Engineering Economics: 16-Civ-B8 Management of Construction, December 2015, Q4

Question 4: Engineering Economics (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. Two mutually exclusive projects with identical 12-year lives, appraised at a discount rate of 10 per cent per year. No revenue or benefit stream is stated, so the cash flows are costs only.

Given data — cost cash flows (CAD)
Item	Project A	Project B
Initial investment (year 0)	62,000	80,000
Yearly operating cost (years 1–12)	4,500	7,000
Major maintenance	15,000 every 3 years	13,000 every 4 years
Service life	12 years	12 years
Discount rate	10 per cent per year

Find. The present worth of each alternative over the common 12-year study period and, from the comparison, the more economical plan.

Figure 4.1 — Project A. Downward arrows are disbursements: the initial investment at year 0, the uniform annual operating cost in years 1 to 12, and major maintenance in years 3, 6, 9 and 12 (drawn upward only to separate it from the annual series).

Figure 4.2 — Project B, drawn to the same convention, with major maintenance in years 4, 8 and 12.

Approach. Both alternatives already share a 12-year life, so no least-common-multiple or repeatability argument is needed: discount each cost stream to time zero with the uniform-series and single-payment factors and compare the totals directly.

Evaluate the discount factors once. The uniform series runs the full twelve years, so $$(P/A,10\%,12) = \frac{1-(1.10)^{-12}}{0.10} = \frac{1-0.318631}{0.10} = 6.8137$$ and the single-payment factors needed for the maintenance events are $(P/F,10\%,3)=0.7513$, $(P/F,10\%,4)=0.6830$, $(P/F,10\%,6)=0.5645$, $(P/F,10\%,8)=0.4665$, $(P/F,10\%,9)=0.4241$ and $(P/F,10\%,12)=0.3186$.
Present worth of Project A. Major maintenance every three years over a twelve-year life falls in years 3, 6, 9 and 12, four events in all. Their combined discount factor is $0.7513+0.5645+0.4241+0.3186 = 2.0585$, so $$\begin{aligned} PW_A &= 62{,}000 + 4{,}500\,(6.8137) + 15{,}000\,(2.0585)\\ &= 62{,}000 + 30{,}662 + 30{,}878 = 123{,}539 \end{aligned}$$ It is worth noticing that the maintenance programme is as expensive in present-worth terms as twelve years of routine operation, which is not obvious from the table of raw figures.
Present worth of Project B. Major maintenance every four years falls in years 4, 8 and 12, three events, with combined factor $0.6830+0.4665+0.3186 = 1.4682$: $$\begin{aligned} PW_B &= 80{,}000 + 7{,}000\,(6.8137) + 13{,}000\,(1.4682)\\ &= 80{,}000 + 47{,}696 + 19{,}086 = 146{,}782 \end{aligned}$$
Compare and choose. Both figures are present worths of cost, so the more economical plan is the smaller of the two: $$\boxed{PW_A = \text{CAD } 123{,}539 \;\lt\; PW_B = \text{CAD } 146{,}782 \;\Rightarrow\; \text{choose Project A}}$$ The advantage is $23,242 in present worth, about 16 per cent of Project B’s total, which is far too large to be overturned by rounding or by a modest change in the discount rate.
Cross-check on an annual basis. Dividing each present worth by the same series factor converts it to an equivalent uniform annual cost: $$\begin{aligned} EUAC_A &= \frac{123{,}539}{6.8137} = 18{,}131\ \text{per year}\\ EUAC_B &= \frac{146{,}782}{6.8137} = 21{,}542\ \text{per year} \end{aligned}$$ The annual-cost comparison reproduces the same ranking with a difference of $3,411 per year, as it must when the two lives are equal; the check is worth one line because it is the comparison that would be required had the lives differed.

The result is easy to explain in engineering terms. Project B costs $18,000 more to build and $2,500 more to run every year, and it buys back only a less frequent maintenance cycle worth about $11,800 in present worth. The higher first cost would have to purchase a benefit — extra capacity, higher availability, longer residual life — and the question states none.

Check: two modelling assumptions worth stating on the answer paper. (1) The question asks for the “most economical plan using present value profit”, but gives no revenue or benefit stream; the appraisal is therefore a present-worth-of-cost comparison, and the “profit” of choosing A is the $23,242 in avoided present-worth cost. Nothing here implies either project is profitable in itself. (2) “Every 3 years” and “every 4 years” both place a maintenance event in year 12, the year the asset is retired. Taken literally — the reading used above — the charge is incurred; if instead the year-12 event is dropped as falling at retirement, $PW_A = 118{,}760$ and $PW_B = 142{,}640$, and Project A still wins by $23,880. The ranking is insensitive to the assumption, so state it and proceed. Neither project is salvaged at end of life, since no salvage value is given.

Final results — Question 4
Quantity	Project A	Project B
Present worth of first cost	$62,000	$80,000
Present worth of operating cost	$30,662	$47,696
Present worth of major maintenance	$30,878	$19,086
Total present worth of cost	$123,539	$146,782
Equivalent uniform annual cost	$18,131/yr	$21,542/yr
Decision at 10 per cent	Project A is the more economical plan, by $23,242 present worth

Topic. Present-worth comparison of mutually exclusive alternatives with equal lives and periodic (non-annual) disbursements.

Key relations. $(P/A,i,n) = \dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,k) = (1+i)^{-k}$; $PW = P + A\,(P/A,i,n) + \sum_k M\,(P/F,i,k)$; $EUAC = PW/(P/A,i,n)$; with unequal lives, compare over the least common multiple of lives or on an annual-worth basis.

Why this works. Discounting converts payments made at different dates into one figure at a common date, which is the only basis on which two spending patterns can be compared. A periodic charge that does not recur every year cannot use the uniform-series factor at all; it must be discounted event by event, or converted with the sinking-fund factor into an equivalent annual charge. Because both alternatives here span the same twelve years and face the same rate, whichever has the lower present worth also has the lower annual worth, and the two comparisons cannot disagree.

Common pitfall. Applying $(P/A)$ to the major-maintenance amount as though it were annual, which roughly triples its weight; and mis-counting the periodic events — a 12-year life at a 3-year interval gives four charges, not three, and at a 4-year interval three, not two. Where the lives differ, the further trap is comparing present worths taken over each project’s own life, which silently compares twelve years of one asset against eight of another.

Source. Fraser et al., Global Engineering Economics, 5th Canadian ed., Ch. 4 and 5 (present worth, annual worth, repeated lives); Halpin & Senior, Construction Management, 4th ed., Ch. 16.

Question 4 of 6: Engineering Economics

Question 4: Engineering Economics (20 marks)