1. Chapter 7 Annual Worth Analysis
Section 7.2: Single
Section 7-3: Multiple
Section 7-4: Least Common Multiple of Lives Assumption
Section 7-6: Capital Recovery Cost
More frequent than FW&CW but not PW

2. Systematic Economic Analysis Technique
1. Identify the investment alternatives
2. Define the planning horizon
3. Specify the discount rate
4. Estimate the cash flows
5. Compare the alternatives
6. Perform supplementary analyses
7. Select the preferred investment

3. Annual Worth Analysis Section 7-2
Single Alternative

4. Annual Worth Method
converts all cash flows to a uniform annual series over the planning horizon using i=MARR
a popular DCF method
If AW>0, then it’s recommended.
Another way?

5. Example 7.1
A $500,000 investment in a surface mount placement machine is being considered. Over a 10-year planning horizon, it is estimated the SMP machine will produce net annual savings of $92,500. At the end of 10 years, it is estimated the SMP machine will have a $50,000 salvage value. Based on a 10% MARR and annual worth analysis, should the investment be made?
AW(10%) = -$500K(A|P 10%,10) + $92.5K
+ $50K(A|F 10%,10)
= $14,262.50>0 it’s recommended

6. Example 7.2
How does annual worth change over the life of the investment? How does annual worth change when the salvage value decreases geometrically and as a gradient series?

7. Linear
It’s concave

8. It’s called Internal rate of return (PW,FW,AW)
AW =0 when MARR = 13.80%
It’s called Internal rate of return (PW,FW,AW)

9. Annual Worth Analysis Section 7-3
Multiple Alternatives
ranking and incremental approaches

10. Example 7.4
Recall the example involving two alternative designs for a new ride at a theme park: Alt. A costs $300,000, has net annual after-tax revenue of $55,000, and has a negligible salvage value at the end of the 10-year planning horizon; Alt. B costs $450,000, has revenue of $80,000/yr., and has a negligible salvage value. Based on an AW analysis and a 10% MARR, which is preferred?
AWA(10%) = -$300,000(A|P 10%,10) + $55,000
= -$300,000( ) + $55,000 = $
AWB(10%) = -$450,000(A|P 10%,10) + $80,000
= -$450,000( ) + $80,000 = $
Analyze the impact on AW based on salvage values decreasing geometrically to 1¢ after 10 years.

12. AW(A) = AW(B) when MARR = 10.56%(IRR)
AW(B)>AW(A)
AW(A)>AW(B)

13. Example 7.5
For The Scream Machine alternatives (A costing $300,000, saving $55,000, and having a negligible salvage value at the end of the 10-year planning horizon; B costing $450,000, saving $80,000, and having a negligible salvage value), using an incremental AW analysis and a 10% MARR, which is preferred?
AWA(10%) = -$300,000(A|P 10%,10) + $55,000
= -$300,000( ) + $55,000 = $ > $0
(A is better than “do nothing”)
AWB-A(10%) = -$150,000(A|P 10%,10) + $25,000
= -$150,000( ) + $25,000 = $587.50> AWA
(B is better than A)
Conclusion: Prefer B

14. Example 7.6
If an investor’s MARR is 12%, which mutually exclusive investment alternative maximizes the investor’s future worth, given the parameters shown below?
Consider 3 scenarios: individual life cycles; least common multiple of lives; and “one-shot” investments

15. Example 7.6 (Continued)
Scenario 1: individual life cycles (Infinitely Long Horizon)
AW1(12%) = -$10,000(A|P 12%,3) + $ $5000(A|F 12%,3)
= $
AW2(12%) = -$14,500(A|P 12%,6) + $5000 +$2500(A|F 5%,6)
= $
AW3(12%) = -$20,000(A|P 12%,6) + $3000(A|G 12%,6)
= $
Is the result different from PW and FW when having diff. cycles?

