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Advanced Financial Management · Advanced Capital Budgeting Decisions

Replacement, Equivalent Annual Cost and Unequal Lives

Updated 5 October 2026 · Fact-checked

Equivalent annual cost (EAC) converts the present value of an asset's costs into a level yearly amount over its life. Equivalent annual benefit (EAB) does the same for NPV. To compare projects with unequal lives, compute EAC or EAB for each, then pick the lowest EAC or highest EAB. A replacement chain over the common life gives the same ranking.

Understand Replacement, Equivalent Annual Cost and Unequal Lives

Plain NPV comparison fails when mutually exclusive projects have different lives. A project that runs 5 years collects more cash flows than one that runs 3 years, so its NPV looks bigger or its cost looks higher. That is not a fair comparison. The real question is: which option is better per year, assuming you can repeat it?

The fix is to put both projects on a yearly basis. Equivalent annual cost (EAC) is the level annual amount whose present value equals the present value of all costs of the asset over its life. Use it when the options only have costs, such as two machines that do the same job. The lower EAC wins.

Equivalent annual benefit (EAB) is the level annual amount whose present value equals the project's NPV. Use it when the projects have revenues and costs. The higher EAB wins.

The other route is the replacement chain. You assume each project is repeated until both end together, at the least common multiple of the lives (for example 6 years for lives of 2 and 3). You then add the NPVs of every repetition, discounted to today, and compare. Both methods assume the project can be replaced at the same cost and cash flows. If the question says costs will rise or technology will change, say so and adjust.

Replacement decisions ask whether to keep an old asset or replace it now. Treat it as a choice between two options. Include the sale value of the old asset (and any tax on it), the cost of the new asset, and the operating costs of each. Compare on an annual basis if the remaining life of the old asset differs from the life of the new one.

Key rules to remember

Annuity factor
PVAF(r, n) = [1 − (1 + r)^−n] ÷ r
Use the table value if the question gives one. Use the same rate throughout.
Equivalent annual cost (EAC)
EAC = PV of all costs over the life ÷ PVAF(r, n)
PV of costs = initial outlay + PV of operating costs + PV of any later repairs − PV of salvage. Choose the lowest EAC.
Equivalent annual benefit (EAB)
EAB = NPV ÷ PVAF(r, n)
Use when the project has inflows. Choose the highest EAB, only among projects with positive NPV.
Replacement chain NPV (finite)
Chain NPV = NPV × [1 + (1 + r)^−n + (1 + r)^−2n + ...] up to the common life
Each repetition's NPV is discounted back by the number of years elapsed before it starts.
Infinite replacement NPV
NPV∞ = NPV × (1 + r)^n ÷ [(1 + r)^n − 1]
Applies when the project is repeated forever with the same cash flows. It ranks projects the same way as EAB.

How to solve Replacement, Equivalent Annual Cost and Unequal Lives questions

Use this method for any question on unequal lives or replacement. It works for cost-only and benefit-based cases.

  1. 1Read what is being compared: cost-only options (use EAC) or projects with inflows (use EAB or NPV chain). Note each life and the discount rate.
  2. 2Build the after-tax cash flows for each option. Include the initial outlay, operating costs, tax shield on depreciation (if tax is given), and salvage value in the final year.
  3. 3For the replacement of an old asset, include the sale proceeds of the old asset and the tax effect of any gain or loss on sale.
  4. 4Discount the cash flows at the given rate. Get the PV of costs (for EAC) or the NPV (for EAB).
  5. 5Divide by the annuity factor for that option's own life. This gives EAC or EAB.
  6. 6Rank the options: lowest EAC or highest EAB wins. If a replacement chain is asked for, compute the chain NPV over the common life and compare.
  7. 7State the decision in one sentence and list the assumptions: the project can be repeated with the same cash flows, and the discount rate stays constant.

Quickest way: Annual-equivalent shortcut

When to use it: Use it when the question has two or three options with different lives and gives annuity factors. It is faster than building a chain over the common life.

  1. Write the PV of costs (or NPV) for each option in one line.
  2. Look up the annuity factor for each life and divide.
  3. Compare the annual figures. Lowest EAC or highest EAB wins.
  4. If the examiner asks for a chain, cross-check with one line: NPV × (1 + discount factor of repeat years).
  5. Write the assumption sentence, which carries marks.

Common mistakes in Replacement, Equivalent Annual Cost and Unequal Lives

  • Comparing raw NPVs or total PV of costs of projects with different lives.

    It is the method used for equal lives, and students apply it without checking the lives.

    Fix: Check the lives first. If they differ and the options are mutually exclusive and repeatable, convert to EAC or EAB.

  • Dividing by the annuity factor of the wrong life, for example using the longer life for both projects.

    Students pick one table row and reuse it.

    Fix: Divide each project by PVAF for its own life, at the same rate.

  • Choosing the highest EAC or lowest EAB.

    Students mix up the cost and benefit rule.

    Fix: EAC is a cost, so lower is better. EAB is a benefit, so higher is better.

  • Ignoring salvage value, or adding it to costs instead of subtracting it.

    Salvage appears at the end and is forgotten, or its sign is confused.

    Fix: Salvage is an inflow in the final year. Subtract its present value from the PV of costs.

