★ COMPARISONS · REPAIR VS REPLACE · X-VS-Y #241
REPAIR wins by $410.00 — repair $300.00, replace $710.00. A $300 repair beats a $900 new washing machine when the old one has 4 years left — but over 10 years the new machine wins by $540.00, because its savings run the full decade. Enter your numbers to find your break-even.
Should you repair or replace your washing machine? Compare the repair cost against a new washing machine, with break-even repair cost, new price, remaining life and savings — and see how the answer flips over 10 years.
A washing machine repair is a bet on the old machine: you pay the repair cost, and the machine gives you its remaining life. Replacing is a bet the other way: you pay the new price, minus the water and energy saving the new machine 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 washing machine $900, years left 4, saving $60 a year, disposal $50 — the repair total is $300.00 and the replace total is $710.00, so REPAIR wins by $410.00. Cost per year: repair $75.00 vs replace $177.50. But look at the 10-year lens: repair now, then replace at the end of the remaining life, costs $890.00; replace now costs $350.00 — REPLACE wins by $540.00 over 10 years. The new machine's savings run the full decade, while the repaired machine's savings stop when it dies. The 10-year rule: after the repair the old machine 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.
| Scenario | Repair | Replace | Winner | By | 10-year winner |
|---|---|---|---|---|---|
| Default | $300.00 | $710.00 | REPAIR wins by $410.00 | $410.00 | REPLACE wins by $540.00 over 10 years |
| Repair $1,000 | $1,000.00 | $710.00 | REPLACE wins by $290.00 | $290.00 | REPLACE wins by $1,240.00 over 10 years |
| Repair $1,200 | $1,200.00 | $710.00 | REPLACE wins by $490.00 | $490.00 | REPLACE wins by $1,440.00 over 10 years |
| 15 years left | $300.00 | $50.00 | REPLACE wins by $250.00 | $250.00 | REPLACE wins by $1,200.00 over 10 years |
| 16 years left | $300.00 | -$10.00 | REPLACE wins by $310.00 | $310.00 | REPLACE wins by $1,260.00 over 10 years |
| Saving $200/year | $300.00 | $150.00 | REPLACE wins by $150.00 | $150.00 | REPLACE wins by $1,100.00 over 10 years |
| Saving $0/year | $300.00 | $950.00 | REPAIR wins by $650.00 | $650.00 | REPLACE wins by $300.00 over 10 years |
| New machine $800 | $300.00 | $610.00 | REPAIR wins by $310.00 | $310.00 | REPLACE wins by $540.00 over 10 years |
| New machine $2,000 | $300.00 | $1,810.00 | REPAIR wins by $1,510.00 | $1,510.00 | REPLACE wins by $540.00 over 10 years |
| Disposal $0 | $300.00 | $660.00 | REPAIR wins by $360.00 | $360.00 | REPLACE wins by $540.00 over 10 years |
| Disposal $200 | $300.00 | $860.00 | REPAIR wins by $560.00 | $560.00 | REPLACE wins by $540.00 over 10 years |
At the defaults — repair $300, new washing machine $900, years left 4, saving $60 a year, disposal $50 — the repair total is $300.00, the replace total is $710.00, REPAIR wins by $410.00, the cost per year is $75.00 for the repair vs $177.50 for the replacement, the break-even repair cost is $710.00, the break-even new price is $490.00, the break-even remaining life is 10.83 yrs (exact 10.8333), the break-even annual saving is $162.50/yr, the break-even disposal is none (-$360.00), and the 10-year lens is $890.00 vs $350.00 — REPLACE wins by $540.00 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.
Setup: repair $300, new washing machine $900, years left 4, saving $60 a year, disposal $50.
Line 1 — the repair total. Just the repair: $300.00.
Line 2 — the replace total. $900 − $60 × 4 + $50 = $900 − $240 + $50 = $710.00.
Line 3 — the difference. $710.00 − $300.00 = $410.00 in favour of repairing.
