1. Engineering Economic Analysis
Chapter 5
PRESENT WORTH ANALYSIS
We will use all the stuff in Ch 1-4 from here on
Now we need to know how to convert a series of cash flows to another form that is desired. What form do want to compare mutually exclusive alternatives? That depends on our analysis technique. In this chapter we will compare the present worth's of alternatives. In the next chapter we will compare the equivalent uniform annual benefit of alternatives. And in chapter 7 we will compare alternatives based on their rate of return.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

2. Economic Criteria Projects are judged against an economic criterion.
Situation
Criterion
Fixed input
Maximize output
Fixed output
Minimize input
Neither fixed
Maximize difference
(output-input)
Remember in chapter one we talked about the different situations you could have to choose between when comparing alternatives
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

3. Economic Criteria Restated Present Worth Techniques
Situation
Criterion
Fixed input
Amount of capital available fixed
Maximize present worth of benefits
Fixed output
$ amount of benefit is fixed
Maximize present worth of costs
Neither fixed
Neither capital nor $ benefits are fixed
Maximize net present worth
(NPV)
For present worth analysis this translates into the following situations and criterion
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

4. Economic Criteria - Example Purchasing Building Space
Alt
Situation
Example
Criterion A
Fixed input
$150,000 max
Maximum square feet of building for the price B
Fixed output
20,000 ft2 building available
Negotiate for minimum cost/ft2 C
Neither fixed
15-20,000 ft2 required
Simultaneously negotiate for maximum building size & minimum cost/ft2
Here are examples of the three types of situations you could have
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

5. Applying Present Worth Techniques
Analysis period must be considered
Useful life of the alternative equals the analysis period
Alternatives have useful lives different from the analysis period
The analysis period is infinite, n = ¥
Present worth analysis is used to determine the present value of future money receipts and disbursements, especially things like income producing property or stocks and bonds. Choose option with “best” present worth. If you have two or more alternatives to consider, each has a time period associated with it that covers the useful life of the alternative. The problem covers an analysis period over which consequences are considered as the planning horizon. There are 3 situations then:
Useful life of all alternatives are the same
Useful life of alternatives are not the same
The analysis period is infinite
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

6. Useful Lives Equal the Analysis Period
In this type all alternatives have equal useful lives
The length of the useful lives are considered to be the planning horizon since:
Either this length was already the length of the planning horizon or,
This length is when you need to replace all alternatives, thereby starting over
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

7. Example 1
Two devices are being considered. Both cost $10,000 now, have useful lives of 5 years, and have no salvage value. Device A will result in $300 savings per year, for each of the next 5 years. Device B, will result in $400 savings in the 1st year and reduce by $50 each year. If the interest rate is 7%, which device should you choose?
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

8. Example 1
The first thing we need to do is determine what situation we are in
Since both devices cost $1000 now we are in a fixed input situation, so our goal should be to maximize the present worth of benefits.
Next we need to determine the planning horizon.
Since both devices last 5 years, our planning horizon is 5 years
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

9. Example 1
Now we need to look at the two cash flow diagrams so we can find the present worth of each.
We can see that device A is a uniform series of cash flows so we need to use the uniform series present worth equation to find the present worth. We also can see that device B is a derivation of the arithmetic gradient present worth that we saw in the last chapter so we know how to find the present worth of each.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

10. Example 1
So to find the present worth of device A we need to find the present worth of a series of 5 uniform payments at 7% interest. So we find the appropriate uniform series present worth factor, in orange, and find that the present worth of benefits for device A is $1,230
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

11. Example 1
So to find the present worth of device B we need to find the present worth the following CFDs. It’s the combination of a uniform series of 5 payments with an A of $400 with an arithmetic gradient with a G of $50 subtracted from it.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

12. Example 1
So to find the present worth of device B we need to find the present worth the following CFDs. So we find the appropriate uniform series present worth factor, in orange, and the appropriate arithmetic gradient present worth factor, in purple, and find that the present worth of benefits for device A is $1,
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

13. Example 1 Device Present Worth of Benefits A $1,230 B $1,257.65
Since our goal was to maximize the present worth of benefits we choose device B as the best alternative.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

14. Example 1: Excel
I like to make a section for each alternative in PW Analysis
I use NPV because I have no flows in year 0 and do not want to worry about the specific pattern (A or F, etc)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

