5.1 Comparing Alternatives ME 005-ECE21S2 - Engineering Economy PDF

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11/11/21, 2:57 PM

5.1: Comparing Alternatives: ME 005-ECE21S2 - Engineering Economy

5.1: Comparing Alternatives ▶ METHODS OR PATTERNS IN COMPARING ALTERNATIVES The Rate of Return on Additional Investment Method Determine the rate of return for each alternative. Rate of return on additional investment = If the rate of return on additional investment is satisfactory, then, the alternative requiring a bigger investment is more economical and should be chosen. The Annual Cost (AC) Method Determine the annual cost of the alternatives including interest on investment. The alternative with the least annual is chosen. This method applies only to the alternatives which have a uniform cost data for each year and a single investment of capital at the beginning of the first year of the project life. The Equivalent Uniform Annual Cost (EUAC) Method All cash flows (irregular or uniform) must be converted to an equivalent uniform annual cost. The alternative with the least EUAC is chosen. The Present Worth Cost (PWC) Method Determine the present worth of the net cash outflows. The alternative with the least PWC is chosen. The Capitalized Method Determine the capitalized cost of all the alternatives. Capitalized cost = first cost + present worth of all perpetual operation and maintenance + present worth of cost of all perpetual replacement The alternative with the least capitalized cost should be chosen. Payback (Payout) Period Method Determine the payback period of each alternative. Payout period(years) = https://tip.instructure.com/courses/26120/pages/5-dot-1-comparing-alternatives?module item id=1936843

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5.1: Comparing Alternatives: ME 005-ECE21S2 - Engineering Economy

The alternative with the shortest payback period is chosen.

To illustrate the above principles please refer to the sample problems below. Sample Problem no. 35 A company is considering two types of equipment for its manufacturing plant. Pertinent data are as follows: Type A

Type B

First cost

P200,000

Annual operating cost

32,000

Annual labor cost

P300,000 24,000

50,000

Insurance and property taxes

32,000

3%

3%

Payroll taxes

4%

Estimated life

10

4% 10

If the minimum required rate of return is 15%, which equipment should be selected. Solution By the rate of return on additional investment ROR =

(100%)

Type A Annual costs: Depreciation =

=

Operation

=

P9,850 = P32,000

Labor Payroll taxes = P50,000(0.04) Taxes and insurance = P200,000(0.03)

= P50,000 = P2,000 = P6,000

Total annual cost

= P99,850

Type B Annual costs: https://tip.instructure.com/courses/26120/pages/5-dot-1-comparing-alternatives?module item id=1936843

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Depreciation =

5.1: Comparing Alternatives: ME 005-ECE21S2 - Engineering Economy

=

=

P14,776

Operation

= P24,000

Labor

= P32,000

Payroll taxes = P32,000(0.04)

= P1,280

Taxes and insurance = P300,000(0.03)

= P9,000

Total annual cost

= P81,056

Annual savings = P99,850 - P81,056 = P18,794 Additional investment = P300,000 - P200,000 = P100,000 Rate of return on additional investment =

(100%) = 18.79

Since the computed ROR is greater than 15%, Type B should be selected. By the annual cost method Type A Annual costs: Depreciation =

=

=

P9,850

Operation

= P32,000

Labor

= P50,000

Payroll taxes = P50,000(0.04)

= P2,000

Taxes and insurance = P200,000(0.03)

= P6,000

Interest on capital = P200,000(0.15)

= P30,000

Total annual cost

= P129,850

Type B Annual costs: Depreciation = Operation

=

=

P14,776 = P24,000

Labor https://tip.instructure.com/courses/26120/pages/5-dot-1-comparing-alternatives?module item id=1936843

= P32,000 3/6

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5.1: Comparing Alternatives: ME 005-ECE21S2 - Engineering Economy

Payroll taxes = P32,000(0.04)

= P1,280

Taxes and insurance = P300,000(0.03)

= P9,000

Interest on capital = P300,000(0.15)

= P45,000

Total annual cost Since ACB

= P126,056 ACA, type B should be selected

By the present worth cost method Type A Annual costs (excluding depreciation) = P32,000 + P50,000 + P50,000(0.04) + P200,000 (0.03) = P90,000

PWCA = P200,000 + P90,000(P/A, 15%, 10) PWCA = P200,000 + P90,000

= P651,692

Type B Annual costs (excluding depreciation) = P24,000 + P32,000 + P32,000(0.04) + P300,000 (0.03) = P66,280

PWCB = P300,000 + P66,280(P/A, 15%, 10) PWCB = P300,000 + P66,280 Since PWCB

= P632,646

PWC A for the same period of time, type B should be selected.

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5.1: Comparing Alternatives: ME 005-ECE21S2 - Engineering Economy

By the equivalent uniform annual cost method Type A

EUACA = P200,000 (A/P, 15%, 10) + P90,000 EUACA = P200,000

+ P90,000

EUACA = P129,860 Type A

EUACB = P300,000 (A/P, 15%, 10) + P66,280 EUACB = P300,000

+ P66,280

EUACB = P126,070 Since EUACB

EUACA , type B is more economical

Sample Problem no. 36 A plant to provide the company's present needs can be constructed for P2,800,000 with annual operating disbursements of P600,000. It is expected that at the end of 5 years the production requirements could be doubled, which all necessitate the addition of an extension costing P2,400,000. The disbursements after 5 years will likewise double. A plan to provide the entire expected capacity can be constructed for P4,000,000 and its operating disbursements will be P640,000 when operating on half capacity (for the first 5 years) and P900,000 on full capacity. The plants are predicted to have an indeterminately long life. The required rate of return is 20%. What would you recommend? Solution Deferred Expansion

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5.1: Comparing Alternatives: ME 005-ECE21S2 - Engineering Economy

Capitalized cost = P2,800,000 + P600,000 (P/A, 20%, 5) + P2,400,000 (P/F, 20%, 5) + (P/F, 20%, 5) = P7,970,152

Full-size Plant

Capitalized cost = P4,000,000 + P640,000 (P/A, 20%, 5) +

(P/F, 20%, 5) = P7,722,444

The full-size plant should be constructed.

After completing this module, solve the exercise problem ($CANVAS_COURSE_REFERENCE$/modules/items/gb9633a27da6426485a69a433e0aa0334) and work

on your assignment ($CANVAS_COURSE_REFERENCE$/modules/items/g25b8dbe1e736d552a422a02c0a46dd73) .

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