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11-CS-1 Engineering Economics · May 2019

Question 4 of 5: Experienced Grower versus Automated Irrigation

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

Notes on this paper

National Exams — May 2019 — 11-CS-1 Engineering Economics. Three hours; open book; any non-communicating calculator permitted. Any four of the five questions constitute a complete exam paper and each question is of equal value (25 marks). Fully worked solutions to all five follow.

Question 4: Experienced Grower versus Automated Irrigation (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.

Assumptions stated (exam note 1). Two conventions are load-bearing and neither is spelled out in the question, so both are adopted explicitly here.

NOTE 1 — timing of the re-programming events. "Re-programmed every $k$ years" is read as an event at every year that is a multiple of $k$ and no later than the horizon: five events at $t=1,2,3,4,5$ in part (a), and two events at $t=2$ and $t=4$ in part (b). This is the only reading under which parts (a) and (b) come out on opposite sides, which is plainly why both are asked. The $15,000 installation and first-time programming is a separate $t=0$ cost and is not one of these events.

NOTE 2 — scope of each option. The automated system replaces the grower entirely, so the automated option carries no salary or benefits; the human-based option carries no capital cost. Neither option is credited with a salvage value at $t=5$.

Costs of the two options. Human-based: $40,000 salary plus $3,000 benefits, i.e. $43,000 per year for five years, with no capital outlay. Automated: $135,000 purchase plus $15,000 installation and first-time programming, i.e. $150,000 at $t=0$, plus $12,000 for each re-programming event. MARR 7%, horizon 5 years, and $(P/A,7\%,5)=4.100197$.

(a) Re-Programmed Every Year — Present Worth

  1. Human-based system. A five-year uniform annual cost:
    $$PW_{H} = 43{,}000\,(P/A,7\%,5) = 43{,}000(4.100197) = \$176{,}308$$
  2. Automated system. Capital at $t=0$ plus a re-programming cost in each of the five years, which is itself a uniform series:
    $$PW_{A} = 150{,}000 + 12{,}000\,(P/A,7\%,5) = 150{,}000 + 49{,}202 = \$199{,}202$$
  3. Compare. Both are costs, so the lower present worth wins:
    $$PW_{A} - PW_{H} = 199{,}202 - 176{,}308 = \boxed{+\$22{,}894\ \text{in favour of the grower}}$$

No — with yearly re-programming it is not an economic decision to replace the experienced grower. The automated system costs about $22,894 more in present-worth terms over the five-year horizon. The reason is easy to see: re-programming every year costs $12,000 against the grower's $43,000, an annual saving of $31,000 worth $127,106 in present value, which does not cover the $150,000 of capital.

(b) Re-Programmed Every 2 Years — Present Worth

Now only two re-programming events fall inside the horizon, at $t=2$ and $t=4$. They are not a uniform annual series, so discount each singly with $(P/F,7\%,2)=0.873439$ and $(P/F,7\%,4)=0.762895$:

$$PW_{\text{reprog}} = 12{,}000\,[\,0.873439+0.762895\,] = 12{,}000(1.636334) = \$19{,}636$$
$$PW_{A} = 150{,}000 + 19{,}636 = \$169{,}636$$

The human-based option is unchanged at $176,308, so

$$PW_{H} - PW_{A} = 176{,}308 - 169{,}636 = \boxed{\$6{,}672\ \text{saved by automating}}$$

Yes — with re-programming every two years it is an economic decision to replace the experienced grower, though only just: the margin is about $6,672 on a $176,308 commitment, under 4%. Halving the number of re-programming events saved $29,566 of present worth, which is what tips a $22,894 deficit into a $6,672 surplus.

(c) Part (a) Redone with Future Worth

Move the same two cash-flow streams to the end of the horizon instead of the beginning, using $(F/A,7\%,5)=5.750739$ and $(F/P,7\%,5)=1.402552$:

$$FW_{H} = 43{,}000\,(F/A,7\%,5) = 43{,}000(5.750739) = \$247{,}282$$
$$FW_{A} = 150{,}000\,(F/P,7\%,5) + 12{,}000\,(F/A,7\%,5) = 210{,}383 + 69{,}009 = \$279{,}392$$
$$FW_{A} - FW_{H} = 279{,}392 - 247{,}282 = \boxed{+\$32{,}110\ \text{in favour of the grower}}$$

The decision is unchanged: do not replace the grower. This is exactly what should happen, and it is worth saying why. Future worth is present worth multiplied by $(F/P,7\%,5)=1.402552$, a single positive constant applied to both options alike, so it cannot reorder them; indeed $176{,}308.49(1.402552)=247{,}282$ and $199{,}202.37(1.402552)=279{,}392$, and the $22,894 present-worth gap becomes a $32,110 future-worth gap, the same gap valued five years later. Present worth, future worth and annual worth are three views of one comparison and can never disagree.

(d) Ten-Year Service Life and Ten-Year Horizon — Would Part (b) Change?

No. The part (b) decision stands, and stands more firmly — the automated system is still the economic choice. The hint is right that no calculation is needed, and the reasoning is the following.

Confirming the argument (not required by the question): $(P/A,7\%,10)=7.023582$ gives $PW_{H}=43{,}000(7.023582)=\$302{,}014$, while re-programming at $t=2,4,6,8,10$ costs $12{,}000(0.873439+0.762895+0.666342+0.582009+0.508349)=12{,}000(3.393034)=\$40{,}716$, so $PW_{A}=150{,}000+40{,}716=\$190{,}716$. The margin grows from $6,672 to $111,298.

Note that the same question asked about part (a) would not be so safe: yearly re-programming lost by $22,894 at five years, and whether ten years reverses that requires the calculation the hint says is unnecessary here. It is the fact that part (b) had already won that makes the extension argument airtight.