Why a repair quote and a purchase price cannot be compared directly
A $350 repair and a $1,200 replacement look like a four-to-one decision, and people decide on that ratio all the time. It is the wrong comparison, because the two numbers buy different things: the repair buys perhaps four more years on a machine that is already nine years old, and the purchase buys fourteen years on a machine that is new. Divide each by what it buys and the gap narrows sharply — $87.50 a year against $92.86 a year in that example.
Then add running cost, which is the term most people leave out entirely and which is often larger than either capital figure. A refrigerator using 700 kWh a year at $0.17 costs $119 a year to run; a modern equivalent at 350 kWh costs $59.50. That $59.50 difference is a permanent annual advantage to the new machine, and it is why the total comparison in the example above flips from a near-tie on capital to a clear $54 a year in favour of replacing.
So the calculation has exactly three moving parts: annualised capital on each path, running cost on each path, and the number of years the repair honestly buys. The third is the one you are least certain about and the one the answer is most sensitive to, which is why the table in the results sweeps it rather than trusting a single value.
The three rules of thumb, and what each one actually tests
The 50% rule says replace when the repair quote exceeds half the price of a new unit. It is a capital-only test: it compares one lump against another and ignores both the service life and the energy. It is popular because it needs two numbers you already have, and it is a reasonable screen — but it has no view on whether the machine will last another year or another eight, which is the actual question.
The 50-50 rule adds age: replace if the machine is more than halfway through its typical life and the repair exceeds half the replacement cost. This is closer to what the annualised comparison does, because age is a proxy for how few years the repair will buy.
The annualised comparison on this page is the general form both rules approximate. Divide each one-off cost by the years it purchases, add each machine's annual electricity, and compare. It reduces to the 50% rule when the two machines use the same energy and the repair buys exactly half the new unit's life — which is roughly the case the rule was built around, and precisely why the rule works as often as it does.
Payback answers a different question again: how long the energy saving alone takes to repay the extra you spend by buying rather than repairing. Gap divided by annual saving. A payback longer than the new machine's life means the energy saving never repays the difference, and the purchase has to be justified on reliability or features instead. In the default example the payback is 16 years against a 14-year life — so replacing is still the cheaper path per year, but not because of the energy.
The break-even repair quote inverts the whole thing. Set the two annual costs equal and solve for the repair price: (replace cost per year − old machine's annual electricity) × years the repair buys. It is the most useful single number to take to a repair technician, because it is the price above which you would rather buy. If the old machine's running cost alone already exceeds the entire annual cost of a new one, the expression goes negative and no repair price wins — the calculator reports that plainly rather than printing a meaningless negative dollar figure.
Worked example: a nine-year-old refrigerator with a $350 quote
The compressor relay and start capacitor have failed on a nine-year-old refrigerator. The technician quotes $350 including the call-out and thinks it should run another four years. A comparable new unit is $1,200 plus $100 delivery and haul-away, with no rebate, and you expect fourteen years from it. The old unit's EnergyGuide label says 700 kWh a year; the new one says 350. Electricity is $0.17.
- Old running cost. 700 × $0.17 = $119.00 a year.
- New running cost. 350 × $0.17 = $59.50 a year, so the saving is $59.50.
- Repair path. $350 ÷ 4 = $87.50 of capital, plus $119.00 of electricity = $206.50 a year.
- Net cost of new. $1,200 + $100 − $0 = $1,300.
- Replace path. $1,300 ÷ 14 = $92.86 of capital, plus $59.50 of electricity = $152.36 a year.
- Advantage. $206.50 − $152.36 = $54.14 a year in favour of replacing.
- 50% rule. $350 ÷ $1,200 = 29.2%, comfortably below the threshold — the rule says repair.
- Break-even quote. ($152.36 − $119.00) × 4 = $133.43.
The two methods disagree, and the disagreement is instructive. The 50% rule sees a cheap repair on an expensive machine. The annualised comparison sees that the repair buys only four years while the purchase buys fourteen, and that the old machine burns $59.50 a year more in electricity for every one of those four years. At $350 the repair is a worse deal per year; it would have to come in under $133 to compete. If the technician thought the machine had eight good years left rather than four, the repair path would fall to $162.75 a year and the gap would nearly close.
