FE Electrical Engineering Economics Practice Problems

Answer: D) $13,382

The Handbook's Single Payment Compound Amount factor on p. 235, \(F = P(1 + i)^n\):

$$F = 10{,}000\,(1.06)^5 = 10{,}000 \times 1.3382 = 13{,}382$$

so the investment grows to $13,382. The factor \((F/P, 6\%, 5) = 1.3382\) is also tabulated in the 6% table on p. 240. Compound interest earns interest on the interest: the first year adds $600, the second $636 on the new balance of $10,600, and so on, which is why five years of 6% yield 33.8% rather than the 30% of simple interest. The difference, $382 here, grows with both the rate and the number of periods, and every other factor in the Handbook's table is built from this one: the present worth factor is its reciprocal, and the series factors are sums of it over the periods. The period and the rate must match, annual compounding with an annual rate and five annual periods, as they do here.

$13,000 is simple interest, \(10{,}000(1 + 5 \times 0.06)\), with no interest on the accumulated interest.

$12,625 compounds for four periods, \((1.06)^4\).

$10,600 is a single year's growth.

Answer: C) 26.1 hours

Handbook: FE Reference Handbook 10.6, Industrial and Systems Engineering → Learning Curves, p. 428 (PDF viewer: page 434). Note the chapter — it is not in Engineering Economics, which is where this item's category points. The Handbook writes it as

$$T_N = K N^{s}, \qquad s = \frac{\ln(\text{learning rate as a decimal})}{\ln 2}$$

with \(K\) the time for the first unit — the 50 hours the figure supplies.

Step 1 – Exponent.

$$s = \frac{\ln 0.85}{\ln 2} = \frac{-0.16252}{0.69315} = -0.23446$$

Step 2 – Evaluate at N = 16.

$$T_{16} = 50 \times 16^{-0.23446} = 50 \times 0.52200 = \mathbf{26.1 \text{ hours}}$$

Step 3 – Answer: (C).

The 20-second route. A learning rate applies per doubling of cumulative units, and \(16 = 2^4\) — four doublings from unit 1:

$$1 \to 2 \to 4 \to 8 \to 16, \qquad T_{16} = 50(0.85)^4 = 50(0.52201) = 26.1 \text{ h}$$

No logarithms needed. Count the doublings carefully: 16 is the fourth doubling, not the third.

Step 4 – Eliminate the distractors:

(A) 32.5 h is the average time per unit across all 16 panels, from the second formula printed on the same Handbook page:

(B) 36.1 h \(= 50(0.85)^2\) — two doublings, the time for panel 4. Stopping two rungs early.

(D) 30.7 h \(= 50(0.85)^3\) — three doublings, which is the time for panel 8. Miscounting the doublings by one is the most common error, and 8 is a tempting stopping point.

$$T_{avg} = \frac{K}{N}\left[\frac{(N+0.5)^{1+s} - (0.5)^{1+s}}{1+s}\right] = \frac{50}{16}\left[\frac{(16.5)^{0.76554} - (0.5)^{0.76554}}{0.76554}\right] = 32.5$$

That is a real quantity, but it is not what was asked. The question wants the time for one specific unit, so use \(T_N\), not \(T_{avg}\) — the two formulas sit side by side on p. 428, so check which one you have copied.

Answer: C) About 20.9%

The rate of return (p. 236 of the Handbook) makes the present worth of the savings equal to the cost, through the Uniform Series Present Worth factor of p. 235:

$$32{,}000 = 7{,}440\,(P/A, i^*, 12) \quad\Rightarrow\quad (P/A, i^*, 12) = \dfrac{32{,}000}{7{,}440} = 4.301$$

The 18% table on p. 242 gives \((P/A, 18\%, 12) = 4.7932\), so the rate lies above the tables; from the factor formula, \((P/A, 20\%, 12) = 4.4392\) and \((P/A, 22\%, 12) = 4.1274\), and interpolating,

$$i^* = 20\% + 2\% \times \dfrac{4.4392 - 4.301}{4.4392 - 4.1274} = 20.9\%$$

which the direct solution of the formula confirms to the same decimal. The investment returns about 21% a year. That is below the simple ratio of saving to cost, 23.3%, because the $32,000 must be recovered out of the savings over the 12 years as well as earning the return; the two figures converge only for a very long life. A demand penalty is a charge for reactive power, and capacitors remove it at a cost that is small next to the penalty, which is why such projects routinely show returns above 20% and paybacks under five years, here 4.3 years undiscounted.

