Lecture No.15
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Chapter Opening
Story – GE’s
Healthymagination
Project
GE Unveils $6 Billion
Health-Unit Plan:
• Goal: Increase the
market share in the
healthcare sector.
• Strategies: Develop
products that will lower
costs, increase access
and improve health-care
quality.
• Investment required:
$6 billion over six years
• Desired project
outcome: Would help
GE’s health-care unit
grow at least twice as
fast as the broader
economy.
Contemporary Engineering Economics, 5th edition, © 2010
Ultimate Questions
 GE’ s Point of View:
 Would there be enough demand for their
products to justify the investment required in
new facilities and marketing?
 What would be the potential financial risk if the
actual demand is far less than its forecast or
adoption of technology is too slow?
 If everything goes as planned, how long does it
take to recover the initial investment?
Contemporary Engineering Economics, 5th edition, © 2010
Bank Loan vs. Project Cash Flows
Contemporary Engineering Economics, 5th edition, © 2010
Example 5.1 Describing Project Cash Flows –
A Computer-Process Control Project
Year
(n)
Cash Inflows
(Benefits)
0
0
1
Cash Outflows
(Costs)
Net
Cash Flows
$650,000
-$650,000
215,500
53,000
162,500
2
215,500
53,000
162,500
…
…
…
…
8
215,500
53,000
162,500
Contemporary Engineering Economics, 5th edition, © 2010
Cash Flow Diagram for the Computer Process
Control Project
Contemporary Engineering Economics, 5th edition, © 2010
Payback Period
 Principle:
How fast can I recover my initial investment?
 Method:
Based on the cumulative cash flow (or accounting
profit)
 Screening Guideline:
If the payback period is less than or equal to some
specified bench-mark period, the project would be
considered for further analysis.
 Weakness:
Does not consider the time value of money
Contemporary Engineering Economics, 5th edition, © 2010
Example 5.3 Payback Period
N
Cash Flow
Cum. Flow
0
1
2
3
4
5
6
-$105,000+$20,000
$15,000
$25,000
$35,000
$45,000
$45,000
$35,000
-$85,000
-$70,000
-$45,000
-$10,000
$35,000
$80,000
$115,000
Payback period should occurs somewhere
between N = 3 and N = 4.
Contemporary Engineering Economics, 5th edition, © 2010
Annual cash flow
$45,000
$45,000
$35,000
$35,000
$25,000
$15,000
0
1
2
Years
3
4
5
6
Cumulative cash flow ($)
$85,000
150,000
3.2 years
Payback period
100,000
50,000
0
-50,000
-100,000
0
1
2
3
4
Contemporary Engineering Economics, 5th edition, © 2010
5
Years (n)
6
Practice Problem
 How long does it take to recover the initial
investment for the computer process control system
project in Example 5.1?
Initial Cost
Payback Period =
Uniform annual benefit
$650,000

$162,500
 4 years
Contemporary Engineering Economics, 5th edition, © 2010
Discounted Payback Period
 Principle:
How fast can I recover my initial investment plus
interest?
 Method:
Based on the cumulative discounted cash flow
 Screening Guideline:
If the discounted payback period (DPP) is less than
or equal to some specified bench-mark period, the
project could be considered for further analysis.
 Weakness:
Cash flows occurring after DPP are ignored
Contemporary Engineering Economics, 5th edition, © 2010
Discounted Payback Period Calculation
Period
(n)
Cash Flow
(An)
Cost of Funds
(15%)*
Ending Cash
Balance
0
-$85,000
0
1
15,000
-$85,000(0.15) = -$12,750
-82,750
2
25,000
-$82,750(0.15) = -12,413
-70,163
3
35,000
-$70,163(0.15) = -10,524
-45,687
4
45,000
-$45,687(0.15) =-6,853
-7,540
5
45,000
-$7,540(0.15) = -1,131
36,329
6
35,000
$36,329(0.15) = 5,449
76,778
* Cost of funds = (Unrecovered beginning balance) X (interest rate)
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-$85,000
Illustration of Discounted Payback Period
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Payback periods can be used as a screening tool for
liquidity, but we need a measure of investment worth for
profitability.
Contemporary Engineering Economics, 5th
edition, © 2010
Lecture No.16
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th
edition, © 2010
Net Present Worth Measure
 Principle: Compute the equivalent net surplus at n = 0 for a given
interest rate of i.
