Count at the process exit
Use completed parts at one defined exit. Do not count the same part again as it moves between workstations. A part that passes through rework still counts as one output part.
Hypothetical arithmetic scenarios, without a performance promise. The time scenario holds good output constant. It does not identify downtime causes or establish how to shorten the period.
Define the process boundary and keep the time basis consistent across shifts. Compare gross rate and good-part throughput. Count reworked parts only once at the final exit.
Use completed parts at one defined exit. Do not count the same part again as it moves between workstations. A part that passes through rework still counts as one output part.
Shift throughput includes downtime within the observed shift. A rate calculated only during machine running time is a different operating measure. Keep those time bases separate when comparing results.
Subtract defective parts from total output, then divide by the observation time. Gross rate uses total output over the same period. The good-part share is not first pass yield unless first-pass results were recorded.
G = total output − defective partsUse one consistent quality decision at the defined process exit.
G ÷ observation time in hoursGood parts per hour, including downtime within the chosen time basis.
Total output ÷ observation timeProduction rate before excluding defective parts.
3,600 ÷ good-part throughputThe reciprocal average is not an individual part processing-time measurement.
Illustrative shift data: 1,000 completed parts, of which 40 are defective. The eight-hour observation period includes downtime within the shift.
| Item | Calculation | Result |
|---|---|---|
| Good parts | 1,000 − 40 | 960 parts |
| Good-part throughput | 960 ÷ 8 | 120 parts/h |
| Gross rate | 1,000 ÷ 8 | 125 parts/h |
| Good-part share | 960 ÷ 1,000 × 100 | 96% |
| Average time | 3,600 ÷ 120 | 30 s/part |
Divide the hourly result by 60. In the example, 120 parts per hour equals 2 parts per minute.
The reciprocal of an output rate gives average time per output part. With parallel processes, this differs from the processing time at one machine.
Use the same process boundary, quality counting method and time basis. A changing product mix can alter unit throughput even when machines operate in the same way.
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