Equivalent annual cost

The equivalent annual cost (EAC) of an asset is its total cost over a holding period $N$, expressed as the equivalent uniform annual cost — the cost-side counterpart of the annual worth. Compute it for several candidate holding periods and the minimum identifies the economic life; compare it across alternatives and the lower-EAC option wins (assuming the asset has to be held indefinitely).

The reason EAC works for unequal-life comparisons where PW does not: converting a 6-year asset’s total cost into a per-year figure implicitly assumes that the asset will be replaced by an identical copy at the end of year 6, and again after year 12, and so on indefinitely. Under that assumption every year of operation costs EAC, regardless of which “copy” you’re on. A 10-year competitor likewise turns into a per-year figure under the same assumption. Year-for-year comparison is then apples-to-apples. PW lacks this trick because the dollar totals being compared cover different numbers of years — $50k over 6 years isn’t directly comparable to $60k over 10 years.

EAC has two standard components:

Capital cost EAC. Captures the cost of buying, holding, and eventually disposing of the asset:

${\text{EAC}}_{\text{capital}}=P\cdot (A/P,i,N)-S\cdot (A/F,i,N).$

Amortise the first cost $P$ over $N$ years (that’s $P\cdot (A/P)$), then subtract the equivalent annual amount that the salvage $S$ contributes via a sinking-fund accumulation (that’s $S\cdot (A/F)$). $P$ is the asset’s first cost (purchase price + installation), $S$ is the salvage value at the end of $N$ years, $i$ is the MARR.

Using the identity $(A/P,i,N)=(A/F,i,N)+i$ (Capital recovery factor), the same expression can be rewritten as

${\text{EAC}}_{\text{capital}}=(P-S)\cdot (A/P,i,N)+S\cdot i.$

The $+S\cdot i$ term reflects the foregone interest on the salvage value — the capital $S$ is tied up in the asset until disposal and could otherwise have been earning $i$ each year.

O&M EAC. Captures the operating-and-maintenance cash flows. If O&M is a uniform $C$ per year, it’s just $C$. If O&M follows an arithmetic gradient (starting at $C_1$ and increasing by $G$ each year), use \text{EAC}_{\text{O&M}} = C_1 + G \cdot (A/G, i, N). If it’s a geometric gradient or an irregular pattern, compute the PW directly and then multiply by $(A/P,i,N)$.

Total EAC:

\text{EAC}_{\text{total}} = \text{EAC}_{\text{capital}} + \text{EAC}_{\text{O&M}}.

This is the headline number for keep-or-replace decisions.

A worked example. A $50,000 truck has $10,000 salvage after 6 years and $8,000/year in O&M costs. At $i=7\%$:

• Capital recovery: $(A/P,7\%,6)=0.2098$, sinking fund: $(A/F,7\%,6)=0.1398$.
• EAC_capital: 50{,}000(0.2098) - 10{,}000(0.1398) = 10{,}490 - 1{,}398 = \9{,}092/\text{year}$.Equivalently$(50{,}000 - 10{,}000)(0.2098) + 10{,}000(0.07) = 8{,}392 + 700 = $9{,}092/\text{year}$.
• EAC_O&M: $8,000/year
• EAC_total: 9{,}092 + 8{,}000 = \17{,}092/\text{year}$

To find the economic life, recompute EAC_total for several values of $N$ and pick the lowest.

For the underlying Capital recovery factor and Annual worth method, see those notes. For the keep-or-replace decision context see Replacement decision and Economic life.