Chapter 5 Present Worth Analysis - Help-A-Bull
Download
Report
Transcript 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
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:
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.
Engineering Economics
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
Engineering Economics
6
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
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
USEFUL LIFE
DEVICE A $1000
$300 Annually
5 year
DEVICE B $1350
$300 The first year and
increase $50 annually
5 year
Engineering Economics
8
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
Engineering Economics
9
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
Engineering Economics
10
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
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
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
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 INVESTMENT1
Engineering Economics
15
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
Engineering Economics
16
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
Engineering Economics
17
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
Engineering Economics
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)
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.
Engineering Economics
19
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
Engineering Economics
20
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
Engineering Economics
21
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
Engineering Economics
22
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.
Engineering Economics
23
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
Engineering Economics
A
P
i
(5-2)
24
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
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
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
Engineering Economics
26
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
Engineering Economics
27
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
28
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
Engineering Economics
29
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
Engineering Economics
30
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)
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
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!
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)
(b)
Which option to choose?
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
End of Chapter 5
Engineering Economics
37