1. Time Value of Money Lecture No.2 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005

2. Time Value of Money Interest: The Cost of Money Economic Equivalence Interest Formulas – Single Cash Flows Equal-Payment Series Dealing with Gradient Series Composite Cash Flows.

3. What Do We Need to Know? To make such comparisons, we must be able to compare the value of money at different point in time. To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis.

4. Time Value of Money Money has a time value because it can earn more money over time (earning power). Money has a time value because its purchasing power changes over time (inflation). Time value of money is measured in terms of interest rate. Interest is the cost of money-a cost to the borrower and an earning to the lender

5. Delaying Consumption

7. Elements of Transactions involve Interest 1. Initial amount of money in transactions involving debt or investments is called the principal (P). 2. The interest rate (i) measures the cost or price of money and is expressed as a percentage per period of time. 3. A period of time, called the interest period (n), determines how frequently interest is calculated. 4. A specified length of time marks the duration of the transactions and thereby establishes a certain number of interest periods (N). 5. A plan for receipts or disbursements (An) that yields a particular cash flow pattern over a specified length of time. [monthly equal payment] 6. A future amount of money (F) results from the cumulative effects of the interest rate over a number of interest periods.

8. Which Repayment Plan? End of YearReceiptsPayments Plan 1Plan 2 Year 0$20,000.00 $200.00 Year 15,141.850 Year 25,141.850 Year 35,141.850 Year 45,141.850 Year 55,141.8530,772.48 The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)

9. Cash Flow Diagram Represent time by a horizontal line marked off with the number of interest periods specified. Cash flow diagrams give a convenient summary of all the important elements of a problem.

10. Methods of Calculating Interest Simple interest: the practice of charging an interest rate only to an initial sum (principal amount). Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

11. Simple Interest P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 8% N = 3 years End of Year Beginning Balance Interest earned Ending Balance 0$1,000 1 $80$1,080 2 $80$1,160 3 $80$1,240

12. Simple Interest Formula

13. Compound Interest the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn. P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 8% N = 3 years End of Year Beginning Balance Interest earned Ending Balance 0$1,000 1 $80$1,080 2 $86.40$1,166.40 3 $93.31$1,259.71

14. Compounding Process $1,000 $1,080 $1,166.40 $1,259.71 0 1 2 3

15. 0 $1,000 $1,259.71 1 2 3

16. Compound Interest Formula

17. Practice Problem Problem Statement If you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10?

18. Solution 0 1 2 3 4 5 6 7 8 9 10 $100 $200 F

19. Practice problem Problem Statement Consider the following sequence of deposits and withdrawals over a period of 4 years. If you earn 10% interest, what would be the balance at the end of 4 years? $1,000 $1,500 $1,210 0 1 2 3 4 ? $1,000

20. $1,500 $1,210 01 2 3 4 ? $1,000 $1,100 $2,100$2,310 -$1,210 $1,100 $1,210 + $1,500 $2,710 $2,981 $1,000

21. Solution End of Period Beginning balance Deposit made WithdrawEnding balance n = 0 0$1,0000 n = 1 $1,000(1 + 0.10) =$1,100 $1,0000$2,100 n = 2 $2,100(1 + 0.10) =$2,310 0$1,210$1,100 n = 3 $1,100(1 + 0.10) =$1,210 $1,5000$2,710 n = 4 $2,710(1 + 0.10) =$2,981 00$2,981

22. Economic Equivalence What do we mean by economic equivalence? Why do we need to establish an economic equivalence? How do we establish an economic equivalence?

23. Economic Equivalence (EE) Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another. EE refers to the fact that a cash flow-whether a single payment or a series of payments-can be converted to an equivalent cash flow at any point in time. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.

24. Economic Equivalence (EE)

25. If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N. N F P 0 Equivalence from Personal Financing Point of View

26. F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i. N F P 0 Alternate Way of Defining Equivalence N0 =

27. Practice Problem 012 3 345 $2,042 5 F 0 At 8% interest, what is the equivalent worth of $2,042 now 5 years from now? If you deposit $2,042 today in a savings account that pays 8% interest annually. how much would you have at the end of 5 years? =

28. Solution

29. Example 2.2 $3,000$2,042 50 At what interest rate would these two amounts be equivalent? i = ?

30. Equivalence Between Two Cash Flows Step 1: Determine the base period, say, year 5. Step 2: Identify the interest rate to use. Step 3: Calculate equivalence value. $3,000$2,042 50

31. Example - Equivalence Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%. 0 1 2 3 4 5 P F $2,042 $3,000 $2,205 $2,382 $2,778$2,572

32. Example 2.3 012 3 4 5 $100 $80 $120 $150 $200 $100 012 3 4 5 V = Compute the equivalent lump-sum amount at n = 3 at 10% annual interest.

