1. Chapter 5 Present Worth Analysis
EGN 3615
ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1

2. Three Economic Analysis Methods
There are three major economic analysis techniques:
Present Worth Analysis
Annual Cash Flow Analysis
Rate of Return Analysis
This chapter discusses the first techniques

3. Chapter Contents Economic Criteria Considering Project Life
Net Present Worth
Applying Present Worth Techniques
Useful Lives Equal the Analysis Period
Useful Lives Different from the Analysis Period
Infinite Analysis Period: Capitalized Cost
Multiple Alternatives
Spreadsheet Solution

4. Economic Criteria Situation Criterion
Depending on situation, the economic criterion should be chosen from one of the following 3:
Situation
Criterion
Neither input nor output fixed
Maximize (Output – Input)
Fixed input
Maximize output
Fixed output
Minimize input
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5. Analysis Period
Specific time period, same for each alternative, called the analysis period, planning horizon, or project life
Three different analysis-period situations may be considered:
All alternatives have the same useful life: Set it as the analysis period.
Alternatives have different useful lives: Let the analysis period equal the least common multiple, or some realistic time (based on needs).
Infinite analysis period, n=∞
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6. Net Present Worth (NPW or PW)
Here is the basic NPW formula:
PW = PW of benefits – PW of cost
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7. Present Worth Techniques
Mutually exclusive alternatives:
Resolve their consequences to the present time.
Situation
Criterion
Neither input nor output fixed
Maximize net present worth
Amount of money or other input resources are fixed
Maximize present worth of benefits or other outputs
Fixed task, benefit, or other outputs
Minimize present worth of costs or other inputs
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8. Present Worth—Equal Useful Lives
Example: Consider two mechanical devices to install to reduce cost. Expected costs and benefits of machines are shown in the following table for each device. If interest rate is 6%, which device should be purchased?
DEVICE
COST
COST SAVING
USEFUL LIFE
DEVICE A
$1000
$300 Annually
5 year
DEVICE B
$1350
$300 The first year and increase $50 annually
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9. Example Continues 1 2 3 4 5
A=$300
P= $1000
i=6%
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10. Example Continues $500 $450 $400 $350 $300 i=6% P= $1350 1 2 3 4 51 2 3 4
$300
P= $1350
i=6%
$350
$400
$450
$500 5
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11. Example Continues A=$300 i=6% P= $1000 $500 $450 $400 $350 $300 i=6%1 2 3 4 5
A=$300
P= $1000
i=6% 1 2 3 4
$300
P= $1350
i=6%
$350
$400
$450
$500 5
Work 5-4
DEVICE B has the larger present worth & is the preferred alternative
Engineering Economics