How to read the answer without fooling yourself
Read the sensitivity table before the headline figure. The replace column does not move, because it does not depend on the repair; the repair column falls as you assume more years. Find the row where the verdict flips and ask whether you genuinely believe the machine will last that long. That is the whole decision expressed as one honest question, and it is much easier to answer than a dollar comparison.
Be sceptical of your own repair-years number when the machine is already past its typical life. A failed component that has been replaced is now the newest part in the machine, but everything around it is the same age as the part that just failed. Repairing the compressor on a fifteen-year-old refrigerator does not reset the door seals, the defrost timer or the fan motor.
Treat energy figures with the same care. EnergyGuide labels are test-cycle estimates under a standard protocol, not measurements of your kitchen, and an old machine with degraded seals or a dirty condenser can use considerably more than its original label claimed. If the decision is close, a $25 plug-in meter left on the machine for a week gives a far better number than the label.
Finally, note what the arithmetic cannot price. A second failure risk on an out-of-warranty machine, the inconvenience of another week without a fridge, a match to an existing cabinet run, water damage risk from a failing washer hose, and the residual value at sale are all real and none of them are dollars per year. Use the calculator to establish the size of the financial gap, then decide whether the unpriced factors are worth that gap.
The highest repair quote worth paying, by new-unit price
| Net cost of new unit | If both use the same electricity | If the new unit saves 350 kWh/yr |
|---|---|---|
| $600 | $150.00 | none — replacing wins at any quote |
| $900 | $225.00 | $46.50 |
| $1,200 | $300.00 | $121.50 |
| $1,800 | $450.00 | $271.50 |
| $2,500 | $625.00 | $446.50 |
The 350 kWh column subtracts 350 × $0.17 × 3 = $178.50 from the first. At $600 that turns the break-even negative, which the calculator reports as no winning repair price rather than a negative dollar figure.
What this comparison leaves out
- The risk of a second failure. A machine that has failed once out of warranty is more likely to fail again than a new one. If you want to price it, shorten the repair-years figure rather than inflating the quote.
- Water, gas and vent costs. A washer or dishwasher also uses water and often hot water; a gas dryer or range uses therms. Only electricity is modelled here, so add those separately if they differ between the machines.
- Time value of money. Both paths use simple averages with no discounting. Over three to fourteen years at ordinary rates the effect is real but small compared with the uncertainty in the repair-years assumption.
- Disposal and recycling. Refrigerant-bearing appliances must be handled properly, and some utilities pay for old working units under recycling programmes. Enter any such credit in the rebate field.
- Capacity or feature changes. A new machine that is larger, quieter or gentler on clothes delivers value the arithmetic does not see. So does one that no longer fits the opening.
- Rental and landlord economics. If a tenant pays the electricity, the running-cost saving accrues to them and not to the person paying for the machine, which inverts the whole comparison for the buyer.
Where this decision sits among household running costs
The structure of this calculation — annualise every one-off cost, then add the running cost — is the right frame for almost every keep-or-replace question in a house, and for most keep-or-switch ones too. The appliance energy cost calculator gives you the running-cost half for any individual device, and the same logic drives the pool heating cost calculator, where the equipment choice and the fuel price interact the same way.
For vehicles the identical arithmetic decides tyres and maintenance: a consumable with a measurable life and a cost per mile. The tire tread life remaining calculator works out the life half of that ratio, and the monthly parking pass break-even calculator is the same break-even algebra applied to a subscription.
One more household use: emergency preparedness. A failing appliance is a scheduling problem as well as a money problem, and if the machine in question is a well pump, a water heater or a freezer, the emergency water storage calculator is worth running before you decide to wait for a part.
Get the break-even number before the technician arrives
The most useful output on this page is the highest repair quote that still wins, because it turns an open-ended judgement into a threshold you can hold in your head. Work it out from the new-unit price you would actually buy, then when the diagnosis comes back you already know whether to authorise the work. It also protects against the common trap of authorising a small diagnostic fee and then feeling committed to the repair it recommends.