Matakuliah : D0762 – Ekonomi Teknik
Tahun
: 2009
Present Worth Analysis
Course Outlined 4
Outline
• Definition
next
• Present Worth Comparisonnext
• Assumption on Using PWA
next
References :
- Engineering Economy – Leland T. Blank, Anthoy J.
Tarquin p.152-163
- Engineering Economic Analysis, Donald G. Newman, p.
111-127
- Engineering Economy, William G. Sulivan, p.137-194, p.
212-284
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Present Worth Analysis (PWA)
•
•
•
Present worth analysis (PWA) can resolve alternatives into equivalent present
consequences.
“Present worth analysis is most frequently used to determine the present value of
future money receipts and disbursements.”
We can restate the above three criteria in terms of PWA as follows:
Case
Situation
Criterion
Fixed input
Amount of money or other input
resources fixed
Maximize PW of benefits or
other outputs.
Fixed output
A fixed task, benefit, or other output
must be accomplished.
Minimize PW of costs or other
inputs
Free input &
output
Amounts of money, other inputs,
amounts of benefits, other outputs
can vary.
Maximize Net PW, which is PW
of benefits less PW of costs.
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Present Worth Analysis
• Careful consideration must be given to the time period covered by the
analysis.”
The time period is usually called the analysis period, or the planning
horizon.
Three different analysis-period situations occur:
1.
2.
3.
The useful life of each alternative equals the analysis period.
The alternatives have useful lives different from the analysis
period (and from each other).
The analysis period is effectively infinite.
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Present Worth
Comparison
Same-Length Analysis Periods
Example 5-1. GatorCo is considering buying device A or B. Each device can reduce
costs. Each device has a useful life of five years, and no salvage value. Device A
saves $300 a year, device B saves $400 the first year, but savings in later years
decrease by $50 a year. Interest is 7%. Which device should they choose?
•
•
Device A:
NPW = 300 (P/A,7%,5) = 300 (4.1000) = $1230
•
•
Device B:
NPW = 400 (P/A,7%,5) - 50 (P/G,7%,5) = 400(4.1000) - 50 (7.647) = $1257.65
•
•
Device B has the largest NPW of benefits.
Device B gives more of its benefits in the earlier years.
•
Note, If we ignore the time value of money (we should not), both devices have a NPW of benefits of $1500
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Present Worth Comparison
Same-Length Analysis Periods
•
Example Wayne County plans to build an aqueduct to carry water. The county
can:
a) spend $300 million now, and enlarge the aqueduct in 25 years for $350 million
more,
b) construct a full-size aqueduct now for $400 million.
The analysis period is 50 years. We ignore maintenance costs. Interest is 6%.
There is no salvage value.
a) NPW = $300 million + $350 million (P/F,6%,25) = $381.6 million
b) NPW = $400 million
This is an example of stage construction. The two-stage construction appears
preferable.
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Present Worth
Comparison
Different-Length Analysis Periods
Sometimes the useful lives of projects differ from the analysis period.
Example The mailroom needs new equipment. Alternative choices are
as follows:
Make
Cost
Useful life
EOL salvage value
Speedy
$1500
5 years
$200
Allied
$1600
10 years
$325
We no longer have a situation where either choice will provide the same
desired level of (fixed) output.
Speedy equipment for five years is not equivalent to Allied equipment
for ten years.
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Present Worth
Comparison
Different-Length Analysis Periods
Allied
200
200
Allied
5 years
10 years
1500
1500
Speedy
5 years
200
1500
325
325
5 years
Speedy
10 years
1600
1600
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Present Worth Comparison
Different-Length Analysis Periods
•
•
Solution : Compare one Allied with two Speedy’s
We buy a Speedy for $1500, use it for 5 years, get $200 salvage, buy a second
Speedy for $1500, use it for the second 5 years, and again get $200 salvage.
Two Speedy’s:
PW = 1500 + (1500 – 200) (P/F,7%,5) – 200 (P/F,7%,10)
= 1500 + 1300 (0.7130) – 200 (0.508)
= 1500 + 927 – 102 = $2325.
•
Allied for 10 years:
PW = 1600 – 325 (P/F,7%,10) = 1600 – 325 (0.5083)
= 1600 – 165 = $1435.
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Present Worth
Analysis
Different-Length Analysis Periods
Generalization. “The analysis period for an economy study should be determined from
the situation.”
The period can be:
• short: PC manufacture,
• intermediate length: steel manufacture
• indefinite length: national government
“Least common multiple” idea.
In the above example, it made some sense to use 10 years as the analysis period.
If one piece of equipment had a life of 7 years, and the other a life of 13 years, and we
followed the same approach, we would need to use
7 (13) = 91 years. But an analysis period of 91 years is not too realistic.
Terminal Value Idea.
We estimate terminal values for the alternatives at some point prior to the end of their
useful lives.
