MTBF and MTTR: how to calculate them and where they mislead
Definitions, formulas and a worked example for MTBF and MTTR, plus the pitfalls: what counts as a failure, what counts as repair time and why averages hide long outages.
· 4 minThe shift we will calculate
Numbers make OEE concrete, so this guide follows one eight-hour shift from raw counters to a minute-by-minute loss breakdown. These are the same values the OEE calculator loads as its example, so you can check every step.
| Input | Value | Where it comes from |
|---|---|---|
| Planned production time | 480 min | Shift length minus breaks and unscheduled time |
| Run time | 432 min | Planned time minus the 48 minutes the machine was stopped |
| Ideal cycle time | 10 s per part | The agreed line standard, not the average actually achieved |
| Total count | 2,400 parts | Machine counter, good and bad together |
| Good count | 2,280 parts | Parts that passed inspection first time |
Availability: was the machine running?
Availability compares run time with planned production time: 432 ÷ 480 = 90.0 %. Every stop during planned production reduces it, whether it was a breakdown, a changeover or a wait for material. Breaks and unscheduled time are already excluded from the 480 minutes, so they do not count against the machine.
Performance: did it run at the intended speed?
Performance compares the time the parts should have needed with the time the machine was actually running. At an ideal 10 seconds per part, 2,400 parts need 24,000 seconds. The machine ran for 432 × 60 = 25,920 seconds. Performance is 24,000 ÷ 25,920 = 92.6 %.
Performance absorbs everything that makes the machine slower than its ideal cycle: reduced speed, micro-stops and a cycle time that was set too optimistically. If this figure exceeds 100 %, the ideal cycle time, the counter or the time base is wrong. The calculator asks you to fix the inputs rather than capping the value.
Quality: how many parts were good?
Quality is the share of good parts: 2,280 ÷ 2,400 = 95.0 %. The 120 rejected parts each consumed a full cycle of machine time that produced nothing sellable.
OEE: the product of the three
OEE = 0.900 × 0.926 × 0.950 = 79.2 %. There is a shortcut that avoids the three intermediate numbers: good count × ideal cycle time ÷ planned time = (2,280 × 10 s) ÷ (480 × 60 s) = 22,800 ÷ 28,800 = 79.2 %. Both routes must agree. If they do not, one of your inputs is inconsistent.
The loss waterfall: where the minutes went
A percentage tells you how large the gap is. A waterfall in minutes tells you where it is. Start with 480 planned minutes and subtract each loss in turn:
| Step | Calculation | Minutes | Remaining |
|---|---|---|---|
| Planned production time | 480 | ||
| Stop-time loss | 480 − 432 | 48 | 432 |
| Speed loss | 432 − (2,400 × 10 s ÷ 60) | 32 | 400 |
| Quality loss | 120 rejected × 10 s ÷ 60 | 20 | 380 |
| Productive time | 2,280 × 10 s ÷ 60 | 380 |
380 productive minutes out of 480 is the same 79.2 %. In this shift the largest single loss is stopped time, followed by speed and then quality. That ordering is a starting point for investigation, not a conclusion: a 20-minute quality loss may cost more than a 48-minute stop if the rejected parts carry expensive material.
What the numbers do not say
- Why the machine stopped. OEE shows the size of the loss, not its cause. Pair it with stop-reason codes before choosing an improvement.
- Whether 79.2 % is good. There is no universal target. Compare the line with its own history, and compare plants only when the definitions of planned time and ideal cycle time match.
- Whether the counters are right. A wrong good-count or a counter that resets mid-shift produces a confident but false answer. The OEE pilot acceptance guide explains how to validate counters before trusting the result.
Reproduce it and go further
Open the calculator with this shift preloaded, change one input at a time and watch the waterfall move. Then read about the six hidden losses behind OEE and how MTBF and MTTR relate to the 48 stopped minutes.
