KPI Example
Calculate efficiency hours for a hydraulic excavator.
- 50 week/a, 5 day/week, 1-8h shift
- 85% Availability
- 92% Utilization
- ½ h lunch break, 12 min. travel time
- Average conditions
| Conditions | Front-End Loaders | Excavators, Hydraulic |
|---|---|---|
| Favourable | 50 | 55 |
| Average | 45 | 50 |
| Unfavourable | 40 | 45 |
Efficient working minutes per hour (after International 1975).
How many tons are produced, if the cycle time is 30 seconds and 28 t per cycle are loaded?
How many total hours would we need to achieve 2000 h/a efficiency hours?
Total hours
Up-time (with 85% Availability)
Working hours (with 92% Utilization)
Working efficiency (3 ways to get there)
a) 0.5 h lunch + 12/60 h travel = 0.7 h per shift → 7.3h of 8h → 7.3/8 = 0.9125 (91.25%)
b) 0.7 h per shift → 175 h/a with 0.92*0.85 (availability and utilization) → 136.85 h/a → (1564-136.85)/1564 = 0.9125
c) (2000-175)/2000 = 0.9125
Operating hours
Operating efficiency and Efficiency hours
Value taken from Table → 50 min per hour → 50/60 = 0.83
With 1184 efficiency hours, cycle time 30 sec and 28 t/cycle → Production [t]?
Reversing: How many total hours to get 2000 h/a efficiency hours?
