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23-CS-1 Engineering Economics · December 2016

Question 5 of 5: Four Projects — IRR Table

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 5: Four Projects — IRR Table (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) Independent Projects, MARR = 17%

Independent projects do not compete with one another — the budget is assumed sufficient for any combination — so each is judged on its own merits and no incremental analysis is needed. Accept every project whose overall IRR exceeds the MARR:

ProjectFirst costOverall IRRvs MARR 17%Decision
1$100,00019%19 > 17Accept
2$175,00015%15 < 17Reject
3$200,00018%18 > 17Accept
4$250,00016%16 < 17Reject

Select Projects 1 and 3, a total commitment of $300,000. Note that the incremental columns are simply not used here; they answer a question — "which one?" — that independence does not pose.

(b) Mutually Exclusive, MARR = 15% (Incremental Analysis)

Now only one project may be chosen, so the overall IRRs cannot decide the matter: a percentage says nothing about how much money is working at that percentage. Order the alternatives by increasing first cost, take do-nothing as the initial "current best" (it is permitted here), and advance to a costlier alternative only when the extra investment required to get there earns at least the MARR:

  1. P1 vs do-nothing: the relevant rate is P1's own overall IRR, 19% > 15% → P1 is acceptable and becomes the current best (defender).
  2. P2 vs P1: $\text{IRR}_{2\text{-}1} = 9\% < 15\%$ → the extra $75,000 earns only 9%, below the MARR. Reject P2; P1 remains the defender.
  3. P3 vs P1: the challenger is compared against the surviving defender, not against the rejected P2. $\text{IRR}_{3\text{-}1} = 17\% > 15\%$ → the extra $100,000 earns 17%. Accept the increment; P3 becomes the defender.
  4. P4 vs P3: $\text{IRR}_{4\text{-}3} = 13\% < 15\%$ → the extra $50,000 earns only 13%. Reject P4; P3 stands.

Select Project 3. The 23% in the $\text{IRR}_{3\text{-}2}$ cell is never used, because P2 was eliminated before P3 was considered — comparing a challenger against an already-rejected alternative is the classic error in this method.

(c) When Is a ROR Method Recommended?

A rate-of-return method is the right choice when:

Where cash flows are non-conventional (more than one sign change) the IRR may not be unique, and a present- or annual-worth method should be used instead or as a check.

(d) Is the Highest-ROR Alternative Always Best?

No. Project 1 has the highest overall IRR (19%), yet the mutually-exclusive choice in (b) is Project 3, whose overall IRR is only 18%. The reason is scale: a rate of return is a ratio and carries no information about the size of the investment earning it. Moving from P1 to P3 costs an extra $100,000 that earns 17%, comfortably above the 15% MARR, so that extra money is worth committing even though it drags the average return down from 19% to 18%. Maximizing a percentage is not the objective; maximizing total value at the MARR is. The correct procedure is therefore always incremental analysis — which, applied properly, gives the same answer as maximizing present worth.

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