1. TM 661 Engineering Economics for Managers
Investment Worth

2. Investment Worth
MARR
Suppose a company can earn 12% / annum in U. S. Treasury bills
No way would they ever invest in a project earning < 12%
Def: The Investment Worth of all projects are measured at the Minimum Attractive
Rate of Return (MARR) of a company.

3. MARR MARR is company specific MARR based on
utilities - MARR = %
mutuals - MARR = %
new venture - MARR = %
MARR based on
firms cost of capital
Price Index
Treasury bills

4. Investment Worth Alternatives
NPW(MARR) > 0 Good Investment

5. Investment Worth Alternatives
NPW(MARR) > 0 Good Investment
EUAW(MARR) > 0 Good Investment

6. Investment Worth Alternatives
NPW(MARR) > 0 Good Investment
EUAW(MARR) > 0 Good Investment
IRR > MARR Good Investment

7. Present Worth
Example: Suppose you buy and sell a piece of equipment.
Purchase Price $16,000 Sell Price (5 years) $ 4,000 Annual Maintenance $ 3,000 Net Profit Contribution $ 6,000 MARR %
Is it worth it to the company to buy the machine?

8. Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5) 4,000 4,000 6,000
16,000
6,000
3,000 5
4,000
3,000 5
4,000
16,000
NPW = (P/A,12,5) + 4(P/F,12,5)

9. Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)
16,000
6,000
3,000 5
4,000
3,000 5
4,000
16,000
NPW = (P/A,12,5) + 4(P/F,12,5)
= (3.6048) + 4(.5674)

10. Present Worth NPW = -16 + 3(P/A,12,5) + 4(P/F,12,5)
16,000
6,000
3,000 5
4,000
3,000 5
4,000
16,000
NPW = (P/A,12,5) + 4(P/F,12,5)
= (3.6048) + 4(.5674) =
= -$2,916

11. Annual Worth  Annual Worth (AW or EUAW) AW(i) = PW(i) (A/P, i%, n)
= [ At (P/F, i%, t)](A/P, i%, n)
AW(i) = Annual Worth of Investment
AW(i) > 0 **OK Investment** 

12. Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000 5
4,000
16,000
Repeating our PW example,
we have
AW(12) = -16(A/P,12,5) (A/F,12,5)

13. Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000 5
4,000
16,000
Repeating our PW example,
we have
AW(12) = -16(A/P,12,5) (A/F,12,5)
= -16(.2774) (.1574)

14. Annual Worth; Example AW(12) = -16(A/P,12,5) + 3 + 4(A/F,12,5)
3,000 5
4,000
16,000
Repeating our PW example,
we have
AW(12) = -16(A/P,12,5) (A/F,12,5)
= -16(.2774) (.1574)
= -.808
= -$808

15. Alternately AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774)
3,000 5
4,000
16,000
AW(12) = PW(12) (A/P, 12%, 5)
= (.2774)
= - $810 < 0 NO GOOD

16. Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0
i* = IRR
i* > MARR **OK Investment**

17. Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0
i* = IRR
i* > MARR **OK Investment**
Alt:
FW(i*) = 0 = At(1 + i*)n - t 

18. Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0
i* = IRR
i* > MARR **OK Investment**
Alt:
FW(i*) = 0 = At(1 + i*)n - t
PWrevenue(i*) = PWcosts(i*) 

19. Internal Rate of Return
Example
PW(i) = (P/A, i, 5) + 4(P/F, i, 5)
3,000 5
4,000
16,000

20. Internal Rate of Return
Example
PW(i) = (P/A, i, 5) + 4(P/F, i, 5)
3,000 5
4,000
16,000

21. Internal Rate of Return
Example
PW(i) = (P/A, i, 5) + 4(P/F, i, 5)
i* = 5 1/4 %
i* < MARR
3,000 5
4,000
16,000

22. Public School Funding

23. Public School Funding
216%
16 yrs

24. School Funding F = P(F/P,i*,16) (F/P,i*,16) = F/P = 2.163 16
100
216
F = P(F/P,i*,16)
(F/P,i*,16) = F/P = 2.16
(1+i*)16 = 2.16

