The repair wins today — the replacement wins the decade
An air conditioner repair is cheap and a new air conditioner 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: $600.00. The replace total is the new price, minus the saving over the old unit’s remaining life, plus disposal: $5,500 − $180 × 5 + $100 = $4,700.00. On the defaults — repair $600.00, new air conditioner $5,500.00, remaining life 5 years, saving $180.00/year, disposal $100.00 — REPAIR wins by $4,100.00; cost per year repair $120.00 vs replace $940.00; break-even repair $4,700.00, break-even price $1,400.00, break-even life 27.78 yrs (exact 27.7778), break-even saving $1,000.00/yr, break-even disposal none (-$4,000.00); 10-year repair-then-replace $5,300.00 vs replace-now $3,800.00 — REPLACE wins by $1,500.00 over 10 years.
| Scenario | Repair | Replace | Winner | By | 10-year winner |
|---|
| Default | $600.00 | $4,700.00 | REPAIR wins by $4,100.00 | $4,100.00 | REPLACE wins by $1,500.00 over 10 years |
| Repair $1,500 | $1,500.00 | $4,700.00 | REPAIR wins by $3,200.00 | $3,200.00 | REPLACE wins by $2,400.00 over 10 years |
| Repair $5,000 | $5,000.00 | $4,700.00 | REPLACE wins by $300.00 | $300.00 | REPLACE wins by $5,900.00 over 10 years |
| 10 years left | $600.00 | $3,800.00 | REPAIR wins by $3,200.00 | $3,200.00 | REPLACE wins by $2,400.00 over 10 years |
| 12 years left | $600.00 | $3,440.00 | REPAIR wins by $2,840.00 | $2,840.00 | REPLACE wins by $2,760.00 over 10 years |
| Saving $400/year | $600.00 | $3,600.00 | REPAIR wins by $3,000.00 | $3,000.00 | REPLACE wins by $2,600.00 over 10 years |
| Saving $0/year | $600.00 | $5,600.00 | REPAIR wins by $5,000.00 | $5,000.00 | REPLACE wins by $600.00 over 10 years |
| New air conditioner $3,000 | $600.00 | $2,200.00 | REPAIR wins by $1,600.00 | $1,600.00 | REPLACE wins by $1,500.00 over 10 years |
| New air conditioner $8,000 | $600.00 | $7,200.00 | REPAIR wins by $6,600.00 | $6,600.00 | REPLACE wins by $1,500.00 over 10 years |
| Disposal $0 | $600.00 | $4,600.00 | REPAIR wins by $4,000.00 | $4,000.00 | REPLACE wins by $1,500.00 over 10 years |
| Disposal $500 | $600.00 | $5,100.00 | REPAIR wins by $4,500.00 | $4,500.00 | REPLACE wins by $1,500.00 over 10 years |
At the defaults — repair $600.00, new air conditioner $5,500.00, remaining life 5 years, saving $180.00/year, disposal $100.00 — the repair total is $600.00, the replace total is $4,700.00, REPAIR wins by $4,100.00; cost per year repair $120.00 vs replace $940.00; break-even repair $4,700.00, break-even price $1,400.00, break-even life 27.78 yrs (exact 27.7778), break-even saving $1,000.00/yr, break-even disposal none (-$4,000.00); 10-year repair-then-replace $5,300.00 vs replace-now $3,800.00 — REPLACE wins by $1,500.00 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 $600.00, new air conditioner $5,500.00, remaining life 5 years, saving $180.00/year, disposal $100.00.
Line 1 — the repair total. The repair total is the repair cost: $600.00.
Line 2 — the replace total. $5,500 − $180 × 5 + $100 = $5,500 − $900 + $100 = $4,700.00. The five years of saving shave $900 off the new unit’s price.
Line 3 — the winner. $4,700.00 − $600.00 = $4,100.00, so REPAIR wins by $4,100.00. Cost per year: $600 ÷ 5 = $120.00 for the repair, $4,700 ÷ 5 = $940.00 for the replacement.
