Question 3 of 5: GoldBrick Engineering — A Non-Conventional Cash Flow

Question 3: GoldBrick Engineering — A Non-Conventional Cash Flow (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) Cash-Flow Diagram

(b) Should GoldBrick Accept at a MARR of 25%?

The question asks for a rate-of-return method, so start where such a method must start — by asking whether an internal rate of return exists at all.

1. Count the sign changes. The series $+250{,}000,\;-200{,}000,\;-450{,}000,\;0,\;+600{,}000$ changes sign twice. By the rule of signs the equation $PW(i)=0$ may therefore have as many as two positive real roots, so a single internal rate of return is not guaranteed.
2. Look for the roots. Writing $v=1/(1+i)$,
   $$PW(i)=250{,}000-200{,}000v-450{,}000v^{2}+600{,}000v^{4}$$
   Evaluating this across the whole range $-90\% \lt i \lt 310\%$ produces no sign change whatever: the present worth is positive everywhere, bottoming out at about $33,551 near $i=42.3\%$ and rising towards $250,000 as $i$ grows without bound. This project has no internal rate of return — there is no discount rate at which it breaks even, because the $250,000 received at $t=0$ is never discounted at all and the later net outlay can never overtake it.
3. Use an external (approximate) rate of return instead. The standard remedy is to stop requiring intermediate balances to earn the project's own unknown rate, and instead reinvest net receipts at the MARR. Carry the receipts forward to $t=4$ at 25%:
   $$FW_{\text{receipts}} = 250{,}000(1.25)^{4}+600{,}000 = 610{,}352+600{,}000 = \$1{,}210{,}352$$
   then find the rate $i^{*}$ at which the disbursements accumulate to that same amount:
   $$200{,}000(1+i^{*})^{3}+450{,}000(1+i^{*})^{2} = 1{,}210{,}352 \;\Longrightarrow\; i^{*} = \boxed{30.48\%}$$
   Since $30.48\% > 25\%$, the project clears the MARR.
4. Confirm with the worth test. At $i=25\%$ the discount factors are $1,\;0.8,\;0.64,\;0.512,\;0.4096$:
   $$PW_{25\%} = 250{,}000-200{,}000(0.8)-450{,}000(0.64)+600{,}000(0.4096)$$
   $$= 250{,}000-160{,}000-288{,}000+245{,}760 = \boxed{+\$47{,}760}$$

Both tests agree: GoldBrick Engineering should accept the project.

It is worth checking that the accept test points the right way, because the project opens with a cash inflow, and a stream that only ever lends money must be judged by the reverse rule. Tracking the project balance at the MARR settles it:

From the end of year 2 onward GoldBrick has money committed to the project, so over the stretch that governs the decision this is an investment rather than a loan — and for an investment a rate above the MARR means accept. The terminal balance of $116,602 is the future worth; discounted four years at 25% it returns $47,760, matching the present worth above exactly.

(c) Is the Highest Rate of Return Always the Best Alternative?

No. A rate of return is a percentage earned on whatever capital an alternative happens to tie up, and percentages on different amounts cannot be ranked against one another. A small project can post a spectacular rate while adding very little value; a larger one can earn a lower percentage and still be worth more in absolute terms. The only valid rate-of-return test between mutually exclusive alternatives is the incremental one: order the alternatives by first cost, and accept each additional increment of capital only if that increment itself earns at least the MARR.

This question presents a single project, so the paper supplies no counter-example of its own; one is easy to construct at the same 25% MARR. Let alternative S cost $100,000 now and return $150,000 in one year, and alternative L cost $400,000 now and return $540,000 in one year:

S has by far the higher rate, yet L is the better alternative. The incremental test says so directly: the increment L − S is $300,000 extra now for $390,000 extra at $t=1$, a return of 30.0%, which exceeds the 25% MARR, so the extra capital is justified and L should be chosen. That is also the alternative with the larger present worth, as it must be.

(d) Do Worth-Based and Rate-of-Return Methods Always Agree?

Yes — provided each is applied correctly. Present worth, annual worth and future worth are positive multiples of one another, so they can never disagree; and the incremental rate-of-return test is mathematically equivalent to maximising present worth, because an increment's rate exceeds the MARR precisely when that increment's present worth at the MARR is positive. Applied properly, the two families give the same accept/reject decision and the same ranking.

They appear to disagree only when the rate-of-return method is misapplied, and this paper illustrates both classic ways of doing so:

• Comparing standalone rates instead of incremental ones, which reverses the ranking in part (c).
• Insisting on an internal rate of return for a non-conventional cash flow, as in part (b), where there is no internal rate of return to find. A worth calculation still returns a clear answer there, which is why present worth — or a properly defined external rate — is the safe arbiter whenever the sign pattern is not simple.