Engineering Economics GENG 315 Engineering Economics GENG 315

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Engineering Economics- GENG 315 Engineering Economics GENG 315 CHAPTER 6 a Comparing Alternatives for

Engineering Economics- GENG 315 Engineering Economics GENG 315 CHAPTER 6 a Comparing Alternatives for an Engineering Project

Engineering Economics- GENG 315 Objectives To develop and demonstrate the economic analysis and comparison

Engineering Economics- GENG 315 Objectives To develop and demonstrate the economic analysis and comparison of mutually exclusive design alternatives for an engineering project.

Engineering Economics- GENG 315 Comparing Alternatives for an Engineering Project q In engineering, the

Engineering Economics- GENG 315 Comparing Alternatives for an Engineering Project q In engineering, the selection of one feasible design alternative excludes the choice of any of the other alternatives. These alternatives are called mutually exclusive alternatives. q Typically, the alternatives under considerations require investing different amounts of capital, revenues and costs and it also may have different useful lives. q Because economic outcomes should be proportional to the to determine the preferred alternative (step 6). amount of investment, an analysis must be performed among the alternatives (step 5 of the seven step procedure of an engineering economy study) (refer to Chapter 1)

Engineering Economics- GENG 315 Engineering Economy and The Design Process Procedure of the Engineering

Engineering Economics- GENG 315 Engineering Economy and The Design Process Procedure of the Engineering Economic analysis Activities of the engineering design process

Engineering Economics- GENG 315 q In this study, our comparison between Mutually Exclusive Projects

Engineering Economics- GENG 315 q In this study, our comparison between Mutually Exclusive Projects is based on economical consideration alone. Basic concepts for comparing alternatives The decision of choosing the best Mutually Exclusive projects is based on the following rule: The alternative that requires the lowest amount of initial investment of capital and produces satisfactory functional results will be chosen and will be called (Base Alternative) unless the incremental capital associated with the other alternative that have a higher initial investment can be justified with respect to its incremental benefits. q. Therefore, before additional initial investment of capital is invested in the other investment, it must be shown that the extra benefits obtained by investing additional capital are better than those that can be obtained from investment of the same capital elsewhere at the MARR.

Engineering Economics- GENG 315 Investment and cost alternatives projects q Investment Alternative projects are

Engineering Economics- GENG 315 Investment and cost alternatives projects q Investment Alternative projects are those with initial investment of capital that produce positive cash flows (inflow) from increased revenue, savings through reduced costs or both. Ø For Investment Alternatives projects, the PW of all cash flows must be positive, at the MARR, to be attractive. Select the alternative with the largest PW. q Cost Alternative projects are those with initial investment of capital that produce negative cash flows (Outflow), except for a possible cash flow from disposal of assets at the end of the project useful life (Market value). Ø For Cost Alternatives projects, the PW of all cash flows must be negative, at the MARR, to be attractive. Select the alternative with the largest PW (smallest in absolute value in terms of Cost).

Engineering Economics- GENG 315 Example: Investment Alternative projects q The following example involves an

Engineering Economics- GENG 315 Example: Investment Alternative projects q The following example involves an Investment project, where A and B are two mutually exclusive investment alternatives with estimated net cash flows as shown. The useful life of each alternative in this example is four years. The MARR =10% per year. Alternative A Alternative B A = $ 22, 000 0 1 2 3 Alternative A = $ 26, 225 0 4 $60, 000 1 2 3 4 $73, 000 Alternative B – Alternative A (year by year) A = $ 4, 225=(26, 225 -22, 000) 0 1 2 3 4 $13, 000=(73, 000 -60, 000) Capital investment Annual revenues less expenses A B -$ 60, 000 -$ 73, 000 22, 000 26, 225 Cash flow diagrams for alternatives A and B and their difference The cash flow diagrams for alternatives A and B and for the year-by-year differences between them are shown above.

