Repair vs Replace Refrigerator: Break-Even Calculator 2026

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

A refrigerator repair is a bet on the old machine: you pay the repair cost, and the fridge gives you its remaining life. Replacing is a bet the other way: you pay the new price, minus the energy the new fridge saves over that same life, plus the cost of getting rid of the old one. Repair total = R. Replace total = P + Dd − S × Lr. The difference is replace − repair: positive means REPAIR wins, negative means REPLACE wins. On the defaults — repair $300, new refrigerator $1,400, remaining life 4 years, saving $60 a year, disposal $50 — the repair total is $300.00 and the replace total is $1,210.00, so REPAIR wins by $910.00. Cost per year: repair $75.00 vs replace $302.50. But look at the 10-year lens: repair now, then replace at the end of the remaining life, costs $1,390.00; replace now costs $850.00 — REPLACE wins by $540.00 over 10 years. The new fridge's energy savings run the full decade, while the repaired fridge's savings stop when it dies.

At the defaults — repair $300, new refrigerator $1,400, remaining life 4 years, saving $60 a year, disposal $50 — the repair total is $300.00, the replace total is $1,210.00, REPAIR wins by $910.00, the cost per year is $75.00 for the repair vs $302.50 for the replacement, the break-even repair cost is $1,210.00, the break-even new price is $490.00, the break-even remaining life is 19.17 yrs (exact 19.1667), the break-even annual saving is $287.50/yr, the break-even disposal is none (-$860.00), and the 10-year lens is $1,390.00 vs $850.00 — REPLACE wins by $540.00 over 10 years. Check the identities: repair = R; replace = P + Dd − S × Lr; difference = replace − repair; break-even repair cost = replace total; break-even new price = R − Dd + S × Lr; break-even remaining life = (P + Dd − R) / S; break-even annual saving = (P + Dd − R) / Lr; break-even disposal = R − P + S × Lr; 10-year repair-then-replace = R + P + Dd − S × (10 − Lr); 10-year replace-now = P + Dd − 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 refrigerator $1,400, remaining life 4 years, saving $60 a year, disposal $50.

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

Line 2 — the replace total. $1,400 + $50 − $60 × 4 = $1,400 + $50 − $240 = $1,210.00.

Line 3 — the difference. $1,210.00 − $300.00 = $910.00 in favour of repairing.

Line 4 — cost per year. Repair: $300.00 ÷ 4 = $75.00 a year. Replace: $1,210.00 ÷ 4 = $302.50 a year.

Line 5 — the break-evens. The repair would have to cost $1,210.00 to tie. The new refrigerator would have to cost $490.00 to tie. The old fridge would have to last 19.17 yrs (exact 19.1667) for the saving to pay for the new one. The new fridge would have to save $287.50/yr to tie. There is no break-even disposal cost — it would be -$860.00, so repairing wins even with free disposal.

Line 6 — the 10-year lens. Repair now, then replace after 4 years: $300 + $1,400 + $50 − $60 × 6 = $1,390.00. Replace now: $1,400 + $50 − $60 × 10 = $850.00. REPLACE wins by $540.00 over 10 years — the verdict flips.

Reverse: the break-even repair cost is $1,210.00; a $1,210 repair costs $1,210.00, exactly tying the replace total. The break-even new price is $490.00; a $490 fridge costs $300.00 to replace, tying the repair. The break-even remaining life is 19.17 yrs (exact 19.1667); at 19.17 yrs the replace total is $299.80 (REPLACE wins by $0.20), at 19.16 yrs it is $300.40 (REPAIR wins by $0.40) — 19.17 is the nearest. The break-even annual saving is $287.50/yr; at $287.50/yr saving the replace total is $300.00, tying the repair. The break-even disposal is none — the disposal cost that would tie is -$860.00; repairing wins even with free disposal. The 10-year lens: repair-then-replace $1,390.00 vs replace-now $850.00 — REPLACE wins by $540.00.

Every branch, checked

Repair $100: repair $100.00, replace $1,210.00 — REPAIR wins by $1,110.00. A $100 repair is a bargain: the repair costs $100.00 against $1,210.00 to replace, so REPAIR wins by $1,110.00.

Repair $1,000: repair $1,000.00, replace $1,210.00 — REPAIR wins by $210.00. Even a $1,000 repair still wins: $1,000.00 against $1,210.00, so REPAIR wins by $210.00 — the repair is close to the break-even of $1,210.00.

Repair $1,300: repair $1,300.00, replace $1,210.00 — REPLACE wins by $90.00. At $1,300 the repair costs more than replacing: $1,300.00 against $1,210.00, so REPLACE wins by $90.00 — the verdict flips once the repair passes the $1,210.00 break-even.

New refrigerator $800: repair $300.00, replace $610.00 — REPAIR wins by $310.00. An $800 new refrigerator costs $610.00 to replace, so REPAIR wins by $310.00 — a cheap new fridge narrows the gap but does not close it.

