### Chapter 5 Present Worth Analysis - Help-A-Bull

```Chapter 5
Present Worth Analysis
EGN 3615
ENGINEERING ECONOMICS
WITH SOCIAL AND GLOBAL
IMPLICATIONS
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
2
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
3
Economic Criteria
 Depending on situation, the economic criterion
should be chosen from one of the following 3:
Situation
Neither input nor
output fixed
Fixed input
Fixed output
Criterion
Maximize (Output – Input)
Maximize output
Minimize input
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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.
2. Alternatives have different useful lives: Let the analysis
period equal the least common multiple, or some
realistic time (based on needs).
3. Infinite analysis period, n=∞
1.
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5
Net Present Worth (NPW or PW)
 Here is the basic NPW formula:
 NET P RESENT  P resent wort h   P resent wort h 

 
 

W ORT H  
of
of

-


  Benefit s  

NP
W
or
P
W
Cost
s

 
 

PW = PW of benefits – PW of cost
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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 Minimize present worth of
outputs
costs or other inputs
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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
5 year
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Example Continues
A=\$300
0
1
2
3
4
5
i=6%
P= \$1000
PW A  1000 300(P/A,6%,5)
 1000 300 (4.212)
 263.6
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Example Continues
\$300
0
1
\$450
\$400
\$350
2
3
4
\$500
5
i=6%
P= \$1350
PW B  1350  300 (P/A,6%,5)  50 (P/G,6%,5)
 1350  300 (4.212)  50 (7.934)
 310.3
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Example Continues
A=\$300
0
1
2
3
4
5
P WA  263.6
i=6%
P= \$1000
\$300
0
1
\$450
\$400
\$350
\$500
P WB  310.3
2
3
4
i=6%
P= \$1350
5
Work 5-4
DEVICE B has the larger present worth & is the preferred alternative
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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
Investment 1 \$2000 \$450
Annually
Investment 2 \$3000 \$600
Annually
Engineering Economics
Useful
Salvage Value
Life
(End of Useful Life)
5 year
\$100
5 year
\$700
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Example Continues
Investment 1 :
PW of Benefits  PW of Costs
 450 (P/A, 10%, 5)  [200  100 (P/F, 10%,5)]
 450 (3.791)  [200  100 (0.6209) ]
 231.96
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Example Continues
Investment 2 :
PW of Benefits  PW of Costs
 600(P/A, 10%, 5)  [3000  700 (P/F, 10%,5) ]
 600 (3.791)  [3000  700 (0.6209) ]
 290.77
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Example Continues
Investment 1 :
PW of Benefits - PW of Costs  450 (P/A, 10%, 5)  [200  100 (P/F, 10%,5) ]
 450 (3.791)  [200  100 (0.6209) ]
 231.96
Investment 2 :
PW of Benefits - PW of Costs  600(P/A, 10%, 5)  [3000  700 (P/F, 10%,5)]
 600 (3.791)  [3000  700 (0.6209)]
 290.77
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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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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Example Continues
 One method to select an analysis period is the least
common multiple of useful lives.
EQUIPMENT A
0
1
\$200
Original Equipment A
Investment
2
3
4
\$1500
5
\$200
Replacement Equipment A
Investment
6
7
8
9
10
\$1500
PW of cost
 1500 (1500 200)(P/F,6%, 5)  200(P/F,6%,10)
 1500 1300(0.7473)  200(0.5584)
 \$2,359 .81
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Question Continues
\$350
EQUIPMENT B
0
\$1600
1
2
3
Original Equipment B
Investment
4
5
6
7
8
9
10
PW of cost
 1600  350 (P/F,6%,10)
 1600  350 (0.5584)
 \$1,404.56
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Question Continues
PW of cost  1500  (1500 - 200) (P/F,6%, 5)
 200 (P/F,6%,10)
 1500  1300 (0.7473)  200 (0.5584)
EQUIPMENT A
 \$2,359 .81
PW of cost  1600  325 (P/F,6%,10)
 1600  325(0.5584)
EQUIPMENT B
 \$1,404.56
For fixed output of 10 years of service of equipments, Equipment B
is preferred because it has a smaller cost.
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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) 10year analysis period (by using the present worth method)?
