PRESENT WORTH ANALYSIS
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PRESENT WORTH ANALYSISAnastasia L. Maukar
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Formulating Mutually Exclusive Alternatives
• One of the important functions of financial management and engineering is the creation of “alternatives”
• If there are no alternatives to consider then there really is no problem to solve!
• Given a set of “feasible” alternatives, engineering economy attempts to identify the “best” economic approach to a given problem
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Types of Economic Projects• Mutually exclusive alternatives
– From a set of feasible alternatives, pick one and only one to execute
– Mutually exclusive alternatives compete against each other
• Independent projects– From a set of feasible alternatives select as many as can be
funded in the current period
• The “Do Nothing” (DN) alternative should always be considered
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Economic Criteria
• The decision process requires that the outcomes of feasible alternatives be arranged so that they may be judged for economic efficiency in terms of the selection criterion.
• Equivalence provides the logic by which we may adjust the cash flow for a given alternative into some equivalent sum or series
• How should they be compared?• Learn how analysis can resolve alternatives into equivalent
present consequences, referred to simply as present worth analysis.
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APPLYING PRESENT WORTH TECHNIQUES
• One of the easiest ways to compare mutually exclusive alternatives is to resolve their consequences to the present time
• The three criteria for economic efficiency are restated in terms of present worth analysis :
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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.
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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?
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Bank Loan vs. Project Cash Flows
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Example 3.1 Describing Project Cash Flows – A Computer-Process Control Project
Year(n)
Cash Inflows(Benefits)
Cash Outflows(Costs)
NetCash Flows
0 0 $650,000 -$650,000
1 215,500 53,000 162,500
2 215,500 53,000 162,500
… … … …
8 215,500 53,000 162,500
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Cash Flow Diagram for the Computer Process Control Project
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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.
2 3 4 50 1
Inflow
Outflow
0PW(i) inflow
PW(i) outflow
Net surplus
PW(i) > 0
0
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Example 3.2 - Tiger Machine Tool Company
inflow
outflow
(12%) $35,560( / ,12%,1) $37,360( / ,12%,2)$31,850( / ,12%,3) $34,400( / ,12%,4)$
(12%) $76,000(12%) $106,065 $76,000
$30,065 0, Accept
PW P F P FP F P F
PWPW
$76,000
$35,560 $37,360$34,400
01 2 3
outflow
inflow
$31,850
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Present Worth Amounts at Varying Interest Rates
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Can you explain what $30,065 really means?
1. Project Balance Concept
2. Investment Pool Concept
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Project Balance Concept
• Suppose that the firm has no internal funds to finance the project, so will borrow the entire investment from a bank at an interest rate of 12%.
• Then, any proceeds from the project will be used to pay off the bank loan.
• Then, our interest is to see if how much money would be left over at the end of the project period.
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Calculating Project Balances
End of Year (n ) 0 1 2 3 4
Beginning Project Balance (76,000)$ (49,560)$ (18,147)$ 11,525$
Interest Charged (12%) (9,120)$ (5,947)$ (2,178)$ 1,383$
Payment (76,000)$ 35,560$ 37,360$ 31,850$ 34,400$
Ending Project Balance (76,000)$ (49,560)$ (18,147)$ 11,525$ 47,308$
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Project Balance Diagram – Four Pieces of Information
The exposure to financial risk
The discounted payback period
The profit potential
The net future worth
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Useful Lives Equal the Analysis Period
Examples 3.3:
• A firm is considering which of two mechanical devices to
install to reduce costs in a particular situation. Both devices
cost $1000 and have useful lives of 5 years and no salvage
value. Device A can be expected to result in $300 savings
annually. Device B will provide cost savings of $400 the first
year but will decline $50 annually, making the second-year
savings $350, the third-year savings $300, and so forth. With
interest at 7%, which device should the firm purchase?
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Useful Lives Equal the Analysis Period
• Answer:
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Useful Lives Equal the Analysis Period
Examples 3.4:• A purchasing agent is considering the purchase of some new
equipment for the mail room. Two different manufacturers have provided quotations. An analysis of the quotations indicates the following:
• The equipment of both manufacturers is expected to perform at the desired level of (fixed)output. For a 5-year analysis period, which manufacturer's equipment should be selected? Assume 7% interest and equal maintenance costs
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Useful Lives Equal the Analysis Period
• answer
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Useful Lives Equal the Analysis Period
• A firm is trying to decide which of two weighing scales it should install to check a package-filling operation in the plant. The ideal scale would allow better control of the filling operation and result in less overfilling. If both scales have lives equal to the 6-year analysis period, which one should be selected? Assume an 8% interest rate
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Useful Lives Equal the Analysis Period
• answer
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Useful Lives Different from the Analysis Period
• A diesel manufacturer is considering the two alternative production machines which are specific data are as follows:
• The manufacturer uses an interest rate of 8% and wants to use the PW method to compare these alternatives over an analysis period of 10 years.
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Useful Lives Different from the Analysis Period
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Useful Lives Different from the Analysis Period
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Infinite Analysis Period: Capitalized Cost
• Another difficulty in present worth analysis arises when we encounter an infinite analysis period (n= ~)
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Infinite Analysis Period: Capitalized Cost
Example:• A city plans a pipeline to transport water from a
distant watershed area to the city. The pipeline will cost $8 million and will have an expected life of 70 years. The city anticipates it will need to keep the water line in service indefinitely. Compute the capitalized cost, assuming7% interest
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Infinite Analysis Period: Capitalized Cost
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Infinite Analysis Period: Capitalized Cost
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Multiple Alternatives
• The minimum required interest rate for invested money is called the minimum attractive rate of return, or MARR
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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.
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If $76,000 were left in the investment poolfor 4 years
$76,000(F/P,12%,4) $119,587
If $76,000 withdrawalfrom the investmentpool were invested inthe project
• AmountYear• $35,6501• $37,3602
• $31,8503• $34,4004
$35,650(F/P,12%,3)
$37,360(F/P,12%,2)
$31,850(F/P,12%,1)
$34,400(F/P,12%,0)
1 • $49,959
2 • $46,864
3 • $35,672
4• $34,400
$166,896
$166,896
$119,587
$47,309
Option A
Option B
The net benefit of investing in the project
Investment Pool
PW(12%) = $47,309(P/F,12%,4) = $30.065
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What Factors Should the Company Consider in Selecting a MARR in Project Evaluation?
• Cost of capital– The required return necessary to
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
MA
RR
Cos
t of c
apita
lR
isk
prem
ium
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WEEK 4
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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.
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Solution
1HP=0.7457kW 15HP = 15 0.7457 = 11.1855kW Required input power at 85% efficiency rating:
11.1855 13.15940.85
Required total kWh per year 13.1594kW 4,000 hours/year =52,638
kW kW
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
(10%) $1,000 $4,211( / ,10PW P A %,10) = $26,875
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Exercise 1
• A piece of land may be purchased for $610,000 to be strip-mined for the underlying coal. Annual net income will be $200,000 per year for 10 years. At the end of the 10 years, the surface of the land will be restored as required by a federal law on strip mining. The reclamation will cost $1.5 million more than the resale value of the land after it is restored. Using a 10%interest rate, determine whether the project is desirable
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Exercise
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Exercise 2
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Exercise 3
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Exercise 4
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Exercise 5