1. Example 1 Ms. Smith loans Mr. Brown $10,000 with interest compounded at a rate of 8% per year. How much will Mr. Brown owe Ms. Smith if he repays the loan at the end of 5 years?

2. Example 1 - Solution Since the problém is of the form "find F when given P" the formula to use is F = P(F/P, 8%, 5) = $10,000(1.4693) = $14,693.

3. Example 2 Mr. Lee wishes to accumulate $10,000 in a savings account in 10 years. If the bank pays 5% compounded annually on deposits of this size, how much should Mr. Lee deposit ir the account?

4. Example 2 - Solution This problem is of the form "find P when given F," and the formula to use is P = F(P/F,5%,10) = $10,000(0.6139) = $6,139.

5. Example 3 An individual has been making equal annual payments of $2,000 to repay a loan. The individual wishes to pay off the loan immediately after having made an annual payment. Four payments remain to be paid. With an interest rate of 8%, how much should be paid in the final payment?

6. Example 3 - Solution P = A(P/A,8%,4) = $2,000(3.3121) = $6,624

7. Example 4 A person borrows $10,000 at 6% compounded annually. If the loan is repaid in ten equal annual payments, what will be the size of the payments if the first payment is made l year after borrowing the money?

8. Example 4 - Solution A = P(A/P,6%,10) = $10,000(0.1359) = $1,359

9. Example 5 If $800 is deposited annually for 10 years in an account that pays 6% compounded annually, how much money will be in the fund immediately after the tenth deposit?

10. Example 5 - Solution F = A(F/A,6%,10) = $800(13.1808) = $10,545

11. Example 6 An individual wishes to accumulate $1,000,000 in 30 years. If 30 end-of-year deposits are made into an account that pays interest at a rate of 10% compounded annually, what size deposit is required to meet the stated objective?

12. Example 6 - Solution A = F(A/F, 10%, 30) = $1,000,000(0.0061) = $6,100

13. Example 7 It is expected that a machine will incur operatíng costs of $4,000 the first year and that these costs will increase by $500 each year thereafter for the 10-year life of the machine. If money is worth 15% per year to the firm, what is the equivalent annual worth of the operating costs?

14. Example 7 - Solution The problem reduces to a constant $4,000 per year plus the $500 gradient. A = $4,000 + G(A/G, i%, N) = $4,000 + $500(A/G,15%,10) = $4,000 + $500(3.3832) = $5,692

15. Example 8 Work Problem 7 if the timing of the costs is reversed as shown in the following diagram:

16. Example 8 - Solution The problem reduces to a constant $8,500 per year minus a $500 gradient of payments. A = $8,500 - 500(A/(7,15%, 10) = $8,500 - 500(3.3832) = $6,808

17. Example 9 In Problem 7 suppose the operating cost the first year is $4,000 and that each year thereafter for the 10-year life of the machine operating costs increase by 6% per year. If money is worth 15% per year to the finn, what is the equivalent annual worth of the operatíng costs?

18. Example 9 - Solution Since j = 6% and i = 15%, F = A j [ (F/P,15%,10) - (F/P, 6%, 10)] / 0.15 - 0.06 = $4,000[4.0456 - 1.7908] / 0.09 = $100,213 and A = F(A/F, 15%, 10) = $100,213(0.0493) = $4,941

19. Example 10 A person makes six end-of-year deposits of $1,000 in an account paying 5% compounded annually. If the accumulated rund is withdrawn 4 years after the last deposit, how much money will be withdrawn?

20. Example 10 - Solution Since F does not occur at the ume of the last A, it is necessary that the solution proceed in two steps. The amount of money in the account at the ume of the last deposit may be computed as F 6 = A(F/A,5%,6) = $1,000(6.8019) = $6,802. The problem now is to find F 10 given F 6, = P 6 = $6,802. F 10 = F 6 (F/P,5%,4) = $6,802(1.2155) = $8,268

21. Example 11 - Solution What is the effective interest rate for 4.75% compounded annually and 4.60% compounded quarterly?

22. Example 11 - Solution Effectíve rate = (1+ rIM) M - 1 = (1+ 0.0475) 1 - 1 = 4.75% Effective rate = (1+ rIM) M - 1 = (1+ 0.0476/4) 4 - 1 = 4.68%

23. Example 12 A loan company advertises that it will loan $1,000 to be repaid in 30 monthly installments of $44.60. What is the cffective interest rate?