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Present Worth
Analysis
Different-Length Analysis Periods
Alternative 1
C1 = initial cost
S1 = salvage value
R1
= replacement cost
T1
= terminal value at the end of 10th
year
Alternative 2
C2 = initial cost
T2 = terminal value at the end of 10th
year
S1
C1
T1
S1
R1
S2
T2
C2
7 years
3 years
3 years
1 year
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Present Worth
Analysis
Infinite-Length Analysis Periods – Capitalized Cost
Present worth of costs with 10-yr. analysis period:
PW1 = C1 + (R1 – S1) (P/F,i%,7) – T1 (P/F,i%,10)
PW2 = C2 – T2 (P/F,i%, 10)
Infinite Analysis Period – Capitalized Cost.
Sometimes the analysis period is of indefinite length.
The need for roads, dams, pipelines, etc. is sometimes considered permanent.
The authors refer to this situation as an infinite analysis period.
Present worth analysis in this case is called capitalized cost.
Capitalized cost is the present sum of money that would need to be set aside now, at
some know interest rate, to yield the funds needed to provide a service indefinitely.
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Present Worth
Analysis
Infinite-Length Analysis Periods – Capitalized Cost
Example LA plans a pipeline to transport water from a distant watershed area to the city.
The pipeline will cost $8 million and have an expected life of 70 years. The water line
needs to be kept in service indefinitely. We estimate we need $8 million every 70 years.
Compute capitalized cost (compounding 7% yearly).
8M
8M
8M
P=?
To find the capitalized cost, we first compute an annual disbursement A that is equivalent
to $8 million every seventy years.
8M
A A A A
A A
A = F (A/F,i,n) = 8 million (A/F,7%,70) = 8 million (0.00062) = $4960
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Present Worth Analysis
Infinite-Length Analysis Periods – Capitalized Cost
Motivating Example.
Ima Rich wants to set up a scholarship fund to provide $20,000 yearly to deserving
undergraduate women engineering students at UF. UF will invest her donation, and
expects it to earn 10% a year. How much will Ima need to donate in one lump sum
so that $20,000 is available every year?
Observation. If Ima donates $200,000, 10% of it is $20,000. The money grows in
one year to $220,000, a scholarship is funded, $200,000 remains, and grows in
another year again to $220,000, another scholarship is funded, etc.
With P = $200,000, i = 10%, A = $20,000, we see that:
P=A/i A=Pi
In fact this approach works generally. To make an amount A available every year
beginning with an initial present sum P and given an interest rate i, just take P = A/i.
P is called the capitalized cost.
Go Back
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Example. Spreadsheet
• A City engineer is consideration two alternatives for the local water
supply. The first alternative involves construction of an earthen dam
on a nearby river, which has a highly variable flow. The dam will
form a reservoir so the city may have a dependable source of water.
The initial cost of the dam is expected to be $8 million and will
require annual upkeep cost of $25,000. The dam Is expected to last
indefinitely
• Alternatively the city can drill wells as needed and construct
pipelines for transportation the water to the city. The engineer
estimates that an average 10 wells will be required initially at a cost
of $45,000 per well, including the pipelines. The average life of a
well is expected to be 5 years with an annual operating cost of
$12,000 per well. If the City uses an interest rate of 15% per year,
determine which alternative should be selected on the basis of their
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capitalized cost
Solution - Spreadsheet
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Concepts and Assumptions
End-of-Year Convention
Textbooks in this area usually follow an end-of-year convention. For each time period, all the series
of receipts and disbursements occur at the end of the time period. If, in fact, they do not, you can
replace them by their equivalent values at the end of the year. Multiply each by the appropriate
factor (F/P,i %,n) to move it to the end of the year.
Viewpoint of Economic Analysis Studies.
Usually we take the point of view of an entire firm when doing an industrial economic analysis.
What is best for the entire firm may not be best for smaller groups in the firm. It is easy to make a
bad decision if we ignore part of the problem.
Sunk Costs
It is the differences between alternatives that are relevant to economic analysis. Events that have
occurred in the past have no bearing on what we should do in the future. What is important are the
current and future differences between alternatives. Past costs, like past events, have no bearing on
deciding between alternatives unless the past costs somehow actually affect the present or future
costs. Usually, past costs do not affect the present or the future costs, so we call them sunk costs and
disregard them.
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Concepts and Assumptions
Borrowed Money Viewpoint
Economic analysis involve spending money. It is thus natural to ask the source
of the money. There are two aspects of money to determine:
Financing – obtaining the money
Investment – spending the money
Experience shows it is important to distinguish between these two aspects. Failure
to separate them sometimes leads to confusing results and poor decision making.
The conventional assumption in economic analysis is that the money required to
finance alternatives and/or solutions in problem solving is considered to be
borrowed at interest rate i.
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Concepts and Assumptions
Effect of Inflation and Deflation
For the time being we assume prices are stable. We deal with inflation and
deflation later in the course.
Income Taxes
We defer our introduction of income taxes into economic analyses until later.
Stability
The economic situation is stable.
Determinism (This is a strong assumption. It is almost never satisfied.)
All data of interest are known
deterministically (no randomness)
and can be accurately predicted.
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