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Engineering Economics Questions and Answers

Answer: A

The opportunity to put the money to work is what makes a rupee today worth more than a rupee next year, and it holds quite apart from inflation or any risk of default. Interest is simply the price paid for that opportunity to change hands.

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Answer: B

Correct answer: the return earned for parting with money for a time

Interest is the price of time. The lender gives up the use of the money and the chance to invest it elsewhere, and interest is the compensation for that. Default risk and administrative cost are added to the rate charged, but they are not what interest fundamentally is - a riskless loan still commands interest.

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Answer: C

Equivalence at a stated interest rate is what makes comparison possible, and every technique in the subject amounts to moving cash flows to a common point. Adding them as they stand is the commonest error a beginner makes.

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4.

On a cash flow diagram, an arrow pointing upward conventionally represents a

Answer: D

Upward for money coming in and downward for money going out is the usual convention, though the matters is only that it be applied consistently throughout. The diagram is drawn from the point of view of one party, so a lender's and a borrower's diagrams are mirror images.

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5.

The end-of-period convention assumes that a cash flow within a period occurs at

Answer: A

Correct answer: the end of that period

The convention places every cash flow at a period boundary so that the standard factors apply directly. It costs a little accuracy where flows arrive throughout the period, but the alternative - discounting each flow at its own date - adds a great deal of arithmetic for a difference usually smaller than the estimates themselves.

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Answer: B

Correct answer: the original principal only

\[ F = P(1 + in). \] Because the interest never itself earns interest, the balance grows linearly with time. Simple interest survives in short-term instruments and in some statutory calculations, but it is not how money actually accumulates when the interest is left invested.

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Answer: C

Correct answer: the accumulated balance including past interest

\[ F = P(1+i)^n. \] Each period’s interest is added to the balance and earns interest thereafter, so growth is exponential rather than linear. This is the assumption behind every factor in engineering economy, and it is why long horizons are so sensitive to the rate chosen.

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Answer: D

Correct answer: the same amount

Over a single period there is no accumulated interest for compound interest to work on, so \( P(1+i) \) and \( P(1 + i \times 1) \) are identical. The two methods diverge only from the second period onward, and the divergence then grows without limit.

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Answer: A

Correct answer: a present sum into its future equivalent

The single-payment compound amount factor is \( (F/P, i, n) = (1+i)^n \), and the notation is worth reading as it is written: the quantity sought over the quantity known. Its reciprocal, \( (P/F, i, n) \), performs the opposite conversion.

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Answer: B

Correct answer: dividing it by one plus the rate raised to the number of periods

\[ P = \frac{F}{(1+i)^n}. \] Discounting is compounding run backwards, and it asks what sum invested today would grow into the future amount. Multiplying instead of dividing is the standard slip, and it gives an answer larger than the future sum rather than smaller.

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