24. Example 12 - Solution A = P(A/P,i%,30) $44.60 / $1,000 =(A/P,i%,30) =0.0446 By inspection (with interpolation in tables), i = 2%, and Effective rate = (F/P,2%,12) - 1 = 0.2682 = 26.82%.

25. Example 13 Annual deposite of $1,000 are made in an account that pays 4% compounded quarterly. How much money should be in the account immediately after the fifth deposit?

26. Example 13 - Solution Effective rate = (F/P, 1%, 4) – 1 = 4.06% F = A(F/A,4.06%,5) = $1,000 [(1+ 0.0406) 5 - 1] / 0.0406 = $5 423 or A = $1,000(A/F,1%,4) = $1,000(0.2463) = $246.30 F 20 = $246.30(F/A,1%,20) = 246.30(22.019) = $5,423 An alternate solution method is to treat the five annual deposits as single sums of money.

27. Example 13 - Solution Therefore, F = $1,000(A/F,1%,16)+(F/P,1%,12)+(F/P,1%,8)+(F/P,1%,4)+1] = $1,000[1.1726 + 1.1268 + 1.0829 + 1.0406 + 1.000] = $5,423

28. Example 14 Given the payments shown in the following cash flow diagram, what is the equivalent worth in 2005 with interest at 6%?

29. Example 14 - Solution X = [$500(F/A, 6%, 5) + $100(F/A, 6%, 3) + $200](F/P, 6%, 1) = [$500(5.6371) + $100(3.1836) + $200](1.060) = $3,537

30. Example 15 With interest at 8% compounded annually, how long does it take for a certain amount to double in magnitude? (F/P, 8%, N) = 2.00 By inspection of 8% interest tables, N= 9 years.

31. Example 16 An individua! approaches the Loan Shark Agencv for $ 1,000 to be repaid in 24 monthly installments. The agency advertises interest at 1.5% per month. They proceed to calcu-ate the slze of his payment in the following manner: Amount requested: $1,000 Credit investigation: 25 Credit risk insurance: 5 Total: $1,030 Interest: (1030) (24) (0.015) = $371 Total owed: $1,030 + $371 = $1,401 Payment: $1,401/ 24 = $58.50 What effective interest rate is the individua! paying?

32. Example 16 - Solution A =$58.50 A = P(AIP, 1%, 24) $58.50 = $1,000(A/P, í%, 24) By interpolation in tables, i = 2.9% per month, and Effective rate = (F/P,2.9%,12) - 1 = 1.41 - 1= 41%

33. Example 17 Money is to be invested for a chilďs college expenses. Annual deposits of $2,000 are made in a fund that pays 5% compounded annually. If the first deposit is made on the chilďs 5th birthday and the last on the chilďs 15th biithday, what is the size of4 equal withdrawals on the chilďs 18th, 19th,20th, and 21st birthdays that will just deplete the account?

34. Example 17 - Solution Amount in fund at t = 15: F 15 = A(F/A,5%,11) = $2,000(14.2068) = $28,414. Amount in fund at t = 17: F 17 = P 15 (F/P, 5%, 2) = $28,414(1.1025) = $31,326 Amount of withdrawals: A = P 15 (A/P,5%,4) = $31,326(0.2820) = $8,834

35. Example 18 An university student borrows money in her senior year to buy a car. She defers payments for 6 months and makes 36 beginning-of- month payments thereafter. If the original note is for $12,000 and interest is 0.5% per month on the unpaid balance, how much will her payments be?

36. Example 18 - Solution Amount owed at t = 5: F 5 = P o (F/P,0.5%,5) = $12,000(1.0253) = $12,304 Amount of monthly payments: A = P,(A/P, 0.5%,36) = $12,304 [i(1 + i) N / (1 + i) N - 1] = $12,304 [(1,005)(1,005) 36 / (1,005) 36 - 1] = $372.45

37. Example 19 What is the present wonh of $100,000 ten years hence if interest is a) 15% compounded annually? b) 15% compounded continuously?

38. Example 19 - Solution a. P = F(P/F,15%10) = $100,000(0.2472) = $24,720 b. P= F(P/F,15%,10) = $100,000(0.2231) = $22,310