Two Pumps, Unequal Lives, and a Payback Argument
Your plant is replacing a boiler feed pump at end of life. Two proposals are on the table, evaluated at a 9% MARR in constant dollars. Option A, a standard pump: $180,000 installed, 8-year life, $22,000/year O&M, $15,000 salvage. Option B, a premium alloy pump: $260,000 installed, 12-year life, $14,000/year O&M, $30,000 salvage. Production demand is expected to continue indefinitely, and either unit can be replaced like-for-like at end of life at today's real costs. The finance manager recommends Option A because it is $80,000 cheaper up front and 'pays back in under four years.' The plant manager is inclined to agree because capital is tight this year.
1. Compute the annual equivalent cost of each option and state which is preferred.
Read the worked answer
Annualize with capital recovery plus salvage interest.
Option A
- (A/P, 9%, 8) = 0.18067
- CR = (180,000 - 15,000)(0.18067) + 15,000(0.09) = 29,810 + 1,350 = $31,160
- O&M = $22,000
- EAC(A) = $53,160/year
Option B
- (A/P, 9%, 12) = 0.13966
- CR = (260,000 - 30,000)(0.13966) + 30,000(0.09) = 32,122 + 2,700 = $34,822
- O&M = $14,000
- EAC(B) = $48,822/year
B is cheaper by about $4,340 per year, every year, indefinitely. On a 24-year common horizon that is roughly $4,340 x (P/A, 9%, 24) = about $43,000 in present value in B's favor - which is also the answer you get by repeating each option over a 24-year common horizon and discounting. Both methods agree; EAC is simply faster.
2. Evaluate the finance manager's payback reasoning.
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The payback claim is arithmetically wrong, and the metric is the wrong tool.
- The number: incremental first cost is $260,000 - $180,000 = $80,000; incremental O&M saving is $22,000 - $14,000 = $8,000/year. Simple payback on the increment is 10 years, not under four. The four-year claim appears to compare the $80,000 to total O&M rather than to the saving.
- Payback ignores everything after the cutoff - salvage, the remaining life, and the 4 extra years of savings B delivers.
- Payback penalizes long-lived assets by construction. It systematically selects the option you will replace soonest.
- Comparing raw first costs across unequal lives is the same error in a different suit. A 12-year asset is not comparable to an 8-year asset without annualizing or using a common horizon.
The honest framing for the plant manager: capital is tight this year, but B costs about $4,340 less per year to own and operate, forever. That is a financing question (can we fund the extra $80,000 now?), not an economic one. If capital is genuinely rationed, say so explicitly and treat the $4,340/year as the cost of that rationing - then compare it against other uses of the $80,000.
3. What non-economic factors could legitimately overturn this ranking?
Read the worked answer
A defensible reversal needs a stated mechanism, not a preference:
- Expected remaining plant life. If the plant is expected to close or be reconfigured in 6-7 years, the 12-year life of B is never realized and the annualization is invalid. Recompute over the actual horizon.
- Downtime cost. If the standard pump fails more often and unplanned outage costs $25,000/day, a modest reliability difference swamps the $4,340. Quantify MTBF and mean time to repair for both and price the difference.
- Alloy suitability. If the service is genuinely corrosive, the standard pump's 8-year life may be optimistic and its O&M may escalate - which strengthens B. If the alloy is over-specified, the reverse.
- Capital rationing. As above: a real constraint, but it should be disclosed and priced, not used to justify a metric error.
- Spares and lead time. One unit may need a stock spare ($30k-$60k tied up) or a 40-week lead time that threatens availability.
Recommendation: keep B as the economic answer, present the sensitivity in one table (life horizon, downtime cost, O&M escalation), and let the plant manager make a constraint decision with the price of the constraint on the page.