Transcript CHAPTER VI
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ENGINEERING ECONOMY Fifth Edition Blank and Tarquin M Hill c Graw
CHAPTER VI
Annual Worth Analysis
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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Learning Objectives: AW over one project life cycle AW calculations Selecting Alternatives by Annual Worth AW analysis for a permanent investment
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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ENGINEERING ECONOMY Fifth Edition Blank and Tarquin M Hill c Graw
CHAPTER VI
6.1 AW over one project life cycle
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 Advantages and Uses of Annual Worth Popular Analysis Technique Easily understood – results are reported in $/time period Eliminates the LCM problem associated with the present worth method
Only have to evaluate one life cycle of a project
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 AW Calculations
General in nature such that:
AW = PW(A/P,i%,n) AW = FW(A/F,i%,n)
Convert all cash flows to their equivalent annual amounts
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 Repeatability Assumption
Given alternatives with unequal lives The assumptions are
:
1. The services so provided are needed forever 2. The first cycle of cash flows is repeated for successive cycles 3. All cash flows will have the same estimated values in every life cycle
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6. 1. One or More Cycles Cycle 1 Cycle 2 Cycle K Find the annual worth of any given cycle ($/period) Annualize any one of the cycles AW assumes repeatability of CF’s
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 6-year & 9-year Problem (Ex. 6.1)
See Example 6.1
Location A Location B First cost,$ -15,000 -18,000 Annual Lease cost -3,500 -3,100 Deposit return,$ 1,000 2,000 Lease term, years 6 9 Determine which location should be selected, if the MARR is 15% per year.
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 6-year & 9-year Problem (Ex. 6.1)
Need an 18-year study period for both 6-year Project 6-year Project 6-year Project
9-year Project 9-year Project
Means a lot of calculation time!
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 Example Consider a project with $3,000 annual operating cost and a $5,000 investment required each 5 years. i = 10% 0 1 2 3 4 5 A 1-5 = $3,000 $5,000 For one cycle
EAC = 3,000 + 5,000 (A|P, 10%, 5) = $4,319/yr Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 Multiple cycle..same result!
0 1 2 3 4 5 6 7 8 9 10 $5,000 A 1-10 = $3,000 $5,000
For two cycles EAC = 3000 + 5000 (1+(P|F, .10, 5))(A|P, .10, 10) = 3000 + 1319 = $4319/yr Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.1 AW Requirements
Similar to the Present Worth Method, AW analysis requires:
A discount rate before the analysis is started Estimates of the future cash flows
Estimate of the time period(s) involved
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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ENGINEERING ECONOMY Fifth Edition Blank and Tarquin M Hill c Graw
CHAPTER VI
6.2 AW Calculation
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.2 Capital Recovery and AW Values
Assume the potential purchase of any productive asset One needs to know or estimate:
Initial Investment P Estimated Future Salvage Value Estimated life of the asset N S Estimated operating costs and timing Operative interest rate – i%
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.2 Capital Recovery Cost
•Thus, management is concerned about the
equivalent annual cost of owing a productive asset.
•This cost is termed “Capital Recovery Cost” •CR is a function of {P, S, i%, and “n” } Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.2 Capital Recovery Cost (CR)
•CR = the equivalent annual worth of the
asset given: F N S
……….
0 1 2 3 N-1 N P 0 Capital Recovery (CR) is the annualized equivalent of the initial investment P 0 and the annualized amount of the future salvage value F n
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.2 Capital Recovery Cost – CR S
•Given:
F N
……….
0 1 2 3 N-1 N P 0
•Convert to:
F N
……….
0 1 2 3 N-1 N P 0 $A per year (CRC)
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.2 Capital Recovery Cost
COMPUTING CR FOR INVESTMENTS WITH SALVAGE VALUES: Method I - compute EAC of the original cost and subtract the EAC of the salvage value EAC = P (A|P, i, n) S (A|F, i, n) Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.2 More Traditional CR Approach
Method II – Subtract the salvage value from the original cost and compute the annual cost of the difference. Add to the interest that the salvage value would return each year, SV (i).
CR(i%)= (
P
-
S
) (A|P,
i
,
n
) +
S
(
i
)
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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ENGINEERING ECONOMY Fifth Edition Blank and Tarquin M Hill c Graw
CHAPTER VI
6.3 Selecting Alternatives by Annual Worth
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.3 Alternatives by Annual Worth Mutually Exclusive Analysis
If pure cost situation – select min. cost alternative
If mixed costs and revenues – select the max. AW (i%) alternative Single Alternative
Accept if AW positive, else reject
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.3 Example 6.3
Cash Flow Diagram is: S = +$1,500 A + = $1,200/yr 1 2 3 4 5 P=-23,000 -$650 -$700 -$750 -$800 -$850
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6.3 Example 6.3
The Capital Recovery component is: S = +$1,500 1 2 3 4 5 P=-23,000 CR(10%) = -23,000(A/P,10%,5) + 1,500(A/F,10%,5) = -$5,822
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.3 Example 6.3
Revenue – Op Costs are: A + = $1,200/yr 1 2 3 4 5 $650 $700 $750 $800 $850
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.3 Example 6.3
Cost Revenue component is seen to equal (costs treated as positive values): = 550 – 90.50
= $459.50
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.3 Example 6.3
Total Annual worth (CR + Cost/Rev)
CR(10%) = -$5,822
Revenue/Cost Annual amount: $459.50
AW(10%) = -$5,822+$459.50
AW(10%) = $5,362.50
This amount would be required to recover the investment and operating costs at the 10% rate on a per-year basis
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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ENGINEERING ECONOMY Fifth Edition Blank and Tarquin M Hill c Graw
CHAPTER VI
6.4 AW of a Perpetual Investment
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.4 AW of a Perpetual Investment
EAC of a perpetual investment If an investment has no finite cycle, it is called a perpetual investment then EAC is P times i.
. If “P” is the present worth of the cost of that investment,
Remember: P = A/i From the previous chapter
EAC=A =
P* i
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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6.4 AW of a Perpetual Investment
Example 6.6
One person received a bonus of $10,000. If she deposits it now at an interest rate of 8% per year, how many years must the money accumulate before she can withdraw $2000 per year forever?
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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End of Chapter 6 Lecture Set
Blank & Tarquin: 5th Edition. Ch. 6 Authored by: Dr. Don Smith, Texas A&M University.
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