25. School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .7701 1 2 3 16
100
216
(1+i*)16 = 2.16
16 ln(1+i*) = ln(2.16) = .7701

26. School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .77013 16
100
216
(1+i*)16 = 2.16
16 ln(1+i*) = ln(2.16) = .7701
ln(1+i*) = .0481

27. School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .77013 16
100
216
(1+i*)16 = 2.16
16 ln(1+i*) = ln(2.16) = .7701
ln(1+i*) = .0481
(1+i*) = e.0481 =

28. School Funding (1+i*)16 = 2.16 16 ln(1+i*) = ln(2.16) = .77013 16
100
216
(1+i*)16 = 2.16
16 ln(1+i*) = ln(2.16) = .7701
ln(1+i*) = .0481
(1+i*) = e.0481 =
i* = = 4.93%

29. School Funding We know i = 4.93%, is that significant growth? 1 2 3 16
100
216
We know i = 4.93%, is that significant growth?

30. School Funding We know i = 4.93%, is that significant growth?1 2 3 16
100
216
We know i = 4.93%, is that significant growth?
Suppose inflation = 3.5% over that same period.

31. School Funding We know i = 4.93%, is that significant growth?1 2 3 16
100
216
We know i = 4.93%, is that significant growth?
Suppose inflation = 3.5% over that same period. d i j    1
0493
0350 .
d= 1.4%

32. School Funding We know that d, the real increase in school funding1 2 3 16
100
216
We know that d, the real increase in school funding
after we discount for the effects of inflation
is 1.4%.
So schools have experienced a real increase in
funding?

33. School Funding We know that d, the real increase in school funding1 2 3 16
100
216
We know that d, the real increase in school funding
after we discount for the effects of inflation
is 1.4%.
So schools have experienced a real increase in
funding?
Rapid City growth rate 3% / yr. 

34. Summary
NPW > Good Investment

35. Summary
NPW > Good Investment
EUAW > Good Investment

36. Summary NPW > 0 Good Investment EUAW > 0 Good Investment
IRR > MARR Good Investment

37. Summary NPW > 0 Good Investment EUAW > 0 Good Investment
IRR > MARR Good Investment
Note: If NPW > EUAW > 0
IRR > MARR

38. Internal Rate of Return
IRR - internal rate of return is that return for which NPW(i*) = 0
i* = IRR
i* > MARR **OK Investment**
Alt:
FW(i*) = 0 = At(1 + i*)n - t
PWrevenue(i*) = PWcosts(i*) 

39. Internal Rate of Return
Example
PW(i) = (P/A, i, 5) + 4(P/F, i, 5)
3,000 5
4,000
16,000

40. Internal Rate of Return
Example
PW(i) = (P/A, i, 5) + 4(P/F, i, 5)
i* = 5 1/4 %
i* < MARR
3,000 5
4,000
16,000

41. Spreadsheet Example

42. Spreadsheet Example

43. Spreadsheet Example

44. Spreadsheet Example

45. Spreadsheet Example

46. IRR Problems Consider the following cash flow diagram. We
1,000
4,100
5,580
2,520 n 1 2 3
Consider the following
cash flow diagram. We
wish to find the Internal
Rate-of-Return (IRR).

47. IRR Problems Consider the following cash flow diagram. We
1,000
4,100
5,580
2,520 n 1 2 3
Consider the following
cash flow diagram. We
wish to find the Internal
Rate-of-Return (IRR).
PWR(i*) = PWC(i*)
4,100(1+i*)-1 + 2,520(1+i*)-3 = 1, ,580(1+i*)-2

48. IRR Problems NPV vs. Interest ($5) $0 $5 $10 $15 $20 $25 0% 10% 20%
30%
40%
50%
60%
Interest Rate
Net Present Value

49. External Rate of Return
Purpose: to get around a problem of multiple roots in IRR method
Notation:
At = net cash flow of investment in period t
At , At > 0
0 , else
-At , At < 0
0 , else
rt = reinvestment rate (+) cash flows (MARR)
i’ = rate return (-) cash flows 
Rt = 
Ct =

