EGN 3615
ENGINEERING ECONOMICS
WITH SOCIAL AND GLOBAL
IMPLICATIONS

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
 Spreadsheet Solution
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
Engineering Economics 4

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:
1.
2.
3.
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=∞
Engineering Economics 5

Net Present Worth (NPW or PW)
 Here is the basic NPW formula:


NET PRESENT
WORTH
NPW or PW


 

Present wo rth of
Benefits




Present wo rth of
Costs


PW = PW of benefits – PW of cost
Engineering Economics 6

Present Worth Techniques
 Mutually exclusive alternatives:
 Resolve their consequences to the present time .
Situation
Neither input nor output fixed
Amount of money or other input resources are fixed
Criterion
Maximize net present worth
Maximize present worth of benefits or other outputs
Fixed task, benefit, or other outputs
Minimize present worth of costs or other inputs
Engineering Economics 7

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
DEVICE A $1000 $300 Annually
DEVICE B $1350 $300 The first year and increase $50 annually
USEFUL LIFE
5 year
5 year
Engineering Economics 8

Example Continues
A=$300
0 1 2 i=6%
3 4 5
P= $1000
PW
A



1000

300 (P/A,6%,5)

1000

300 (4.212)

263.6
Engineering Economics 9

Example Continues
$350 $400
$450
$500
$300
0 1 2 i=6%
3 4 5
P= $1350
PW
B



1350

300 (P/A,6%,5)

50 (P/G,6%,5)

1350

300 (4.212)

50 (7.934)

310.3
Engineering Economics 10

Example Continues
A=$300
0 1 2 3 4 5 i=6%
P= $1000
$350 $400
$450
$500
$300
0 1
P= $1350
PW
A

263.6
PW
B

310.3
2 i=6%
3 4 5
Work 5-4
DEVICE B has the larger present worth & is the preferred alternative
Engineering Economics 11

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
Investment 1 $2000 $450
Annually
Investment 2 $3000 $600
Annually
5 year
5 year
Engineering Economics
Salvage Value
(End of Useful Life)
$100
$700
12

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
Engineering Economics 13

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
Engineering Economics 14

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 INVESTMENT 1
Engineering Economics 15

 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
EQUIPMENT A $1500
EQUIPMENT B $1600
SALVAGE
VALUE
$200
$350
USEFUL
LIFE
5 year
10 year
Engineering Economics 16

0
Example Continues
 One method to select an analysis period is the least common multiple of useful lives.
EQUIPMENT A
$200 $200
Original Equipment A
Investment
2 3
Replacement Equipment A
Investment
6 7 8 9 1 4 5 10
$1500
PW of cost
$1500

1500

(1500

200) (P/F, 6%, 5)

200 (P/F, 6%, 10)

1500

1300 ( 0 .
7473 )

200 ( 0 .
5584 )

$ 2 , 359 .
81
Engineering Economics 17

0
Question Continues
EQUIPMENT B
3
Original Equipment B
Investment
4 5 6 1 2
$1600
7
PW of cost

1600

350 (P/F, 6%, 10)

1600

350 (0.5584)

$ 1,404.56
8
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9
$350
10
18

Question Continues
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 EQUIPMENT A
PW of cost

1600

325 (P/F, 6%, 10)

1600

325 (0.5584)

$1,404.56
EQUIPMENT B
For fixed output of 10 years of service of equipments, Equipment B is preferred because it has a smaller cost.
Engineering Economics 19

 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
Value at the
End Of
Useful Life
Terminal Value at the end of
10-year analysis period
MACHINE A $40,000 $8,000 $15,000
MACHINE B $65,000 $10,000 $10,000
USEFUL
LIFE
7 year
13 year
Engineering Economics 20

Example Continues
$8,000
MACHINE A
$ 15,000
0 1 2 5 6 7 8 9 12 13 14 3 4
7-year life
10 11
7-year life
$40,000 $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
Engineering Economics 21

Example Continues
$10,000
MACHINE B
0 1 2 3 4 5 9 10 11 6 7
13-year life
8
$65,000
PW of cost

6 5 , 000

10,000 (P/F, 10%, 10)

6 5 , 000

10,000 ( 0.3855)

$ 61,145.0
12 13 14
Engineering Economics 22

Example Continues
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
MACHINE A
PW of cost

6 5 , 000

10,000 (P/F, 10%, 10)

6 5 , 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.
Engineering Economics 23

 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 P

A i
(5-2)
Engineering Economics 24

 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 )?
Capitalize d Cost, P

Annual disburseme
Interest rate (i) nt (A)
P

1000
0.025

$ 40 , 000
Engineering Economics 25

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 Equipment 2 Equipment 3
0
3
4
1
2
5
6
7
8
-$2000
+1000
+850
+700
+550
+400
+400
+400
+400
Engineering Economics
-$1500
+700
+300
+300
+300
+300
+400
+500
+600
-$3000
+500
+500
+550
+600
+650
+700
+500
+500
26

Question Continues
$1000 $850 $700 $550 $400 $400 $400 $400
0 1 2 3 4 5 6 7 8
EQUIPMENT 1
$2000
PW of Benefits

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
PW of Cost

2000
Net Present Wo rth

3379 .
17

2000

$ 1 , 379 .
17
Engineering Economics 27

Question Continues
$700 $300 $300 $300 $300 $400 $500 $600
0 1 2 3 4 5 6 7
EQUIPMENT 2
$1500
PW of Benefits

300(P/A, 10%, 8)

( 7 00 300)(P/F, 10%, 1)

100(P/G, 10%, 4)(P/F, 10%, 4)

300 ( 5 .
335 )

400 ( 0 .
9091 )
PW of Cost


1500
100(4.378) (0.6830)

2263 .
15
Net Present Wo rth

2263
Engineering Economics
.
15

1500

$ 763 .
15
8
28

Question Continues
$500 $500 $550 $600 $650 $700 $500 $500
0 1 2 3 4 5 6 7
EQUIPMENT 3
$3000
PW 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
PW of Cost

3000
Net Present Wo rth

2979 .
36

3000
 
$ 20 .
64
To maximize NPW, choose EQUIPMENT 1
Engineering Economics
8
29

Question Continues (MS EXCEL)
Use function:
- 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
Engineering Economics 30

Question Continues (MS EXCEL)
5
6
7
8
3
4
Year
0
1
2
Interest
10%
Equipment 1
($2,000)
1000
850
700
550
400
400
400
400
Equipment
($1,500)
700
300
300
300
300
400
500
600
2 Equipment
($3,000)
500
500
550
600
650
700
500
500
3
For Equip 1: NPW=NPV(A12,B3:B10)+B2
$1,379.17 $763.15 ($20.64)
Engineering Economics 31

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!
Engineering Economics 32

Problem 5-23
Solution i = 18%/12 = 1.5% per month
A = $500 payment/month n = 36
P = ?
payments 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!
Engineering Economics 33

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) Which option to choose?
(b) Will answer to (a) change if 2000 additional lines are needed in 5 years, instead of 10 years?
Engineering Economics 34

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.
Engineering Economics 35

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.
Engineering Economics 36

Engineering Economics 37