CS代考 Introduction to Engineering Economy

Introduction to Engineering Economy

Topic 8: Comparison and Selection Among Alternatives
Engineering Economics

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Types of mutually exclusive alternatives (MEAs)
Internal rate of return (IRR) method
Comparing and selecting MEAs in consideration of
the study (analysis) period
repeatability of the projects

Topic 8: Comparison and Selection of Alternatives

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Mutually Exclusive Alternatives (MEAs)
MEAs are investment projects that serve the same purpose or perform the same tasks with comparable performance or quality. Therefore only one of them will be selected.
The alternatives may have different initial investments and their annual revenues and costs may vary.
The CBA criteria and basic methods introduced in Topics 6 to 8 provide the basis for economic comparison of the alternatives.
The base alternative, an alternative that requires the minimum investment of capital and produces satisfactory functional results, will be chosen unless
another alternative having a larger investment can be justified with respect to its incremental benefits, i.e. its incremental benefits > its incremental capital costs.

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Two Basic Types of Alternatives
Investment Alternatives
Alternatives with initial capital investment that produces positive cash flows from increased revenue, savings through reduced costs, or both
Cost Alternatives
Alternatives with all negative cash flows, except for a possible positive cash flow from disposal of assets (salvage value or market value) at the end of the project’s useful life
For investment alternatives to be attractive, the PW of all cash flows must be positive at the MARR.
Select the alternative with the largest PW.
For cost alternatives, the PW of all cash flows will be negative.
Select the alternative with the largest PW (smallest in absolute value).

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All rights reserved.
Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Example of Investment Alternatives
Use a MARR of 10% and useful life of 5 years to select between the investment alternatives below.
Alternative
Capital investment $100,000 $125,000
Annual revenues less expenses (net revenue) $34,000 $41,000

Both alternatives are attractive, but Alternative B provides a greater present worth (PWB > PWA). Therefore, it is better economically (based on profit maximization).
PWA = ‒ $100,000 + $34,000(P/A, 10%, 5) = $28,894
PWB = ‒ $125,000 + $41,000(P/A, 10%, 5) = $30,431

Copyright ©2012 by , Inc.
Upper Saddle River, 07458
All rights reserved.
Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Interest rate = 10%

Copyright ©2012 by , Inc.
Upper Saddle River, 07458
All rights reserved.
Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Example of Cost Alternatives
Use a MARR of 12% and useful life of 4 years to select between the cost alternatives below.
Alternative
Capital investment $80,000 $60,000
Annual expenses $25,000 $30,000

Alternative D has a greater PW (PWD > PWC) because it costs less than Alternative C. Therefore, Alternative D is better economically (based on cost minimization).
PWC = ‒ $80,000 ‒ $25,000(P/A, 12%, 4) = ‒ $155,925
PWD = ‒ $60,000 ‒ $30,000(P/A, 12%, 4) = ‒ $151,110

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Interest rate = 12%

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Figure 6-1 in textbook Cash-Flow Diagrams for Investment Alternatives A and B and Their Difference

Alternative A is the base alternative and would be selected unless the additional capital associated with Alternative B ($13,000) is justified.
= ‒$73,000 + $26,225 (P/A, 10%, 4)
The PW of the extra net benefits
obtained by investing the
additional $13,000 of capital
$10,133 ‒ $9,740 = $393
PWB-A(10%)
= ‒$13,000 + $4,225 (P/A, 10%, 4)
= ‒$60,000 + $22,000 (P/A, 10%, 4)
Alternative B should be chosen as it maximizes profit.

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.

Figure 6-2 in textbook Cash-Flow Diagrams for Cost Alternatives C and D and Their Difference

Differences in cash flows (Alternatives D – C):
N=0: ‒$(415,000‒380,000) = ‒$35,000
N=1: ‒$(27,400‒$38,100) = $10,700
N=2: ‒$(27,400‒$39,100) = $11,700
N=3: ‒$(27,400‒$40,100) + $26,000 = $38,700
PW(10%)C = ‒$477,077
Alternative D should be chosen as it minimizes costs.
Alternative C is the base alternative and would be selected unless the additional capital associated with Alternative D ($35,000) is justified.
PW(10%)D = ‒$463,607
PWD-C(10%)
= ‒$463,607 ‒ (‒$477,077)
You may derive PWD-C(10%) based on the annual cash flow differences. But an easier method is the one below:
Discrepancy may arise because the results are derived from a different interest table.

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.

