Engineering Finance
Lessons


Solution Alternatives
Comparisons with Net Worth

At least in my family there is a difference between me and my wife with respect to shopping. Say we have an immediate need, or want, to purchase some item for the home. My strategy, as the husband, is to enter a store that we frequent, look around at the alternatives on display and pick the one, perhaps the first one, that I think satisfies the need. My wife, on the other hand, likes to shop so visiting just one store is often insufficient. Rather, we travel from store to store to find the best one. Although cost plays a role, it is often not the deciding factor, especially with the easy availability of credit. For me, convenience is the issue. I would rather be done with the project and go on with other activities. For my wife, shopping is the driving force. The original need is the excuse to shop. Her decision criterion is a mystery to me, but at least in my family, it is usually the deciding criterion.

For engineers and managers the decision criteria may be similarly obscure. An engineer faced with a hard technical problem when finally discovering the answer may breath a sigh of relief and present the answer as the solution to the problem. When the decision maker asks about alternatives, the engineer may reply that one is enough. It is difficult and perhaps expensive to come up with another. In other cases, there may be alternatives, but the deciding criteria may involve politics, personal gain, or simply personal preference based on unknown or unexpressed factors.

In most cases of personal or organizational decision making there should be alternatives. Projects should include specifications and project teams should have the discipline and resources to present more than one solution that meets the specifications. Although there may be several measures that differ among solutions, the economic or financial aspects of the solutions should be carefully examined. Then an explicit analysis of the solutions can provide an objective comparison. Whether or not the decision maker uses the objective analysis as the deciding factor, at least the information has been generated and supposedly considered.

Two methods for comparing alternatives using economic criteria are presented in this lesson. Here we describe the NPW and NAW methods. A later lesson considers the rate of return (ROR) method. Most businesses should, and often do, use economic-based methods for making decisions.

Goals
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  • Use the NPW Method to compare two or more alternatives and select the best
  • Use the NAW Method to compare two or more alternatives and select the best
  • Select and use a study period when alternative lives are different
  • Do incremental analysis for two alternatives
  • Select challenger and defender for incremental analysis
  • Use the Economics add-in for comparisons.
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Section 3.4 describes the NPW, NAW and ROR methods for comparing alternatives.

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3.4 Equivalent Worth Methods
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Net Present Worth Method
For the NPW Method compute the net present worth of the cash flows for the alternatives and choose the best. Be careful when the lives of the alternatives are not equal.

An important issue when evaluating economic alternatives is the interval of time for the evaluation. Usually we say that the interval begins when the first cash is expended (or received) and identify this time as time 0. The interval ends when expenditures or revenues can no longer be associated with the project. This is the life of the project. For some projects, there is a revenue associated with the disposal of assets called the salvage value. For other projects there may be considerable expenditures associated with termination and the final payment might be a cost rather than a revenue. We have considered the project life cycle in an earlier lesson, and in most instances the life will correspond to the length of the life cycle.

Click the QuickTime symbol to see how to use the present worth method when the alternatives have the same lives.

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NPW Comparison
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Summary of Example
Two machines are being considered to manufacture some product.  The first machine costs $1100 initially and has a $100 salvage value after ten years.  It costs $200 a year to operate.  The second costs $1300 initially.  It has a salvage value of $100 after ten years and costs $150 per year to operate.  At the minimum rate of return of 15%, which machine should you select?

First Alternative (F)
Second Alternative (S)
NPW costs = 1100 – 100(P/F, 0.15, 10) + 200(P/A, 0.15, 10) = 2079
NPW costs = 1300 – 100(P/F, 0.15, 10) + 150(P/A, 0.15, 10) = 2028
Select the second alternative with the smaller NPW of costs.

Incremental Analysis

Component Challenger (S) Defender (F) Increment (S-F)
Investment
–1300
–1100
–200
Operations Cost
–150
–200
+50
Salvage
100
100
0
For the Incremental Investment
NPW = –200 + 50(P/A, 0.15, 10) = 51
NPW > 0, so accept the challenger. The incremental investment is justified.
Note that when performing incremental analysis, the defender is the less expensive alternative and the challenger is the more expensive alternative. When there are more than two alternatives, we always start with the least expensive and perform pairwise comparisons, discarding the defender if the incremental cost is beneficial (NPW > 0). If not, the challenger is discarded, the next candidate becomes the new challenger, and the analysis is repeated until only one alternative remains.
Different Lives
When the alternatives have different lives, they must be compared over a common study period. Often the choice is the least common multiple of the lives.

