Repair vs Replace Your Oven: Break-Even Calculator 2026

The repair wins today — but the new oven wins the decade

An oven repair is a bet on the old oven: you pay the repair cost, and the oven gives you its remaining life. Replacing is a bet the other way: you pay the new price, minus the energy and service-call saving the new oven gives you over that same life, plus the cost of getting rid of the old one. Repair total = R. Replace total = P − S × L + D. The difference is replace − repair: positive means REPAIR wins, negative means REPLACE wins. On the defaults — repair $300, new oven $1,200, years left 6, saving $55 a year, disposal $60 — the repair total is $300.00 and the replace total is $930.00, so REPAIR wins by $630.00. Cost per year: repair $50.00 vs replace $155.00. But look at the 10-year lens: repair now, then replace at the end of the remaining life, costs $1,340.00; replace now costs $710.00 — REPLACE wins by $630.00 over 10 years. The new oven's savings run the full decade, while the repaired oven's savings stop when it dies. The 10-year rule: after the repair the old oven runs its remaining L years, then a new one runs the rest of the decade. The rule is unclamped: (10 − L) is used as-is, so for L over 10 the repair-then-replace total grows with L.

At the defaults — repair $300, new oven $1,200, years left 6, saving $55 a year, disposal $60 — the repair total is $300.00, the replace total is $930.00, REPAIR wins by $630.00, the cost per year is $50.00 for the repair vs $155.00 for the replacement, the break-even repair cost is $930.00, the break-even new price is $570.00, the break-even remaining life is 17.45 yrs (exact 17.4545), the break-even annual saving is $160.00/yr, the break-even disposal is none (-$570.00), and the 10-year lens is $1,340.00 vs $710.00 — REPLACE wins by $630.00 over 10 years. Check the identities: repair = R; replace = P − S × L + D; difference = replace − repair; break-even repair cost = replace total; break-even new price = R + S × L − D; break-even remaining life = (P + D − R) / S; break-even annual saving = (P + D − R) / L; break-even disposal = R − P + S × L; 10-year repair-then-replace = R + P + D − S × (10 − L); 10-year replace-now = P + D − S × 10. Rounding convention: totals are computed from unrounded values, so rounded parts may differ by a cent from their rounded sum.

The worked example, line by line

Setup: repair $300, new oven $1,200, years left 6, saving $55 a year, disposal $60.

Line 1 — the repair total. Just the repair: $300.00.

Line 2 — the replace total. $1,200 − $55 × 6 + $60 = $1,200 − $330 + $60 = $930.00.

Line 3 — the difference. $930.00 − $300.00 = $630.00 in favour of repairing.

Line 4 — cost per year. Repair: $300.00 ÷ 6 = $50.00 a year. Replace: $930.00 ÷ 6 = $155.00 a year.

Line 5 — the break-evens. The repair would have to cost $930.00 to tie. The new oven would have to cost $570.00 to tie. The old oven would have to last 17.45 yrs (exact 17.4545) for the saving to pay for the new one. The new oven would have to save $160.00/yr to tie. There is no break-even disposal cost — it would be -$570.00, so repairing wins even with free disposal.

Line 6 — the 10-year lens. Repair now, then replace after 6 years: $300 + $1,200 + $60 − $55 × 4 = $1,340.00. Replace now: $1,200 + $60 − $55 × 10 = $710.00. REPLACE wins by $630.00 over 10 years — the verdict flips.

Reverse: the break-even repair cost is $930.00; a $930 repair costs $930.00, exactly tying the replace total. The break-even new price is $570.00; a $570 oven costs $300.00 to replace, tying the repair. The break-even remaining life is 17.45 yrs (exact 17.4545); at 17.45 yrs the replace total is $300.25 (REPAIR wins by $0.25), at 17.46 yrs it is $299.70 (REPLACE wins by $0.30) — 17.45 is the nearest. The break-even annual saving is $160.00/yr; at $160.00/yr saving the replace total is $300.00, tying the repair. The break-even disposal is none — the disposal cost that would tie is -$570.00; repairing wins even with free disposal. The 10-year lens: repair-then-replace $1,340.00 vs replace-now $710.00 — REPLACE wins by $630.00.

Every branch, checked

Repair $100: repair $100.00, replace $930.00 — REPAIR wins by $830.00. A $100 repair is a bargain: the repair costs $100.00 against $930.00 to replace, so REPAIR wins by $830.00.

Repair $600: repair $600.00, replace $930.00 — REPAIR wins by $330.00. Even a $600 repair still wins: $600.00 against $930.00, so REPAIR wins by $330.00 — the repair is close to the break-even of $930.00.

Repair $1,000: repair $1,000.00, replace $930.00 — REPLACE wins by $70.00. At $1,000 the repair costs more than replacing: $1,000.00 against $930.00, so REPLACE wins by $70.00 — the verdict flips once the repair passes the $930.00 break-even.

New oven $700: repair $300.00, replace $430.00 — REPAIR wins by $130.00. A $700 new oven costs $430.00 to replace, so REPAIR wins by $130.00 — a cheap new oven narrows the gap but does not close it.

New oven $1,800: repair $300.00, replace $1,530.00 — REPAIR wins by $1,230.00. A $1,800 new oven costs $1,530.00 to replace, so REPAIR wins by $1,230.00 — the pricier the replacement, the stronger the case for repairing.

3 years left: repair $300.00, replace $1,095.00 — REPAIR wins by $795.00. With only 3 years of life left, replacing costs $1,095.00 against a $300.00 repair, so REPAIR wins by $795.00 — but the 10-year lens is closer, because the new oven's savings run longer.