Your local foundry is adding a new furnace. There are several different styles and types of furnaces, and the foundry will select from three mutually exclusive alternatives. Initial capital investment and annual expenses for each alternative are given in the table below. None have any market value at the end of its useful life. Using a MARR of 15%, which furnace will you choose?
Investment (I) $110,000 $125,000 $138,000
Useful life (N) 10 years 10 years 10 years
Total annual expenses (A) $53,800 $51,625 $45,033

Exercise on Cost Alternative
PW = ‒I ‒ A(P/A, 15%, 10)
= ‒I ‒ A(5.0188)
For cost alternatives, select the alternative with the largest PW (smallest in absolute value)  F3

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Internal Rate of Return (IRR) Method
When we introduced MARR, the rate of profit of a project is usually measured by the Internal Rate of Return (IRR).
The IRR method is the most widely used method for conducting rate of return analyses.
The IRR is the interest rate that equates the equivalent worth of the cash inflows (revenue, R) to the equivalent worth of the cash outflows (expenses, E) of an alternative.
It is sometimes referred to as the breakeven interest rate.
IRR is i’% that satisfies the following equation in which the subscript t stands for year t and N is the project life or study period .

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.

Evaluation criterion for accepting or rejecting an investment project:
If a project’s IRR ≥ MARR, then the project is acceptable.
If a project’s IRR < MARR, then it should be rejected. Refer to Section 7.3.1 and Example 6 in Self-study Notes Unit 7 for the derivation of IRR using Excel and linear interpolation. Internal Rate of Return (IRR) Method Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. Challenges in Applying the IRR Method It is computationally difficult with only a simple calculator. Multiple IRRs can be found in some rare cases in which apart from period 0, some subsequent periods also have negative net cash flows. Possible inconsistency between the IRR and PW criteria, i.e. a project with a higher IRR may have a lower PW than another project with a lower IRR, and vice versa.  IRR method must be carefully applied and interpreted.  We should not directly compare the IRRs of two or more mutually exclusive alternatives (MEAs). Refer to Figure 7.6 and its explanation in Self-study Notes Unit 7 for possible inconsistency between IRR and PW criteria. Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. Using IRR Method in the Presence and Absence of Capital Rationing In the absence of capital rationing, all investment projects whose IRRs ≥ MARR are economically justified and worth-investing. When there is capital rationing, or when we are considering mutually exclusive alternatives that serve the same purpose, then we should use the following method to compare MEAs. Incremental investment analysis using the IRR method Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. This method includes the following steps: Include only the alternatives that satisfy the IRR criterion: IRR ≥ MARR of 12%  economically justified. Arrange the alternatives in ascending order of their capital investment: starting from the one with the smallest amount  project A is the base alternative. Given that the base alternative is acceptable, we conduct the incremental analysis to decide whether moving on to the next larger investment alternative is economically justified. IRR 12.99% 12.04% 13.48% 14.99% 14.61% MARR = 12% Incremental Investment Analysis Using the IRR Method Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. On an Excel spreadsheet: 4. Repeat the previous step to decide whether it is justified to move from Alternative A to Alternative C. Accept the move to the next larger investment alternative as long as IRR of the incremental analysis ≥ MARR. Otherwise, drop that alternative and consider the next larger one. The largest-capital alternative that satisfies this IRR rule will be selected. = IRR(B5:B13) Incremental Investment Analysis Using the IRR Method The IRR derived for incremental cash flows (from Alternative A to Alternative B) is ‒20.11%  Alternative B should be rejected. Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. Exercise on Incremental Analysis Alt. A Alt. B Alt. B-Alt. A Initial cost $25,000 $35,000 $10,000 Net annual income $7,500 $10,200 $2,700 IRR on total cash flow 15% 14% 11% Which alternative is preferred using a 5-year study period and MARR=10%? Q1) Are both Alternatives A and B acceptable? Why? Q2) Will it be a correct decision to choose Alternative A based on its larger IRR value? Q3) Which of the MEAs will you choose based on the incremental analysis using IRR? Verify your answer with the PW method. Alternative B should be chosen.  choose B PWA = ‒$25,000 + $7,500 (P/A, 10%, 5) = $3,433 PWB = ‒$35,000 + $10,200 (P/A, 10%, 5) = $3,668 Their IRR > MARR

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.

Interest rate = 10%

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
The Study (Analysis) Period
A study period (or planning horizon) is the time period over which MEAs are compared, and it must be appropriate for the decision situation.
MEAs that have equal lives: Their study period = their useful lives
MEAs that have unequal lives: At least one life span does not match the study period. Such MEAs are handled in the following 2 ways:
With repeatability assumption:
The study period is either indefinitely long or equal to the LCM of the life spans of the MEAs.
When each MEA is repeated, its cash flows will remain the same in succeeding life spans (replacements).
With Co-terminated assumption:
We use a finite and identical study period for all MEAs.
Necessary adjustments will be made to the cash flows of individual MEA to make them consistent with the study period. (The necessary adjustments will be specified in the case you analyse.)

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Comparing MEAs With Equal Lives
When their useful lives are equal, the MEAs can be compared directly based on their equivalent worth (PW, FW, or AW) calculated using the MARR. The decision will be the same regardless of which method you use.
For example, select from among the following MEAs using a MARR of 12%.
Investment Alternatives
Capital investment (I) $150,000 $85,000 $75,000 $120,000
Annual revenues ( ) $28,000 $16,000 $15,000 $22,000
Annual expenses ( ) $1,000 $550 $500 $700
Salvage Value (S)
(EOL Market Value) $20,000 $10,000 $6,000 $11,000
Life (years) 10 10 10 10

$27,000 $15,450 $14,500 $21,300

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Interest rate = 12%

Calculations in this example are based on a different interest table in the textbook where (P/A, 12%, 10) = 5.6502.