 Decision Rule for Single Project Evaluation: Accept the project if the
net surplus is positive.
Decision Rule for Comparing Multiple Alternatives: Select the
alternative with the largest net present worth.
Inflow
0
1
2
3
4
5
Net surplus
Outflow
PW(i) > 0
PW(i)inflow
0
PW(i)outflow
Contemporary Engineering Economics, 5th
edition, © 2010
0
Example 5.5 - Tiger Machine Tool Company
$35,560
$37,360
$31,850
$34,400
inflow
0
1
2
3
outflow
$76,000
PW (12%)inflow  $35,560(P / F ,12%,1)  $37,360(P / F ,12%,2)
$31,850(P / F ,12%,3)  $34,400(P / F ,12%,4)
PW (12%)outflow
$
 $76,000
PW (12%)  $106,065  $76,000
 $30,065  0, Accept
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edition, © 2010
Present Worth Amounts at Varying Interest Rates
Contemporary Engineering Economics, 5th
edition, © 2010
Can you explain
what $30,065
really means?
1.
2.
Project
Balance
Concept
Investment
Pool
Concept
Contemporary Engineering Economics, 5th
edition, © 2010
Project Balance Concept
 Suppose that the firm has no internal funds to
finance the project, so will borrow the entire
investment from a bank at an interest rate of
12%.
 Then, any proceeds from the project will be used
to pay off the bank loan.
 Then, our interest is to see if how much money
would be left over at the end of the project
period.
Contemporary Engineering Economics, 5th
edition, © 2010
Calculating Project Balances
End of Year (n )
0
1
2
3
4
Beginning Project Balance
$ (76,000) $ (49,560) $ (18,147) $ 11,525
Interest Charged (12%)
$ (9,120) $
(5,947) $
(2,178) $
1,383
Payment
$ (76,000) $ 35,560 $
Ending Project Balance
$ (76,000) $ (49,560) $ (18,147) $ 11,525 $ 47,308
Contemporary Engineering Economics, 5th
edition, © 2010
37,360 $ 31,850 $ 34,400
Project Balance Diagram – Four Pieces of Information
 The exposure to
financial risk
The discounted
payback period
The profit
potential
 The net future
worth
Contemporary Engineering Economics, 5th
edition, © 2010
Investment Pool Concept
 Suppose the company has $76,000. It has two
options. (1)Take the money out and invest it in the
project or (2) leave the money in the pool and
continue to earn a 12% interest.
 If you take Option 1, any proceeds from the project
will be returned to the investment pool and earn
12% interest yearly until the end of the project
period.
 Let’s see what the consequences are for each
option.
Contemporary Engineering Economics, 5th
edition, © 2010
Option A
If $76,000 were left
in the investment pool
for 4 years
$76,000(F/P,12%,4)
$119,587
Option B
If $76,000 withdrawal
from the investment
pool were invested in
the project
3)
1
• $49,959
$37,360(F/P,12%
,2)
2
• $46,864
3
• $35,672
$35,650(F/P,12%,
$31,850(F/P,12%,
1)
Year • Amount
1
• $35,650
2
• $37,360
3
• $31,850
4
• $34,400
4
• $34,400
$166,896
$34,400(F/P,12%,
0)
$166,896
Investment Pool
$47,309
$119,587
PW(12%) = $47,309(P/F,12%,4) = $30.065
Contemporary Engineering Economics, 5th
edition, © 2010
The net benefit of
investing in the project
What Factors Should the Company Consider in
Selecting a MARR in Project Evaluation?
 Cost of capital
Contemporary Engineering Economics, 5th
edition, © 2010
Cost of capital
MARR
make an investment project
worthwhile.
 Viewed as the rate of return that a
firm would receive if it invested its
money someplace else with a
similar risk
 Risk premium
 The additional risk associated with
the project if you are dealing with a
project with higher risk than normal
project
Risk
premium
 The required return necessary to
Practice Problem
 An electrical motor rated at 15HP needs to be
purchased for $1,000.
 The service life of the motor is known to be 10 years
with negligible salvage value.
 Its full load efficiency is 85%.
 The cost of energy is $0.08 per Kwh.
 The intended use of the motor is 4,000 hours per
year.
 Find the total present worth cost of owning and
operating the motor at 10% interest.
