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Present Worth and Annual Worth Analysis Questions and Answers

Answer: A

Correct answer: what the asset would actually fetch

Market value is an economic quantity established by what a buyer would pay; book value is an accounting one, being cost less accumulated depreciation under whatever method was chosen. The two are related only by coincidence, and it is the market value that enters a replacement decision because only it represents money forgone.

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22.

Raising the discount rate in a present worth comparison tends to favour the alternative with

Answer: B

A high rate penalises distant cash flows, so the alternative whose burden lies in the future is treated more leniently and the capital-intensive one loses its advantage. The ranking can reverse at some rate, which is the crossover point.

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Answer: C

Correct answer: crossover or incremental rate of return

At the crossover rate the two alternatives have identical present worths and the preference between them reverses. It is the internal rate of return of the incremental cash flow, and knowing where it lies relative to the minimum attractive rate is what makes an incremental comparison decisive.

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Answer: D

First costs and operating costs can often be estimated closely, whereas the appropriate rate is a matter of judgement. Showing that the ranking holds over a plausible range of rates is what makes a recommendation robust.

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25.

A project costing 100 000 returns 30 000 a year for 5 years. At 10 per cent its present worth is about

Answer: A

Correct answer: 13 730

\[ PW = -100{,}000 + 30{,}000 \times (P/A, 10\%, 5) = -100{,}000 + 30{,}000 \times 3.7908 = 13{,}724. \] The undiscounted receipts total 150,000, so discounting removes some 36,000 of apparent value - which is the whole point of the calculation.

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26.

For that project the annual worth at 10 per cent is about

Answer: B

Multiplying the present worth of 13 730 by the capital recovery factor of 0.2638 gives 3622 a year. The positive sign agrees with the present worth, as it always must, since the two differ only by a positive multiplier.

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27.

A machine costs 50 000, has annual costs of 8000 and a salvage of 5000 after 10 years. At 10 per cent its equivalent uniform annual cost is about

Answer: C

Correct answer: 15 820

\[ EUAC = 50{,}000(A/P, 10\%, 10) + 8{,}000 - 5{,}000(A/F, 10\%, 10) \] \[ = 50{,}000 \times 0.16275 + 8{,}000 - 5{,}000 \times 0.06275 = 8{,}138 + 8{,}000 - 314 = 15{,}824. \] Note the salvage enters through the sinking fund factor and reduces the annual cost, since it is a receipt at the end rather than a payment.

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28.

A perpetual project with an equivalent annual cost of 12 000 at 6 per cent has a capitalised cost of

Answer: D

Dividing 12 000 by 0.06 gives 200 000, which is the principal whose interest alone would meet the cost for ever without touching the capital. Halving the interest rate would double the capitalised cost, so the figure is highly rate-sensitive.

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29.

Alternative A costs 20 000 with annual costs of 6000 over 10 years; B costs 35 000 with annual costs of 3500. At 10 per cent, A's equivalent annual cost is about

Answer: A

Correct answer: 9255

\[ EUAC_A = 20{,}000 \times (A/P, 10\%, 10) + 6{,}000 = 20{,}000 \times 0.16275 + 6{,}000 = 3{,}255 + 6{,}000 = 9{,}255. \] The two components pull in opposite directions between the alternatives: A is cheap to buy and dear to run, and B the reverse.

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30.

For that comparison, B's equivalent annual cost is about

Answer: B

Correct answer: 9196

\[ EUAC_B = 35{,}000 \times 0.16275 + 3{,}500 = 5{,}696 + 3{,}500 = 9{,}196. \] B is cheaper by 59 a year, which on figures of this size is well inside the accuracy of the estimates. The honest conclusion is that the two are effectively equal and that other considerations should decide.

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