MACHINE
INITIAL
COST
Salvage
Terminal Value
Value at the
at the end of
End Of
10-year analysis
Useful Life
period
USEFUL
LIFE
MACHINE A \$40,000
\$8,000
\$15,000
7 year
MACHINE B \$65,000
\$10,000
\$10,000
13 year
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Example Continues
\$8,000
MACHINE A
0
1
2
3
4
5
6
7
\$15,000
8
7-year life
\$40,000
9
10
11
12
13
14
7-year life
\$40,000
PW of cost
 40,000 (40,000 8,000)(P/F, 10%,7)
 15,000(P/F,10%,10)
 40,000 32,000 (0.5132)  15,000(0.3855)
 \$50,639.90
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Example Continues
\$10,000
MACHINE B
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
13-year life
\$65,000
PW of cost
 65,000  10,000(P/F,10%,10)
 65,000  10,000( 0.3855)
 \$61,145.0
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Example Continues
P W of cost  40,000 (40,000 - 8,000) (P /F,10%,7)
 15,000 (P /F,10%,10)
 40,000  32,000 (0.5132)  15,000(0.3855)
 \$50,639.90 MACHINE A
P W of cost  65,000  10,000(P /F,10%,10)
 65,000  10,000( 0.3855)
 \$61,145.00 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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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.
Capitalize d Cost
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A
P
i
(5-2)
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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)?
Annualdisbursement (A)
Capitalized Cost, P 
Interestrate(i)
1000
P
 \$40,000
0.025
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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
0
-\$2000
-\$1500
-\$3000
1
+1000
+700
+500
2
+850
+300
+500
3
+700
+300
+550
4
+550
+300
+600
5
+400
+300
+650
6
+400
+400
+700
7
+400
+500
+500
+400
+600
+500
8
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Question Continues
\$1000 \$850 \$700 \$550 \$400 \$400 \$400 \$400
0
\$2000
1
2
3
4
5
6
7
8
EQUIPMENT 1
P W of Benefit s 400(P /A,10%,8)  600(P /A,10%, 4)
 150(P /G,10%, 4)
 400(5.335)  600(3.170)
 150(4.378) 3379.17
P W of Cost  2000
Net P resent W ort h  3379.17  2000 \$1,379.17
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Question Continues
\$700 \$300 \$300 \$300 \$300 \$400 \$500 \$600
0
\$1500
1
2
3
4
5
6
7
8
EQUIPMENT 2
PW of Benefits 300(P/A,10%,8)
 (700 - 300)(P/F,10%,1)
 100(P/G,10%,4)(P/F,10%,4)
 300(5.335)  400(0.9091)
 100(4.378)(0.6830) 2263.15
PW of Cost  1500
Net PresentEngineering
Wo
rthEconomics
 2263.15  1500 \$763.15
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Question Continues
\$500 \$500 \$550 \$600 \$650 \$700 \$500 \$500
0
\$3000
1
2
3
4
5
6
7
8
EQUIPMENT 3
P W of Benefits 500(P /A,10%,8)
 50(P/G,10%,5)(P /F,10%,1)
 500(5.335)
 100(6.862)
(0.9091) 2979.36
P W of Cost  3000
Net P resent Worth  2979.36  3000 \$20.64
To maximize NPW, choose EQUIPMENT 1
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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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Question Continues (MS EXCEL)
Year
Equipment 1
Equipment2
Equipment3
0
(\$2,000)
(\$1,500)
(\$3,000)
1
1000
700
500
2
850
300
500
3
700
300
550
4
550
300
600
5
400
300
650
6
400
400
700
7
400
500
500
8
400
600
500
Interest
10%
For Equip 1: NPW=NPV(A12,B3:B10)+B2
\$1,379.17
Engineering Economics
\$763.15
(\$20.64)
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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)
= –980000 + 200000(6.424) + 20000(0.2292)
= \$309,384
As PW > 0, purchase the machine.
Or using MS EXCEL
PW = -P + pv(0.12, 13, -200000, -20000) = \$309,293.17
Terms A(P/A, 0.12, 13) and F(P/F, 0.12, 13) are combined!
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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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Problem 5-41
Outputs:
2000 lines for years 1~10
4000 lines for years 21~30
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)
(b)
Which option to choose?
Will answer to (a) change if 2000 additional lines are
needed in 5 years, instead of 10 years?
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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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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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End of Chapter 5
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```