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Repair or replace calculator

Repair or replace comes down to one number for each option: the equivalent annual cost (EAC), what the option costs per year once the price or repair bill, the running costs and the end value are spread evenly over its life at your discount rate. Enter the old machine's sell value, repair bill and rising running costs, and the new machine's price, life and costs, to see which is cheaper per year, each one's economic life and the break-even figures.

EACequalsPV × A/P

PVequalsI₀ + Σ Cₜ ÷ (1 + r)t − S ÷ (1 + r)n

A/Pequalsr divided by 1 − (1 + r)−n

P/Aequals1 − (1 + r)−n divided by r

I₀ is the cost at the start: for the old machine, the sell value given up plus the repair now; for the new one, price plus installation. Cₜ is year t's operating and maintenance (O&M) cost plus any other yearly cost, summed over years 1 to n, S the end value (negative for a removal cost), n the years and r the discount rate. A/P is the capital recovery factor and P/A, its inverse, the annuity factor; at r = 0, A/P = 1 ÷ n. Source: NIST Handbook 135, Life Cycle Costing Manual (2022), sections 3.2 and 17.4. Costs fall at the end of each year.

From Finance: the return your company requires, before tax.

Keep and repair the old machine

What it would fetch as is, net of removal. Negative if removal costs more.

The overhaul or repair needed to keep it running.

How long it can run after the repair.

Maintenance labour and parts, energy, consumables.

Added each year. Blank is 0.

Optional: downtime, scrap, overtime you can price. See below.

Resale or scrap value at the end. Negative for a removal cost.

Replace it with a new machine

Delivered price of the new machine.

Rigging, foundation, wiring, trials. Blank is 0.

Its full expected life, not the old one's years left.

Maintenance labour and parts, energy, consumables.

Added each year. Blank is 0.

Optional: downtime, scrap, overtime you can price. See below.

Resale or scrap value at the end. Negative for a removal cost.

Pre-filled with the worked example (illustrative). Before tax: income tax and depreciation are left out. What the old machine cost and its book value are sunk and do not go in.

Replacing saves

$4,826/yr

Equivalent annual cost at 10%: keep $57,289, replace $52,463 a year.

Keep: EAC

$57,289

a year over 4 years

Replace: EAC

$52,463

a year over 10 years

Equivalent annual cost of each option, split into capital, O&M and other costs
A yearKeepReplace
Capital, net of end value$9,764$26,737
O&M$47,525$25,725
Other yearly costs$0.00$0.00
EAC$57,289$52,463

Each at its economic life

Lowest EAC: old machine $57,203 if kept 3 years, new machine $52,463 if kept 10 years. Same answer.

Resale values assumed to fall in a straight line to the end value.

One more year with the old machine?

Next year costs $66,300 (repair, value given up, interest and O&M) against the new machine's lowest EAC of $52,463. On this test, replace now.

Break-even O&M next year (old)

$37,174

Keeping wins below this, all else the same.

Break-even price (new)

$189,654

Replacing wins below this price.

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The working, step by step

Each year's cost is discounted to today with 1 ÷ (1 + 10%)t, added up with the time 0 costs and less the end value, then spread evenly over the life with the capital recovery factor A/P = r ÷ (1 − (1 + r)−n).

Keep and repair: 4 years

Keep and repair: yearly costs, discount factor and present value
YearCost1 ÷ (1+r)^tPresent value
0: Sell value given up + repair now$33,0001.0000$33,000
1$42,0000.9091$38,182
2$46,0000.8264$38,017
3$50,0000.7513$37,566
4$54,0000.6830$36,883
4: end value-$3,0000.6830-$2,049

Present value of costs = $181,598
Annuity factor P/A (10%, 4) = 3.1699, so A/P = 1 ÷ 3.1699 = 0.31547
EAC = $181,598 × 0.31547 = $57,289 a year

If kept fewer years: the economic life
Keep and repair: year-by-year cost and equivalent annual cost if kept that many years
Years keptO&MResaleThat yearEACNo discount
1$42,000$12,000$66,300$66,300$63,000
2$46,000$9,000$50,200$58,633$56,000
3$50,000$6,000$53,900$57,203$55,000
4$54,000$3,000$57,600$57,289$55,500

Economic life: 3 years, the lowest EAC ($57,203). Resale is a straight line from $15,000 to the end value. "That year" is the cost of keeping it one more year: value at the start × (1 + r), less the value at the end, plus that year's costs. "No discount" is total cost ÷ years, the EAC at 0%.

