Comparing economic alternatives
Download
1 / 23

Comparing Economic Alternatives - PowerPoint PPT Presentation


  • 153 Views
  • Uploaded on

Comparing Economic Alternatives. Using time value of money to choose between two or more options. Examples of Engineering Decisions. Is it better to invest in an automated process or a manual process? Is it better to buy a high efficiency appliance or a low efficiency appliance?

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Comparing Economic Alternatives' - adara


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Comparing economic alternatives l.jpg

Comparing Economic Alternatives

Using time value of money to choose

between two or more options


Examples of engineering decisions l.jpg
Examples of Engineering Decisions

  • Is it better to invest in an automated process or a manual process?

  • Is it better to buy a high efficiency appliance or a low efficiency appliance?

  • Should we buy a new machine or keep the old one?


Economically different alternatives l.jpg
Economically Different Alternatives

  • One may require more initial investment than the other.

  • One may have smaller operating cost than the other.

  • One may last longer than the other.

  • One may have a different salvage than the other.

    They differ because they have different cash flows.

    Yet we must make a choice. Sometimes, we are allowed the choice of neither, but not always.


Example l.jpg
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?


Example cont d l.jpg
Example (cont’d)

  • How do we make a choice?

  • Recall the steps seen earlier.

    • Describe the cash flow of each alternative

    • Specify a Minimal Acceptable Rate of Return

    • Compute the equivalent NPW, NAW or FW

    • Select the alternative that has the greatest NPW, NAW or FW


Cash flows of first machine f l.jpg
Cash Flows of First Machine (F)

  • NPW costs =

  • 1100 - 100 (P/F, 0.15, 10) + 200 (P/A, 0.15, 10)

  • = 2079


Cash flows of second machine s l.jpg
Cash Flows of Second Machine (S)

  • NPW costs =

  • 1300 - 100(P/F, 0.15, 10) + 150(P/A, 0.15, 10)

  • = 2028


Which do we select l.jpg
Which do we select?

  • First machine:

    NPW costs =

    1100 - 100 (P/F, 0.15, 10) + 200 (P/A, 0.15, 10)

    = 2079

  • Second machine:

    NPW costs =

    1300 - 100(P/F, 0.15, 10) + 150(P/A, 0.15, 10)

    = 2028

  • Which do we select?

    • Select the second machine,with the lowest NPW of the costs.


Incremental analysis l.jpg
Incremental Analysis

Defender

Challenger

-

  • NPW =

  • -200 + 50(P/A, 0.15, 10)

  • = 51

NPW > 0, so accept the extra investment


Annual worth comparison l.jpg
Annual Worth Comparison

Defender

Challenger

-

  • NAW costs =

  • 1300 (A/P, 0.15, 10)

  • - 100(A/F, 0.15, 10) + 150

  • = 404.10

  • NAW costs =

  • 1100 (A/P, 0.15, 10)

  • - 100(A/F, 0.15, 10) + 200

  • = 414.25

Choose the Challenger (Second) with the Smaller Net Annual Cost


Naw of the increment l.jpg
NAW of the Increment

  • We can also use incremental analysis with NAW

  • Incremental Analysis: S - F:

    • NAW S-F: -200(A/P, 0.15, 10) + 50 = 10.15

  • Accept the extra investment of S over F?

    Accept the extra investment of S over F. Choose S.


Example broom versus vacuum l.jpg
Example: Broom versus Vacuum

  • 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%.

    • Do a present worth analysis.


Present worth broom versus vacuum l.jpg
Present Worth Broom versus Vacuum

  • NP Cost= 90 + 10(P/A, .15, 4)

  • = 118.55

  • NP Cost = 35


Example broom versus vacuum cont d l.jpg
Example: Broom versus Vacuum (cont’d)

  • Can we use the Present Worth Analysis?

  • If so, which is alternative best?

    • NPW of Broom = 35

    • NPW of Vacuum = 90 + 10(P/A, .15, 4) = 118.55

    • Select the broom?

    • No. The analysis here is wrong because the alternatives are compared over unequal periods.


Present worth broom versus vacuum15 l.jpg
Present Worth: Broom versus Vacuum

  • Select a common study period for each alternative. Use a study period of 4 years.

  • NPW = 90 +10(P/A,0.15,4)

  • = 118.55

NPW = 35 + 35(P/A,0.15,3) = 114.91


Npw when useful lives are not equal l.jpg
NPW when Useful Lives are Not Equal

  • We can also use the NPW method when lives are not equal.

    • Select a common study period for each alternative. Often we use the least common multiple of the lives.

    • Compute the NPW Benefits or NPW of Costs for each alternative for the study period.

    • Select the alternative that has the greatest NPW of Benefits or smallest NPW of Costs


Annual worth broom versus vacuum l.jpg
Annual Worth: Broom versus Vacuum

  • Select the broom, the alternative with the smallest net annual cost.

NAC = 90(A/P,0.15,4)+ 10

= 41.52/year

NAC = 35(A/P,0.15,1)

= 40.25 per yr.


Useful lives l.jpg
Useful Lives

  • Differences among alternatives occur in many forms.

  • The useful life of an asset is the time period during which it is kept in productive use in a trade or business.

  • In many cases, alternatives will not have the same useful lives.

  • In this case, the Annual Worth Method is often the best way to compare them.


Example with incomes l.jpg
Example with Incomes

  • A proposed project has three possible levels of investment with the annual costs and incomes given in the table below.

  • Select the best with the present worth and annual worth methods. MARR = 10%.


Income example with present worth l.jpg
Income Example with Present Worth

  • Comparing with the NPW method. Select the one with the greatest NPW over a 45-year study period.

  • NPW(A) =

    [-100(A/P, i, 9) + 20](P/A, i, 45) = 26.037

  • NPW(B) =

    [-85(A/P, i, 5) + 24](P/A, i, 45) = 15.554

  • NPW(C) =

    [-60(A/P, i, 5) + 18](P/A, i, 45) = 21.422

    Select A since it is the alternative with the greatest NPW


Example with incomes cont d l.jpg
Example with Incomes (cont’d)

  • Comparing with the NAW method

  • NAW(A) =

    -100(A/P, i, 9) + 20 = 2.64/year

  • NAW(B) =

    -85(A/P, i, 5) + 24 = 1.57/year

  • NAW(C) =

    -60(A/P, i, 5) + 18 = 2.17/year

  • Select A since it is the alternative with the greatest NAW


Summary l.jpg
Summary

  • Equal Useful Lives

    • Use the MARR for equivalence computations

    • Pick the best alternative using the NPW or NAW. Both methods give the same result.

  • Unequal Useful Lives

    • Use the MARR for equivalence computations

    • Select a common study period.

    • Pick the best alternative using the NPW or NAW. Both methods give the same result.

  • Incremental Analysis

    • Common components cancel out. Necessary for ROR comparisons.


Incremental analysis23 l.jpg
Incremental Analysis

1. Arrange (rank order) the feasible alternatives based on increasing initial investment.

2. Select the alternative with the smallest investment to be the defender (D).

3. From the remaining alternatives (not including the defender or those already eliminated) chose the challenger (C) as the alternative with the least initial investment.

4. Evaluate the increment of investment C – D by computing

NPW(C –D), NAW(C – D), ROR(C – D).

Accept of reject (C – D) based on the evaluation.

If (C – D) is accepted, delete D from the list, and make C the defender.

If (C – D) is rejected, delete C from the list.

If any alternatives remain, return to step 3, otherwise quit. The current defender is the best alternative.


ad