  • Treating the old asset's original cost as a relevant cash flow in a replacement decision.

    The book value is given in the question, so it looks important.

    Fix: Original cost is sunk. Only the sale proceeds and the tax on sale of the old asset matter, not the book value itself.

  • Forgetting to state the assumption of repeatability.

    Students stop after the numbers.

    Fix: Add one line: the method assumes the asset can be replaced at the same cost and cash flows, or it may not hold if costs or technology change.

Worked examples

Example 1

A firm needs a machine for a long period and compares two options at a cost of capital of 10%. Machine A costs ₹10,00,000, lasts 3 years, and has operating costs of ₹2,00,000 a year. Machine B costs ₹15,00,000, lasts 5 years, and has operating costs of ₹1,50,000 a year. Neither has salvage value. Ignore tax. Which machine should the firm choose? Given PVAF at 10%: 3 years = 2.4869; 5 years = 3.7908.

Show the solution
  1. Cost-only options with different lives, so use EAC.
  2. Machine A: PV of operating costs = 2,00,000 × 2.4869 = ₹4,97,380. Total PV of costs = 10,00,000 + 4,97,380 = ₹14,97,380.
  3. EAC of A = 14,97,380 ÷ 2.4869 ≈ ₹6,02,107. Check: the outlay spread over 3 years is 10,00,000 ÷ 2.4869 ≈ 4,02,107, and adding 2,00,000 gives 6,02,107.
  4. Machine B: PV of operating costs = 1,50,000 × 3.7908 = ₹5,68,620. Total PV of costs = 15,00,000 + 5,68,620 = ₹20,68,620.
  5. EAC of B = 20,68,620 ÷ 3.7908 ≈ ₹5,45,696. Check: 15,00,000 ÷ 3.7908 ≈ 3,95,696, and adding 1,50,000 gives 5,45,696.
  6. Compare: B has the lower EAC (₹5,45,696 against ₹6,02,107). Its total PV of costs is higher, but it is spread over more years.

Answer: Choose Machine B. Its EAC is about ₹5,45,696 a year against about ₹6,02,107 for Machine A. This assumes either machine can be replaced at the same cost and operating cost.

Example 2

A company can take up only one of two mutually exclusive projects, each repeatable with the same cash flows. Project X has NPV ₹1,20,000 and a life of 2 years. Project Y has NPV ₹1,50,000 and a life of 3 years. The cost of capital is 10%. Compare them using EAB and verify with a replacement chain over 6 years. Given PVAF at 10%: 2 years = 1.7355; 3 years = 2.4869. Discount factors at 10%: year 2 = 0.8264; year 3 = 0.7513; year 4 = 0.6830.

Show the solution
  1. Project X: EAB = 1,20,000 ÷ 1.7355 ≈ ₹69,144.
  2. Project Y: EAB = 1,50,000 ÷ 2.4869 ≈ ₹60,316.
  3. X has the higher EAB, so X is better per year.
  4. Verify with a 6-year chain (the LCM of 2 and 3). X is repeated 3 times, starting at years 0, 2 and 4: chain NPV = 1,20,000 × (1 + 0.8264 + 0.6830) = 1,20,000 × 2.5094 = ₹3,01,128.
  5. Y is repeated 2 times, starting at years 0 and 3: chain NPV = 1,50,000 × (1 + 0.7513) = 1,50,000 × 1.7513 = ₹2,62,695.
  6. The chain also favours X, which agrees with the EAB ranking. Y's higher single-cycle NPV does not make it better.

Answer: Choose Project X. Its EAB is about ₹69,144 against about ₹60,316 for Y. Its 6-year chain NPV is about ₹3,01,128 against about ₹2,62,695 for Y.

Exam tips

  • Look for the signal words 'mutually exclusive', 'different lives' or 'replace'. They tell you to move away from a plain NPV comparison.
  • Show the annuity factor division clearly. Marks are given for the method even if the table value differs slightly.
  • Write the repeatability assumption in one line at the end. Examiners often reward it.
  • In replacement cases, list relevant cash flows first (sale proceeds, tax on sale, new outlay, savings) and cross out sunk costs.
  • If both EAC and a chain are possible, do EAC first. Use the chain only if the question asks for it.

Practice questions from Advanced Capital Budgeting Decisions

Replacement, Equivalent Annual Cost and Unequal Lives: frequently asked questions

What is the difference between EAB and EAC?

EAC spreads the present value of costs into a yearly amount and is used for cost-only options, where lower is better. EAB spreads the NPV into a yearly amount and is used for projects with inflows, where higher is better. Both divide by the annuity factor for the project's own life.

When should I use the equivalent annual cost method?

Use it when you compare mutually exclusive assets that do the same job but have different lives, such as two machines. It also suits the replacement decision when the old and new assets have different remaining lives. It assumes the asset can be replaced on the same terms.

How do I compare projects with unequal lives?

Either convert each NPV to EAB and pick the highest, or build a replacement chain to the common life (the LCM) and compare the total NPVs. Both give the same ranking if the same assumptions are used. EAB is quicker when the common life is long.

Do I include the old machine's book value in a replacement decision?

No. Book value is a sunk figure. Include the sale proceeds of the old machine, and any tax on the gain or loss on sale, as relevant cash flows.