q The present worth values for projects A and B are: PW = Inflow

q The present worth values for projects A and B are: PW = Inflow - Outflow PW (10%) of project A = 22, 000 ( P/A, 10%, 4) - 60, 000 PW (10%) of project A = (22, 000 X 3. 1699) - 60, 000 = 9, 738 PW (10%) of project B = 26, 225 ( P/A, 10%, 4) - 73, 000 PW (10%) of project B = (26, 225 X 3. 1699) - 73, 000 = 10, 131 q PW of B - PW of A = $ 10, 131 - $ 9, 738 = $ 393 q Since the initial investment of A = 60, 000 is less than the initial investment of B = 73, 000 and its PW > 0 at i = MARR, So A is the (Base Alternative) and would be selected, unless Incremental/additional Capital associated with B ($ 13, 000) is justified.

Engineering Economics- GENG 315 q In this case, B is preferred to A, because

Engineering Economics- GENG 315 q In this case, B is preferred to A, because B has a greater PW value. Hence, the extra benefits obtained by increasing the additional $ 13, 000 of capital in B have a PW of $ 393. This also can be obtained from the following: PW (10%) of B-A = Inflow - Outflow PW (10%) of B-A = (26, 225 -22, 000) (P/A, 10%, 4) - (73, 000 -60, 000) = PW (10%) of B-A = 4, 225 (3. 1699) - 13, 000 = $ 393 Thus the additional capital invested in B is justified.

Engineering Economics- GENG 315 Example: Cost Alternative projects q The following example involves a

Engineering Economics- GENG 315 Example: Cost Alternative projects q The following example involves a Cost project, where A and B are two mutually exclusive cost alternatives with estimated net cash flows as shown. The useful life of each alternative in this example is three years. The MARR =10% per year. Alternative C 1 0 $38, 100 $380, 000 2 Alternative D 3 1 0 $39, 100 $ 26, 000 2 A = $27, 400 $40, 100 $415, 000 Alternative D – Alternative C $10, 700 (year by year) $ 12700 + 26, 000 = 38700 $11, 700 0 1 $35, 000 2 3 Alternative 3 End of year C D 0 -$ 380, 000 -$ 415, 000 1 -38, 100 -27, 400 2 -39, 100 -27, 400 3 -40, 100 -27, 400 0 26, 000 3 (Market value) Cash flow Diagrams for alternatives C and Alternative D and their difference The cash flow diagrams for alternatives C and D and for the year-by-year differences between them are shown above.

q The present worth values for projects A and B are: PW = Cash

q The present worth values for projects A and B are: PW = Cash flow – Initial Investment PW (10%) of project C = - 38, 100(P/F, 10%, 1) - 39, 100(P/F, 10%, 2) - 40, 100(P/F, 10%, 3) - 380, 000 = PW (10%) of project C = - 38, 100(0. 9901) - 39, 100(0. 9804) - 40, 100(0. 9709) - 380, 000 = PW (10%) of project C = - 34, 636. 71 - 32, 312. 24 - 30, 127. 13 - 380, 000 = - 477, 077 PW (10%) of project D = - 27, 400(P/A, 10%, 3) + 26, 000(P/F, 10%, 3) - 415, 000 = PW (10%) of project D = - 27, 400(2. 4869) + 26, 000(0. 7513) - 415, 000 = PW (10%) of project D = - 68, 141. 06 + 19, 533. 8 - 415, 000 = - 463, 607. 26 q PW of D – PW of C = - 463, 607. 27 – (- 477, 077) = 13, 470 q C has lower capital investment (380, 000<415, 000) so it is the Base Alternative. However alternative D is preferred to C because it has the less negative PW (minimizes cost). q The lower annual expenses obtained by investing the additional 415, 000 -380, 000=35, 000 in D have PW = 13, 470. This can also be obtained from the cash flow diagram of the difference between the

Engineering Economics- GENG 315 q Alternative C which has the lesser capital investment is

Engineering Economics- GENG 315 q Alternative C which has the lesser capital investment is automatically the (Base Alternative) and would be selected unless the additional capital associated with alternative D (415, 000 -380, 000=$35, 000) is justified. q With the greater capital investment 35, 000, alternative D has lower annual expenses. Otherwise D will not be a feasible alternative. q Note that the difference between two feasible cost alternatives is an investment alternative.