Line 4 — the break-evens. The break-even repair cost is the replace total, $4,700.00. The break-even new price is $600 + $180 × 5 − $100 = $1,400.00. The break-even remaining life is ($5,500 + $100 − $600) ÷ $180 = 27.78 yrs (exact 27.7778). The break-even annual saving is ($5,500 + $100 − $600) ÷ 5 = $1,000.00/yr. The break-even disposal is $600 − $5,500 + $180 × 5 = -$4,000.00 — none, because the disposal cost that would tie is negative.
Line 5 — the 10-year lens. Repair now, then replace in 5 years: $600 + $5,500 + $100 − $180 × 5 = $5,300.00. Replace now: $5,500 + $100 − $180 × 10 = $3,800.00. So REPLACE wins by $1,500.00 over 10 years — the verdict flips over 10 years.
Reverse: the break-even repair cost is $4,700.00 (exact $4,700.00); a $4,700.00 repair costs $4,700.00, exactly tying the replace total — verdict: TIE. The break-even new price is $1,400.00 (exact $1,400.00); a $1,400.00 air conditioner costs $600.00 to replace, tying the repair — verdict: TIE. The break-even remaining life is 27.78 yrs (exact 27.7778); at 27.78 yrs the replace total is $599.60 (REPLACE wins by $0.40), at 27.77 yrs it is $601.40 (REPAIR wins by $1.40), at 27.79 yrs it is $597.80 (REPLACE wins by $2.20) — 27.7778 is the exact tie. The break-even annual saving is $1,000.00/yr (exact $1,000.00); at $1,000.00/yr saving the replace total is $600.00, tying the repair — verdict: TIE. The break-even disposal is none — the disposal cost that would tie is -$4,000.00; repairing wins even with free disposal — verdict: NONE. The 10-year lens: repair-then-replace $5,300.00 vs replace-now $3,800.00 — REPLACE wins by $1,500.00 over 10 years.
Every branch, checked
Repair $1,500: Repair $1,500.00, replace $4,700.00 — REPAIR wins by $3,200.00. 10-year: $6,200.00 vs $3,800.00 — REPLACE wins by $2,400.00 over 10 years. A $1,500 repair is still far cheaper than replacing — REPAIR wins by $3,200.00.
Repair $5,000: Repair $5,000.00, replace $4,700.00 — REPLACE wins by $300.00. 10-year: $9,700.00 vs $3,800.00 — REPLACE wins by $5,900.00 over 10 years. At $5,000 the repair costs more than the $4,700.00 replacement — REPLACE wins by $300.00, a flip from the default verdict.
10 years left: Repair $600.00, replace $3,800.00 — REPAIR wins by $3,200.00. 10-year: $6,200.00 vs $3,800.00 — REPLACE wins by $2,400.00 over 10 years. Ten years of $180 savings are worth $1,800, so the replace total falls to $3,800.00 — but REPAIR wins by $3,200.00 today; the 10-year lens is still REPLACE by $2,400.00.
12 years left: Repair $600.00, replace $3,440.00 — REPAIR wins by $2,840.00. 10-year: $6,560.00 vs $3,800.00 — REPLACE wins by $2,760.00 over 10 years. Twelve years of saving drop the replace total to $3,440.00 — REPAIR wins by $2,840.00 today. And the 10-year repair-then-replace is $6,560.00 — the unclamped (10 − L) = −2 means the replacement you buy in 12 years is charged with negative saving, so the path costs more, not less.
Saving $400/year: Repair $600.00, replace $3,600.00 — REPAIR wins by $3,000.00. 10-year: $4,200.00 vs $1,600.00 — REPLACE wins by $2,600.00 over 10 years. A $400-a-year saving pays for the new unit fast — the replace total is $3,600.00 and REPAIR wins by $3,000.00; over 10 years the replace-now total is $1,600.00.
Saving $0/year: Repair $600.00, replace $5,600.00 — REPAIR wins by $5,000.00. 10-year: $6,200.00 vs $5,600.00 — REPLACE wins by $600.00 over 10 years. With no saving the new unit costs its full price plus disposal — $5,600.00 — so REPAIR wins by $5,000.00 and the break-even remaining life is — (no saving, so no life ever ties).
