How to Calculate OEE: One Shift, Worked Through Step by Step
T. Krause
OEE in one sentence
Overall equipment effectiveness is the share of a machine's planned production time in which it made good parts at its ideal speed.
You can calculate it directly from that sentence:
OEE = (good count × ideal cycle time) ÷ planned production time
That gives you the score and nothing else. To see where the lost time went, split it into three factors.
- Availability. Was the machine running when it was meant to run?
- Performance. While it ran, did it run as fast as it can?
- Quality. Were the parts right the first time?
OEE = availability × performance × quality
Here is one shift with every figure filled in.
The shift we are going to calculate
The plant and the figures are made up for this example. They are chosen so that you can follow every line with a pocket calculator.
A CNC lathe turns bushings. The early shift finishes the tail of one job, then the setter changes the machine over to a second bushing. Both parts have the same ideal cycle time.
| Input | Value |
|---|---|
| Shift length | 480 min |
| Scheduled break, machine stopped | 30 min |
| Breakdowns from the stop log | 35 min |
| Changeover to the second job | 55 min |
| Ideal cycle time | 45 s per part |
| Total count from the machine counter | 420 parts |
| of which scrap | 9 |
| of which rework | 12 |
Percentages are shown to one decimal place. Round only the result of a line and carry the unrounded value forward.
Step 1: Planned production time
Planned production time = shift length − scheduled breaks
480 min − 30 min = 450 min
These 450 minutes are the denominator for everything that follows. Any of them that doesn't end in a good part is a loss.
Step 2: Run time
Run time = planned production time − stop time
Stop time is 35 min of breakdowns plus 55 min of changeover, 90 min in total.
450 min − 90 min = 360 min
The changeover counts as a stop even though it was on the schedule. Mistake 2 below explains why.
Step 3: Availability
Availability = run time ÷ planned production time
360 min ÷ 450 min = 0.8 = 80.0%
Step 4: Performance
Performance = (total count × ideal cycle time) ÷ run time
420 parts × 45 s = 18,900 s = 315 min
315 min ÷ 360 min = 0.875 = 87.5%
Put another way, 360 minutes at 45 seconds a part had room for 480 parts. The counter shows 420. The 60 missing parts are worth 45 minutes, and no stop log has a line for them.
Step 5: Quality
Good count = total count − scrap − rework
420 − 9 − 12 = 399 parts
Quality = good count ÷ total count
399 ÷ 420 = 0.95 = 95.0%
A good part is one that was right the first time through. A reworked bushing may still ship, but it counts as a loss here.
Step 6: OEE
OEE = availability × performance × quality
0.8 × 0.875 × 0.95 = 0.665 = 66.5%
Here all three factors come out exact. Your own numbers rarely will, so multiply the unrounded factors and round only the OEE.
The cross-check
Now the one-line formula from the top.
399 parts × 45 s = 17,955 s = 299.25 min
299.25 min ÷ 450 min = 0.665 = 66.5%
The two results have to match. If they don't, a minute or a part has been counted twice or left out.
Out of 450 planned minutes, the lathe spent 299.25 making good parts at ideal speed. The other 150.75 minutes went here:
| Item | Loss | Remaining |
|---|---|---|
| Planned production time | 450 min | |
| Breakdowns | 35 min | 415 min |
| Changeover | 55 min | 360 min |
| Small stops and slow cycles | 45 min | 315 min |
| Scrap and rework, 21 parts × 45 s | 15.75 min | 299.25 min |
Six ways to count it wrong
Mistakes 2 to 5 raise the score on the example shift without one extra part being made. Mistake 6 adds a Saturday run and reports the two runs together six points too high.
1. Putting breaks and planned downtime in the wrong place
Planned production time is the time the machine is scheduled to produce. If it stops during the break because nobody runs it, the break is not part of it. If a bar feeder keeps it running, the break is. Time with no order stays out. Plants differ on planned maintenance. Write the rule down once and apply it to every machine the same way.
Calculate against the calendar instead and you get a different metric. It is called TEEP, total effective equipment performance, and it divides the same productive time by every minute of the day. On the example shift that is 299.25 min ÷ 1,440 min = 20.8%. OEE tells you how well the machine uses its scheduled time. TEEP tells you how much capacity the machine has left across the whole day. Both figures are correct, but they don't belong in the same column of a report.
2. Treating changeover as planned time
Changeovers are on the schedule, so some plants subtract them like a break. On the example shift, planned production time then drops from 450 to 395 minutes. Availability rises to 360 ÷ 395 = 91.1%, and OEE to 299.25 ÷ 395 = 75.8%. That is a good nine points more for the same number of parts.
