1. Chapter 4 More Interest Formulas
EGN 3615
ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1

2. Chapter Contents Uniform Series Compound Interest Formulas
Uniform Series Compound Amount Factor
Uniform Series Sinking Fund Factor
Uniform Series Capital Recovery Factor
Uniform Series Present Worth Factor
Arithmetic Gradient
Geometric Gradient
Nominal Effective Interest
Continuous Compounding

3. Uniform Series Compound Amount Factor
F1+F2+F3+F4 =F 1 2 3 4 A 1 2 3 4 A 1 2 3 4 F1 A 1 2 3 4 F2 A 1 2 3 4 F3
A=F4 1 2 3 4

4. Uniform Series Compound Amount Factor
That is, for 4 periods, F = F1 + F2 + F3 + F4 = A(1+i)3 + A(1+i)2 + A(1+i) + A = A[(1+i)3 + (1+i)2 + (1+i) + 1]

5. Uniform Series Compound Amount Factor
For n periods with interest (per period),
F = F1 + F2 + F3 + … + Fn-1 + Fn
= A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3 + … + A(1+i) + A
= A[(1+i)n-1 + (1+i)n-1 + (1+i)n-3 + …+ (1+i) + 1] A 1 2 3 4 A A
n-1 n

6. Uniform Series Compound Amount Factor
Notation
i = interest rate per period n = total # of periods

7. Uniform Series Formulas (Compare to slide 25)
(1) Uniform series compound amount: Given A, i, & n, find F F = A{[(1+i)n – 1]/i} = A(F/A, i, n) (4-4) (2) Uniform series sinking fund: Given F, i, & n, find A A = F{i/[(1+i)n – 1]} = F(A/F, i, n) (4-5) (3) Given F, A, & i, find n n = log(1+Fi/A)/log(1+i) (4) Given F, A, & n, find i There is no closed form formula to use. But rate(nper, pmt, pv, fv, type, guess)

8. Uniform Series Compound Amount Factor
Question: If starting at EOY1, five annual deposits of $100 each are made in the bank account, how much money will be in the account at EOY5, if interest rate is 5% per year? 1 2 3 4 5
$100
F= 552.6
i=0.05

9. QUESTION CONTINUES (USING INTEREST TABLE)1 2 3 4 5
$100
F= 552.6
i=0.05

10. QUESTION CONTINUES(SPREADSHEET)
Go to XL --Chap 4 extended examples-A1
Use function: FV(rate, nper, pmt, pv, type)

11. Uniform Series Compound Amount Factor
Question: Five annual deposits of $100 each are made into an account starting today. If interest rate is 5%, how much money will be in the account at EOY5?
F2= 580.2 F1 1 2 3 4 5
$100
i=5%

12. QUESTION CONTINUES(INTEREST TABLE)
F2= 580.2 F1 1 2 3 4 5
$100
i=5%

13. QUESTION CONTINUES(SPREADSHEET)

14. Uniform Series Compound Amount Factor
Question: If starting at EOY1, five annual deposits of $100 each are made in the bank account, how much money will be in the account at EOY5, if interest rate is 6.5% per year? 1 2 3 4 5
$100
F= ?
i=6.5%

15. INTERPOLATION
6.0
7.0
6.5
5.6371
5.7507
0.5 X 1
0.1136
Interpolation

16. Uniform Series Sinking Fund Factor1 2 3 4 5
F=Given
i=Given
A=?
n=Given
A= Equal Annual Dollar Payments
F= Future Some of Money
i = Interest Rate Per Period
n= Number of Interest Periods

17. Uniform Series Sinking Fund Factor
Notation

18. Uniform Series Sinking Fund Factor
Question: A family wishes to have $12,000 in a bank account by the EOY 5. to accomplish this goal, five annual deposits starting at the EOF year 1 are to be made into a bank account paying 6% interest. what annual deposit must be made to reach the stated goal?
F=$12,000 1 2 3 4 5
i=5%
A =$2172
n=5

19. QUESTION CONTINUES(INTEREST TABLE)
F=$12,000 1 2 3 4 5
i=5%
A = $2172
n=5

20. Uniform Series Sinking Fund Factor
Question: A family wishes to have $12,000 in a bank account by the EOY 5. to accomplish this goal, six annual deposits starting today are to be made into a bank account paying 5% interest. What annual deposit must be made to reach the stated goal? 1 2 3 4 5
F=$12,000
i=5%
A = $1764
n=6