New air conditioner $3,000: Repair $600.00, replace $2,200.00 — REPAIR wins by $1,600.00. 10-year: $2,800.00 vs $1,300.00 — REPLACE wins by $1,500.00 over 10 years. A cheaper new unit narrows the gap to $2,200.00 — but REPAIR wins by $1,600.00, the closest call on the page.
New air conditioner $8,000: Repair $600.00, replace $7,200.00 — REPAIR wins by $6,600.00. 10-year: $7,800.00 vs $6,300.00 — REPLACE wins by $1,500.00 over 10 years. An $8,000 unit is thirteen times the repair — REPAIR wins by $6,600.00.
Disposal $0: Repair $600.00, replace $4,600.00 — REPAIR wins by $4,000.00. 10-year: $5,200.00 vs $3,700.00 — REPLACE wins by $1,500.00 over 10 years. Free haul-away drops the replace total to $4,600.00 — REPAIR wins by $4,000.00.
Disposal $500: Repair $600.00, replace $5,100.00 — REPAIR wins by $4,500.00. 10-year: $5,700.00 vs $4,200.00 — REPLACE wins by $1,500.00 over 10 years. A $500 disposal fee pushes the replace total to $5,100.00 — REPAIR wins by $4,500.00.
What the defaults assume
The model assumes the repair restores the old air conditioner 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 energy 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: $180 a year of lower energy and fewer service calls. With a $0 repair the total is $0.00 and REPAIR wins by $4,700.00, but the 10-year lens is still REPLACE by $900.00. 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 ($4,700.00, $1,400.00, 27.7778, $1,000.00, -$4,000.00) are stated wherever the rounded figures appear.
Output · formula · meaning
| Output | Formula | What it means |
|---|
| Repair total | the repair cost R | $600.00. |
| Replace total | new price − saving × life + disposal | $4,700.00. |
| Difference | replace total − repair total | $4,100.00. |
| Winner | REPAIR when the difference is positive, REPLACE when negative | REPAIR wins by $4,100.00. |
| Cost per year (repair) | repair cost ÷ remaining life | $120.00. |
| Cost per year (replace) | replace total ÷ remaining life | $940.00. |
| Break-even repair cost | the replace total | $4,700.00. |
| Break-even new price | repair + saving × life − disposal | $1,400.00. |
| Break-even remaining life | (new price + disposal − repair) ÷ saving | 27.7778 yrs → 27.78 yrs. |
| Break-even annual saving | (new price + disposal − repair) ÷ life | $1,000.00/yr. |
| Break-even disposal | repair − new price + saving × life | -$4,000.00 → none. |
| 10-year: repair-then-replace | repair + new price + disposal − saving × (10 − life) | $5,300.00. |
| 10-year: replace-now | new price + disposal − saving × 10 | $3,800.00. |
| 10-year winner | the smaller 10-year total | REPLACE wins by $1,500.00 over 10 years. |
| Break-even repair cost | the replace total | $4,700.00 → $4,700.00. |
| Break-even new price | repair + saving × life − disposal | $1,400.00 → $1,400.00. |
| Break-even remaining life | (new price + disposal − repair) ÷ saving | 27.7778 → 27.78 yrs. |
| Break-even annual saving | (new price + disposal − repair) ÷ life | $1,000.00/yr → $1,000.00/yr. |
| Break-even disposal | repair − new price + saving × life | -$4,000.00 → none. |
| 10-year lens | repair-then-replace vs replace-now | $5,300.00 vs $3,800.00. |
