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Equivalent Annual Cost

Also known as EAC · equivalent uniform annual cost · EUAC · annualised cost · annual owning and operating cost · life cycle cost per year

EAC=P i (1+i)n(1+i)n−1+M\mathit{EAC} = P\,\frac{i\,(1+i)^{n}}{(1+i)^{n} - 1} + M
$
$
yr
$

Worked example: $100,000 over 5 years at 10% plus $12,000/yr → $38,379.75 — press Try an example to run it live, then adjust anything.

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See also
Capital Recovery Factor — The fraction multiplying P here is exactly the capital recovery factor.Net Present Value of a Uniform Annual Cash Flow — NPV and EAC rank projects identically; EAC is the one that compares unequal lives fairly.
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UniversityApplied Field Engineering

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Equivalent Annual Cost explained

PMEACin

Equivalent annual cost puts the capital charge and the running cost on the same yearly footing: EAC=P×CRF+M\mathit{EAC} = P \times \mathit{CRF} + M. A $100,000 machine over five years at 10% carries a capital charge of $26,380 a year, and with $12,000 of energy and maintenance the true cost of owning it is $38,380 a year. That is the number to put beside a rental quote or a subcontract price, and it is usually a shock the first time.

EAC exists mainly to solve one problem that NPV handles badly, which is comparing assets with unequal lives. A cheap pump lasting six years and an expensive one lasting eighteen cannot be compared by present worth without inventing a replacement chain out to a common horizon. Annualise both and the comparison is immediate and fair, because EAC already carries the life in its denominator.

Two cautions. The comparison only holds if you genuinely intend to replace like with like at the end, since EAC quietly assumes an indefinite chain of identical replacements. And the answer is very sensitive to the interest rate on long-lived assets: run the same $100,000 over twenty years and the capital charge is $11,750 a year at 10% but $8,020 at 5%. Where the discount rate came from deserves as much scrutiny as the equipment quote.

Equivalent Annual Cost formula

EAC=P i (1+i)n(1+i)n−1+M\mathit{EAC} = P\,\frac{i\,(1+i)^{n}}{(1+i)^{n} - 1} + M
Where
  • EAC\mathit{EAC}= Equivalent annual cost ($)
  • PP= Capital cost ($)
  • ii= Interest rate per year
  • nn= Service life in years (yr)
  • MM= Operating cost per year ($)

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