Replace: 10 years

Replace: yearly costs, discount factor and present value
YearCost1 ÷ (1+r)^tPresent value
0: Price + installation$172,0001.0000$172,000
1$22,0000.9091$20,000
2$23,0000.8264$19,008
3$24,0000.7513$18,032
4$25,0000.6830$17,075
5$26,0000.6209$16,144
6$27,0000.5645$15,241
7$28,0000.5132$14,368
8$29,0000.4665$13,529
9$30,0000.4241$12,723
10$31,0000.3855$11,952
10: end value-$20,0000.3855-$7,711

Present value of costs = $322,361
Annuity factor P/A (10%, 10) = 6.1446, so A/P = 1 ÷ 6.1446 = 0.16275
EAC = $322,361 × 0.16275 = $52,463 a year

If kept fewer years: the economic life
Replace: year-by-year cost and equivalent annual cost if kept that many years
Years keptO&MResaleThat yearEACNo discount
1$22,000$146,000$65,200$65,200$48,000
2$23,000$132,000$51,600$58,724$42,500
3$24,000$118,000$51,200$56,451$41,000
4$25,000$104,000$50,800$55,233$40,500
5$26,000$90,000$50,400$54,442$40,400
6$27,000$76,000$50,000$53,866$40,500
7$28,000$62,000$49,600$53,416$40,714
8$29,000$48,000$49,200$53,048$41,000
9$30,000$34,000$48,800$52,735$41,333
10$31,000$20,000$48,400$52,463$41,700

Economic life: 10 years, the lowest EAC ($52,463). Resale is a straight line from $160,000 to the end value. "That year" is the cost of keeping it one more year: value at the start × (1 + r), less the value at the end, plus that year's costs. "No discount" is total cost ÷ years, the EAC at 0%.

Next step

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Before you enter numbers

Numbers that maintenance and finance will both accept

Sell value today. Get a quote from a used machinery dealer or an auction house for the machine as it stands, less rigging and removal. Keeping the machine means not taking that money, so it is the first cost of keeping it. The calculator charges it to keeping, so do not also take it off the new machine's price.

O&M from the repair history. Add up two or three years of work orders for this asset (labour, parts, contractors), plus energy and consumables, and look at how fast the total is rising. The rise matters as much as the level: it is what usually tips the answer.

Each machine's own life. The old machine gets the years it can run after the repair. The new one gets its full expected service life, from the maker or your own history with similar machines.

The rate from Finance. Use the rate your company applies to capital spending, and keep it consistent with the cost figures: a real rate with costs in today's money, a nominal rate with costs that include inflation (NIST Handbook 135, section 3.1).

Before tax. The calculator leaves out income tax and depreciation. They can change the answer, so treat this as the agreed operating case and ask Finance for the after-tax version before a large purchase.

Worked example

Illustrative numbers, not a benchmark

A 12-year-old mechanical press needs a clutch and brake rebuild quoted at $18,000. A used machinery dealer would pay $15,000 for it as is. The maintenance records point to $42,000 of O&M next year, rising about $4,000 a year, and the team expects 4 more years from it after the rebuild, with $3,000 of scrap value at the end. A new press costs $160,000 plus $12,000 to install, should last 10 years, and costs $22,000 a year to run and maintain at first, rising $1,000 a year, with $20,000 resale at the end. Finance uses 10%. The press was bought for $140,000 and the books still carry it at $28,000; neither figure goes in.

  1. 1Keep, at the start: $15,000 sell value given up + $18,000 rebuild = $33,000.
  2. 2Keep, present value: 33,000 + 42,000 ÷ 1.1 + 46,000 ÷ 1.1² + 50,000 ÷ 1.1³ + 54,000 ÷ 1.1⁴ − 3,000 ÷ 1.1⁴ = 33,000 + 150,646.81 − 2,049.04 = $181,597.77.
  3. 3Keep, EAC: A/P = 0.1 ÷ (1 − 1.1⁻⁴) = 0.31547, so 181,597.77 × 0.31547 = $57,289 a year.
  4. 4Replace: 172,000 + O&M of 22,000 to 31,000 discounted (158,071.82) − 20,000 ÷ 1.1¹⁰ (7,710.87) = $322,360.95. A/P over 10 years = 0.16275, so the EAC = $52,463 a year.
  5. 5Replacing is $4,826 a year cheaper. At each machine's economic life the answer holds: the old press's lowest EAC is $57,203 if kept 3 years, the new one's $52,463 at 10. One more year of the old press would cost $66,300, more than $52,463.
  6. 6Break-even: keeping wins only if next year's O&M on the old press were under $37,174 with the same $4,000 rise, or if the new press cost more than $189,654.
Replace: $52,463 against $57,289 a year, $4,826 a year cheaper before tax. The calculator above is filled in with this example. The numbers are illustrative.