50. External Rate of Return
Method
find i = ERR such that
Rt (1 + rt) n - t = Ct (1 + i’) n - t
Evaluation
If i’ = ERR > MARR
Investment is Good  

51. External Rate of Return
Example MARR = 15%
Rt ( ) n - t = Ct (1 + i’) n - t
4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1
i’ = .1505 1 2 3
1,000
4,100
5,580
2,520  

52. External Rate of Return
Example MARR = 15%
Rt ( ) n - t = Ct (1 + i’) n - t
4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1
i’ = .1505
ERR > MARR 1 2 3
1,000
4,100
5,580
2,520  

53. External Rate of Return
Example MARR = 15%
Rt ( ) n - t = Ct (1 + i’) n - t
4,100(1.15)2 + 2,520 = 1,000(1 + i’)3 + 5,580(1 + i’)1
i’ = .1505
ERR > MARR Good Investment 1 2 3
1,000
4,100
5,580
2,520  

54. Critical Thinking IRR < MARR a. IRR < MARR < ERR
b. IRR < ERR < MARR
c. ERR < IRR < MARR

55. Critical Thinking IRR > MARR a. IRR > MARR > ERR
b. IRR > ERR > MARR
c. ERR > IRR > MARR

56. Savings Investment Ratio
Method 1 Let i = MARR
SIR(i) = Rt (1 + i)-t
Ct (1 + i)-t
= PW (positive flows) -
PW (negative flows)  

57. Savings Investment Ratio
Method #2
SIR(i) = At (1 + i) -t
Ct (1 + i) -t
SIR(i) = PW (all cash flows)
PW (negative flows)  

58. Savings Investment Ratio
Method #2
SIR(i) = At (1 + i) -t
Ct (1 + i) -t
SIR(i) = PW (all cash flows)
PW (negitive flows)
Evaluation:
Method 1: If SIR(t) > 1 Good Investment
Method 2: If SIR(t) > 0 Good Investment  

59. Savings Investment Ratio16 1 2 3 4 5 7
Example
SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16
= 3(3.6048) + 4(.5674)
= < 1.0

60. Savings Investment Ratio16 1 2 3 4 5 7
Example
SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16
= 3(3.6048) + 4(.5674)
= < 1.0

61. Payback Period   Method Find smallest value m such that where
Co = initial investment
m = payback period investment m
t = 1 R t C o  

62. Payback Period R t n å
Example c n o 16 1 2 3 4 5 7
m = 5 years

63. Capitalized Costs Example: $10,000 into an account @ 20% / year
A = P(A/P, i, )
P = A / i A
10,000  

64. Class Problem
A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.

65. Class Problem Pc = 10,000 + A/i = 10,000 + 100/.08 = 11,250
10,000
A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.
Pc = 10,000 + A/i
= 10, /.08
= 11,250
Capitalized Cost = $11.25 million

66. Class Problem 2
Suppose that the flood control project has major repairs of $1 million scheduled every 5 years. We now wish to re-compute the capitalized cost.

67. Class Problem 2 Compute an annuity for the 1,000 every 5 years:
10,000
1,000
Compute an annuity for the 1,000 every 5 years:

68. Class Problem 2 Compute an annuity for the 1,000 every 5 years:
10,000
1,000
Compute an annuity for the 1,000 every 5 years:
A = ,000(A/F,8,5)
= ,000(.1705)
= 270.5

69. Class Problem 2
10,000
1,000
10,000
Pc = 10, /.08
= 13,381

70. How Many to Change a Bulb?
How does Bill Gates change
a light bulb?

71. How Many to Change a Bulb?
How does Bill Gates change
a light bulb?
He doesn’t, he declares darkness
a new industry standard!!!

72. How Many to Change a Bulb?
How many Industrial
Engineers does it take to
change a light bulb?

73. How Many to Change a Bulb?
How many Industrial
Engineers does it take to
change a light bulb?
None, IE’s only change
dark bulbs!!!!!