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11-CS-1 Engineering Economics · May 2018

Question 4 of 5: Gasoline versus All-Electric Car

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2018 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any four of the five questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all five follow; standard compound-interest factors are used and minor rounding is immaterial.

Question 4: Gasoline versus All-Electric Car (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Annual energy cost: Gas $=20{,}000\times\tfrac{6.5}{100}\times0.85 = \$1{,}105$/yr; Electric $=20{,}000\times\tfrac{12}{100}\times0.20 = \$480$/yr. Resale after $n$ years: $MV_n = P(1-d)^n$. At 0% interest, total cost = purchase − resale + $\sum$ energy.

(a) Total Cost if Sold After 3 Years

Gasoline: $MV_3 = 24{,}000(0.88)^3 = \$16{,}355$. Cost $= 24{,}000 - 16{,}355 + 3(1{,}105) = \$10{,}960$.

Electric: $MV_3 = 36{,}000(0.90)^3 = \$26{,}244$. Cost $= 36{,}000 - 26{,}244 + 3(480) = \$11{,}196$.

$$\text{Gasoline } \$10{,}960 \;<\; \text{Electric } \$11{,}196 \Rightarrow \boxed{\text{Gasoline is more economic (by }\$236)}$$

(b) Break-Even Gas Price at a 4-Year Resale

Gas depreciation loss $=24{,}000-24{,}000(0.88)^4 = 24{,}000-14{,}393 = \$9{,}607$; annual litres $=1{,}300$. Electric total $=36{,}000-36{,}000(0.90)^4 + 4(480) = 12{,}380 + 1{,}920 = \$14{,}300$. Setting gasoline cost equal:

$$9{,}607 + 4(1{,}300)p = 14{,}300 \;\Rightarrow\; 5{,}200\,p = 4{,}693 \;\Rightarrow\; p \approx \boxed{\$0.90/\text{L}}$$

Gasoline is justified only while the gas price stays at or below ≈$0.90/L. (At today's $0.85/L it is justified.)

(c) Years of Usage that Justify the Electric Car

Define $f(n) = \text{Gas cost} - \text{Electric cost}$; the electric car is justified when $f(n)>0$:

$$f(n) = -12{,}000 - 24{,}000(0.88)^n + 36{,}000(0.90)^n + 625n$$

Evaluating: $f(3)=-236$, $f(8)=-134$, $f(9)=-23$, $f(10)=+118$. The sign change is between 9 and 10 years; interpolating:

$$n \approx 9 + \frac{23}{23+118} \approx \boxed{9.2\ \text{years}}$$

The all-electric car becomes the economic choice only if kept about 9.2 years or longer—its higher purchase price and slower depreciation take that long to be repaid by its lower running cost.