10 years left: repair $300.00, replace $710.00 — REPAIR wins by $410.00. With 10 years of life left, replacing costs $710.00 against a $300.00 repair, so REPAIR wins by $410.00 — a longer remaining life makes the new oven's saving worth more, so the gap shrinks.

18 years left: repair $300.00, replace $270.00 — REPLACE wins by $30.00. With 18 years of life left, replacing costs $270.00 against a $300.00 repair, so REPLACE wins by $30.00 — the verdict flips once the old oven has more than 17.45 years left. The 10-year lens is $2,000.00 vs $710.00, so REPLACE wins by $1,290.00 over 10 years.

19 years left: repair $300.00, replace $215.00 — REPLACE wins by $85.00. With 19 years of life left, replacing costs $215.00 against a $300.00 repair, so REPLACE wins by $85.00 — the saving has had long enough to pay for the new oven.

Saving $0/year: repair $300.00, replace $1,260.00 — REPAIR wins by $960.00. With no saving at all, replacing costs $1,260.00 against a $300.00 repair, so REPAIR wins by $960.00 — and there is no break-even remaining life, because no amount of life makes the new oven pay for itself.

Saving $100/year: repair $300.00, replace $660.00 — REPAIR wins by $360.00. At $100 a year of saving, replacing costs $660.00 against a $300.00 repair, so REPAIR wins by $360.00 — the saving has to reach $160.00 a year before the two tie.

Saving $200/year: repair $300.00, replace $60.00 — REPLACE wins by $240.00. At $200 a year of saving, replacing costs $60.00 against a $300.00 repair, so REPLACE wins by $240.00 — the savings alone nearly pay for the new oven, and the verdict flips. The 10-year lens is $760.00 vs -$740.00, so REPLACE wins by $1,500.00 over 10 years.

Disposal $0: repair $300.00, replace $870.00 — REPAIR wins by $570.00. With free disposal, replacing costs $870.00 against a $300.00 repair, so REPAIR wins by $570.00 — free disposal is not enough to flip the verdict, because the break-even disposal cost is -$570.00.

Disposal $150: repair $300.00, replace $1,020.00 — REPAIR wins by $720.00. With $150 of disposal, replacing costs $1,020.00 against a $300.00 repair, so REPAIR wins by $720.00 — a pricier haul-away only strengthens the case for repairing.

What the defaults assume

The model assumes the repair restores the oven to its full remaining life — if the repair only buys a year, use 1 year of remaining life, not 6. It assumes the new oven's saving is steady every year; the $55 a year covers energy and fewer service calls — a modern oven holds its temperature better and wastes less heat. It assumes the disposal fee is what you actually pay — many retailers haul the old oven away free, so set it to $0 when that applies (at $0 disposal the replace total is $870.00 and REPAIR still wins by $570.00). It assumes the 10-year lens buys exactly one replacement oven in each path — an oven that lasts longer than 10 years would change the second decade, not the first. The 10-year rule is unclamped: (10 − L) is used as-is, so for L over 10 the repair-then-replace total grows with L — at 18 years left the repair-then-replace total is $2,000.00. With a repair cost of $0 the verdict is REPAIR wins by $930.00. With 1 year of life left and no saving, the break-even life is — (no saving, so no life ever ties — a model-level edge; at the calc level a saving of $0 is valid input and the break-even-life field renders —). Rounding convention: totals are computed from unrounded values, so rounded parts may differ by a cent from their rounded sum; the reverses are nearest-value solutions, and the exact unrounded figures (17.4545) are stated wherever the rounded figures appear.

Output · formula · meaning

Who uses this

• The homeowner quoted $300 for a heating-element repair: they learn the repair wins by $630.00 today, because the old oven still has 6 years of life — and they check the $1,000 branch (REPLACE wins by $70.00) before authorising a bigger repair.
• The cook whose oven died with 3 years left before a move: they learn the repair wins by $795.00 on the short lens, but the 10-year lens favours replacing — so they weigh whether they will still be in the same home in 10 years.
• The energy-conscious buyer eyeing a $200-a-year saving: they learn the new oven nearly pays for itself outright — the replace total is $60.00 and REPLACE wins by $240.00 — so they check the $100-a-year branch (REPAIR wins by $360.00) to see how much saving the flip actually needs.
• The landlord replacing ovens across several units: they learn the break-even new price is $570.00 — any oven under that price beats repairing — and they check the disposal-$150 branch (REPAIR wins by $720.00) before budgeting haul-away.

Accuracy & limitations

The math is exact arithmetic on your five inputs: repair = R; replace = P − S × L + D; difference = replace − repair; break-even repair cost = replace total; break-even new price = R + S × L − D; break-even remaining life = (P + D − R) / S; break-even annual saving = (P + D − R) / L; break-even disposal = R − P + S × L; 10-year repair-then-replace = R + P + D − S × (10 − L); 10-year replace-now = P + D − S × 10. The 4-decimal exact is shown where the rounding matters: the break-even remaining life is 17.4545 (shown as 17.45 yrs); at 17.45 yrs the replace total is $300.25 (REPAIR wins by $0.25) and at 17.46 yrs it is $299.70 (REPLACE wins by $0.30). What the model cannot do is predict how long your oven will actually last — the remaining-life input is your best guess, and the break-even life of 17.45 years tells you how long it would have to last for the new oven's saving to pay for it. For the refrigerator version, see the repair vs replace refrigerator calculator; for the dishwasher version, see the repair vs replace dishwasher calculator; for the washer version, see the repair vs replace washer calculator; for the dryer version, see the repair vs replace dryer calculator; for gas versus electric, see the gas vs electric oven comparison. This is a calculator, not repair advice.