Repair vs Replace Garbage Disposal: Cost Calculator 2026

The repair wins today — the replacement wins the decade

A disposal repair is cheap and a new disposal is not — but a new unit runs cheaper every year, and over a decade that saving adds up. The repair total is just the repair cost: $180.00. The replace total is the new price, minus the saving over the old unit’s remaining life, plus disposal: $450 − $40 × 3 + $25 = $355.00. On the defaults — repair $180.00, new disposal $450.00, remaining life 3 years, saving $40.00/year, disposal $25.00 — REPAIR wins by $175.00; cost per year repair $60.00 vs replace $118.33; break-even repair $355.00, break-even price $275.00, break-even life 7.38 yrs (exact 7.3750), break-even saving $98.33/yr, break-even disposal none (-$150.00); 10-year repair-then-replace $375.00 vs replace-now $75.00 — REPLACE wins by $300.00 over 10 years.

At the defaults — repair $180.00, new disposal $450.00, remaining life 3 years, saving $40.00/year, disposal $25.00 — the repair total is $180.00, the replace total is $355.00, REPAIR wins by $175.00; cost per year repair $60.00 vs replace $118.33; break-even repair $355.00, break-even price $275.00, break-even life 7.38 yrs (exact 7.3750), break-even saving $98.33/yr, break-even disposal none (-$150.00); 10-year repair-then-replace $375.00 vs replace-now $75.00 — REPLACE wins by $300.00 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 $180.00, new disposal $450.00, remaining life 3 years, saving $40.00/year, disposal $25.00.

Line 1 — the repair total. The repair total is the repair cost: $180.00.

Line 2 — the replace total. $450 − $40 × 3 + $25 = $450 − $120 + $25 = $355.00. The three years of saving shave $120 off the new unit’s price.

Line 3 — the winner. $355.00 − $180.00 = $175.00, so REPAIR wins by $175.00. Cost per year: $180.00 ÷ 3 = $60.00 for the repair, $355.00 ÷ 3 = $118.33 for the replacement.

Line 4 — the break-evens. The break-even repair cost is the replace total, $355.00. The break-even new price is $180 + $40 × 3 − $25 = $275.00. The break-even remaining life is ($450 + $25 − $180) ÷ $40 = 7.38 yrs (exact 7.3750). The break-even annual saving is ($450 + $25 − $180) ÷ 3 = $98.33/yr. The break-even disposal is $180 − $450 + $40 × 3 = -$150.00 — none, because the disposal cost that would tie is negative.

Line 5 — the 10-year lens. Repair now, then replace in 3 years: $180 + $450 + $25 − $40 × 7 = $375.00. Replace now: $450 + $25 − $40 × 10 = $75.00. So REPLACE wins by $300.00 over 10 years — the verdict flips over 10 years.

Reverse: the break-even repair cost is $355.00 (exact $355.00); a $355.00 repair costs $355.00, exactly tying the replace total — verdict: TIE. The break-even new price is $275.00 (exact $275.00); a $275.00 disposal costs $180.00 to replace, tying the repair — verdict: TIE. The break-even remaining life is 7.38 yrs (exact 7.3750); at 7.37 yrs the replace total is $180.20 (REPAIR wins by $0.20), at 7.38 yrs it is $179.80 (REPLACE wins by $0.20), at 7.39 yrs it is $179.40 (REPLACE wins by $0.60) — 7.3750 is the exact tie. The break-even annual saving is $98.33/yr (exact $98.3333); at $98.33/yr saving the replace total is $180.01 (REPAIR wins by $0.01), at $98.34/yr it is $179.98 (REPLACE wins by $0.02) — verdict: NEAREST. The break-even disposal is none — the disposal cost that would tie is -$150.00; repairing wins even with free disposal — verdict: NONE. The 10-year lens: repair-then-replace $375.00 vs replace-now $75.00 — REPLACE wins by $300.00 over 10 years.

Every branch, checked

Repair $300: Repair $300.00, replace $355.00 — REPAIR wins by $55.00. 10-year: $495.00 vs $75.00 — REPLACE wins by $420.00 over 10 years. A $300 repair is still cheaper than replacing — REPAIR wins by $55.00.

Repair $400: Repair $400.00, replace $355.00 — REPLACE wins by $45.00. 10-year: $595.00 vs $75.00 — REPLACE wins by $520.00 over 10 years. At $400 the repair costs more than the $355.00 replacement — REPLACE wins by $45.00, a flip from the default verdict.

5 years left: Repair $180.00, replace $275.00 — REPAIR wins by $95.00. 10-year: $455.00 vs $75.00 — REPLACE wins by $380.00 over 10 years. Five years of $40 savings are worth $200, so the replace total falls to $275.00 — but REPAIR wins by $95.00 today; the 10-year lens is still REPLACE by $380.00.

8 years left: Repair $180.00, replace $155.00 — REPLACE wins by $25.00. 10-year: $575.00 vs $75.00 — REPLACE wins by $500.00 over 10 years. Eight years of saving drop the replace total to $155.00 — REPLACE wins by $25.00 today, a flip: a long remaining life lets the savings pay for the new unit. And the 10-year repair-then-replace is $575.00 — the unclamped (10 − L) = 2 means the replacement you buy in 8 years only saves for 2 of the 10 years, so the path costs more, not less.

