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11-CS-1 Engineering Economics · May 2018

Question 3 of 5: Labour versus Collaborative Robot

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 3: Labour versus Collaborative Robot (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.

Assumption (stated, as the paper’s Note 1 invites). The $15,000 installation line already covers the first programming at $t=0$, so a re-programming event falls at every year $k$ that is a multiple of the stated interval with $k\le 5$: five events at $t=1,\dots,5$ in part (a) and two events at $t=2,4$ in part (b). Nothing else is charged to the robot—no salvage is given, so the horizon-end value is taken as zero for both alternatives, and the second labour worker is common to both options and therefore cancels.

(a) Reprogrammed Every Year — Present Worth

Labour: $PW = 43{,}000(P/A,7\%,5) = 43{,}000(4.100197) = \$176{,}308$. Robot: $150,000 plus $12,000/yr reprogramming:

$$PW_{\text{robot}} = 150{,}000 + 12{,}000(4.100197) = 150{,}000 + 49{,}202 = \$199{,}202$$

The robot ($199,202) costs more than the labourer ($176,308), so with yearly reprogramming replacement is not economic.

(b) Reprogrammed Every 2 Years — Present Worth

Reprogramming only at years 2 and 4: $12{,}000[(P/F,7\%,2)+(P/F,7\%,4)] = 12{,}000(0.87344+0.76290)=\$19{,}636$.

$$PW_{\text{robot}} = 150{,}000 + 19{,}636 = \$169{,}636 \;<\; \$176{,}308$$

Now the robot is cheaper—replacement is economic (saves ≈$6,700 in present worth).

(c) Reprogrammed Every Year — Future Worth (t = 5)

$$FW_{\text{labour}} = 43{,}000(F/A,7\%,5) = 43{,}000(5.750739) = \$247{,}282$$
$$FW_{\text{robot}} = 199{,}202(F/P,7\%,5) = 199{,}202(1.402552) = \$279{,}391$$

The robot's future worth is higher (more costly), so it is not economic—the same conclusion as (a), as it must be since FW = PW × (F/P).

(d) 10-Year Life and 10-Year Horizon — Different from (b)?

No—the decision in (b) still holds (replace), even more strongly. Doubling the horizon spreads the robot's one-time $150,000 capital cost over twice as many years while the labourer's $43,000/yr cost simply doubles. This shifts the balance further in the robot's favour, so a case that already favoured the robot at 5 years favours it by an even wider margin at 10 years. No calculation is needed—the capital cost is diluted while the labour cost is not.