Shoe Repair: Cost Per Month Of Wear, Repair Vs Replace

Excel Formulas › Shoe Repair

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The customer at the counter is doing one calculation in their head: is $45 for a resole worth it when new shoes are $160? Put it in months. A resole that adds 18 months costs $2.50 a month; the new pair that lasts three years costs $4.44 a month. Two divisions and a comparison, and the cobbler has a number on the ticket that makes the case for them.


Quick formula: Repair per month, new per month, and which is lower:
=B2/C2 and =D2/E2

$45 over 18 months is $2.50 a month; $160 over 36 months is $4.44. The repair wins by $1.94 a month.

Functions used (tap for the full reference guide):

The example

Three jobs. The cheap flats are the one where replacement wins — a $30 repair that only buys six months is more expensive per month than the $90 replacement.

ABCDEFGH
1JobRepairMonths addedNew priceNew life (mo)Repair /moNew /moBetter
2Full resole, boots$4518$16036$2.50$4.44Repair
3Goodyear welt rebuild$8524$24048$3.54$5.00Repair
4Heel tips, flats$306$9024$5.00$3.75Replace

The formula

Two divisions and one IF:

=B2/C2 =D2/E2 =IF(F2<=G2,"Repair","Replace") // per-month costs, then the verdict

How it works

Both sides are the same shape: price over months of wear.

  1. B2/C2 is the repair price over the months of wear it buys. A resole on a welted boot adds a year and a half; a heel tip adds a few months.
  2. D2/E2 is the new-shoe price over its expected life. Be honest here — a $160 boot that gets resoled twice lasts far longer than 36 months, and that argument is a second repair.
  3. IF(F2<=G2,"Repair","Replace") compares the two. Ties go to the repair because it is less waste and less money up front.
  4. Round both per-month figures to cents for the ticket. The comparison should use the unrounded values.

The comparison assumes the repaired shoe is as good as new for the months it adds. For a well-made shoe that is true; for a glued sneaker it is not, and the honest thing is to shorten the months-added figure until it is.

Try it: interactive demo

Interactive

Enter the repair price and months it adds, and the new-shoe price and expected life.

Variations

Savings over the repair period

The per-month gap times the months the repair adds is the dollar saving the customer actually keeps.

=(D2/E2-B2/C2)*C2

Break-even months

How many months the repair has to last to match the new pair per month. If your honest months-added estimate is above this, repair.

=B2/(D2/E2)

Pitfalls & errors

Months added is a judgment call and it is the whole answer. A cobbler who inflates it to win the job loses the customer six months later. Quote the low end.

Add a third column for a second repair. A welted boot that can be resoled twice at $45 has a real life of 36 + 18 + 18 months on a $160 base, which is the comparison that sells premium footwear.

Do not compare the repair price to the new price directly. $45 versus $160 is the customer's instinct and it is the wrong comparison — the months are what make it an answer.

Practice workbook

📊
Download the free Shoe Repair: Cost Per Month Of Wear, Repair Vs Replace practice workbook
Edit the yellow price and month cells; per-month costs and the verdict recalculate.

Frequently asked questions

What is a fair life for new shoes?
It depends entirely on construction and use. A cemented fashion shoe worn daily is 12 to 18 months; a welted leather boot is 3 years before its first resole and much longer with care. Use the customer's own history if they have it — how long did the last pair last?
Should I include the cost of the customer's time?
You can add a flat amount to the repair side for the trip and the wait, but most shops leave it out. The comparison is already lopsided in favour of repair for anything well-made, and adding soft costs makes it look like you are arguing.

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Function references: IFROUND