approaches infinity, (1+r/m)mn approaches: ern Under continuous compounding, F = P ern 28

29. 29 Changing interest rates F = P(1+i1) (1+i2) (1+i3) ... (1+in-1) (1+in)

30. 30 How much money would be accumulated in 9 years with an initial deposit of $1000, if the account earned interest at 8% per year for the first 3 years, 10% per year for the next 4 years, and 12% per year for the last 2 years? F = 1000(F/P,8%,3) (F/P,10%,4) (F/P,12%,2) = $2313.53 Class Pb. 8

31. 31 Class Pb. 9 Find, FW,PW and AW of this cash flow series. Solution: P = 200(P/F,10%,1) -200(P/F,10%,2) +300(P/F,8%,1)(P/F,10%,2) +200(P/F,12%,1)(P/F,8%,2)(P/F,10%,2) =$372.62

32. 32 Class Pb. 9 Find, FW,PW and AW of this cash flow series. Solution: F = 200 +300(F/P,8%,1)(F/P,12%,1) -200(F/P,8%,2)(F/P,12%,1) +200(F/P,10%,1)(F/P,8%,2)(F/P,12%,1) = $589.01, or use the P found earlier F = 372.62(F/P,10%,2)(F/P,8%,2)(F/P,12%,1)

33. 33 Class Pb. 9 $372.62 = A(P/A,10%,2) +A(P/A,8%,2)(P/F,10%,2) +A(P/F,12%,1)(P/F,8%,2)(P/F,10%,2) Solve for A. ? A = $96.99

34. Principal and interest in loan payments Suppose, $10,000 is borrowed at t=0 at 1% per month. It is to be paid of in 6 equal monthly payments, A. Then, A = 10000(A/P, 1%, 6) = $1725.48 Then, 6 x 1725.48 -10,000 = 10,352.88 -10,000= $352.88, is the total interest payment made But how much of each month’s payment goes towards the principal and how much towards interest? 34

35. Principal and interest in loan payments 35

36. Limiting and special cases of factors (P/A, 0% , n) = n (P/A,i,n) ? 1/i as n ? 8 when i ? 0 (A/P,i,n) ? i as n ? 8 when i ? 0 (F/A,i,n) ? 8 as n ? 8 when i ? 0 (A/F,i,n) ? 0 as n ? 8 when i ? 0 36