Saving $100/year: Repair $180.00, replace $175.00 — REPLACE wins by $5.00. 10-year: -$45.00 vs -$525.00 — REPLACE wins by $480.00 over 10 years. A $100-a-year saving pays for the new unit fast — the replace total is $175.00 and REPLACE wins by $5.00. And both 10-year totals are negative — -$45.00 vs -$525.00: 10 years of $100/yr savings ($1,000) exceed the $450 price, so the new unit has paid for itself and you are ahead by $525.00. A negative 10-year total means the savings have paid back more than the price.

Saving $0/year: Repair $180.00, replace $475.00 — REPAIR wins by $295.00. 10-year: $655.00 vs $475.00 — REPLACE wins by $180.00 over 10 years. With no saving the new unit costs its full price plus disposal — $475.00 — so REPAIR wins by $295.00 and the break-even remaining life is — (no saving, so no life ever ties).

New disposal $300: Repair $180.00, replace $205.00 — REPAIR wins by $25.00. 10-year: $225.00 vs -$75.00 — REPLACE wins by $300.00 over 10 years. A cheaper new unit narrows the gap to $205.00 — but REPAIR wins by $25.00. And the replace-now total over 10 years is -$75.00 — negative, because 10 years of $40/yr savings ($400) exceed the $300 price plus $25 disposal, so the new unit has paid for itself and you are ahead by $75.00.

New disposal $600: Repair $180.00, replace $505.00 — REPAIR wins by $325.00. 10-year: $525.00 vs $225.00 — REPLACE wins by $300.00 over 10 years. A $600 unit is more than three times the repair — REPAIR wins by $325.00.

Disposal $0: Repair $180.00, replace $330.00 — REPAIR wins by $150.00. 10-year: $350.00 vs $50.00 — REPLACE wins by $300.00 over 10 years. Free haul-away drops the replace total to $330.00 — REPAIR wins by $150.00.

Disposal $100: Repair $180.00, replace $430.00 — REPAIR wins by $250.00. 10-year: $450.00 vs $150.00 — REPLACE wins by $300.00 over 10 years. A $100 disposal fee pushes the replace total to $430.00 — REPAIR wins by $250.00.

What the defaults assume

The model assumes the repair restores the old disposal for its full remaining life, and that the new unit’s saving is the same every year — it does not model a repair that fails again, a new unit that breaks early, or water and power prices changing. It assumes the disposal fee is paid when you replace, not when you repair — you keep the old unit either way. It assumes the saving is real money: $40 a year of fewer clogs and service calls, lower water and power use. With a $0 repair the total is $0.00 and REPAIR wins by $355.00, but the 10-year lens is still REPLACE by $120.00. With 1 year left and no saving the break-even life is — — at the model level there is no life that ties, because the saving is $0; at the calc level S = 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 ($355.00, $275.00, 7.3750, $98.3333, -$150.00) are stated wherever the rounded figures appear.

Output · formula · meaning

Who uses this

• The homeowner with a $180 repair quote: they learn REPAIR wins by $175.00 — and they check the $400 branch (REPLACE wins by $45.00) to see the repair price where replacing wins.
• The owner whose unit has 8 years left: they learn REPLACE wins by $25.00 — eight years of saving pay for the new unit — and they check the 5-year branch (REPLACE wins by $380.00 over 10 years) to see the decade margin.
• The shopper eyeing a $300 disposal: they learn REPAIR wins by $25.00 — the closest call on the page — and they check the $600 branch (REPAIR wins by $325.00) to see how fast a dearer model loses.
• The planner thinking in decades: they learn the 10-year lens flips the verdict — REPLACE wins by $300.00 over 10 years — and they check the $100-a-year branch (REPLACE wins by $480.00 over 10 years) to see the new unit pay for itself.

Accuracy & limitations

The math is first-principles comparison: repair = R; replace = P − S × L + D; difference = replace − repair; cost per year = total ÷ L; 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 7.3750 (shown as 7.38 yrs); at 7.37 yrs the replace total is $180.20 (REPAIR wins by $0.20), at 7.38 yrs it is $179.80 (REPLACE wins by $0.20), at 7.39 yrs it is $179.40 (REPLACE wins by $0.60). The break-even annual saving is $98.3333 (shown as $98.33/yr); at $98.33/yr the replace total is $180.01 (REPAIR wins by $0.01), at $98.34/yr it is $179.98 (REPLACE wins by $0.02). What the model cannot do is predict a disposal’s real life: a repair can fail again, a new unit can break early, and the saving depends on how much you run it — the break-even life of 7.38 yrs is the number to compare against your honest estimate of the old unit’s remaining life. A negative 10-year total means the new unit’s savings have paid back more than its price — you are ahead by that amount. For the same question on the heating side, see the repair vs replace furnace calculator and the repair vs replace water heater calculator; for the cooling side, see the repair vs replace air conditioner calculator. This is a calculator, not a repair manual.