Contemporary Engineering Economics, 5th
edition, © 2010
Solution
1HP=0.7457kW
15HP = 15  0.7457 = 11.1855kW
Required input power at 85% efficiency rating:
11.1855kW
 13.1594kW
0.85
Required total kWh per year
13.1594kW  4,000 hours/year =52,638 kWh/yr
Total annual energy cost to operate the motor
52,638kWh  $0.08/kWh =$4,211/yr
The total present worth cost of owning and operating the motor
PW (10%)  $1,000  $4,211(P / A,10%,10)
= $26,875
Contemporary Engineering Economics, 5th
edition, © 2010
Cash Flow Series Associated with Owning and
Operating the Motor
0
1
2
3
4
5
6
7
8
9
10
$1,000
$4,211
PW (10%)  $1,000  $4,211(P / A,10%,10)
 $26,875
Contemporary Engineering Economics, 5th
edition, © 2010
Lecture No.17
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th
edition, © 2010
Future Worth Criterion
 Given: Cash flows and
MARR (i)
 Find: The net
equivalent worth at a
specified period other
than “present”,
commonly the end of
project life
 Decision Rule: Accept
the project if the
equivalent worth is
positive.
$47,309
$35,560 $37,360 $31,850 $34,400
0
1
$76,000
Contemporary Engineering Economics, 5th
edition, © 2010
2
3
Project life
Example 5.6 Future Equivalent at an
Intermediate Time
Contemporary Engineering Economics, 5th
edition, © 2010
Example 5.8 Project’s Service Life is Extremely
Long
•
Q1: Was Bracewell's
$800,000 investment
a wise one?
•Q2: How long does
he have to wait to
recover his initial
investment, and will
he ever make a
profit?
Contemporary Engineering Economics, 5th
edition, © 2010
Ex. 5.8: Mr. Bracewell’s Hydroelectric Project
V1  V2  $1,101K  $1, 468 K
 $367 K  0
V2  120 K ( P / A,8%,50)
 $1, 468K
V1  $50 K ( F / P,8%,9)  $50 K ( F / P,8%,8)
 $100 K ( F / P,8%,1)  60 K
 $1,101K
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edition, © 2010
How Would You Find P for a Perpetual Cash
Flow Series, A?
Contemporary Engineering Economics, 5th
edition, © 2010
Capitalized Equivalent Worth
 Principle: PW for a
project with an
annual receipt of A
over infinite service
life
A

0
Equation:
CE(i) = A(P/A, i, )
= A/i
P =CE(i)
Contemporary Engineering Economics, 5th
edition, © 2010
Practice Problem
Given: i = 10%, N = ∞
Find: P or CE (10%)
$2,000
$1,000
0
10
P = CE (10%) = ?
Contemporary Engineering Economics, 5th
edition, © 2010
∞
Solution:
$2,000
$1,000
0
10
P = CE (10%) = ?
∞
$1,000 $1,000
CE(10%) 

(P / F ,10%,10)
0.10
0.10
 $10,000(1  0.3855)
 $13,855
Contemporary Engineering Economics, 4th
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A Bridge Construction Project
 Construction cost = $2,000,000
 Annual Maintenance cost = $50,000
 Renovation cost = $500,000 every 15 years
 Planning horizon = infinite period
 Interest rate = 5%
Contemporary Engineering Economics, 5th
edition, © 2010
Cash Flow Diagram for the Bridge
Construction Project
15
Years
30
45
60
$500,000
$500,000
$500,000
$500,000
0
$50,000
$2,000,000
Contemporary Engineering Economics, 5th
edition, © 2010
Solution:
 Construction Cost
P1 = $2,000,000
 Maintenance Costs
P2 = $50,000/0.05 = $1,000,000
 Renovation Costs
P3 = $500,000(P/F, 5%, 15)
+ $500,000(P/F, 5%, 30)
+ $500,000(P/F, 5%, 45)
+ $500,000(P/F, 5%, 60)
.