Click the Quicktime symbol to see how to use the present worth method when the alternatives have the different lives.

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Different Lives
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Summary of Example
To sweep the floor of a machine shop a company is considering two alternatives.  An automatic broom is available that costs $35.  The broom must be replaced every year.  The old broom is thrown away.  A vacuum cleaner is available for $90 with an annual expense of $10.  This machine will last four years with no salvage value.  The company’s minimum rate of return is 15%.  Which alternative should be chosen?

Select a common study period for each alternative. Use a study period of 4 years.
Broom (B)
Vacuum (V)
NPW costs = 35 + 35(P/A, 0.15, 3) = 114.91 
NPW costs = 90 + 10(P/A, 0.15, 4) = 118.55
Select the broom with the smaller NPW of costs.
Net Annual Worth Method
For the NAW Method compute the equivalent uniform annual worth of the cash flows for the alternatives and choose the best.

Click the QuickTime symbol to see how to use the annual worth method when the alternatives have the same lives.

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NAW Comparison
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Summary of Example with both Alternatives having the Same Life
Two machines are being considered to manufacture some product.  The first machine costs $1100 initially and has a $100 salvage value after ten years.  It costs $200 a year to operate.  The second costs $1300 initially.  It has a salvage value of $100 after ten years and costs $150 per year to operate.  At the minimum rate of return of 15%, which machine should you select?

First Alternative (F)
Second Alternative (S)
NAC(F) = 1100(A/P, 0.15, 10) – 100(A/F, 0.15, 10) + 200 = $414.25
NPC(S) = 1300(A/P, 0.15, 10) – 100(A/F, 0.15, 10) + 150 = $404.1
Select the second alternative with the smaller NAW of costs.

Incremental Analysis

Component Challenger (S) Defender (F) Increment (S-F)
Investment
–1300
–1100
–200
Operations Cost
–150
–200
+50
Salvage
100
100
0
For the Incremental Investment
NAW = –200(A/P, 0.15, 10) + 50 = $10.15
NAW > 0, so accept the challenger. The incremental investment is justified.

Summary of Example with the Alternatives having Different Lives
To sweep the floor of a machine shop a company is considering two alternatives.  An automatic broom is available that costs $35.  The broom must be replaced every year.  The old broom is thrown away.  A vacuum cleaner is available for $90 with an annual expense of $10.  This machine will last four years with no salvage value.  The company’s minimum rate of return is 15%.  Which alternative should be chosen?

A study period need not be selected. The implicit assumption is the least common multiple of the lives, or four years.
Broom (B)
Vacuum (V)
NAC(B) = 35(A/P, 0.15, 1) = 40.25/year
NAC(V) = 90(A/P, 0.15, 4) + 10 = 41.52/year
Select the broom with the smaller NPW of costs.
The Economics Add-in
You should be able to perform NPW or NAW analyses by hand with equivalence expressions, but for problem solving, the Economics add-in can handle problems of arbitrary complexity.

The Economics add-in is easy to use and it should be handy for homework and computer-based exams.

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Economics Add-in
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The figure below shows the two alternatives defined for the machine example introduced above. Column J computes information necessary for the comparison.

The add-in has a Compare command that explicitly compares two alternatives. The Compare Multiple command compares more than two. Click the icon for the documentation of the two methods.

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Economics: Compare
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The results for the example are below. When two projects are compared the add-in automatically prepares an incremental analysis.

Project Comparison
Incremental Cash Flow

The figure shows the MARR for comparison as 15%. The study period is the 10 year life of the alternatives. The Second is the better of the two. Near NPW and NAW values are computed for the 15% return. At the bottom of the figure we see the NPW and NAW of the incremental investment of the second alternative over the first. The IRR reported is the rate of return on the increment of investment.

Also produced by the comparison, is the cash flow of the incremental investment. The payback period for the incremental investment is 4 years.

Summary

 

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PW & AW Comparison Summary
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Problems

The links below open problem sets. The first uses the the ROR method as well as the methods of this lesson. The ROR method is covered in later lessons.

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Comparison Problems
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Comparison Exercise 1
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Comparison Exercise 2
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Division

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Engineering Finance
by Paul A. Jensen
Copyright 2005 - All rights reserved

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