21. QUESTION CONTINUES(INTEREST TABLE)
$1764 1 2 3 4 5
F=$12,000
i=5%
A = $1764
n=6

22. Uniform Series Sinking Fund Factor
Example: the current balance of a bank account is $2,500. starting EOY 1 six equal annual deposits are to be made into the account. The goal is to have a balance of $9000 by the EOY 6. if interest rate is 6%, what annual deposit must be made to reach the stated goal? 1 2 3 4 5
F=$9000
i=6%
A = $796.23 6
P=$2,500

23. Uniform Series Capital Recovery Factor1 2 3 4 5
P=Given
i=Given
A=? n
P= Present Sum of Money
A= Equal Annual Dollar Payments
i = Interest Rate
n= Number of Interest Periods

24. Uniform Series Capital Recovery Factor
Notation

25. Uniform Series Formulas (Compare to slide 7)
(1) Uniform series present worth: Given A, i, & n, find P P = A{[(1+i)n – 1]/[i(1+i)n]} = A(P/A, i, n) (4-7) (2) Uniform series capital recovery: Given P, i, & n, find A A = P{[i(1+i)n]/[(1+i)n – 1]} = P(A/P, i, n) (4-6) (3) Given P, A, & i, find n n = log[A/(A-Pi)]/log(1+i) (4) Given P, A, & n, find i (interest/period) There is no closed form formula to use. But rate(nper, pmt, pv, fv, type, guess)

26. Uniform Series Capital Recovery Factor
Example: A person borrows $100,000 from a commercial bank. The loan is to be repaid with five equal annual payments. If interest rate is 10%, what should the annual payments be? 1 2 3 4 5
A =26,380
P= $100,000
i=10%

27. Example CONTINUES(INTEREST TABLE)1 2 3 4 5
A =26,380
P= $100,000
i=10%

28. Uniform Series Capital Recovery (MS EXCEL)
Use function: PMT(rate, nper, pv, fv, type)
rate = interest rate/period
nper = # of periods
pv = present worth
fv = balance at end of period n (blank means 0).
type = 1 (payment at beginning of each period) or
0 (payment at end of a period)(blank means 0)
See spreadsheet

29. Uniform Series Capital Recovery Factor
Example: At age 30, a person begins putting $2,500 a year into account paying 10% interest. The last deposit is made on the man’s 54th birthday (25 deposits). Starting at age 55, 15 equal annual withdrawals are made. How much should each withdrawal be?
Solution
Step 1: First A will be converted into F.
Step2: F will be considered as P.
Step3: P will be converted into Second A

30. EXAMPLE CONTINUES F = 245,868 i=10% A=$2500 i=10% A =$32,3321 2 3 21 22
F = 245,868
i=10% 23 24
A=$2500
i=10% 1 2 3 12 13 14 15
A =$32,332
P= $245,868

31. Uniform Series Present Worth Factor
A= Equal Annual Dollar Payments
P= Present Sum of Money (at Time 0)
i = Interest Rate/Period
n= Number of Interest Periods 1 2 3 4 5
P=?
i=Given
A=Given
n=Given

32. Uniform Series Present Worth Factor
Notation

33. Uniform Series Present Worth Factor
Example: A special bank account is to be set up. Each year, starting at EOY 1, a $26,380 withdrawal is to be made. After five withdrawals the account is to be depleted. if interest rate is 10%, how much money should be deposited today? 1 2 3 4 5
A=26,380
P= 100,001
i=10%

34. EXAMPLE CONTINUES (USING INTEREST TABLE)1 2 3 4 5
A=26,380
P= 100,001
i=10%

35. Uniform Series Present Worth (Using MS EXCEL)
Use function: PV(rate, nper, pmt, fv, type)
rate = interest rate/period
nper = total # of periods (payments)
pmt = constant payment/period
fv = balance at end of period n (blank means 0)
type = 1 or 0
PV(0.1, 5, ) = $100,000.95
See spreadsheet

36. Uniform Series Present Worth Factor
Example: Eight annual deposits of $500 each are made into a bank account beginning today. Up to EOY 4, the interest rate is 5%. After that, the interest rate is 8%. What is the present worth of these deposits? 1 2 3 4 5
A=500 6 7
i=5%
i=8%

37. EXAMPLE CONTINUES 1 2 3 4 5
A=500 6 7
i=5%
i=8%

38. EXAMPLE CONTINUES (Using MS EXCEL)
P1 = PV(0.08, 3, -500)(1+0.05)–4 = ( )(0.8227) = $1, P2 = PV(0.05, 4, -500) = $1,772.98

