1. Chapter 2 Time Value of Money
Interest: The Cost of Money
Economic Equivalence
Development of Interest Formulas
Unconventional Equivalence Calculations

2. Decision Dilemma—Take a Lump Sum or Annual Installments
A suburban Chicago couple won the Power-ball.
They had to choose between a single lump sum $104 million, or $198 million paid out over 25 years (or $7.92 million per year).
The winning couple opted for the lump sum.
Did they make the right choice? What basis do we make such an economic comparison?

3. Option A
(Lump Sum)
Option B
(Installment Plan) 1 2 3 25
$104 M
$7.92 M

4. What Do We Need to Know?
To make such comparisons (the lottery decision problem), we must be able to compare the value of money at different points 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.

5. 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

6. Delaying Consumption

7. Delaying Consumption

8. Notation for interest calculations
An: A discrete payment or receipt occurring at the end of some interest period
i: Interest rate per period
N: Total number of interest periods
P: A sum of money at a time chosen for purposes of analysis as time zero – present value/worth
F: A future sum of money at the end of the analysis period

9. Example: Paying back a loan
You get a loan of $20000 from a bank at a 9% annual interest rate. You also pay a $200 loan origination fee when the loan commences (begins). The bank offers two repayment plans.
Plan 1: Equal payments at the end of every year for the next 5 years
Plan 2: Single payment at the end of the loan period (5 years)

10. Repayment Plans End of Year Receipts Payments Plan 1 Plan 2 Year 0
$20,000.00
$200.00
Year 1
A=?
Year 2
Year 3
Year 4
Year 5
F=?

11. Repayment Plans End of Year Receipts Payments Plan 1 Plan 2 Year 0
$20,000.00
$200.00
Year 1
5,141.85
Year 2
Year 3
Year 4
Year 5
30,772.48
P = $20,000, A = $5,141.85, F = $30,772.48

12. Cash Flow Diagram

13. End-of-Period Convention
Beginning of
Interest period
End of interest
period 1
Interest Period
Important simplifying
assumption: All cash flows are placed at the end of an interest period.

14. Methods of Calculating Interest
Simple interest: the practice of charging an interest rate only to an initial sum (principal amount) – even though you do not withdraw it
Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.

15. 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
$1,000 1
$80
$1,080 2
$1,160 3
$1,240

16. Compound 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
$1,000 1
$80
$1,080 2
$86.40
$1,166.40 3
$93.31
$1,259.71

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

18. Compound Interest The Fundamental Law of Engineering Economy
End of 1st period:
End of 2nd period:
End of 3rd period:
At the end of N periods:
The Fundamental Law of Engineering Economy

19. Comparing Simple to Compound Interest

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

21. Economic Equivalence The compound interest formula
expresses the equivalence between some present amount P and future amount F for given i and N.
Equivalent cash flows are equivalent at any common point in time.

22. Economic Equivalence N F P
If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N.
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.

23. Typical Repayment Plans for a Bank Loan of $20,000
Repayments
Plan 1
Plan 2
Plan 3
Year 1
$5,141.85
$1,800.00
Year 2
5,141.85
1,800.00
Year 3
Year 4
Year 5
$30,772.48
21,800.00
Total of payments
$25,709.25
$29,000.00
Total interest paid
$5,709.25
$10,772.48
$9,000.00

24. Equivalence Between Two Cash Flows
You are offered the alternative of receiving either $3000 at the end of 5 years or P dollars today. You have no current need for the money, you would deposit the P dollars in an account paying 8% interest. What value of P would make you indifferent to your choice between P dollars today and the promise of $3000 at the end of 5 years?

25. 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 5

26. Equivalent Cash Flows are Equivalent at Any Common Point In Time

27. Equivalence Calculations with Multiple Payments

28. Practice problem
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? ?
$1,210 1 4 2 3
$1,500
$1,000
$1,000

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

30. Solution n = 0 n = 1 n = 2 n = 3 n = 4 End of Period Beginning balance
Deposit made
Withdraw
Ending
n = 0
$1,000
n = 1
$1,000( ) =$1,100
$2,100
n = 2
$2,100( ) =$2,310
$1,210
$1,100
n = 3
$1,100( ) =$1,210
$1,500
$2,710
n = 4
$2,710( ) =$2,981
$2,981

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

32. Solution P 2P
N = ?

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

34. Approach
$1,000
$500
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. A
$ $ $502 B

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

36. Types of Cash Flows Single cash flow Equal (uniform) payment series
Linear gradient series
Geometric gradient series
Irregular payment series

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

39. Uneven Payment Series
Wilson Technology wishes to set aside money now to invest over the next 4 years. The company can earn 10% on a lump sum deposited now. The money will be withdrawn in the following increments:
Year 1: $25000 to purchase computer hardware and software
Year 2: $3000 for additional hardware
Year 3: no expenses
Year 4: $5000 for software upgrades
How much must be deposited now?