About 23.3% is the saving over the cost, \(7{,}440/32{,}000\), with no capital recovery.

About 4.3% reads the factor value 4.30 as a rate.

About 1.9% is the monthly saving over the cost, \(620/32{,}000\).

Answer: B) \(\$186{,}842\)

Handbook: FE Reference Handbook 10.6, Engineering Economics:

• p. 235 (PDF viewer: page 241) — the factor table lists Uniform Series Present Worth (P/A, i%, n) and Uniform Gradient Present Worth (P/G, i%, n).
• p. 241 (PDF viewer: page 247) — the i = 10.00% table. Its n = 8 row gives P/A = 5.3349 and P/G = 16.0287 directly, so no factor has to be evaluated by hand.

Step 1 – Split the series in two. This is the move the whole problem turns on. An arithmetic gradient is not handled by one factor; it is a uniform series plus a gradient laid on top:

• a base annuity \(A = \$20{,}000\) running in all eight years, and
• a gradient \(G = \$5{,}000\) that starts at zero in year 1 and adds \(\$5{,}000\) more in each subsequent year.

Check the split against the figure: year 1 is \(20{,}000 + 0 = \$20{,}000\); year 8 is \(20{,}000 + 7(5{,}000) = \$55{,}000\). Both match.

Step 2 – Look up both factors at i = 10%, n = 8:

$$(P/A, 10\%, 8) = 5.3349, \qquad (P/G, 10\%, 8) = 16.0287$$

Step 3 – Combine.

$$P = 20{,}000(5.3349) + 5{,}000(16.0287)$$
$$P = 106{,}699 + 80{,}144 = \mathbf{\$186{,}842}$$

Step 4 – Answer: (B).

Note the n in the gradient factor. Even though the gradient contributes nothing in year 1, you still use \(n = 8\), not \(n = 7\). The \((P/G)\) factor is defined with that zero first term already built in. Using \(n = 7\) gives \((P/G, 10\%, 7) = 12.7631\) and an answer of \(\$170{,}514\) — a plausible-looking number that is simply wrong.

Sanity check. The eight payments total \(\$300{,}000\) undiscounted. At 10% over eight years the answer should land well below that but above the \(\$106{,}699\) that the flat \(\$20{,}000\) base alone is worth. \(\$186{,}842\) sits between them.

Step 5 – Eliminate the distractors:

(A) \(\$106{,}699 = 20{,}000(5.3349)\) is the base annuity alone, with the gradient dropped. It treats the contract as eight flat payments of \(\$20{,}000\) and throws away every dollar of escalation.

(C) \(\$200{,}060 = 37{,}500(5.3349)\) averages the series to \(\$37{,}500\) per year and discounts that as a uniform annuity. The average is right, but averaging is not: it moves money from the late years to the early ones, where it is discounted less, so the result is too high.

(D) \(\$300{,}000\) is the plain sum \(20{,}000 + 25{,}000 + \ldots + 55{,}000\), with no discounting at all. It is the ceiling the present worth can never reach.

Answer: A) SPD-A $71,000; SPD-B $64,000; select SPD-B

Each system's cost for the year is the purchase of the 50 units, the expected repair cost, failures weighted by their probability as in the Handbook's expected value on p. 237, and the maintenance:

$$\text{SPD-A: } 50 \times 1{,}200 + 50 \times 0.015 \times 8{,}000 + 5{,}000 = 60{,}000 + 6{,}000 + 5{,}000 = 71{,}000$$

$$\text{SPD-B: } 50 \times 900 + 50 \times 0.040 \times 8{,}000 + 3{,}000 = 45{,}000 + 16{,}000 + 3{,}000 = 64{,}000$$

SPD-B is expected to cost $64,000 against $71,000 for SPD-A, so SPD-B is selected. Here the cheaper unit wins even though it fails nearly three times as often: the extra 1.25 expected failures a year cost $10,000, less than the $15,000 saved on the purchase and the $2,000 saved on maintenance. The outcome turns on the repair cost; at $21,600 per failure the two systems would break even, and above that the reliable unit would be preferred, so the sensitivity of the decision to that estimate is worth checking before ordering. Because the comparison is over a single year, the purchase price is a one-year cost; for a multi-year life it would be spread by the capital recovery factor of p. 235 instead.

SPD-A $71,000; SPD-B $64,000; select SPD-A has the costs right and picks the dearer system, presumably for its reliability alone.

SPD-A $65,000; SPD-B $48,000; select SPD-B omits the expected repair costs.

SPD-A $66,000; SPD-B $61,000; select SPD-B omits the maintenance.