33. 012 3 4 5 $100 $80 $120 $150 $200 $100 V Approach

34. 012 3 4 5 $100 $80 $120 $150 $200 $100 V

35. Practice Problem How many years would it take an investment to double at 10% annual interest? P 2P2P 0 N = ?

36. Solution P 2P2P 0 N = ?

37. Given: i = 10%, Find: C that makes the two cash flow streams to be indifferent Practice Problem $500 $1,000 0 1 2 3 A B CC

38. Approach Step 1: Select the base period to use, say n = 2. Step 2: Find the equivalent lump sum value at n = 2 for both A and B. Step 3: Equate both equivalent values and solve for unknown C. $500 $1,000 0 1 2 3 A B CC

39. Solution For A: For B: To Find C: $500 $1,000 0 1 2 3 A B CC

40. At what interest rate would you be indifferent between the two cash flows? $500 $1,000 0 1 2 3 $502 $502 $502 A B Practice Problem

41. Approach Step 1: Select the base period to compute the equivalent value (say, n = 3) Step 2: Find the net worth of each at n = 3. $500 $1,000 0 1 2 3 $502 $502 $502 A B

42. Establish Equivalence at n = 3 Find the solution by trial and error, say i = 8%

43. Single Cash Flow Formula Single payment compound amount factor (growth factor) Given: Find: F P F N 0

44. Practice Problem If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years? $2,000 F = ? 8 0 i = 10%

45. Solution

46. Single Cash Flow Formula Single payment present worth factor (discount factor) Given: Find: P F N 0

47. Practice Problem You want to set aside a lump sum amount today in a savings account that earns 7% annual interest to meet a future expense in the amount of $10,000 to be incurred in 6 years. How much do you need to deposit today?

48. Solution 0 6 $10,000 P

49. Multiple Payments ….. (Example 2.8) How much do you need to deposit today (P) to withdraw $25,000 at n =1, $3,000 at n = 2, and $5,000 at n =4, if your account earns 10% annual interest? 0 1 2 3 4 $25,000 $3,000 $5,000 P

50. 0 1 2 3 4 $25,000 $3,000 $5,000 P 0 1 2 3 4 $25,000 P1P1 0 1 2 3 4 $3,000 P2P2 0 1 2 3 4 $5,000 P4P4 ++ Uneven Payment Series

51. Check 01234 Beginning Balance 028,6226,484.204,132.624,545.88 Interest Earned (10%) 02,862648.42413.26454.59 Payment+28,622-25,000-3,0000-5,000 Ending Balance $28,6226,484.204,132.624,545.880.47 Rounding error It should be 0.

52. Equal Payment Series: Find equivalent P or F A 0 1 2 3 4 5 N-1 N F P

53. Equal Payment Series – Compound Amount Factor 012 N 0 12N AAA F 012 N AAA F

54. Compound Amount Factor 012 N 0 12N AAA F A(1+i) N-1 A(1+i) N-2

55. Equal Payment Series Compound Amount Factor (Future Value of an annuity) Example 2.9: Given: A = $5,000, N = 5 years, and i = 6% Find: F Solution: F = $5,000(F/A, 6%, 5) = $28,185.46 0 1 2 3 N F A

56. Validation

57. Finding an Annuity Value Example: Given: F = $5,000, N = 5 years, and i = 7% Find: A Solution: A = $5,000(A/F, 7%, 5) = $869.50 0 1 2 3 N F A = ?

58. Example 2.10 Handling Time Shifts in a Uniform Series F = ? 0 1 2 3 4 5 $5,000 $5,000 $5,000 $5,000 $5,000 i = 6% First deposit occurs at n = 0

59. 59 Sinking fund (1) A fund accumulated by periodic deposits and reserved exclusively for a specific purpose, such as retirement of a debt. (2) A fund created by making periodic deposits (usually equal) at compound interest in order to accumulate a given sum at a given future time for some specific purpose.