| Repair $1,500 | same model | Repair $1,500.00, replace $4,700.00 — REPAIR wins by $3,200.00. 10-year: $6,200.00 vs $3,800.00 — REPLACE wins by $2,400.00 over 10 years. |
| Repair $5,000 | same model | Repair $5,000.00, replace $4,700.00 — REPLACE wins by $300.00. 10-year: $9,700.00 vs $3,800.00 — REPLACE wins by $5,900.00 over 10 years. |
| 10 years left | same model | Repair $600.00, replace $3,800.00 — REPAIR wins by $3,200.00. 10-year: $6,200.00 vs $3,800.00 — REPLACE wins by $2,400.00 over 10 years. |
| 12 years left | same model | Repair $600.00, replace $3,440.00 — REPAIR wins by $2,840.00. 10-year: $6,560.00 vs $3,800.00 — REPLACE wins by $2,760.00 over 10 years. |
| Saving $400/year | same model | Repair $600.00, replace $3,600.00 — REPAIR wins by $3,000.00. 10-year: $4,200.00 vs $1,600.00 — REPLACE wins by $2,600.00 over 10 years. |
| Saving $0/year | same model | Repair $600.00, replace $5,600.00 — REPAIR wins by $5,000.00. 10-year: $6,200.00 vs $5,600.00 — REPLACE wins by $600.00 over 10 years. |
| New air conditioner $3,000 | same model | Repair $600.00, replace $2,200.00 — REPAIR wins by $1,600.00. 10-year: $2,800.00 vs $1,300.00 — REPLACE wins by $1,500.00 over 10 years. |
| New air conditioner $8,000 | same model | Repair $600.00, replace $7,200.00 — REPAIR wins by $6,600.00. 10-year: $7,800.00 vs $6,300.00 — REPLACE wins by $1,500.00 over 10 years. |
| Disposal $0 | same model | Repair $600.00, replace $4,600.00 — REPAIR wins by $4,000.00. 10-year: $5,200.00 vs $3,700.00 — REPLACE wins by $1,500.00 over 10 years. |
| Disposal $500 | same model | Repair $600.00, replace $5,100.00 — REPAIR wins by $4,500.00. 10-year: $5,700.00 vs $4,200.00 — REPLACE wins by $1,500.00 over 10 years. |
Who uses this
- The homeowner with a $600 repair quote: they learn REPAIR wins by $4,100.00 — and they check the $5,000 branch (REPLACE wins by $300.00) to see the repair price where replacing wins.
- The owner whose unit has 12 years left: they learn REPAIR wins by $2,840.00 — twelve years of saving still lose to the cheap repair today — and they check the 10-year branch (REPLACE wins by $2,400.00 over 10 years) to see the decade margin.
- The shopper eyeing a $3,000 air conditioner: they learn REPAIR wins by $1,600.00 — the closest call on the page — and they check the $8,000 branch (REPAIR wins by $6,600.00) 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 $1,500.00 over 10 years — and they check the $400-a-year branch (REPLACE wins by $2,600.00 over 10 years) to see the decade margin widen.
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 27.7778 (shown as 27.78 yrs); at 27.78 yrs the replace total is $599.60 (REPLACE wins by $0.40), at 27.77 yrs it is $601.40 (REPAIR wins by $1.40), at 27.79 yrs it is $597.80 (REPLACE wins by $2.20). What the model cannot do is predict an air conditioner’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 27.78 yrs is the number to compare against your honest estimate of the old unit’s remaining life. For the same question on a smaller appliance, see the repair vs replace washing machine calculator and the repair vs replace refrigerator calculator; for the size of the unit itself, see the air conditioner size chart. This is a calculator, not a repair manual.