16. Example 7.6 (Continued)- PW
Scenario 1: individual life cycles
PW1(12%) = -$10,000+ $5000 (P|A,12,3) + $5000(P|F 12%,3)
= -$10,000+ $5000 ( ) + $5000( )
= $5,568.05
PW2(12%) = -$14,500+ $5000 (P|A,12,6) + $2500(P|F 12%,6) = $7,321.57
PW3(12%) = -$20,000+ $3000(P|G 12%,6)
= $6,790.51
PW(LCML) = AW(LC)(P|A MARR,LCML)

17. Example 7.6 (Continued)- FW
Scenario 1: individual life cycles
FW1(12%) = -$10,000(F|P 12%,3)+ $5000 (F|A,12,3) + $5000
= -$10,000 ( )+ $5000 ( ) + $ = $7,822.7
FW2(12%) = -$14,500 (F|P 12%,6) + $5000 (F|A,12,6) + $ = $14,455.56
PW3(12%) = -$20,000 (F|P 12%,6) + $3000(P|G 12%,6) (F|P 12%,6) = $13,403.24

18. Example 7.6 (Continued) Identical results!
Scenario 2: least common multiple of lives
AW1(12%) = -$10,000(A|P 12%,6) + $ $5000(A|F 12%,6)
- $5000(A|P 12%,3)(A|P 12%,6)
= $
AW2(12%) = $
AW3(12%) = $
Identical results!

19. Observations (from Ch.4)
Evaluating alternatives based on a LCML planning horizon will yield the same recommendations as for an infinitely long planning horizon under the assumption of identical replacements over the planning horizon.
When we consider capitalized worth and capitalized cost in the next chapter, we will explicitly consider an infinitely long planning horizon, because we will be interested in knowing the magnitude of the present worth for an infinitely long planning horizon.
PW(LCML) = AW(LC)(P|A MARR,LCML)
The present worth over a LCML horizon equals the product of the annual worth for a life cycle (LC) and the P|A factor for a period of time equal to the LCML. (See the text for the calculations for Example 4.1.)
LC = life cycle

20. Example 7.6 (Continued)
Scenario 3: “one-shot” investments: take the longest one and assume zeros for the remaining years
AW1(12%) = {-$10,000 + [$5000(P|A 12%,3)
+ $5000(P|F 12%,3)]}(A|P 12%,6) = $
AW2(12%) = $
AW3(12%) = $
What is the difference between the one-shot and individual investments when calculating AW? (individual life= least common multiple approach only with AW)

21. Considering scenarios 1 and 2, is it reasonable to assume an investment alternative equivalent to Alt. 1 will be available in 3 years? If so, why was the MARR set equal to 12%?

22. Future Worths Assuming Investment 1 Is Not Repeated
One shot investment

23. Example 7.7
Three industrial mowers (Small, Medium, and Large) are being evaluated by a company that provides lawn care service. Determine the economic choice, based on the following cost and performance parameters:
Use AW analysis (multiple shots) to determine the preferred mower, based on a MARR of 12%.

24. Example 7.7 (Continued)
annually
(55-35)
AWsmall = -$1500(A|P 12%,2) + $20(1000) = $19,112.45
AWmed = -$2000(A|P 12%,3) + $25(1100) = $26,667.30
AWlarge = -$5000(A|P 12%,5) + $24(1200) = $27,412.95
What did we assume when solving the example?
Identical costs for the coming 30 years
Service will be needed for the coming 30 years.

25. Example 7.8
20,000 + SV 4 5
1500 AW

26. Example 7.8
Two years of service of the medium mower will have the following salvage value:
SVmed = $26,667.30(F|A 12%,2) - $27,500(F|A 12%,2)
+ $2000(F|P 12%,2) = $ (37.17% of its purchase price)
Given the length of the cycle is 5 years then,
SVlarge = $0

27. Compare investment alternatives over a common period of time
Principle #8
Compare investment alternatives over a common period of time

28. Capital Recovery Cost تكلفة الاستخدام السنوية
 is the annual equivalent of the capital cost. CR is often described from the bank's point of view and is a function of the MARR, i%, no. of years, cost price, and selling price.
e.g. selling a car

29. CFD for Capital Recovery Cost (CR).
P(A|P i,n)
F(A|F i, n) n
CR = P(A|P i,n) – F(A|F i, n) main formula
The relationship between the siniking fund factor and the capital recovery cost
Remember: (A|F i, n)=(A|P i,n) - i

