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23-CS-1 Engineering Economics · December 2016

Question 3 of 5: Gasoline versus All-Electric Car

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

Notes on this paper

National Exams — December 2016 — 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 questions follow; standard compound-interest factors are used and minor rounding is immaterial.

Question 3: 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 Costs and the Cost Model

Gasoline: $20{,}000\times\frac{6.5}{100}\times0.85 = \$1{,}105$/yr (1,300 L/yr). Electric: $20{,}000\times\frac{12}{100}\times0.14 = \$336$/yr (2,400 kWh/yr). The electric car therefore saves $769/yr in energy.

At 0% interest no discounting is needed, so the net cost of ownership over $n$ years is simply the purchase price less the resale value, plus $n$ years of energy cost. With declining-balance depreciation at 10%/yr the resale value is $P(0.9)^n$, so the depreciation borne by the owner is $P\,[1-(0.9)^n]$ and

$$C(n) = P\left[1-(0.9)^n\right] + (\text{annual energy cost})\,n$$

(a) Net Cost After 3 Years

$(0.9)^3 = 0.729$, so each car loses $1-0.729 = 27.1\%$ of its price:

$$\text{Gasoline} = 20{,}000(0.271) + 1{,}105(3) = 5{,}420 + 3{,}315 = \boxed{\$8{,}735}$$
$$\text{Electric} = 30{,}000(0.271) + 336(3) = 8{,}130 + 1{,}008 = \boxed{\$9{,}138}$$

The gasoline car is more economic after 3 years, by $403. Three years of energy savings ($2,307) do not cover the extra depreciation the $10,000 purchase premium carries ($2,710).

(b) Gas Price to Justify the Electric Over 4 Years

Only the $10,000 price premium depreciates differently, so the electric's extra net cost over 4 years is $10{,}000(1-0.9^{4})=10{,}000(0.3439)=\$3{,}439$. Its yearly saving at gasoline price $p$ is (gasoline energy cost) $-$ (electric energy cost) $= 1{,}300p - 336$. Setting the 4-year saving equal to the extra cost:

$$4(1{,}300p - 336) = 3{,}439 \;\Rightarrow\; 5{,}200p - 1{,}344 = 3{,}439 \;\Rightarrow\; 5{,}200p = 4{,}783$$
$$p = \boxed{\$0.92/\text{L}}$$

Gasoline would have to be about 8% dearer than the stated $0.85/L for the electric car to break even over a 4-year hold.

(c) Years of Usage to Justify the Electric

At the given prices the electric saves $769/yr, while its extra net cost is $10{,}000(1-0.9^{n})$. The electric is justified when

$$769\,n \;\ge\; 10{,}000\left(1-0.9^{n}\right)$$

This has no closed-form solution, so bracket it: at $n=6$ the left side is $4{,}614$ against $4{,}686$ (not yet justified); at $n=7$ it is $5{,}383$ against $5{,}217$ (justified). Interpolating and refining gives

$$n \approx \boxed{6.3\ \text{years}}$$

So the all-electric car pays for itself once the customer keeps it for about six years and four months or longer. Note the premium term saturates — $10{,}000(1-0.9^n)$ can never exceed $10,000 — while the savings term grows without limit, so beyond the break-even the electric car's advantage keeps widening.