How to calculate machine capacity, step by step

• Theoretical capacity: every hour on the calendar, 24 hours a day, 7 days a week. Useful only as an upper bound.
• Available capacity: the hours the machine is scheduled to work, meaning shifts minus breaks, holidays and planned maintenance.
• Effective capacity: available hours adjusted for how the machine actually performs, meaning efficiency losses from slow running, minor stops and unplanned downtime.

Planners should schedule against effective capacity. Theoretical capacity makes every week look possible, and available capacity makes it look slightly more possible than it is.

Start from the shift pattern.

• Hours per shift, minus paid breaks during which the machine stops
• Times shifts per day
• Times working days in the period
• Minus planned maintenance and holidays in the period

Example: a mill runs two 8 hour shifts with a 30 minute break in each, on 5 days a week, and has 4 hours of planned maintenance this week. Hours per shift after breaks: 7.5. Per day: 15. Per week: 75. Minus maintenance: 71 available hours.

If breaks are staggered so the machine keeps running, do not subtract them. That alone is one of the cheapest capacity gains in most plants.

No machine runs at its book rate every minute it is scheduled. Measure the ratio between the standard hours of work it completed and the hours it was scheduled, over a few weeks. That ratio is its efficiency.

Example: in the last four weeks, the mill was scheduled for 284 hours and completed work worth 241 standard hours. Efficiency: 241 divided by 284, about 85 percent.

Effective capacity this week: 71 available hours times 0.85, about 60 hours.

If you do not have this data yet, start with a conservative estimate, such as 80 to 85 percent for a machine with regular setups, and correct it as actual times come in from the floor.

Load is the work you plan to send to the machine in the same period, in hours.

• For each order, quantity times run minutes per unit on this machine
• Plus setup minutes for each changeover
• Add them up and convert to hours

Example: this week the mill has three orders. 50 housings at 13 minutes each, 65 housings at 13 minutes each, and 80 flanges at 4 minutes each, plus three setups of 30 minutes. Run time: 650 plus 845 plus 320 minutes, 1,815 minutes. Setups: 90 minutes. Total load: 1,905 minutes, about 31.8 hours.

Utilisation is load divided by effective capacity.

Example: 31.8 hours of load against 60 hours of effective capacity is 53 percent. The mill has room this week.

• Below about 85 percent, a machine normally absorbs the variation of a real week.
• Between 85 and 100 percent, it will run late whenever anything goes wrong.
• Above 100 percent, some orders will be late for certain, and the only question is which.

Do this for every work center and the one with the highest utilisation is your likely production bottleneck.

A machine can be at 70 percent for the week and still make orders late, because the work does not arrive evenly. If Monday and Tuesday hold most of the load and the last operations of an order have to follow, the order can miss its date while the machine sits idle on Friday. Weekly utilisation tells you whether the week fits in total. It does not tell you whether each order fits in time.

That second question needs finite capacity scheduling: every operation placed on the machine in sequence, inside its shifts, after the operation before it has finished. Read more on advanced planning and scheduling software.

• Available hours = (shift hours minus breaks) times shifts per day times days, minus maintenance and holidays
• Efficiency = standard hours completed divided by hours scheduled
• Effective capacity = available hours times efficiency
• Load = sum of quantity times run time, plus setup time, in hours
• Utilisation = load divided by effective capacity