Line 4 — cost per year. Repair: $300.00 ÷ 4 = $75.00 a year. Replace: $710.00 ÷ 4 = $177.50 a year.
Line 5 — the break-evens. The repair would have to cost $710.00 to tie. The new machine would have to cost $490.00 to tie. The old machine would have to last 10.83 yrs (exact 10.8333) for the saving to pay for the new one. The new machine would have to save $162.50/yr to tie. There is no break-even disposal cost — it would be -$360.00, so repairing wins even with free disposal.
Line 6 — the 10-year lens. Repair now, then replace after 4 years: $300 + $900 + $50 − $60 × 6 = $890.00. Replace now: $900 + $50 − $60 × 10 = $350.00. REPLACE wins by $540.00 over 10 years — the verdict flips.
Reverse: the break-even repair cost is $710.00; a $710 repair costs $710.00, exactly tying the replace total. The break-even new price is $490.00; a $490 machine costs $300.00 to replace, tying the repair. The break-even remaining life is 10.83 yrs (exact 10.8333); at 10.83 yrs the replace total is $300.20 (REPAIR wins by $0.20), at 10.84 yrs it is $299.60 (REPLACE wins by $0.40) — 10.83 is the nearest. The break-even annual saving is $162.50/yr; at $162.50/yr saving the replace total is $300.00, tying the repair. The break-even disposal is none — the disposal cost that would tie is -$360.00; repairing wins even with free disposal. The 10-year lens: repair-then-replace $890.00 vs replace-now $350.00 — REPLACE wins by $540.00.
Repair $1,000: repair $1,000.00, replace $710.00 — REPLACE wins by $290.00. A $1,000 repair is well past the $710.00 break-even — the verdict flips once the repair costs more than replacing. The 10-year lens is $1,590.00 vs $350.00, so REPLACE wins by $1,240.00 over 10 years.
Repair $1,200: repair $1,200.00, replace $710.00 — REPLACE wins by $490.00. At $1,200 the repair costs nearly twice the replace total — replacing is the clear winner. The 10-year lens is $1,790.00 vs $350.00, so REPLACE wins by $1,440.00 over 10 years.
15 years left: repair $300.00, replace $50.00 — REPLACE wins by $250.00. With 15 years of life left, fifteen years of $60 savings beat the $300 repair outright — the replace total is only $50.00. The 10-year lens is $1,550.00 vs $350.00, so REPLACE wins by $1,200.00 over 10 years.
16 years left: repair $300.00, replace -$10.00 — REPLACE wins by $310.00. With 16 years of life left, the savings alone pay for the new machine — the replace total is negative, because $960 of saving beats the $900 price plus $50 disposal. The 10-year lens is $1,610.00 vs $350.00, so REPLACE wins by $1,260.00 over 10 years.
Saving $200/year: repair $300.00, replace $150.00 — REPLACE wins by $150.00. At $200 a year of saving, replacing costs $150.00 against a $300.00 repair — the saving has to reach $162.50 a year before the two tie, and $200 clears it. The 10-year lens is $50.00 vs -$1,050.00, so REPLACE wins by $1,100.00 over 10 years — over a decade the big saving more than pays for the new machine.
Saving $0/year: repair $300.00, replace $950.00 — REPAIR wins by $650.00. With no saving at all, replacing costs $950.00 against a $300.00 repair — and there is no break-even remaining life, because no amount of life makes the new machine pay for itself. The 10-year lens is $1,250.00 vs $950.00, so REPLACE still wins by $300.00 over 10 years, because the repair-then-replace path buys a second machine.
New machine $800: repair $300.00, replace $610.00 — REPAIR wins by $310.00. A cheap $800 machine costs $610.00 to replace, so REPAIR still wins by $310.00 — the new machine would have to cost $490.00 or less to tie. The 10-year lens is $790.00 vs $250.00, so REPLACE wins by $540.00 over 10 years.