60. 60 Sinking Fund Factor is an interest-bearing account into which a fixed sum is deposited each interest period; The term within the colored area is called sinking-fund factor. 0 1 2 3 N F A Example 2.11 Example 2.11 – College Savings Plan: Given: F = $100,000, N = 8 years, and i = 7% Find: A Solution: A = $100,000(A/F, 7%, 8) = $9,746.78

61. OR Given: F = $100,000 i = 7% N = 8 years Find: A Solution: A = $100,000(A/F, 7%, 8) = $9,746.78

62. 62 Capital Recovery Factor (Annuity Factor) Annuity: (1) An amount of money payable to a recipient at regular intervals for a prescribed period of time out of a fund reserved for that purpose. (2) A series of equal payments occurring at equal periods of time. (3) Amount paid annually, including reimbursement of borrowed capital and payment of interest. Annuity factor: The function of interest rate and time that determines the amount of periodic annuity that may be paid out of a given fund.

63. Capital Recovery Factor is the colored area which is designated (A/P, i, N). In finance, this A/P factor is referred to as the annuity factor. Example 2.12: Paying Off Education Loan Given: P = $21,061.82, N = 5 years, and i = 6% Find: A Solution: A = $21,061.82(A/P,6%,5) = $5,000 1 2 3 N P A = ? 0

64. P =$21,061.82 0 1 2 3 4 5 6 A A A A A i = 6% 0 1 2 3 4 5 6 A A A A A i = 6% P = $21,061.82(F / P, 6%, 1) Grace period Example 2.13 Deferred (delayed) Loan Repayment Plan

65. Two-Step Procedure

66. Present Worth of Annuity Series The colored area is referred to as the equal-payment-series present-worth factor Example 2.14: Powerball Lottery Given: A = $7.92M, N = 25 years, and i = 8% Find: P Solution: P = $7.92M(P/A, 8%, 25) = $84.54M 1 2 3 N P = ? A 0

67. 0 1 2 3 4 5 6 7 8 9 10 11 12 44 Option 2: Deferred Savings Plan $2,000 Example 2.15 Early Savings Plan – 8% interest 0 1 2 3 4 5 6 7 8 9 10 44 Option 1: Early Savings Plan $2,000 ? ?

68. Option 1 – Early Savings Plan 0 1 2 3 4 5 6 7 8 9 10 44 Option 1: Early Savings Plan $2,000 ? 6531Age

69. Option 2: Deferred Savings Plan 0 11 12 44 Option 2: Deferred Savings Plan $2,000 ?

70. 0 1 2 3 4 5 6 7 8 9 10 44 0 1 2 3 4 5 6 7 8 9 10 11 12 44 Option 1: Early Savings Plan Option 2: Deferred Savings Plan $2,000 $396,644 $317,253

71. 71 Linear Gradient Series Engineers frequently meet situations involving periodic payments that increase or decrease by a constant amount (G) from period to period. P

72. Gradient Series as a Composite Series

73. $1,000 $1,250 $1,500 $1,750 $2,000 1 2 3 4 5 0 P =? How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure? Example – Present value calculation for a gradient series

74. Method 1: $1,000 $1,250 $1,500 $1,750 $2,000 1 2 3 4 5 0 P =? $1,000(P/F, 12%, 1) = $892.86 $1,250(P/F, 12%, 2) = $996.49 $1,500(P/F, 12%, 3) = $1,067.67 $1,750(P/F, 12%, 4) = $1,112.16 $2,000(P/F, 12%, 5) = $1,134.85 $5,204.03

75. Method 2:

76. 0 1 2 3 4 5 6 7 25 26 $3.44 million 0 1 2 3 4 5 6 7 25 26 Cash Option $175,000 $189,000 $357,000 $196,000 G = $7,000 Annual Payment Option Example 2.16 Supper Lottery

77. Equivalent Present Value of Annual Payment Option at 4.5%

78. Geometric Gradient Series Many engineering economic problems, particularly those relating to construction costs, involve cash flows that increase over time, not by a constant amount, but rather by a constant percentage (geometric), called compound growth.

79. Present Worth Factor

80. Alternate Way of Calculating P

81. Example 2.17: Find P, Given A 1, g, i, N Given: g = 5% i = 7% N = 25 years A 1 = $50,000 Find: P

82. Required Additional Savings

83. $50 $100 $150 $150 $200 0 1 2 3 4 5 6 7 8 9 Composite Cash Flows