30. Capital Recovery Cost Formulas
CR = P(A|P i,n) – F(A|F i, n)
CR = (P-F)(A|F i, n) + Pi
CR = (P-F)(A|P i,n) + Fi
Note:
CR VS Depreciation – Ch.9
The CR is used in Ch. 11 when finding the optimal replacement interval

31. Example 7.7
A $500,000 investment in a surface mount placement machine is being considered. Over a 10-year planning horizon, it is estimated the SMP machine will produce net annual savings of $92,500. At the end of 10 years, it is estimated the SMP machine will have a $50,000 salvage value. Based on a 10% MARR, what is the capital recovery cost?
P = $500, F = $50, i = 10% n = 10 yrs
CR = $500,000( ) - $50,000( ) = $78,237.50/year
CR = $450,000( ) + $500,000(0.10) = $78,237.50/year
CR = $450,000( ) + $50,000(0.10) = $78,237.50/year
Note: To be profitable, revenues must equal or exceed the capital recovery costs, so annual revenues ≥ CR.
Here, 92,5000>78,237.5 so it’s profitable!

32. Why not keeping the machine forever? Maintenance & operating costs
CR decreases with increasing values of n
The longer one keeps an asset, the less its overall annual cost of ownership
Why not keeping the machine forever?
Maintenance & operating costs
Less competitive

33. Example (source: The University of Texas at Austin)
Your company is planning to manufacture a new product which requires a machine that the company does not now own. The cost of the machine is $10,000, its life is 4 years, and its salvage value (at the end of the 4th year) is $1000. With this machine, the new profit from each product is $12. What annual production makes the investment worthwhile? The MARR is 10%.
I = $10,000 S = $1, Useful life =4 MARR= 10%
= $10,000(A/P,10%,4) – $1000(A/F, 10%, 4)
= $10,000(0.3155) – $1000(0.2155)
= $2939.5
To be profitable, revenues must equal or exceed the capital recovery costs. Let X be the number of products produced per year.
Revenues – CR = $12X – $ ( 0. Solving for X, we get X ( 245 units.

34. Pit Stop #7—No Time to Coast
True or False: Annual worth analysis is the most popular DCF measure of economic worth.
True or False: Unless non-monetary considerations dictate otherwise, choose the mutually exclusive investment alternative that has the greatest annual worth over the planning horizon.
True or False: The capital recovery cost is the uniform annual cost of the investment less the uniform annual worth of the salvage value.
True or False: If AW > 0, then PW>0, and FW>0.
True or False: If AW(A) > AW(B), then PW(A) > PW(B).
True or False: If AW (A) < AW(B), then AW(B-A) > 0.
True or False: If AW(A) > AW(B), then CW(A) > CW(B) and DPBP(A) < DPBP(B).
True or False: AW can be applied as either a ranking method or as an incremental method.
True or False: To compute capital recovery cost using Excel, enter =PMT(i%,n,-P,F) in any cell in a spreadsheet.
True or False: When using annual worth analysis with mutually exclusive alternatives having unequal lives, always use a planning horizon equal to the least common multiple of lives.

35. Pit Stop #7—No Time to Coast
True or False: Annual worth analysis is the most popular DCF measure of economic worth. FALSE
True or False: Unless non-monetary considerations dictate otherwise, choose the mutually exclusive investment alternative that has the greatest annual worth over the planning horizon. TRUE
True or False: The capital recovery cost is the uniform annual cost of the investment less the uniform annual worth of the salvage value. TRUE
True or False: If AW > 0, then PW > 0, and FW > 0. TRUE
True or False: If AW(A) > AW(B), then PW(A) > PW(B). TRUE
True or False: If AW (A) < AW(B), then AW(B-A) > 0. TRUE
True or False: If AW(A) > AW(B), then CW(A) > CW(B) and DPBP(A) < DPBP(B). FALSE
True or False: AW can be applied as either a ranking method or as an incremental method. TRUE
True or False: To compute capital recovery cost using Excel, enter =PMT(i%,n,-P,F) in any cell in a spreadsheet. TRUE
True or False: When using annual worth analysis with mutually exclusive alternatives having unequal lives, always use a planning horizon equal to the least common multiple of lives. FALSE