New machine $2,000: repair $300.00, replace $1,810.00 — REPAIR wins by $1,510.00. A $2,000 new machine costs $1,810.00 to replace — the pricier the replacement, the stronger the case for repairing. The 10-year lens is $1,990.00 vs $1,450.00, so REPLACE wins by $540.00 over 10 years.
Disposal $0: repair $300.00, replace $660.00 — REPAIR wins by $360.00. With free disposal, replacing costs $660.00 against a $300.00 repair — free disposal is not enough to flip the verdict, because the break-even disposal cost is -$360.00. The 10-year lens is $840.00 vs $300.00, so REPLACE wins by $540.00 over 10 years.
Disposal $200: repair $300.00, replace $860.00 — REPAIR wins by $560.00. With $200 of disposal, replacing costs $860.00 against a $300.00 repair — a pricier haul-away only strengthens the case for repairing. The 10-year lens is $1,040.00 vs $500.00, so REPLACE wins by $540.00 over 10 years.
The model assumes the repair restores the washing machine to its full remaining life — if the repair only buys a year, use 1 year of remaining life, not 4. It assumes the new machine's saving is steady every year; the $60 a year covers water, energy and fewer service calls — a modern machine uses less water per load and spins faster, so the dryer runs less too. It assumes the disposal fee is what you actually pay — many retailers haul the old machine away free, so set it to $0 when that applies (at $0 disposal the replace total is $660.00 and REPAIR still wins by $360.00). It assumes the 10-year lens buys exactly one replacement machine in each path — a machine 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 15 years left the repair-then-replace total is $1,550.00. With a repair cost of $0 the verdict is REPAIR wins by $710.00. 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 (10.8333) are stated wherever the rounded figures appear.
| Output | Formula | What it means |
|---|---|---|
| Repair total | the repair cost R | $300.00. |
| Replace total | new price − saving × life + disposal | $710.00. |
| Difference | replace total − repair total | $410.00. |
| Winner | REPAIR when the difference is positive, REPLACE when negative | REPAIR wins by $410.00. |
| Cost per year (repair) | repair cost ÷ remaining life | $75.00. |
| Cost per year (replace) | replace total ÷ remaining life | $177.50. |
| Break-even repair cost | the replace total | $710.00. |
| Break-even new price | repair + saving × life − disposal | $490.00. |
| Break-even remaining life | (new price + disposal − repair) ÷ saving | 10.8333 yrs → 10.83 yrs. |
| Break-even annual saving | (new price + disposal − repair) ÷ life | $162.50/yr. |
| Break-even disposal | repair − new price + saving × life | -$360.00 → none. |
| 10-year: repair-then-replace | repair + new price + disposal − saving × (10 − life) | $890.00. |
| 10-year: replace-now | new price + disposal − saving × 10 | $350.00. |
| 10-year winner | the smaller 10-year total | REPLACE wins by $540.00 over 10 years. |
| Break-even repair cost | the replace total | $710.00 → $710.00. |
| Break-even new price | repair + saving × life − disposal | $490.00 → $490.00. |
| Break-even remaining life | (new price + disposal − repair) ÷ saving | 10.8333 → 10.83 yrs. |
| Break-even annual saving | (new price + disposal − repair) ÷ life | $162.50/yr → $162.50/yr. |
| Break-even disposal | repair − new price + saving × life | -$360.00 → none. |
| 10-year lens | repair-then-replace vs replace-now | $890.00 vs $350.00. |
| Repair $1,000 | same model | Repair $1,000.00, replace $710.00 — REPLACE wins by $290.00. 10-year: $1,590.00 vs $350.00 — REPLACE wins by $1,240.00 over 10 years. |
| Repair $1,200 | same model | Repair $1,200.00, replace $710.00 — REPLACE wins by $490.00. 10-year: $1,790.00 vs $350.00 — REPLACE wins by $1,440.00 over 10 years. |
| 15 years left | same model | Repair $300.00, replace $50.00 — REPLACE wins by $250.00. 10-year: $1,550.00 vs $350.00 — REPLACE wins by $1,200.00 over 10 years. |
