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23-CS-1 Engineering Economics · December 2018

Question 5 of 5: Grower versus Automated Irrigation System

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

Notes on this paper

National Exams — December 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 5: Grower versus Automated Irrigation System (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.

(a) Reprogrammed Every Year — Present Worth

Assumption (stated under NOTE 1). The $15,000 installation charge covers the first programming at $t=0$, so a re-programming falls in every year $k$ that is a multiple of the stated interval with $k\le 5$: five events at $t=1\ldots5$ in part (a), two events at $t=2,4$ in part (b). Reading part (a) instead as four events at $t=1\ldots4$ gives $190,647, still dearer than the grower, so the conclusion below does not turn on the reading.

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

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

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

(b) Reprogrammed Every 2 Years — Present Worth

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

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

Now the system is cheaper—replacement is economic (saves $6,672 in present worth).

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

$$FW_{\text{grower}} = 43{,}000(F/A,7\%,5) = \$247{,}282;\qquad FW_{\text{system}} = 199{,}202.37(F/P,7\%,5) = 199{,}202.37(1.4025517) = \$279{,}392$$

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

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