13. Two runs, two products, two performances (minor stops)

In this example

• Shifttime:                                         500 min
• Production Run A:                         100 min
  •   Output 1                              80.000 (Set Speed 1000 pcs/min)
  •   Output 2                              70 (Set Speed 1 pcs/min)
• Production Run B:                         10 min
  •   Output 3                              7.000 (Set Speed 1000 pcs/min)
  •   Output 4                              6 (Set Speed 1 pcs/min)
• Waiting:                                            390 min

Result

1. Availability 22% (110min/500min)
2. Performance: 74,1% (NOT: Actual output / Maximum Output= 79,1%)
3. Quality: Good Product / (Good+Rejects) = 100%
4. OEE = 22% x 74,1% x 100% = 16,3%

Calculation

1. The performance of each product in a run is calculated

First run:

• 80.000/100.000 = 80%
• 70/100 = 70%

Second run:

• 7.000/10.000 = 70%
• 6/10 = 60%

2. The average performance for each run is calculated

• 100 min production: (80% + 70%)/2 = 75% 
• 10 min production: (70% + 60%)/2 = 65%

3. A weighted average is calculated for the different runs

• 100 minutes at 75% and 10 minutes at 65%:
• ((100/110) * 75%) + ((10/110) * 65%) = 74,1%

Warning: Just adding all actual outputs and dividing this by all maximum outputs does not give a correct performance!   In this example it would give: 87076 / 110110 = 79 %   This does not take into account that running 100 minutes at a certain speed weighs more than running 10 minutes at a speed!