15. Example 1: Excel
I use Conditional Formatting so that I can reuse my spreadsheet in the future to identify the highest PW without changing too much
Highlight the cells … Click on Conditional Formatting
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

16. Example 1: Excel Go to Top/Bottom Rules, More Rules
Setup the Rule to be Top 1 (to max PW) and change the Format of the Cell to something noticeable
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

17. Example 1: Excel
Since our goal was to maximize the present worth of benefits we choose device B as the best alternative.
Same result as not using Excel …
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

18. Example 2
An aqueduct is needed to provide water for the next 50 years. There are two plans available:
Alternative
Plan A
Build the full sized aqueduct now for $400 million B
Build a smaller aqueduct now for $300 million. Expand it in 25 years to be full sized for a cost of $350 million then.
If the interest rate is 6% which plan do we use?
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

19. Example 2 First we must decide what our goal is.
Since they both provide water for the next 50 years, we have fixed output so our goal should be to minimize the input or cost.
Next we must decide the analysis period
Since both provide water for the next 50 years, the analysis period is 50 years
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

20. Example 2
The present worth of plan A is $400 million, the present cost of the big aqueduct
The present worth of plan B is the $300 million now plus the $350 million in 25 years brought back to present dollars
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

21. Example 2 Alternative Present Worth of Cost A $400 million B
Since our goal is to find the alternative with the minimum present worth of cost we should choose plan B with a present worth of cost of $ million.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

22. Example 2: Excel
We are minimizing cost, so use conditional formatting to return the lowest value
Be careful about the signs of the answers returned from PV
Since PV returns the flow in the opposite direction as the input, we need to use –PV in the formulas
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

23. Uniform Annual Savings
Example 3
A company needs to decide between 2 scales that have useful lives of 6 years, which is the analysis period. Assume an 8% interest rate and we have the following information on the scales:
Scale
Cost
Uniform Annual Savings
Salvage Value 1
$2,000
$450
$100 2
$3,000
$600
$700
Which scale is preferred?
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

24. Example 3 First we need to determine our goal.
Since the alternatives have different costs and benefits our goal should be too maximize net present worth which is the present worth of the benefits minus the present worth of the costs
Next we need to determine the analysis period
The problem stated that the analysis period is 6 years
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

25. Example 3 Here is cash flow diagram for scale 1
We can see that the net present worth of scale 1 is made up of a current cost plus a uniform annual benefit and a salvage value benefit. To find the present worth of benefits scale 1 we need to take the uniform annual benefit back to present worth using the uniform series present worth factor and we need to bring back the salvage value back to present value using the single payment present worth factor. Since the cost is a current cost the present worth of costs is easy its just the current cost. Therefore the net present worth of scale 1 is given by the formula above.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

26. Example 3
Now to find the net present worth of scale 1 we need the appropriate value for the uniform series present worth factor for 6 payments and an interest rate of 8%, which is circled in orange, and the appropriate value for the single payment present worth factor for 6 periods in the future and an interest rate of 8% which is in purple. This gives us a net present worth for scale 1 of $
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

27. Example 3 Here is cash flow diagram for scale 2
We can see that the net present worth of scale 2 is made up of a current cost plus a uniform annual benefit and a salvage value benefit. Since the cash flow diagram looks exactly as it did for scale one other than the values of A, C, and S we can use the same equation and only change those values in it.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

28. Example 3
Now to find the net present worth of scale 2 we do the same things we did to find the net present worth of scale 1. This gives us a net present worth for scale 2 of $
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

29. Example 3 Scale Net Present Worth 1 $143.37 2 $214.94
Since our goal is to maximize net present worth we should choose scale 2 which has a net present worth of $
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

30. Example 3: Excel
I like to set up a spreadsheet with a section for Costs, Benefits and Salvage
Costs are always negative, Benefits and Salvage are always positive
I always Maximize PW
If I only have costs, I end up maximizing –costs, which is minimizing costs
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

31. Example 3: Excel
Then, I create a cash flow for each alternative for each year, using the inputs
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

32. Example 3: Excel
Finally, I can add the year 0 flows to the NPV of years 1 and later to get the present worth of each option
The interest rate is in Cell B4
Use conditional formatting here as well
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

33. Useful Lives Different From the Analysis Period
There could be a situation where one alternative has a useful life of 5 years and the other has a useful life of 10 years
What would our analysis period be in that case?
When alternatives have different useful lives we typically make our analysis period the least common multiple of the alternatives useful lives
Here it would be 10 years
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