39. Arithmetic Gradient
Arithmetic Gradient series (G): each annual amount differs from the previous one by a fixed amount G. + A 1 2 3 4 5 G 2G 3G 4G
A+G
A+2G
A+3G
A+4G =

40. Arithmetic Gradient Present Worth Factor
Given G, i, & n, find P (4-19)
Arithmetic Gradient
Present Worth Factor
Notation

41. Arithmetic Gradient Present Worth Factor
Question: You has purchased a new car. the following maintenance costs starting at EOY 2 will occur to pay the maintenance of your car for the 5 years. EOY2 $30, EOY3 $60, EOY4 $90, EOY5 $120. If interest rate is 5%, how much money you should deposit into a bank account today? 1 2 3 4 5
G=$30
P= $247.11
i=5%
$30
$60
$90
$120

42. QUESTION CONTINUES (INTEREST TABLE)1 2 3 4 5
G=$30
P= $247.11
i=5%
$30
$60
$90
$120

43. Arithmetic Gradient Present Worth Factor
Question: If interest rate is 8%, what is the present worth of the following sums? +
400 1 2 3 4 5 50
100
150
200
450
500
550
600 =
400
400
400
400

44. QUESTION CONTINUES
400 1 2 3 4 5 1 2 3 4 5 50
100
150
200

45. Arithmetic Gradient Uniform Series Factor
Convert an arithmetic gradient series into a uniform series Given G, i, & n, find A (4-20)
Arithmetic Gradient
Uniform Series Factor
Notation

46. Arithmetic Gradient Uniform Series Factor
Question: Demand for a new product will decrease as competitors enter the market. What is the equivalent annual amount of the revenue cash flows shown below? (interest 12%) 1 2 3 4 5
1000
1500
2000
2500
3000 =
3000
3000
3000
3000
3000 1 2 3 4 5 + 1 2 3 4 5
500
1000
1500
2000

47. Geometric Series Present Worth Factor
Geometric series: Each annual amount is a fixed percentage different from the last. In this case, the change is 10%.
We will look at this problem in a few slides. 1 2 3 4 5
P=?
i=5%
$100
$110
$121
$133 6 7 8 9 10
g=10% ? ? ? ? ?

48. Geometric Gradient
Unlike the Arithmetic Gradient where the amount of period-by-period change is a constant, for the Geometric Gradient, the period-by-period change is a uniform growth rate (g) or percentage rate.
First year maintenance cost
Uniform growth rate (g)

49. Geometric Series Present Worth Factor
Given A1, g, i, & n, find P (4-29) & (4-30)
Geometric Series
Present Worth Factor
When Interest rate equals the growth rate,

50. Geometric Series Present Worth Factor
Question: What is the present value (P) of a geometric series with $100 at EOY1 (A1), 5% interest rate (i), 10% growth rate (g), and 10 interest periods (n)? 1 2 3 4 5
P=?
i=5%
$100
$110
$121
$133 6 7 8 9 10
g=10% ? ? ? ? ?

51. Geometric Series Present Worth Factor1 2 3 4 5
P= $
i=5%
$100
$110
$121
$133 6 7 8 9 10
g=10%
$146
$161
$195
$177
$214
$236

52. Time for a Joke!
What is Recession? Recession is when your neighbor loses his or her job. What is Depression? Depression is when you lose yours. By Ronald Reagan

53. Problem 4-7 Purchase a car: $3,000 down payment
$480 payment for 60 months
If interest rate is 12% compounded monthly, at what purchase price P of a car can one buy?
Solution
i = 12.0%/12 = 1.o% per month, n = 60, and A = $480
P = (P/A, 0.01, 60)
= (44.955)
= $24,578
Important: P = $ $480(60) = $31,800, if i = 0.

54. Problem 4-9
$25 million is needed in three years. Traffic is estimated at 20 million vehicles per year. At 10% interest, what should be the toll per vehicle? (a) Toll receipts at end of each year in a lump sum. (b) Traffic distributed evenly over 12 months, and toll receipts at end of each month in a lump sum.

