Engineering Economic Analysis
9th Edition
Chapter 5
PRESENT WORTH ANALYSIS
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
1
Jerome P. Lavelle:
In general I like this type of slide, it provides a map of the material in terms
of past coverage and future materials. My only comment is that if we chose
to use this approach as a first slide, we need to do it for all chapters. Also, I
believe that we need “chapter learning objectives” incorporated into all of
• Equivalence
the chapter slides (however,
we don’t yetconcept
have these for any chapters).
Given all of this, and for consistency sake, I would say for now we need to
• Cash flows
eliminate this slide.
Where we have been:
• Compound interest factors
Where we are going in this chapter:
• Understanding economic criteria
• Applying present worth techniques
• Assumptions in solving economic analysis
problems
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
2
Economic Criteria
Projects are judged against an economic criterion.
Situation
Criterion
Fixed input
Maximize output
Fixed output
Minimize input
Neither fixed
Maximize difference
(output-input)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
3
Economic Criteria Restated
Present Worth Techniques
Fixed input
Fixed output
Neither fixed
Situation
Amount of
capital available
fixed
$ amount of
benefit is fixed
Neither capital
nor $ benefits
are fixed
Criterion
Maximize
present worth of
benefits
Maximize
present worth of
costs
Maximize net
present worth
(NPV)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
4
Economic Criteria - Example
Purchasing Building Space
Alt
A
Situation Example
Fixed
$150,000
input
max
B
Fixed
output
C
Neither
fixed
20,000 ft2
building
available
$150,000 max
15-20,000 ft2
required
Criterion
Maximum square feet of
building for the price
Negotiate for minimum
cost/ft2
Simultaneously negotiate
for maximum building
size & minimum cost/ft2
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
5
Jerome P. Lavelle:
Would it be useful to have a slide that
demonstrates the mechanics of calculating
by hand using the tabled factors and equati
too?
Applying
Present Worth Techniques
• Analysis period must be considered
• Useful life of the alternative equals the analysis
period
• Alternatives have useful lives different from the
analysis period
• The analysis period is infinite, n = 
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
6
Useful Lives
Equal the Analysis Period
1. Require a project to last
five years.
2. The equipment and tooling
will last five years.
3. Calculate the PW or NPW
over a five year span and
junk the equipment and
tooling at the end of the
five years.
Examples 5-1, 2, 3, & 4
Examples
5-1
Applying present worth techniques
Interest rate
Year
0
1
2
3
4
5
7.00 %
Alternative
A
B
-1000
300
300
300
300
300
PW of benefits $1,230.06
NPW
$230.06
5-2
Interest rate
Year
0
25
5-3
$1,257.75
$257.75
6.00 %
Alternative
A
B
-300
-350
-400
0
PW of costs
NPW
($381.55)
($381.55)
($400.00)
($400.00)
Interest rate
7.00 %
Alternative
Speedy
Allied
Year
0
1
2
3
4
5
5-4
-1000
400
350
300
250
200
-1500
0
0
0
0
200
-1600
0
0
0
0
325
PW of costs ($1,500.00)
NPW
($1,357.40)
($1,600.00)
($1,368.28)
Interest rate
Year
8.00 %
Alternative
Atlas
Tom Thumb
<<<< Some cells are hidden.
You need to unhide them to enter
Alternative Speedy
Allied
NPW
($1,357.40) ($1,368.28)
Alternative
NPW
Atlas
Tom Thumb
$143.31
$214.85
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
7
Jerome P. Lavelle:
We need a slide that mentions the
Repeatability Assumption for handling
these problems – definition and
parameters
Useful Lives Different
From the Analysis Period
1. Require a project to last 10
years.
2. The equipment and tooling
will last five years.
3. Calculate the PW or NPW
over a 10 year span.
1. Purchase new equipment
and tooling twice, at the
beginning of year one and
six.
2. Junk the equipment and
tooling at the end of each five
year period.
Based on example 5-3
Based on 5-3
The lives of the equipment differ, one is a multiple of the other and we
can use the equipment as long as it will last.
Interest rate
7.00 %
Alternative
Year
Speedy
Allied
Alternative Speedy
Allied
NPW
($2,325.21) ($1,434.79)
Purchase new Speedy
0
-1500
-1600 Purchase new Allied
1
0
0
2
0
0
3
0
0
4
0
0
Salvage old Speedy
5
-1300
0
Purchase new Speedy
6
0
0
7
0
0
8
0
0
9
0
0
Salvage old Speedy
10
200
325 Salvage old Allied
PW of costs ($2,325.21) ($1,434.79)
NPW
($2,325.21) ($1,434.79)
Based on 5-3
The lives of the equipment differ and the least common multiple of the
lives is longer than we can use the equipment.