| 16 years left | same model | Repair $300.00, replace -$10.00 — REPLACE wins by $310.00. 10-year: $1,610.00 vs $350.00 — REPLACE wins by $1,260.00 over 10 years. |
| Saving $200/year | same model | Repair $300.00, replace $150.00 — REPLACE wins by $150.00. 10-year: $50.00 vs -$1,050.00 — REPLACE wins by $1,100.00 over 10 years. |
| Saving $0/year | same model | Repair $300.00, replace $950.00 — REPAIR wins by $650.00. 10-year: $1,250.00 vs $950.00 — REPLACE wins by $300.00 over 10 years. |
| New machine $800 | same model | Repair $300.00, replace $610.00 — REPAIR wins by $310.00. 10-year: $790.00 vs $250.00 — REPLACE wins by $540.00 over 10 years. |
| New machine $2,000 | same model | Repair $300.00, replace $1,810.00 — REPAIR wins by $1,510.00. 10-year: $1,990.00 vs $1,450.00 — REPLACE wins by $540.00 over 10 years. |
| Disposal $0 | same model | Repair $300.00, replace $660.00 — REPAIR wins by $360.00. 10-year: $840.00 vs $300.00 — REPLACE wins by $540.00 over 10 years. |
| Disposal $200 | same model | Repair $300.00, replace $860.00 — REPAIR wins by $560.00. 10-year: $1,040.00 vs $500.00 — REPLACE wins by $540.00 over 10 years. |
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 10.8333 (shown as 10.83 yrs); at 10.83 yrs the replace total is $300.20 (REPAIR wins by $0.20) and at 10.84 yrs it is $299.60 (REPLACE wins by $0.40). What the model cannot do is predict how long your machine will actually last — the remaining-life input is your best guess, and the break-even life of 10.83 years tells you how long it would have to last for the new machine's saving to pay for it. For the dryer version, see the repair vs replace dryer calculator; for the dishwasher version, see the repair vs replace dishwasher calculator; for the water heater version, see the repair vs replace water heater calculator; for the refrigerator version, see the repair vs replace refrigerator calculator; for the washer version, see the repair vs replace washer calculator. This is a calculator, not repair advice.
Bright Side Kit. "Repair vs Replace Washing Machine: Cost Calculator 2026" https://brightoolbox.com/repair-vs-replace-washing-machine/ Accessed October 25, 2026.Repair it when the repair is cheap and the old machine still has life; replace it when the repair approaches the break-even cost. At a $300 repair against a $900 new machine with 4 years of life left, REPAIR wins by $410.00. The repair would have to cost $710.00 before replacing ties it.
The cost of replacing it. At the defaults the replace total is $710.00, so any repair under $710.00 wins and any repair over it loses. A $1,000 repair loses by $290.00; a $300 repair wins by $410.00.
When the saving is big enough, or the old machine has enough life left. The new machine would have to save $162.50 a year to tie at 4 years of remaining life, or the old machine would have to last 10.83 years for a $60-a-year saving to pay for it. At $200 a year of saving, the replace total is $150.00 — $150.00 less than the repair.
Yes — it flips it. Repairing wins today by $410.00, but over 10 years replacing now wins by $540.00 ($890.00 vs $350.00), because the new machine’s $60-a-year saving runs the full decade while the repaired machine’s saving stops when it dies after 4 years.
Repairing still wins. With $0 disposal the replace total is $660.00 and REPAIR wins by $360.00. The break-even disposal cost is -$360.00 — there is no disposal price that ties the two, because the repair is already cheaper than the new machine before disposal is counted.
Water, energy and fewer service calls. A modern washing machine uses less water per load, spins faster so the dryer runs less, and needs fewer repairs. At $60 a year the saving is modest — that is why the repair wins today — but over 10 years it adds up to $600.00, which is why the 10-year lens flips the verdict.