Worse, the 55 minutes disappear from the loss table, and with them the reason to shorten them.
3. Missing small stops and slow running
A stop log holds what somebody enters or what exceeds the system's threshold. A jammed part or a feed rate turned down a notch never gets there. Both show up in performance, but only if you calculate performance from the total count and the ideal cycle time.
On the example shift they add up to 45 minutes, more than all the logged breakdowns together. Set performance to 100% for lack of data and the result is 0.8 × 1 × 0.95 = 76.0%, almost ten points too high.
4. An ideal cycle time that is too soft
The ideal cycle time is the shortest time in which the machine can make the part, sustained, under good conditions. What often ends up in the calculation is the standard time from the costing, allowances included.
Say that figure were 54 seconds instead of 45. Then 420 parts × 54 s = 22,680 s = 378 min, and 378 min ÷ 360 min gives a performance of 105.0%. OEE would read 0.8 × 1.05 × 0.95 = 79.8%. Performance above 100% tells you the reference time is too long.
5. Counting rework as good
Count the twelve reworked bushings as good and the good count becomes 411 instead of 399. Quality rises to 411 ÷ 420 = 97.9%. For OEE, 411 parts × 45 s = 18,495 s = 308.25 min, and 308.25 min ÷ 450 min = 68.5%.
Two points sounds harmless. But those twelve parts took time twice, and the second time now appears nowhere.
6. Averaging OEE scores
On Saturday the same lathe runs 150 minutes for a rush order, with no changeover and no breakdown, and makes 181 good parts. That is 181 parts × 45 s = 8,145 s = 135.75 min, so OEE is 135.75 ÷ 150 = 90.5%.
The average of 66.5% and 90.5% is 78.5%. Done properly, you add the times before you divide:
(299.25 min + 135.75 min) ÷ (450 min + 150 min) = 435 min ÷ 600 min = 72.5%
The average treats 150 minutes as if they were 450 and comes out six points too high. Averaging several machines, or the daily scores of a week, goes wrong the same way. Add up minutes and parts first.
What to do with the result
Nobody can act on 66.5%. The loss table is different, because it puts every loss in minutes and shows which factor costs the most. On the example shift:
Availability, 90 minutes. The largest single item is the 55-minute changeover. That is where SMED starts, by separating work that needs the machine stopped from work that can be finished beforehand.
For the 35 minutes of breakdowns, read the reasons. If the log says "waiting for material" or "waiting for release", the cause sits upstream of the machine, in the order flow. How to calculate lead time makes that waiting visible on one order followed through six stations.
Performance, 45 minutes. Nothing was logged for this loss. Stand at the machine for an hour and tally every stop under two minutes, with its reason.
Quality, 21 parts. If scrap and rework come straight after the changeover, they are start-up parts. Then the problem belongs to the changeover and to first-article release.
When OEE is the wrong number
The machine is not the bottleneck. Raising OEE on a machine whose output queues at the next station gets you more inventory and no earlier delivery date. Delivery time is decided at the bottleneck and in the waiting between stations. Measure OEE first where the orders pile up.
The score becomes the target. OEE goes up when a machine runs long batches without changeovers, whether or not anyone ordered the parts. Tie a bonus or a shift's reputation to the score and that is what you will get. Batches grow and the finished goods store fills up. Read OEE next to inventory and on-time delivery, never alone. How many parts the customer needs per shift is worked out in how to calculate takt time on one assembly cell.
The 85% yardstick. You will meet this figure as "world-class OEE". It goes back to Seiichi Nakajima, whose book "Introduction to TPM", published in Japanese in 1984 and in English in 1988, gave minimum values for the three factors. They are 90% availability, 95% performance and 99% quality. Multiplied, that is 84.6%, rounded to 85%. We take this from Vorne Industries' page World-Class OEE, as it stood in October 2026. The same page points out that the values come from the Japanese car industry of the 1970s. So 85% is one author's reference value for specific factor levels, not a target for every plant. A job shop with small batches and five changeovers a day can sit well below it without being badly run.
Plants draw the lines around breaks, changeovers and cycle time differently, so compare a machine with itself, week by week, under the same rule.
Try it on one shift
Pick one machine, ideally the bottleneck, and one shift. Write down the eight inputs from the first table, calculate both routes and put the loss table next to them. Paper and the machine counter are enough.
If you want to know whether your losses arise at the machine or in the order flow before it, that is the scope of our process analysis. Fixed scope, fixed duration, fixed price.