Pump Replacement With a Sunk Study and Unequal Lives
A municipal water utility must decide whether to keep a 3-year-old booster pump or replace it. The existing pump needs an immediate overhaul costing $9,000 and then has 5 years of remaining life, a salvage value of $4,000, and annual O&M of $18,000. A new pump costs $60,000 installed, has a 12-year life, a salvage value of $8,000, and annual O&M of $9,500. Last year the utility paid a consultant $15,000 for a vibration study of the existing unit. The finance director has asked for a defensible recommendation using the utility's 9% MARR, and the operations superintendent has argued that "we already have $15,000 in the old pump, so we should run it longer."
1. Which costs belong in the comparison and which do not? State the treatment of the $15,000 study, the $9,000 overhaul, and the salvage values.
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Excluded: the $15,000 vibration study. It is a sunk cost - incurred in the past, identical under both alternatives, and unrecoverable. Its only legitimate role is as information, not as a cash flow. The superintendent's "we have $15,000 in it" reasoning is the classic sunk-cost fallacy; the study does not make the old pump cheaper to own going forward.
Included: the $9,000 overhaul. It is a future cash outflow that is required only if the old pump is kept, so it differentiates the alternatives. Treat it as a year-0 capital cost of the keep alternative.
Included: both salvage values, as future inflows at the end of each alternative's respective life. Salvage of the existing pump ($4,000 at year 5) is a genuine opportunity cost: keeping the pump forgoes selling it. The existing pump's original purchase price and book value are sunk and irrelevant.
Because the lives differ (5 vs 12 years), the comparison must be on an annual worth basis (or over a common multiple of lives), never on present worth of a single cycle.
2. Compute the annual worth of each alternative at 9% and recommend.
Read the worked answer
Keep existing (AW over 5 years):
$$AW_A = -9000(A/P,9%,5) - 18000 + 4000(A/F,9%,5)$$
- $(A/P,9%,5) = 0.2571$ → $-2{,}314$
- $(A/F,9%,5) = 0.2571 - 0.09 = 0.1671$ → $+668$
$$AW_A = -2{,}314 - 18{,}000 + 668 = -$19{,}645/\text{yr}$$
Replace (AW over 12 years):
$$AW_B = -60{,}000(A/P,9%,12) - 9500 + 8000(A/F,9%,12)$$
- $(A/P,9%,12) = 0.1397$ → $-8{,}380$
- $(A/F,9%,12) = 0.0497$ → $+397$
$$AW_B = -8{,}380 - 9{,}500 + 397 = -$17{,}482/\text{yr}$$
Recommendation: replace. The new pump saves about $2,160 per year on an equivalent-annual basis. Because the differences are expressed per year over different lives, no life-matching adjustment is needed beyond using AW.
3. The O&M estimate for the existing pump is uncertain. At what annual O&M for the existing pump does the decision reverse, and how would you present that sensitivity?
Read the worked answer
Set the two annual worths equal. The keep alternative is $AW_A = -1{,}645 - X$, where $X$ is the existing pump's O&M and $-1{,}645$ is the net of overhaul annualized less salvage annualized ($-2{,}314 + 668$).
$$-1{,}645 - X = -17{,}482 \Rightarrow X = $15{,}837/\text{yr}$$
The decision reverses only if the existing pump's annual O&M is below about $15,800 - that is, roughly $2,200 (12%) lower than the $18,000 estimate, plus enough to also absorb the uncertainty.
How to present it: state the recommendation as conditional on the estimate, and show the sensitivity band. "Replace" holds across any O&M estimate above ~$15.8k/yr. If maintenance history suggests the true figure could be $14k–$16k, the decision is genuinely close and the tie-breaker should be non-economic: reliability, outage risk, and whether the 12-year life assumption is realistic for this service. Do not bury this in a single point estimate - give the director the break-even number so the assumption is auditable.