OpenSpecEngineering

Finds how many good units a workstation or cell really turns out. Starting from the cycle time and the number of machines or operators working side by side, it takes off the time lost to downtime and the units lost to scrap. It gives the rate per hour, the good units per period and the effective time per good unit. Production planners and industrial engineers use it to check capacity against demand and to find the bottleneck in a line [1], [2].

Equations

R_gross = m / CT     R_eff = R_gross · A     R_good = R_eff · (1 − s)

capacity = R_good · T_op     CT_good = 1 / R_good

CT is the cycle time of one machine or operator, in minutes per unit. m is the number of parallel resources, each doing the full job. A is the availability, the share of operating time the resources are actually running, as a fraction: 90 % is 0.90. s is the share of units scrapped or sent out for rework. T_op is the operating time in the period, in minutes. The rates are found per minute and shown per hour. CT_good is the average time, in minutes, between good units leaving the station.

Limits and common mistakes

• Parallel resources share the work. Machines in series, each doing a different step, are not parallel; their throughput is set by the slowest one.
• Availability here covers downtime only. A machine that runs below its rated speed has a longer cycle time; enter that real cycle time rather than the rated one.
• Enter planned operating time, after breaks, as T_op. Unplanned stops belong in the availability, so don’t take them out twice.
• The result is the long-run average. With variable cycle times and no buffer, a station can fall short of this rate [1].
• Reworked units that come back as good are not lost. Count only units that leave for good in the scrap rate.

Worked example

A machining cell has three identical CNC lathes, each with a 4.5 min cycle. They run 85 % of the time, and 2 % of the parts are scrapped. A shift has 450 min of operating time. How many good parts will a shift give?

1. Enter Cycle Time per Unit 4.5 and Parallel Resources 3.
2. Enter Availability / Uptime 85 and Scrap / Rework-Out Rate 2.
3. Enter Operating Time per Period 450 and press Solve. The Solution shows: Gross production rate: 40 units/hr
   Effective rate (after availability): 34 units/hr
   Good-unit throughput rate: 33.32 units/hr
   Good-unit capacity per period: 249.9 units
   Effective cycle time per good unit: 1.80072 min

Check by hand. Three lathes at 4.5 min make 3 × 60 / 4.5 = 40 parts an hour. 40 × 0.85 × 0.98 = 33.32, and over 7.5 hr that is 249.9, so plan on 249 good parts a shift. One good part leaves every 60 / 33.32 = 1.80 min.