1. Interest Formulas – Equal Payment Series
Engineering Economy

2. Equal Payment Series F P
Using equal payment series you can find present value or future value 1 2 N A A A P N 1 2 N

3. Compound Amount Factor
A(1+i)N-2 A A A
A(1+i)N-1 N 1 2 1 2 N
This is a geometric series with the first term as A and the constant r = (1+i)
The formula for a geometric series = A (1- r^n)/1-r

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

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

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

7. Annuity Due
Excel Solution
Beginning period
=FV(6%,5,5000,0,1)

8. Example: College Savings Plan
Sinking Fund Factor
The term between the brackets is called the equal-payment-series sinking-fund factor. F N A
Example: 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

9. Excel Solution Given: Find: A Using the equation:
N = 8 years
Find: A
Using the equation:
Using built in Function:
=PMT(i,N,pv,fv,type)
=PMT(7%,8,0,100000,0)
=$9,746.78

10. Capital Recovery Factor
If we need to find A, given P,I, and N
Remember that:
Replacing F with its value

11. Capital Recovery Factor
This factor is called capital recovery factor P N
A = ?
Example: 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

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

13. Two-Step Procedure
Adding the first year interest to the principal then calculating the annuity payment

14. Present Worth of Annuity Series N A
Example:Powerball Lottery
Given: A = $7.92M, N = 25 years, and i = 8%
Find: P
Solution: P = $7.92M(P/A,8%,25) = $84.54M

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

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

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

18. At What Interest Rate These Two Options Would be Equivalent?

20. Using Excel’s Goal Seek Function

21. Result

23. 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

24. Interest Formulas (Gradient Series)

25. Linear Gradient Series
Gradient-series present –worth factor P

26. Gradient Series as a Composite Series
We view the cash flows as composites of two series a uniform with a payment amount of A1 and a gradient with a constant amount of G

27. 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 =?

28. 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 =?

29. Method 2:

30. Example: Supper Lottery
$3.44 million
Cash Option
Annual Payment Option
$357,000
G = $7,000
$196,000
$189,000
$175,000

31. Equivalent Present Value of Annual Payment Option at 4.5%
The gradient series is delayed by one period
To return the calculations to year zero

32. Geometric Gradient Series
Geometric Gradient is a gradient series that is been determined by a fixed rate expressed as a percentage instead of a fixed dollar amount
For example the economic problems related to construction cost which involves cash flows that increase or decrease by a constant percentage

33. Present Worth Factor
Geometric-gradient-series present-worth factor

34. Alternate Way of Calculating P

35. Unconventional Equivalence Calculations
EGN3613
Ch2 Part IV

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

37. Unconventional Equivalence Calculations
Situation:
What value of A would make the two cash flow transactions equivalent if i = 10%?

40. 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

41. Solution

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

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

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

45. Equivalence Relationship

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

47. 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

48. 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

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

50. 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

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