FE Electrical Engineering Economics Practice Problems

Answer: D) $13,382

The Handbook's Single Payment Compound Amount factor on p. 229, \(F = P(1 + i)^n\):

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

so the investment grows to $13,382. The factor \((F/P, 6\%, 5) = 1.3382\) is also tabulated in the 6% table on p. 234. 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: B) $2,652,000

The Handbook's Scaling of Equipment Costs on p. 257 of the Chemical Engineering chapter, to which the Engineering Economics chapter refers on p. 231, scales cost with capacity raised to an exponent \(n\):

$$C_2 = C_1\left(\dfrac{Q_2}{Q_1}\right)^{n} = 2{,}000{,}000 \times \left(\dfrac{80}{50}\right)^{0.6} = 2{,}000{,}000 \times 1.6^{0.6}$$

$$1.6^{0.6} = e^{0.6\ln 1.6} = e^{0.6 \times 0.4700} = e^{0.2820} = 1.3258$$

$$C_2 = 2{,}000{,}000 \times 1.3258 \approx 2{,}652{,}000$$

The 80 MW plant is estimated at $2.65 million, 33% more for 60% more capacity. An exponent below one is the economy of scale: vessels, conductors and structures grow less than proportionally with the throughput they handle, and 0.6 is the classical average, the six-tenths rule, with the Handbook's table on p. 258 giving values from 0.4 to 1.2 for particular equipment. The relationship is an order-of-magnitude tool for early estimates, valid over modest capacity ratios and for equipment of the same type, and a price-level change over the years between the two projects would be applied separately with a cost index. The unit cost falls from $40,000 to $33,150 per megawatt.

$3,200,000 scales linearly, \(2{,}000{,}000 \times 1.6\), with no economy of scale.

$1,920,000 multiplies by the exponent instead of raising to it, \(2{,}000{,}000 \times 1.6 \times 0.6\).

$4,378,000 inverts the exponent, \(1.6^{1/0.6}\).

Answer: D) Premium motor, EUAC $4,652

Each alternative's cash flows are converted to an equivalent uniform annual cost with the Handbook's Capital Recovery factor for the first cost and Sinking Fund factor for the salvage, both on p. 229 and in the 10% table on p. 235, \((A/P, 10\%, 10) = 0.1627\) and \((A/F, 10\%, 10) = 0.0627\):

$$EUAC_{std} = 5{,}000(0.1627) + 4{,}200 - 500(0.0627) = 814 + 4{,}200 - 31 = 4{,}982$$

$$EUAC_{prem} = 8{,}000(0.1627) + 3{,}400 - 800(0.0627) = 1{,}302 + 3{,}400 - 50 = 4{,}652$$

The premium motor costs $4,652 a year against $4,982, so it is the more economical by $331 a year despite costing $3,000 more to buy. The $800 annual energy saving more than covers the extra capital recovery of $488 a year, which is the usual outcome for motors that run many hours, since energy over the life of a motor costs many times its price. The salvage values barely matter, together shifting the comparison by $19 a year. Annual cost comparison is convenient because both alternatives have the same life; with unequal lives it remains valid under repeated replacement, whereas a present worth comparison would need a common study period. The break-even energy saving is $469 a year; below that the standard motor would win.

Standard motor, EUAC $4,982 has the standard motor's cost right and picks it, the dearer alternative.

Premium motor, EUAC $4,982 and Standard motor, EUAC $4,652 attach each cost to the wrong motor.

Answer: D) $105,500

Each risk's expected cost is its probability times its impact, the Handbook's expected value of p. 231 applied to one chance event at a time, and the exposures add because the risks are separate:

$$0.40 \times 120{,}000 = 48{,}000, \quad 0.15 \times 250{,}000 = 37{,}500, \quad 0.25 \times 80{,}000 = 20{,}000$$

$$\text{Risk exposure} = 48{,}000 + 37{,}500 + 20{,}000 = 105{,}500$$

The expected cost of the identified risks is $105,500. Ranking by exposure puts the supply-chain delay first although the permit rejection has the larger impact, because probability counts as much as consequence; that ranking is what directs mitigation effort. The figure is an average over the possible outcomes, and the actual cost will be one of eight combinations ranging from nothing, with probability \(0.6 \times 0.85 \times 0.75 = 38\%\), to all three at $450,000, with probability 1.5%; a contingency reserve set at the expected value therefore covers the average project but not the worst case, and how much above it to hold is a judgement about the tolerance for overrun, not a calculation. The expected value is the starting point of that judgement.

$48,000 is the largest single exposure, the supply-chain delay alone.

$85,500 adds only the two largest exposures.

$37,500 is the permit rejection's exposure alone.

Answer: D) 7 years

The Handbook defines the payback period under Breakeven Analysis on p. 230 as the time for the benefits of an investment to equal its cost; the discounted version values those benefits at the interest rate, so the condition is \(8{,}500\,(P/A, 12\%, n) \ge 35{,}000\), or \((P/A, 12\%, n) \ge 4.118\). From the 12% table on p. 235:

$$n = 6:\ (P/A, 12\%, 6) = 4.1114, \quad 8{,}500 \times 4.1114 = 34{,}947 < 35{,}000$$

$$n = 7:\ (P/A, 12\%, 7) = 4.5638, \quad 8{,}500 \times 4.5638 = 38{,}792 \ge 35{,}000$$

The discounted savings first cover the cost during the seventh year, so the discounted payback period is 7 years. Six years fall short by $53, a near miss that shows why the comparison must be made with the factor and not by eye. The undiscounted payback is \(35{,}000/8{,}500 = 4.1\) years, rounded up to 5 whole years; discounting always lengthens the payback, here by two years, because later savings are worth less. The Handbook's uniform series present worth factor on p. 229 is the tool, and the payback rule, whichever version, ignores everything after the payback year, which is its known weakness as a decision criterion.

4 years truncates the undiscounted ratio \(35{,}000/8{,}500 = 4.1\).

5 years is the undiscounted payback rounded up to a whole year, with no interest.

6 years stops where the discounted savings, $34,947, come close to the cost without reaching it.