= {$500,000(A/F, 5%,15)}/0.05
= $463,423
 Total Present Worth
P = P1 + P2 + P3 = $3,463,423
Contemporary Engineering Economics, 5th
edition, © 2010
Alternate way to calculate P3
 Concept: Find the
effective interest
rate per payment
period
Effective interest rate
for a 15-year period
Effective interest
0
15
30
45
60
rate for a 15-year
cycle
i = (1 + 0.05)15 - 1
= 107.893%
$500,000 $500,000 $500,000 $500,000
Capitalized equivalent worth
P3 = $500,000/1.0789
= $463,423
Contemporary Engineering Economics, 5th
edition, © 2010
Lecture No.18
Chapter 5
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th
edition, © 2010
Comparing Mutually Exclusive Projects – Basic
Terminologies
 Mutually Exclusive Projects
Alternative vs. Project
 Do-Nothing Alternative
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edition, © 2010
Revenue versus Service Projects
Revenue Projects
Service Projects
 Projects whose revenues
 Projects whose revenues do
depend on the choice of
alternatives
 Revenue and cost streams
vary with the choice of
alternatives.
not depend on the choice of
alternatives
 Fixed (or constant) revenues
for all alternatives
Contemporary Engineering Economics, 5th
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Analysis Period versus Required
Service Period
Analysis Period
Required Service Period
 The time span over which
 The time span over which
the economic effects of an
investment will be
evaluated (study period or
planning horizon).
the service of an equipment
(or investment) will be
needed.
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edition, © 2010
Road Map – A Process of Making a Choice
among Mutually Exclusive Alternatives
Contemporary Engineering Economics, 5th
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Comparing Mutually Exclusive Projects
 Principle: Projects must be compared
over an equal time span.
 Rule of Thumb: If the required service
period is given, the analysis period should
be the same as the required service period.
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edition, © 2010
Case 1
 Analysis Period Equals
Project Lives
 What to do: Compute the
NPW for each project over its
life and select the project
with the largest NPW.
Contemporary Engineering Economics, 5th
edition, © 2010
Comparing projects requiring different levels of
investment
Assume that the
unused funds will be
invested at MARR.
This portion
of investment
will earn a 10%
return on
investment.
PW(10%)A  $4,000  $450( P / F ,10%,1)  $600( P / F ,10%, 2)
$4, 493( P / F ,10%,3)
 $283
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Analysis Period Implied in Comparing Mutually
Exclusive Alternatives
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Case 2 – Ex. 5.10
Project Lives Longer than
the Analysis Period
 What to do:
 Estimate the salvage
value at the end of the
required service period.
 Compute the NPW for
each project over the
required service period.
 PW(15%)A = -$362
 PW(15%)B = -$364
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edition, © 2010
Case 3A – Service Projects – Ex. 5.11
 Project Lives Shorter than
the Analysis Period
 What to do:
Come up with
replacement projects
that match or exceed the
required service period.
Compute the NPW for
each project over the
required service period.
 PW(15%)A = -$34,359
 PW(15%)B = -$31,031
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edition, © 2010
Case 3B – Revenue Projects – Ex. 5.12
 Analysis Period Coincides with
the Project with the Longest Life in
the Mutually Exclusive Group
 What to do:
 Compute the NPW of each
project over its analysis
period, assuming that no
cash flows after the service
life for the shorter-lived
project.
 PW(15%)Drill = $2,208,470
 PW(15%)Lease = $2,180,210
Contemporary Engineering Economics, 5th
edition, © 2010
Case 4 – Ex. 5.13
 Analysis Period Is Not
Specified
 What to do:
 Come up with
replacement projects
that serve out the least
common multiple period
(LCM).
 Compute the NPW for
each project over the
LCM.
 PW(15%)A = -$53,657
 PW(15%)B = -$48,534
Contemporary Engineering Economics, 5th
edition, © 2010
Summary




Present worth is an equivalence method of analysis in
which a project’s cash flows are discounted to a lump
sum amount at present time.
The MARR or minimum attractive rate of return is the
interest rate at which a firm can always earn or borrow
money.
MARR is generally dictated by management and is the
rate at which NPW analysis should be conducted.
Two measures of investment, the net future worth and
the capitalized equivalent worth, are variations to the
NPW criterion.
Contemporary Engineering Economics, 5th
edition, © 2010
 The term mutually exclusive means that, when one of several






alternatives that meet the same need is selected, the others
will be rejected.
Revenue projects are those for which the income generated
depends on the choice of project.
Service projects are those for which income remains the same,
regardless of which project is selected.
The analysis period (study period) is the time span over which
the economic effects of an investment will be evaluated.
The required service period is the time span over which the
service of an equipment (or investment) will be needed.
The analysis period should be chosen to cover the required
service period.
When not specified by management or company policy, the
analysis period to use in a comparison of mutually exclusive
projects may be chosen by the individual analyst.
Contemporary Engineering Economics, 5th
edition, © 2010