34. Useful Lives Different From the Analysis Period
This isn’t a problem for the alternative that has a useful life of 10 years, but what do we do for the alternative that only had a useful life of 5 years.
Basically we assume that you can purchase an identical item at the end of the first items useful life
This can be done for multiple alternatives in the case that the least common multiple is not one of the alternatives useful life
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

35. Uniform Annual Savings
Example 4
A company needs to decide between 2 scales. Assume an 8% interest rate and we have the following information on the scales:
Scale
Cost
Uniform Annual Savings
Salvage Value
Useful Life 1
$2,000
$450
$100
10 years 2
$3,000
$600
$700
5 years
Which scale is preferred?
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

36. Example 4 First we need to determine our goal.
Since the alternatives have different costs and benefits our goal should be too maximize net present worth which is the present worth of the benefits minus the present worth of the costs
Next we need to determine the analysis period
Since the useful life of scale 1 is 10 years and the useful life of scale 2 is 5 years, the least common multiple is 10 years
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

37. Example 4
This cash flow diagram is just like the ones in the prior example except that we changed the number of years in the analysis period
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

38. Example 4
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

39. Example 4
For scale 2 we can see that there are an extra cost and an extra salvage value in the CFD as compared to scale 1
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

40. Example 4 .
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

41. Example 4: Excel Extend the template used in Example 3 for these flows
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

42. Useful Lives Different From the Analysis Period
What happens if one alternative has a useful life of 10 years and another has a useful life of 13 years.
The least common multiple is 130 years
Our assumptions for purchasing identical items at the end of the useful life is probably not valid that far in the future.
For this case we set the analysis period for something like 15 years and come up with terminal values for the alternatives at that time.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

43. Infinite Analysis Period Capitalized Cost
Capitalized cost is the present amount of money required at some interest rate to provide the project forever
Use the interest, A, each year to pay for the project
Maintain principal amount so that the fund never runs out
Some projects need to be maintained forever like
Roads
Dams
Graves
Capitalized cost is the present amount of money required to be invested at some interest rate to provide for the project forever. You use the interest each year to pay for the project, so you need to maintain the principal amount in the bank. We assume the cost each year is the same, so each year we withdraw the interest
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

44. Infinite Analysis Period Capitalized Cost
Year
Beginning Year Amount
Interest Earned
Withdrawal Amount 1 P iP 2
P + iP – iP = P …
So from this we can see that A, the interest earned each year, is i times P
429 P iP
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

45. Example 5
You need $50 per year for your grave maintenance after you die. If you earn 4%, how much do you need to set aside by one year before you die.
Since from the time you die on you can’t add money to this fund and you need to take care of the grave forever we have an infinite analysis period
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

46. Example 6
A pipeline is required to carry water for a city. It costs $8 million and is expected to last 70 years. If we assume that we will always need the pipeline, what is the capitalized cost if the interest rate is 7%?
What can we do to find the capitalized cost, P, here: A few things
We could take the current cost then spread all of the other future costs out over the 70 years until we need them
We could spread the current cost of 8 million and spread it over the 70 years until we need it again
We could find the effective interest rate over 70 years, and then we could add the current cost of 8 million and what would be a uniform series after that
We choose to find the effective interest rate.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

47. Example 6
A is
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

48. Multiple Alternatives
For multiple alternatives
Determine analysis period based on useful lives of alternatives
If least common multiple approach doesn’t work, set a reasonable analysis period
Find the present worth of all alternatives
Choose the best for your situation
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.

49. Assumptions in Solving Economic Analysis Problems
End-of-year convention (simplifies calculations)
Viewpoint (generally the firm)
Sunk costs (past has no bearing)
Borrowed money (consider investing only)
Effect of inflation (prices are not stable)
Income taxes (must be considered for realism)
There are some assumptions that we make to solve economic analysis problems
End of year convention
Remember tables and formulas assume cash flows occur at the end of the period
Spreadsheets can handle cash flows at the beginning or end, but be careful!!
Use the viewpoint of the whole company or larger entity
Sunk Costs – Ignore them!!!
Borrowed Money
We assume that money spent in problems is borrowed at the same rate and conditions given in the problem
No Inflation or Deflation…for now. Prices are stable
No income tax effects of consequences…for now
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.