Engineering Economics- GENG 315 Ensuring a comparable basis q Differences among mutual exclusive alternatives

Engineering Economics- GENG 315 Ensuring a comparable basis q Differences among mutual exclusive alternatives may exist. Typical examples of these differences are the useful lives, capital, revenues, costs, performance capabilities, output quality, or others. q To ensure a comparable basis for the analysis of such alternatives, the economic impact of these differences, from the perspective of the firm, must be included in the cash flow estimates and the analysis method. q Two rules were given in Chapter 2 to facilitate the correct analysis of mutually exclusive projects without consideration to the time value of money.

Engineering Economics- GENG 315 Ensuring a comparable basis These two rules are repeated below

Engineering Economics- GENG 315 Ensuring a comparable basis These two rules are repeated below without consideration to the time value of money : Rule 1 (Variable yield) : When revenues and other economic benefits are present and vary among alternatives, consider only the profit, and choose the alternative that maximizes overall profitability, even if the cost is high. That is select the alternative that has the greatest positive equivalent worth at i =MARR and satisfy all project requirements. Rule 2 (Constant yield) : When revenues and other economic benefits are not present or constant among the alternatives, consider only the cost and choose the alternative that minimizes the total cost, even if the profitability is low. That is, select the alternative that has the least equivalent worth at i =MARR and satisfy all project requirements.

Engineering Economics- GENG 315 Investment alternative example Use a MARR of 10% and useful

Engineering Economics- GENG 315 Investment alternative example Use a MARR of 10% and useful life of 5 years to select between the investment alternatives below. Alternative A B Capital investment $100, 000 $125, 000 Annual revenues less expenses $34, 000 $41, 000 PW = Cash flow – Initial Investment Both alternatives are attractive, but Alternative B provides a greater present worth, so is better economically.

Engineering Economics- GENG 315 Cost alternative example Use a MARR of 12% and useful

Engineering Economics- GENG 315 Cost alternative example Use a MARR of 12% and useful life of 4 years to select between the cost alternatives below. Alternative C D Capital investment $80, 000 $60, 000 Annual expenses -$25, 000 -$30, 000 PW = Cash flow – Initial Investment Alternative D costs less than Alternative C, it has a greater PW, so is better economically.

Engineering Economics- GENG 315 Practice Problem Your local foundry is adding a new furnace.

Engineering Economics- GENG 315 Practice Problem Your local foundry is adding a new furnace. There are several different styles and types of furnaces, so the foundry must select from among a set of 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 should be chosen? Furnace Initial Investment F 1 F 2 F 3 $110, 000 $125, 000 $138, 000 Useful life 10 years Total annual expenses $53, 800 $51, 625 $45, 033

(Table A-4 Present Value Interest with Annuity) Engineering Economics- GENG 315 Solution Using a

(Table A-4 Present Value Interest with Annuity) Engineering Economics- GENG 315 Solution Using a MARR of 15%, the PW is shown for each of the three alternatives in the table. PW = Inflow – Outflow PW (F 1) = -53, 800 (P/F, 15%, 10) – 110, 000 = (-53, 800 x 5. 0188) – 110, 000 = -$380, 010 PW (F 2) = -51, 625 (P/F, 15%, 10) – 125, 000 = (- 51, 625 x 5. 0188) – 125, 000 = -$ 384, 094 PW (F 3) = -45, 033 (P/F, 15%, 10) – 138, 000 = (- 45, 033 x 5. 0188) – 138, 000 = -$ 364, 010 Initial Investment Useful life Total annual expenses Present Worth @ 15% F 1 $110, 000 10 years $53, 800 -$380, 010 Furnace F 2 $125, 000 10 years $51, 625 -$384, 094 F 3 $138, 000 10 years $45, 033 -$364, 010 The largest value is -$364, 010, indicating that Furnace F 3 is the best alternative.