Cite this calculatorBright Side Kit. “Repair vs Replace Air Conditioner: Cost Calculator 2026” brightoolbox.com, 2026. Method: repair total = R; replace total = P − S × L + D; difference = replace − repair (> 0 means REPAIR wins, < 0 means REPLACE wins, 0 = tie); cost per year (repair) = R / L; cost per year (replace) = replace 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 lens: repair now then replace at the end of the remaining life = R + P + D − S × (10 − L); replace now = P + D − S × 10. 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. Key idea: the repair wins today because the old unit still has life and the new one saves little — but over 10 years the new unit’s savings run the full decade, so the verdict flips. Defaults: repair $600, new air conditioner $5,500, remaining life 5 years, saving $180/year, disposal $100 → repair $600.00, replace $4,700.00, REPAIR wins by $4,100.00; cost per year repair $120.00 vs replace $940.00; break-even repair $4,700.00, break-even price $1,400.00, break-even life 27.78 yrs (exact 27.7778), break-even saving $1,000.00/yr, break-even disposal none (-$4,000.00); 10-year repair-then-replace $5,300.00 vs replace-now $3,800.00, REPLACE wins by $1,500.00 over 10 years. Branches: Repair $1,500 → Repair $1,500.00, replace $4,700.00 — REPAIR wins by $3,200.00. 10-year: $6,200.00 vs $3,800.00 — REPLACE wins by $2,400.00 over 10 years; Repair $5,000 → Repair $5,000.00, replace $4,700.00 — REPLACE wins by $300.00. 10-year: $9,700.00 vs $3,800.00 — REPLACE wins by $5,900.00 over 10 years; 10 years left → Repair $600.00, replace $3,800.00 — REPAIR wins by $3,200.00. 10-year: $6,200.00 vs $3,800.00 — REPLACE wins by $2,400.00 over 10 years; 12 years left → Repair $600.00, replace $3,440.00 — REPAIR wins by $2,840.00. 10-year: $6,560.00 vs $3,800.00 — REPLACE wins by $2,760.00 over 10 years (10-year repair-then-replace $6,560.00 uses the unclamped (10 − L) = −2); Saving $400/year → Repair $600.00, replace $3,600.00 — REPAIR wins by $3,000.00. 10-year: $4,200.00 vs $1,600.00 — REPLACE wins by $2,600.00 over 10 years; Saving $0/year → Repair $600.00, replace $5,600.00 — REPAIR wins by $5,000.00. 10-year: $6,200.00 vs $5,600.00 — REPLACE wins by $600.00 over 10 years; New air conditioner $3,000 → Repair $600.00, replace $2,200.00 — REPAIR wins by $1,600.00. 10-year: $2,800.00 vs $1,300.00 — REPLACE wins by $1,500.00 over 10 years; New air conditioner $8,000 → Repair $600.00, replace $7,200.00 — REPAIR wins by $6,600.00. 10-year: $7,800.00 vs $6,300.00 — REPLACE wins by $1,500.00 over 10 years; Disposal $0 → Repair $600.00, replace $4,600.00 — REPAIR wins by $4,000.00. 10-year: $5,200.00 vs $3,700.00 — REPLACE wins by $1,500.00 over 10 years; Disposal $500 → Repair $600.00, replace $5,100.00 — REPAIR wins by $4,500.00. 10-year: $5,700.00 vs $4,200.00 — REPLACE wins by $1,500.00 over 10 years. Reverses: break-even repair cost = $4,700.00 (exact $4,700.00; a $4,700.00 repair costs $4,700.00, exactly tying the replace total); break-even new price = $1,400.00 (exact $1,400.00; a $1,400.00 air conditioner costs $600.00 to replace, tying the repair); break-even remaining life = 27.78 yrs (exact 27.7778; at 27.78 yrs the replace total is $599.60, REPLACE wins by $0.40; at 27.77 yrs it is $601.40, REPAIR wins by $1.40); break-even annual saving = $1,000.00/yr (exact $1,000.00; at $1,000.00/yr saving the replace total is $600.00, tying the repair); break-even disposal = none (the disposal cost that would tie is -$4,000.00; repairing wins even with free disposal); 10-year lens: repair-then-replace $5,300.00 vs replace-now $3,800.00, REPLACE wins by $1,500.00 over 10 years. Edges: repair $0 → Repair $0.00, replace $4,700.00 — REPAIR wins by $4,700.00. 10-year: $4,700.00 vs $3,800.00 — REPLACE wins by $900.00 over 10 years; 1 year left, no saving → Repair $600.00, replace $5,600.00 — REPAIR wins by $5,000.00. 10-year: $6,200.00 vs $5,600.00 — REPLACE wins by $600.00 over 10 years (break-even life —). Bright Side Kit. "Repair vs Replace Air Conditioner: Cost Calculator 2026" https://brightoolbox.com/repair-vs-replace-air-conditioner/ Accessed October 25, 2026.
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