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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Selecting the best alternative using present worth method:
($ signs are omitted from the following calculations.)
Based on PW method, select Alternative A to maximize profit.

Do the analysis again using annual worth method.

= $8,995(A/P, 12%, 10)
= $5,516(A/P, 12%, 10)
= $8,860(A/P, 12%, 10)
= $3,891(A/P, 12%, 10)
Comparing MEAs With Equal Lives
Are the results consistent?
Calculations in this example are based on a different interest table in the textbook.
For investment alternatives, select the alternative with the largest PW, AW or FW  A.

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Upper Saddle River, 07458
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Engineering Economy, Fifteenth Edition
By William G. Sullivan, Elin M. Wicks, and C.
Comparing MEAs With Equal Lives
Example 6-2 in the textbook:
A company is choosing among the following plastic-molding presses with the same output capacity. All relevant information of these MEAs is given below. Assume that they do not have market value at the end of their useful life and their output justifies the installation of one of them. For a MARR of 10%, which press should be chosen?
Cost Alternatives
P1 P2 P3 P4
Capital investment $24,000 $30,400 $49,600 $52,000
Useful life (years) 5 5 5 5
Annual expenses
Power 2,720 2,720 4,800 5,040
Labor 26,400 24,000 16,800 14,800
Maintenance 1,600 1,800 2,600 2,000
Property taxes & insurance 480 608 992 1,040
Total annual expenses $31,200 $29,128 $25,192 $22,880

Comparing MEAs With Equal Lives
Calculations in this example are based on a different interest table in the textbook.
For cost alternatives, select the alternative with the largest PW, AW, or FW (smallest in absolute value)  P4

Capital Investment $3,500 $5,000
Annual Net Cash Flow $1,255 $1,480
Useful Life (years) 4 6
MARR = 10%

When the repeatability assumption is applicable, the comparison of MEAs is very simple. We can use the following methods.
Setting the study period equal to the LCM of their useful lives and use PW, FW, or AW method
Setting an infinite study period and use the AW method
Comparing MEAs With Unequal Lives
Example of investment alternatives:

Their cash flow diagrams:

Study period = LCM(4, 6) years =12 years
Comparing MEAs With Unequal Lives

For each useful life of alternative A:
PWA1 = PWA2 = PWA3
= $3,500 + $1,255(P/A,10%,4)
= $3,500 + $1,2553.170 = $478.35
For the study period of 12 years:
PWA(10%) = PWA1 + PWA2(P/F,10%,4) + PWA3(P/F,10%,8)
= $478.35[1+(P/F,10%,4)+(P/F,10%,8)]
= $478.35[1 + 0.6830 + 0.4665]
For each useful life of alternative B:
PWB1 = PWB2
= $5,000 + $1,480(P/A,10%,6)
= $5,000 + $1,4804.355 = $1,445.4
For the study period of 12 years:
PWB(10%) = PWB1 + PWB2(P/F,10%,6)
= $1,445.4[1 + (P/F,10%,6)]
= $ 1,445.4[1 + 0.5645] = $2,261

Calculations in this example are based on a different interest table in the textbook.
PW(10%)A = $1,028 < $2,261 = PW(10%)B Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. Setting an infinite study period assuming that the projects keep repeating forever, use the AW method. Comparing MEAs With Unequal Lives AWA(10%) = $3,500 (A/P,10%,4) + $1,255 = $3,500  0.3155 + $1,255 = $150.75 AWB(10%) = $5,000 (A/P,10%,6) + $1,480 = $5,000  0.2296 + $1,480 AW(10%)A = $151 < $332 = AW(10%)B Which method is better, 1 or 2? Calculations in this example are based on a different interest table in the textbook. Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. When the repeatability assumption is not applicable, we adopt the co-terminated assumption to select an appropriate study period. This is most often used in engineering practice because product life cycles are becoming shorter. Comparing MEAs With Unequal Lives When the useful life of an alternative is shorter than the study period, we may do the following to the cash flows. Cost alternatives Contracting or leasing for remaining years may be appropriate Repeat part of the useful life and use an estimated market value to truncate (i.e. dispose of the asset before the end of its useful life) Investment alternatives Cash flows reinvested at the MARR at the end of the study period Replace with another asset, with possibly different cash flows, to fill the gap between the useful life and the study period Copyright ©2012 by , Inc. Upper Saddle River, 07458 All rights reserved. Engineering Economy, Fifteenth Edition By William G. Sullivan, Elin M. Wicks, and C. Truncate the alternative at the end of the study period and use an estimated market value. When the repeatability assumption is not applicable, we adopt the co-terminated assumption to select an appropriate study period. This is most often used in engineering practice because product 程序代写 CS代考加微信: powcoder QQ: 1823890830 Email: powcoder@163.com

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