Does the 4.2-Month Payback Count the Operating Cost?
Does the 4.2-Month Payback Count the Operating Cost?
Ada re-derives this chapter’s own numbers step by step, at full precision
ADA · CALCULATION AUDIT
Does the 4.2-Month Payback Count the Operating Cost?
The chapter’s ROI table shows a 4.2-month payback: 8 failures a year at $75,000 each is $600,000, and preventing 85% saves $510,000 against a $180,000 investment. But that divides by the gross savings and never subtracts the $36,000/year operating cost. This audit rebuilds the cash flow to ask: does the 4.2-month payback count the operating cost?
Companion to the chapter Predictive Maintenance — every number here comes from that chapter.
See the relationship before changing it
The figure reads from left to right. The blue card is failures prevented. The middle card applies the page rule. The green card is net payback. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 85 %.
- 2
Name the relationship. payback = 180,000 / (8 x 75,000 x prevented fraction - 36,000) x 12
- 3
Substitute with units. 180,000 / (8 x 75,000 x 0.85 - 36,000) x 12 = 4.56 months
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change failures prevented
Try Predict the direction of payback = 180,000 / (8 x 75,000 x prevented fraction - 36,000) x 12. Test another failures prevented, then compare net payback.
Observe Net payback uses annual cash flow after the operating cost is removed. Reset failures prevented to 85 and compare net payback.
Explain Net payback uses annual cash flow after the operating cost is removed.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
Ada: A payback figure is only as honest as the cash flow underneath it, so let me rebuild the Sample ROI table’s 4.2 months from its own inputs and check exactly what that number includes.
First, what does one failure actually cost? The $600,000 annual figure hides a two-part cost per failure — lost production plus a replaced motor:
- Cost per failure:
12 hours x $5,000 + $15,000 = 60,000 + 15,000 = $75,000 - Annual failure cost:
8 x 75,000 = $600,000, matching the table
Now the savings and payback:
- Prevented failures:
8 x 0.85 = 6.8 per year - Gross annual savings:
6.8 x 75,000 = $510,000 - Payback as printed:
180,000 / 510,000 = 0.3529 years = 4.2 months
That reproduces the table exactly — but notice it divides the investment by the gross $510,000 and never subtracts the $36,000/year operating cost the table lists just above it. Net that cost out, the way the chapter’s own chemical-plant example does, and the figure moves:
- Net annual savings:
510,000 - 36,000 = $474,000 - Net payback:
180,000 / 474,000 = 0.3797 years = 4.6 months
The two numbers — 4.2 months gross, 4.6 months net of operating cost — bracket the truth, about ten days apart, and the headline conclusion (payback in under half a year) holds either way. The design lesson is small but load-bearing: a payback figure should always say whether it is gross or net of running costs, because the same investment reads faster or slower depending on that one unstated choice, and the number the operating budget actually feels is the net 4.6 months.
The payback calculation deliberately does not simulate discount rates, tax, ramp-up, false alerts, maintenance escalation, failure-time uncertainty, or variable downtime cost; it uses constant annual expected savings.
Work the audit first, then check the displayed derivation.
Every number above is taken from the chapter’s own material and re-derived step by step.