Question 4 of 6: Engineering Economics — choosing among three highway improvements

Question 4: Engineering Economics — choosing among three highway improvements (20 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.

Given. Three mutually exclusive highway improvements, each with a first cost at time zero, three annual benefit streams grouped under the heading “Yearly Cost Savings”, and an annual maintenance cost. All have a 20-year life and the discount rate is 10 % per year.

Find. Which option, if any, the agency should invest in, on a consistent economic criterion at $i=10\ \%$ over the common 20-year life.

Approach. Because all three options share a 20-year life, present worth, annual worth and benefit–cost ratio can all be applied directly with no least-common-multiple argument. Net the maintenance cost against the savings to get one annual figure per option, discount it over 20 years at 10 %, compare against the first cost, and then test whether the resulting conclusion is an artefact of the discount rate before recommending anything.

1. Compute the series present-worth factor. For a uniform end-of-year series over $n=20$ years at $i=0.10$,

   $$\left(P/A,\,10\%,\,20\right)=\frac{1-(1+i)^{-n}}{i}=\frac{1-1.10^{-20}}{0.10}=8.5136$$

   and its reciprocal, the capital-recovery factor, is $\left(A/P,\,10\%,\,20\right)=0.11746$. Note in passing that the largest value $(P/A)$ can ever take over twenty years is 20, reached only at a zero discount rate — a fact used in step 5.

2. Net the annual flows. The three savings columns are benefits and the maintenance column is a cost, so for each option the net annual benefit is the gross saving less the maintenance:

   $$A_1=8{,}500-1{,}500=\$7{,}000,\qquad A_2=12{,}000-2{,}500=\$9{,}500,\qquad A_3=15{,}800-3{,}000=\$12{,}800$$

   Each is a level end-of-year series over the full twenty years, so a single factor converts it to present worth.

3. Compute net present worth. With $NPW=A\left(P/A,\,10\%,\,20\right)-P$:

   $$NPW_1=7{,}000(8.5136)-185{,}000=59{,}595-185{,}000=-\$125{,}405$$

   $$NPW_2=9{,}500(8.5136)-220{,}000=80{,}879-220{,}000=-\$139{,}121$$

   $$NPW_3=12{,}800(8.5136)-310{,}000=108{,}974-310{,}000=-\$201{,}026$$

   Every present worth is negative, so the immediate conclusion is that

   $$\boxed{\text{none of the three options is economically justified at }i=10\ \%}$$

4. Confirm on the annual-worth and benefit–cost criteria. Because the lives are equal, annual worth must give the same ranking, and it does: $AW=A-P\left(A/P,\,10\%,\,20\right)$ gives −$14,730, −$16,341 and −$23,612 for options 1, 2 and 3. The conventional public-sector benefit–cost ratio, taking the user savings as benefits and the capital plus maintenance as costs,

   $$B/C=\frac{\text{gross saving}\times\left(P/A\right)}{P+\text{maintenance}\times\left(P/A\right)}$$

   gives 0.366, 0.423 and 0.401. All three ratios are far below unity, which is the same verdict. It is worth noticing that the ratio ranks option 2 highest while present worth ranks option 1 highest — the familiar disagreement between a ratio and an absolute measure — but the disagreement is academic here, because incremental analysis is only ever applied among alternatives that have first passed the $B/C\ge1$ test, and none of these does.

5. Test whether the discount rate is to blame. This is the step that turns a bare negative answer into a defensible recommendation. Evaluate each option undiscounted, at $i=0$:

   $$20A_1-P_1=140{,}000-185{,}000=-\$45{,}000,\quad 20A_2-P_2=-\$30{,}000,\quad 20A_3-P_3=-\$54{,}000$$

   Every option loses money even when future dollars are treated as worth as much as present ones, so the shortfall is structural rather than an artefact of discounting. It follows that no positive rate of return exists for any option, and equivalently that the simple payback periods — 26.4, 23.2 and 24.2 years — all exceed the 20-year life. To break even at 10 % the options would need net annual benefits of $21,730, $25,841 and $36,412 respectively; option 1, the closest, delivers only 32 % of what it needs.

6. Run the incremental comparison for completeness. Even taking option 1 as a hypothetical defender, neither increment pays for itself:

   $$\Delta NPW_{2-1}=(9{,}500-7{,}000)(8.5136)-35{,}000=-\$13{,}716$$

   $$\Delta NPW_{3-1}=(12{,}800-7{,}000)(8.5136)-125{,}000=-\$75{,}621$$

   so the extra capital is rejected in both cases and the ordering by present worth is confirmed.

7. State the recommendation. On the numbers as printed the correct answer is to invest in none of them — the do-nothing alternative, always the implicit fourth option in a mutually exclusive set, dominates all three. If the agency is nevertheless committed to improving this corridor for reasons the table does not capture, then

   $$\boxed{\text{Option 1 is the least uneconomic, at }NPW=-\$125{,}405\ \ (AW=-\$14{,}730/\text{yr})}$$