12. Present Worth—Equal Useful Lives
Example: Consider two investments with expected costs and benefits shown below for each investment. If investments have lives equal to the 5-year analysis period, which one should be selected at 10% interest rate?
Investment
Cost
Benefit
Useful
Life
Salvage Value
(End of Useful Life)
Investment 1
$2000
$450 Annually
5 year
$100
Investment 2
$3000
$600 Annually
$700
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13. Example Continues
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14. Example Continues
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15. Example Continues
Salvage value is considered as another positive cash flow. Since criterion is to maximize PW (= present worth of benefits – present worth of costs), the preferred alterative is INVESTMENT1
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16. Alternatives with different Useful Lives
Example: Consider two new equipments to perform desired level of (fixed) output. expected costs and benefits of machines are shown in the below table for each equipment. If interest rate is 6%, which equipment should be purchased?
EQUIPMENT
COST
SALVAGE
VALUE
USEFUL
LIFE
EQUIPMENT A
$1500
$200
5 year
EQUIPMENT B
$1600
$350
10 year
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17. Replacement Equipment A Investment
Example Continues
One method to select an analysis period is the least common multiple of useful lives. 1 2 3 4 5
$200
$1500 6 7 8 9 10
Replacement Equipment A Investment
Original Equipment A
Investment
EQUIPMENT A
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18. Question Continues $350 $1600 EQUIPMENT B 1 2 3 4 5 6 7 8 9 101 2 3 4 5
$1600 6 7 8 9 10
Original Equipment B
Investment
EQUIPMENT B
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19. Question Continues EQUIPMENT A EQUIPMENT B
For fixed output of 10 years of service of equipments, Equipment B is preferred because it has a smaller cost.
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20. Present Worth-Useful Lives are Different from the Analysis Period
Example: Consider two alternative production machines with expected initial costs & salvage values shown below. If interest rate is 10%, compare these alternatives over a (suitable) 10-year analysis period (by using the present worth method)?
MACHINE
INITIAL COST
Salvage
Value at the End Of Useful Life
Terminal Value at the end of 10-year analysis period
USEFUL
LIFE
MACHINE A
$40,000
$8,000
$15,000
7 year
MACHINE B
$65,000
$10,000
13 year
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21. Example Continues $40,000 $8,000 $15,000 MACHINE A 1 2 3 4 5 6 7 8 91 2 3 4 5
$8,000 6 7 8 9 10
$15,000
7-year life
MACHINE A 11 12 13 14
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22. Example Continues $10,000 $65,000 MACHINE B 1 2 3 4 5 6 7 8 9 10 11 121 2 3 4 5 6 7 8 9 10
$10,000
$65,000
MACHINE B 11 12 13 14
13-year life
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23. Example Continues MACHINE A MACHINE B
For fixed output of 10 years of service of equipments, Machine A is preferred because it has a smaller cost.
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24. Infinite Analysis Period (Capitalized Cost)
Capitalized cost is the present sum of money that is set aside now at a given interest rate to yield the funds (future interest earned) required to provide the service indefinitely.
(5-2)
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25. Infinite Analysis Period (Capitalized Cost)
Example: How much should one set aside to pay $1000 per year for maintenance on an equipment if interest rate is 2.5% per year and the equipment is kept in service indefinitely (perpetual maintenance)?
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26. Multiple (3+) Alternatives
Question: Cash flows (costs and incomes) for three pieces of construction equipments are shown below. For 10% interest rate, which alternative should be selected?
Year
Equipment 1
Equipment2
Equipment3
-$2000
-$1500
-$3000 1
+1000
+700
+500 2
+850
+300 3
+550 4
+600 5
+400
+650 6 7 8
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27. Question Continues $2000 $400 $1000 $850 $700 $550 EQUIPMENT 1 1 2 3 41 2 3 4 5
$400 6 7 8
EQUIPMENT 1
$1000
$850
$700
$550
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28. Question Continues $300 $1500 $600 $700 $400 $500 EQUIPMENT 2 1 2 3 41 2 3 4 5
$300
$1500 6 7 8
$600
EQUIPMENT 2
$700
$400
$500
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29. Question Continues $650 $3000 $500 $550 $600 $700 EQUIPMENT 31 2 3 4 5
$650
$3000 6 7 8
$500
EQUIPMENT 3
$550
$600
$700
To maximize NPW, choose EQUIPMENT 1
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30. Question Continues (MS EXCEL)
Use function: npv(rate, value range)
- Return the net present value of a series of
future cash flows “value range” at interest
“rate”/period.
rate = interest rate per period
value range = the cash flow values
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31. Question Continues (MS EXCEL)
Year
Equipment 1
Equipment2
Equipment3
($2,000)
($1,500)
($3,000) 1
1000
700
500 2
850
300 3
550 4
600 5
400
650 6 7 8
Interest
For Equip 1: NPW=NPV(A12,B3:B10)+B2
10%
$1,379.17
$763.15
($20.64)
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32. Problem 5-15 Solution i = 12% P = $980,000 purchase cost
F = $20,000 salvage value after 13 years
A = $200,000 annual benefit for 13 years
PW = –P + A(P/A, 0.12, 13) + F(P/F, 0.12, 13)
= – (6.424) (0.2292)
= $309,384
As PW > 0, purchase the machine.
Or using MS EXCEL
PW = -P + pv(0.12, 13, , ) = $309,293.17
Terms A(P/A, 0.12, 13) and F(P/F, 0.12, 13) are combined!
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33. Problem 5-23 Solution i = 18%/12 = 1.5% per month
A = $500 payment/month
n = 36 payments
P = ? price of a car she can afford
P = A(P/A, 0.015, 36)
= 500(27.661)
= $13,831
What is P, if r = 6%?
i = 6%/12 = 0.5%
P = pv(0.005, 36, -500) = $16,435.51
Do Problems 5-24, 5-25, 5-26!
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34. Problem 5-41 Outputs: 2000 lines for years 1~10
i = 10% per year, cables last for at least 30 yrs
Option 1: 1 cable with capacity of 4000 lines
Cost: $200k with $15k annual maintenance cost
Option 2: 1 cable with capacity of 2000 lines now
1 cable with capacity of 2000 lines in 10 years
Cost: $150k with $10k maintenance cost/year/cable
(a) Which option to choose?
(b) Will answer to (a) change if 2000 additional lines are
needed in 5 years, instead of 10 years?
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35. Problem 5-41 Solution (a) Present worth of cost for option 1
PW 0f cost = $200k + $15k(P/A, 10%, 30)
= $341,400
Present worth of cost for option 2:
PW of cost = $150k + $10k(P/A, 10%, 30)
+ $150k(P/F, 0.1, 10) + $10k(P/A, 0.1, 20)(P/F, 0.1, 10)
= $334,900
Select option 2, as it has a smaller PW of cost.
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36. Problem 5-41 Solution (b) Cost for option 1 will not change.
PW 0f cost = $341,400
Present worth of cost for option 2:
PW of cost = $150k + $10k(P/A, 10%, 30)
+ $150k(P/F, 0.1, 5) + $10k(P/A, 0.1, 25)(P/F, 0.1, 5)
= $394,300
Therefore, the answer will change to option 1.
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37. End of Chapter 5
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