55. Problem 4-9
Solution (a) Let x = the toll/vehicle. Then F = $25,000,000 i = 10%/year, n = 3 years Find A (=20,000,000x). A = F(A/F, 0.1, 3) 20,000,000x = 25,000,000(0.3021) x = $ = $0.38 per vehicle

56. Problem 4-9
Solution (b) Let x = the toll/vehicle. Then F = $25M i = (1/12)10%/month, n = 36 months Find A (=20,000,000x/12). A = F{i/[(1+i)n–1]} 20,000,000x/12 = 25,000,000{(0.1/12)/(1+0.1/12)36–1} x = $0.359 = $0.36 per vehicle

57. Problem 4-32
If i = 12%, for what value of B is the PW = 0? Solution Consider now = time 1. Then PW = B+800(P/A, 0.12, 3) – B(P/A, 0.12, 2) – B(P/F, 0.12, 3) = – 1.758B Letting PW = 0 yields B = $1, For any cash flow diagram, if PW = 0, then its worth at anytime = 0!

58. Problem 4-46
Solution FW = FW [1000(F/A, i, 10)](F/P, i, 4) = By try and error: At i = 12%, LHS = [1000(17.549)](1.574) = $27,622 too low At i = 15%, LHS = [1000(20.304)](1.749) = $35,512 too high Using linear interpolation: i = 12% + 3%[(28000 – 27622)/(35512 – 27622)] = 12.14%

59. Use of MS EXCEL pmt(i, n, P, F, type) returns A, given i, n, P, and F
sinking fund (P=0) A = F{i/[(1+i)n – 1]} (4-5)
capital recovery (F=0) A = P{[i(1+i)n]/[(1+i)n – 1]} (4-6)
or combined (P ≠ 0, and F ≠ 0)
rate(n, A, P, F, type, guess) returns i, given n, A, P, and F

60. Use of MS EXCEL
pv(i, n, A, F, type) returns P, given i, n, A, and F present worth (A=0) P = F/(1+i)n (3-5) series present worth (F=0) P = A{[(1+i)n – 1]/[i(1+i)n]} (4-7) or combined (A ≠ 0, and F ≠ 0) fv(i, n, A, P, type) returns F, given i, n, A, and P compound amount (A=0) F = P(1+i)n (3-3) series compound amount (P=0) F = A{[(1+i)n – 1]/i} (4-4) or combined (A ≠ 0, and P ≠ 0)

61. Use of MS EXCEL
nper(i, A, P, F, type) returns n, given i, A, P, and F. If A = 0, n = log(F/P)/log(1+i) single payment If P = 0, n = log(1+Fi/A)/log(1+i) uniform series If F = 0, n = log[A/(A-Pi)]/log(1+i) uniform series effect(r, m) returns ia, given r and m. effective annual interest rate ia = (1+r/m)m – 1 (3-7) nominal(ia, m) returns r, given ia and m. nominal annual interest rate r = m[(ia – 1)1/m + 1]

62. A Real Life Case
Mr. Goodman set up a trust fund of $1.5M for his 2 children in It is worth more than $300M today (January 2012). What is the effective annual interest rate? Solution P = $1.5M, F = $300M, n = 20 years ia = (F/P)1/n – 1 = (300/1.5)1/20 – 1 = % ia = rate(20, 0, 1.5, –300) = % ia = rate(20, 0, -1.5, 300) = %

63. End of Chapter 4 Uniform Series Compound Interest Formulas
Uniform Series Compound Amount Factor: F/A
Uniform Series Sinking Fund Factor: A/F
Uniform Series Capital Recovery Factor: A/P
Uniform Series Present Worth Factor: P/A
Arithmetic Gradient
Geometric Gradient
Spreadsheet Solutions

64. Interpolation-1 Given: F(X1); F(X2)
What is F(X3) where X1 < X3 < X2?
Assuming linearity so that a linear equation will do:
Basic equation: y = mx + b so
F(X1) = mX1 + b
F(X2) = mX2 + b
Subtract 2 from 1:
F(X1)-F(X2) = m (X1-X2)  m = (F(X1)-F(X2))/(X1-X2)
From 1 we get b = (F(X1) - mX1)

65. Interpolation-2 F(X1)-F(X2) = m (X1-X2)  m = (F(X1)-F(X2))/(X1-X2)
From 1 we get b = (F(X1) - mX1)
Now
F(X3) = m X3 + b
= m X3 + F(X1) - mX1 = m (X3 - X1) + F(X1)
= (X3 - X1) (F(X1)-F(X2))/(X1-X2) + F(X1)
= F(X1) + (F(X1)-F(X2)) (X3 - X1) /(X1-X2)

66. Interpolation-3 F(X3) = F(X1) + (F(X1)-F(X2)) (X3 - X1) /(X1-X2)
Suppose that the Xs are interest rates, i, and the Fs are the functions (F/A,i,n), then
F(i3) = F(i1) + (F(i1)-F(i2)) (i3 - i1) /(i1-i2)
Return