Sunk costs, and why the old machine starts at its sell value

What the press cost when it was bought and the $28,000 still on the books are sunk: they were spent in the past, and nothing decided today changes them. NIST Handbook 135 tells analysts to leave sunk costs out (section 2.4.1.1), and OMB Circular A-94 says sunk costs should be ignored when deciding whether a new investment is worthwhile.

The value that does count is what the old machine would fetch today. Engineering economics texts call this the outsider view (White, Case, Pratt and Agee call it the opportunity cost approach, chapter 11 of Principles of Engineering Economic Analysis): look at the old machine as if you were buying it today at its market value. Both options then start with money spent now, and they can be compared on the same footing.

Book value still matters to Finance. Selling below it books a loss, and after tax that loss changes the tax bill, which is one reason the after-tax answer can differ from this one. It does not change the cash before tax.

Defender and challenger. Replacement studies call the machine you have the defender and the one that might replace it the challenger. The textbook test has two parts, and the calculator shows both: compare each one's EAC, ideally at its economic life, and check whether one more year of the defender costs less than the challenger's lowest EAC. If it does, keep it a year and look again.

Mistakes that flip the answer

  • Comparing the price with the repair bill. $160,000 against $18,000 makes the rebuild look like the easy choice. Per year it is the other way round: the old press's rising O&M makes keeping it $4,826 a year dearer.
  • Ignoring end values. Leave out the new press's $20,000 resale and its EAC rises by $1,255 a year (20,000 × 0.06275). Leave out the old press's $15,000 sell value today and keeping looks $4,732 a year cheaper than it is (15,000 × 0.31547).
  • Using the same remaining life for both. Price the new press over the old one's 4 years with no resale and its EAC comes out at $77,642, so keeping looks $20,353 a year cheaper. Give each machine its own life; that is what EAC is for.
  • Counting the sell value twice. Charge it to keeping, or take it off the new price, not both.
  • Mixing inflation in on one side only. If the O&M figures include future price rises, the rate must too, and the new machine's costs must be inflated the same way.
  • Adding loan interest as a cost. The discount rate is already the charge for the money. Adding interest payments on top counts it twice; a specific loan or lease offer is better compared as its own option.

What the calculator leaves out, and how to price it

EAC only weighs what goes into it. These are the usual gaps; each can go into "Other yearly costs" once someone can back up a figure from the plant's own records. Leave the box at zero rather than guess: a figure nobody can defend is the first thing Finance strikes out.

  • Downtime risk. Expected breakdown hours a year (from MTBF and MTTR) times the cost of an hour of lost output (from the downtime cost calculator).
  • Capacity. If the old machine cannot make the volume, the overtime, outsourcing or lost margin it causes each year.
  • Quality. The scrap and rework this machine causes each year, priced as in the cost of poor quality calculator.
  • Safety. Guarding or controls the old machine needs to stay in service go into the repair cost now. A hazard you cannot put a price on still counts: it belongs in the decision, outside the EAC.
  • Financing, tax and depreciation. Left out on purpose; see the mistakes above and the FAQ.

How maintenance and finance can agree on a decision

  1. Agree the discount rate, and whether it is real or nominal, before anyone sees a result.
  2. Maintenance brings the evidence: the repair quote, two or three years of work order costs for the asset, a dealer's sell value and the new machine's quote with installation. Finance brings the rate and the tax view.
  3. Agree the two lives: what the old machine has left after the repair, and the new one's full service life.
  4. Run it here before tax and read the break-even figures. If the answer only flips when next year's O&M is $5,000 lower, argue about that one number, not the whole model.
  5. Finance reruns it after tax with the depreciation schedule. If the answer changes, the reason is now visible.
  6. Write it up on a CapEx request form with both EACs, the break-even figures and what was left out.

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<iframe src="https://www.theleansuite.com/tools/repair-or-replace-calculator/embed" title="Repair or replace calculator by LeanSuite" width="100%" height="4450" style="border:0;max-width:1000px" loading="lazy" allow="clipboard-write"></iframe>
<p style="font:13px/1.4 sans-serif"><a href="https://www.theleansuite.com/tools/repair-or-replace-calculator">Repair or replace calculator</a> by LeanSuite</p>

How LeanSuite helps

LeanSuite does not do the capital analysis. In LeanSuite, technicians close each work order with the fix and the parts used, building a searchable repair history for every asset, and breakdown work orders record downtime and MTTR. That history is where the old machine's O&M figure and its yearly rise should come from.

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FAQ

Repair or replace calculator: common questions

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