Two machines do the same duty. One is expensive and lasts fifteen years; the other is cheap and lasts eight. Nearly every instinct available to you is wrong here, and there is exactly one honest comparison.

Equivalent annual cost: $\mathit{EAC} = P\,\dfrac{i(1+i)^{n}}{(1+i)^{n} - 1} + M$, or in plain words EAC equals P times CRF, plus M. $P$ is the installed capital cost at time zero, $i$ the rate per year as a decimal, $n$ the service life in years, and $M$ the operating cost per year — energy, maintenance, consumables, labour. The first term annualises the capital; the second is already annual and never touches the factor. The answer is dollars per year of ownership, everything included.

Now the trap, and it is the reason this lesson exists. Discount each machine over its own life and the long-lived one always shows the bigger total — because it is buying more years. Comparing those two totals is comparing a fifteen-year lease with an eight-year lease on the sticker price alone. EAC fixes it by dividing each whole-life cost by the service it delivers, so both bids end up in the same currency: cost per year of duty. Present worth and EAC rank projects identically when the lives MATCH; when the lives differ, only EAC is telling the truth.

One assumption is being made on your behalf, so state it out loud: EAC assumes each machine is replaced by its own kind at the end of its life, indefinitely. That is usually reasonable for a pump and plainly silly for a technology about to be superseded. When it is silly, the fix is a common study period, not a different factor.

The nugget for the field: the cheaper machine wins the tender and loses the decade. First cost is one line of four — the rate, the life and the running cost are the other three, and on any duty that runs continuously the running cost usually dwarfs everything else. That is not an argument for always buying the expensive one. It is an argument for putting all four lines on the page before anybody signs.