Interest rate
7.00 %
Alternative
Year
Speedy
Allied
Alternative Speedy
Allied
Life
7
13
NPW
($1,793.91) ($1,294.45)
Purchase new Speedy
0
-1500
-1600 Purchase new Allied
1
0
0
2
0
0
3
0
0
4
0
0
Salvage old Speedy
5
0
0
Purchase new Speedy
6
0
0
7
-1500
0
Salvage old Speedy
8
1100
525 Salvage old Allied
PW of costs ($1,793.91) ($1,294.45)
NPW
($1,793.91) ($1,294.45)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
8
Infinite Analysis Period
Capitalized Cost
For:
Examples 5 - 5 & 6
n=
A = Pi
Then:
Capitalized cost
P = A/i
This requires
first computing
the future cost
into an
equivalent A.
Infinite analysis period
5-5
Annual disbursement
Interest rate
Capitalized cost
5-6
Disbursement
Periods between disbursements
Interest rate
A/F
PW of perpetual reconstruction
Capitalized cost
Alternate solution A
A/P
Capitalized cost
Alternate solution B
70 year interest rate
PW of perpetual reconstruction
Capitalized cost
$50.00
4%
0.04
$1,250.00
Note: this is
maintenance and
occurs at the end of
each period.
$8,000,000
Note: this is
70
construction and
7%
occurs at the
0.07
beginning of each
$4,956.22
period.
$70,803.11 A/F/Interest rate
$8,070,803.11 Disbursement + Pw of perpetual
reconstruction
$564,956.22
$8,070,803.11 A/P/Interest rate
112.99
$70,803.11 Disbursement/70 year interest rate
$8,070,803.11 Disbursement + Pw of perpetual
reconstruction
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
9
Jerome P. Lavelle:
The text of this table is quite difficult to
read, it seems some of the row widths
need to be readjusted
Multiple Alternatives
Examples 5 - 7, 8, 9 & 10
Where more than one
alternative exists, we
generally want to
compare the
alternatives in a single
table.
The goal is to develop a
single table that
automates as many of
the calculations as
possible.
5-7
Operating years
Operating hours/year
MARR
5
2000
7.00 %
0.07
Alternative
Description
Installed cost of pipeline and pump
Cost per hour for pumping
Annual cost of pumping
PW of pumping costs
PW of pipeline costs
5-8
A
2" Pipe
$22,000.00
$1.20
$2,400.00
($9,840.47)
($31,840.47)
Operating years
MARR
C
4" Pipe
$25,000.00
$0.50
$1,000.00
($4,100.20)
($29,100.20)
A
Description
Total capital investment
Uniform net annual benefit
Terminal value at end of project
Do nothing
$0.00
$0.00
$0.00
B
Vegetable
market
($50,000.00)
$5,100.00
$30,000.00
Gas station
($95,000.00)
$10,500.00
$30,000.00
Small motel
($350,000.00)
$36,000.00
$150,000.00
PW investment
PW of benefits
PW of terminal value
NPW of project
$0.00
$0.00
$0.00
$0.00
($50,000.00)
$43,419.17
$4,459.31
($2,121.52)
($95,000.00)
$89,392.42
$4,459.31
($1,148.27)
($350,000.00)
$306,488.29
$22,296.54
($21,215.16)
Operating years
MARR
Alternative
A
Do nothing
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
B
Strip mining
project
($610,000.00)
$200,000.00
$200,000.00
$200,000.00
$200,000.00
$200,000.00
$200,000.00
$200,000.00
$200,000.00
$200,000.00
$200,000.00
($1,500,000.00)
PW investment
PW of benefits
PW of terminal value
NPW of project
$0.00
$0.00
$0.00
$0.00
($610,000.00)
$1,228,913.42
($578,314.93)
$40,598.49
Operating years
MARR
Alternative Null
Description
Total capital investment
Net annual benefit
1
2
3
4
5
6
7
8
9
10
Terminal value at end of project
C
D
10
10.00 %
0.1
Description
Total capital investment
Net annual benefit
1
2
3
4
5
6
7
8
9
10
Terminal value at end of project
5-10
D
6" Pipe
$30,000.00
$0.40
$800.00
($3,280.16)
($33,280.16)
20
10.00 %
0.1
Alternative
5-9
B
3" Pipe
$23,000.00
$0.65
$1,300.00
($5,330.26)
($28,330.26)
C
D
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
8
8.00 %
0.08
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
$0.00
A
CE-1
($2,000.00)
$1,000.00
$850.00
$700.00
$550.00
$400.00
$400.00
$400.00
$400.00
$0.00
$0.00
$0.00
B
CE-2
($1,500.00)
$700.00
$300.00
$300.00
$300.00
$300.00
$400.00
$500.00
$600.00
$0.00
$0.00
C
$0.00
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
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0
Jerome P. Lavelle:
•Should Assumptions slide come at the beginning of the
chapter versus the end?
Assumptions in Solving Economic
Analysis Problems
•In looking back over the chapter I think we need a few
slides inserted on the theory or pedagogy involved in each
case of the use of Present Worth.
•
•
•
•
•
•
End-of-year convention (simplifies calculations)
Viewpoint (generally the firm)
Sunk costs (past has no bearing)
Borrowed money (consider investing only)
Effect of inflation (prices are not stable)
Income taxes (must be considered for realism)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.
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