New refrigerator $2,000: repair $300.00, replace $1,810.00 — REPAIR wins by $1,510.00. A $2,000 new refrigerator costs $1,810.00 to replace, so REPAIR wins by $1,510.00 — the pricier the replacement, the stronger the case for repairing.

Remaining life 2 years: repair $300.00, replace $1,330.00 — REPAIR wins by $1,030.00. With only 2 years of life left, replacing costs $1,330.00 against a $300.00 repair, so REPAIR wins by $1,030.00 — but the 10-year lens is closer, because the new fridge's savings run longer.

Remaining life 8 years: repair $300.00, replace $970.00 — REPAIR wins by $670.00. With 8 years of life left, replacing costs $970.00 against a $300.00 repair, so REPAIR wins by $670.00 — a longer remaining life makes the new fridge's energy saving worth more, so the gap shrinks.

Energy saving $0/year: repair $300.00, replace $1,450.00 — REPAIR wins by $1,150.00. With no energy saving at all, replacing costs $1,450.00 against a $300.00 repair, so REPAIR wins by $1,150.00 — and there is no break-even remaining life, because no amount of life makes the new fridge pay for itself.

Energy saving $150/year: repair $300.00, replace $850.00 — REPAIR wins by $550.00. At $150 a year of saving, replacing costs $850.00 against a $300.00 repair, so REPAIR wins by $550.00 — the saving has to reach $287.50 a year before the two tie.

Energy saving $300/year: repair $300.00, replace $250.00 — REPLACE wins by $50.00. At $300 a year of saving, replacing costs $250.00 against a $300.00 repair, so REPLACE wins by $50.00 — the energy savings alone nearly pay for the new fridge, and the verdict flips.

Energy saving $350/year: repair $300.00, replace $50.00 — REPLACE wins by $250.00. At $350 a year of saving, replacing costs $50.00 against a $300.00 repair, so REPLACE wins by $250.00 — the savings pay for the new fridge outright.

Disposal $0: repair $300.00, replace $1,160.00 — REPAIR wins by $860.00. With free disposal, replacing costs $1,160.00 against a $300.00 repair, so REPAIR wins by $860.00 — free disposal is not enough to flip the verdict, because the break-even disposal cost is -$860.00.

Disposal $200: repair $300.00, replace $1,360.00 — REPAIR wins by $1,060.00. With $200 of disposal, replacing costs $1,360.00 against a $300.00 repair, so REPAIR wins by $1,060.00 — a pricier haul-away only strengthens the case for repairing.

What the defaults assume

The model assumes the repair restores the fridge to its full remaining life — if the repair only buys a year, use 1 year of remaining life, not 4. It assumes the new fridge's energy saving is steady every year; a modern efficient fridge saves more than an old one, so check the saving figure on the model's energy label. It assumes the disposal cost is what you actually pay — many retailers haul the old fridge away free, so set it to $0 when that applies (at $0 disposal the replace total is $1,160.00 and REPAIR still wins by $860.00). It assumes the 10-year lens buys exactly one replacement fridge in each path — a fridge that lasts longer than 10 years would change the second decade, not the first. With a repair cost of $0 the verdict is REPAIR wins by $1,210.00. 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 an energy 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 (19.1667) are stated wherever the rounded figures appear.

Output · formula · meaning

Who uses this

• The homeowner quoted $300 for a compressor repair: they learn the repair wins by $910.00 today, because the old fridge still has 4 years of life — and they check the $1,300 branch (REPLACE wins by $90.00) before authorising a bigger repair.
• The renter whose fridge died with 2 years left on the lease: they learn the repair wins by $1,030.00 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 $300-a-year saving: they learn the new fridge nearly pays for itself outright — the replace total is $250.00 and REPLACE wins by $50.00 — so they check the $150-a-year branch (REPAIR wins by $550.00) to see how much saving the flip actually needs.
• The landlord replacing fridges across several units: they learn the break-even new price is $490.00 — any fridge under that price beats repairing — and they check the disposal-$200 branch (REPAIR wins by $1,060.00) before budgeting haul-away.

Accuracy & limitations

The math is exact arithmetic on your five inputs: repair = R; replace = P + Dd − S × Lr; difference = replace − repair; break-even repair cost = replace total; break-even new price = R − Dd + S × Lr; break-even remaining life = (P + Dd − R) / S; break-even annual saving = (P + Dd − R) / Lr; break-even disposal = R − P + S × Lr; 10-year repair-then-replace = R + P + Dd − S × (10 − Lr); 10-year replace-now = P + Dd − S × 10. The 4-decimal exact is shown where the rounding matters: the break-even remaining life is 19.1667 (shown as 19.17 yrs); at 19.17 yrs the replace total is $299.80 (REPLACE wins by $0.20) and at 19.16 yrs it is $300.40 (REPAIR wins by $0.40). What the model cannot do is predict how long your fridge will actually last — the remaining-life input is your best guess, and the break-even life of 19.17 years tells you how long it would have to last for the new fridge's saving to pay for it. For the dryer version, see the repair vs replace dryer calculator; for the washer version, see the repair vs replace washer calculator; for the car version, see the repair vs replace car calculator. This is a calculator, not repair advice.