40. Uneven Payment Series

41. Equal Payment Series A
N N F P

42. Equal Payment Series Compound Amount Factor N A
Example 4.13:
Given: A = $3,000, N = 10 years, and i = 7%
Find: F
Solution: F = $3,000(F/A,7%,10) = $41,449.20

43. Given: F = $5,000, N = 5 years, and i = 7% Find: A
Sinking Fund Factor F N A
Example 4.15:
Given: F = $5,000, N = 5 years, and i = 7%
Find: A
Solution: A = $5,000(A/F,7%,5) = $869.50

44. Handling Time Shifts in a Uniform Series
$5,000 $5, $5, $5,000 $5,000
i = 6%
First deposit occurs at n = 0

45. Capital Recovery Factor N A
Example 4.16:
Given: P = $250,000, N = 6 years, and i = 8%
Find: A
Solution: A = $250,000(A/P,8%,6) = $54,075

46. Equal Payment Series Present Worth Factor N A
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

47. Example 2.13 Deferred Loan Repayment Plan
You borrowed $ to finance educational expenses. The loan will be paid with five payments. You want to defer the first payment until the end of year 2, but still desire to make five annual equal installments. With 6% interest, what are the annual payments?

48. Example 2.13 Deferred Loan Repayment Plan
A A A A A
i = 6%
A’ A’ A’ A’ A’
P’ = $21,061.82(F/P, 6%, 1)
Grace period

49. Two-Step Procedure

50. Example 2.15 Early Savings Plan – 8% interest 44
Option 1: Early Savings Plan
$2,000 ? ?
Option 2: Deferred Savings Plan 44
$2,000

51. Option 1 – Early Savings Plan 44
Option 1: Early Savings Plan
$2,000 ?
Age 31 65

52. Option 2: Deferred Savings Plan 44
Option 2: Deferred Savings Plan
$2,000 ?

53. Option 1: Early Savings Plan
$396,644
Option 1: Early Savings Plan 44
$2,000
$317,253
Option 2: Deferred Savings Plan 44
$2,000

54. Present Value of Perpetuities
Perpetuity: stream of cash flows that continues forever

55. Present Value of Perpetuities
N  ∞
i = 10 %
P=10000

56. Linear Gradient SeriesP

57. Linear Gradient Series

58. Linear Gradient Series
Arithmetic-Geometric Series
Let S be the sum

59. Linear Gradient Series
Geometric Series

60. Linear Gradient Series
Letting x=1/(1+i)
Gradient Series Present Worth Factor

61. Gradient Series as a Composite Series

62. Example
$2,000
$1,750
$1,500
$1,250
$1,000
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?
P =?

63. Method 1:
$2,000
$1,750
$1,500
$1,250
$1,000
$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
P =?

64. Method 2:

65. Geometric Gradient Series
=(P|A1,g,i,N)

66. Geometric Gradient Series

67. Geometric Gradient Series

68. Geometric Gradient: Find P, Given A1,g,i,N
N = 5 years
A1 = $54,440
Find: P

69. Geometric Gradient: Find A1, Given F,g,i,N
Planning for retirement: accumulate $ in 20 years time
Local bank account that pays 8% interest per year
Expects that annual income will increase at 6% annually.
Start with a deposit of A1 at the end of year 1.

70. Unconventional Equivalence Calculations
If you make 4 annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount can be withdrawn over 4 subsequent years?

72. Method 3:
A=100(F|P,10%,4)=146.41

73. Composite Cash Flows $200 $150 $150 $150 $150 $100 $100 $100 $50
$150 $150 $150 $150
$100
$100
$100
$50

74. Multiple Interest Rates
F = ?
Find the balance at the end of year 5. 6% 4% 4% 6% 5% 2 4 5 1 3
$400
$300
$500

75. Solution

76. Cash Flows with Missing Payments
$100
i = 10%
Missing payment

77. Solution P = ? i = 10% Add this cash flow to offset the change $100
$100
Pretend that we have the 10th
payment
i = 10%

78. Approach
P = ?
$100
$100
i = 10%
Equivalent Cash Inflow = Equivalent Cash Outflow

79. Equivalence Relationship

80. Unconventional Regularity in Cash Flow Pattern
$10,000
i = 10% 1
C C C C C C C
Payment is made every other year

81. Approach 1: Modify the Original Cash Flows
$10,000
i = 10% 1
A A A A A A A A A A A A A A

82. Relationship Between A and C
$10,000
i = 10% 1
C C C C C C C
$10,000
i = 10% 1
A A A A A A A A A A A A A A

83. Solution
i = 10% C A A
A =$1,357.46

84. Approach 2: Modify the Interest Rate
Idea: Since cash flows occur every other year, let's find out the equivalent compound interest rate that covers the two-year period.
How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%.
(1+0.10)(1+0.10) = 1.21

85. Solution
$10,000
i = 21% 1
C C C C C C C

86. Summary
Money has a time value because of its earning power and inflation rate.
Economic equivalence exists between individual cash flows and/or patterns of cash flows that have the same value. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.
The purpose of developing various interest formulas was to facilitate the economic equivalence computation.