How to Calculate Capacity Utilization: Three Milling Machines, One Week, Worked Through
T. Krause
Capacity utilization in one sentence
Capacity utilization is the share of a machine's or work centre's available capacity that was actually taken up by orders. To calculate it, divide the hours the machine was in use by the hours it was available, both for the same period.
Capacity utilization = hours in use ÷ available capacity × 100%
For a single machine the same ratio is usually called machine utilization. Writing the formula down takes a second. The result depends almost entirely on what you count as "in use" and which capacity you put underneath it.
Which capacity goes in the denominator
Reports use three kinds of capacity, and they rarely say which one they mean.
Theoretical capacity. Every hour the machine could physically run, which is 24 hours a day, 7 days a week, 168 hours. It is also called design capacity. Measured against it, utilization tells you how much a third shift or weekend working could still add.
Available capacity. The hours the shift plan schedules the machine to produce, after taking out breaks when nobody runs it and planned maintenance. It is the same available time the takt time article starts from. For day-to-day utilization, this is the denominator.
Effective capacity. Available capacity minus the losses you plan for anyway, mainly changeovers and the usual breakdowns. It is the number of machining hours production planning can realistically schedule.
The numerator needs a decision too. "In use" can mean the machine was cutting metal. It can also mean an order was on it, changeover included. In this article it means the second, which we will call occupied time:
Occupied time = machining time + changeover time
Breakdowns are left out. During a breakdown an order is assigned to the machine, but nothing happens to it.
How utilization differs from OEE and efficiency
Utilization asks whether the machine had work during its available time. OEE asks how well it ran while it had work. It multiplies availability, performance and quality, and it leaves time with no order out of the calculation. A machine with nothing to do drags utilization down and leaves OEE untouched. Slow running and scrap drag OEE down and never show up in utilization.
Efficiency is the second ratio you will meet. Utilization divides occupied time by available capacity. Efficiency usually divides machining time by effective capacity, the 91.8% in step 5. Plant reports don't always draw the line there, so ask what sits in the denominator before you compare two figures.
On a paced assembly line the related figure is balance efficiency. It compares the work content of the product with the station time available per takt. The line balancing article works one through.
The week we are going to calculate
The plant, the machines and every hour below are invented for this example. They are chosen so that you can check each figure by hand.
A contract machining shop has a milling work centre with three CNC machines, M1, M2 and M3. It mills Monday to Friday in two 8-hour shifts. Each shift has a 30-minute break during which the machines stand, because nobody is there to run them. On Wednesday M2 is down for 5 hours of planned maintenance. The machine data collection system books every hour to a status:
| Machine | Available | Machining | Changeover | Breakdown | No order |
|---|---|---|---|---|---|
| M1 | 75 h | 58 h | 7 h | 6 h | 4 h |
| M2 | 70 h | 49 h | 9 h | 4 h | 8 h |
| M3 | 75 h | 61 h | 6 h | 5 h | 3 h |
| Total | 220 h | 168 h | 22 h | 15 h | 15 h |
"No order" means the machine was staffed and ready, but there was no job waiting or the material for the next one had not arrived. Each row adds up to the machine's available hours.
Step 1: Theoretical capacity
3 machines × 168 h = 504 h
Step 2: Available capacity
Shift plan = 3 machines × 5 days × 2 shifts × 8 h = 240 h
Breaks = 3 machines × 10 shifts × 0.5 h = 15 h
Available capacity = 240 h − 15 h − 5 h maintenance = 220 h
Step 3: Occupied time
Occupied time = 168 h machining + 22 h changeover = 190 h
Step 4: Capacity utilization
Capacity utilization = 190 h ÷ 220 h = 86.4%
Machine by machine:
| Machine | Occupied | Available | Utilization |
|---|---|---|---|
| M1 | 65 h | 75 h | 86.7% |
| M2 | 58 h | 70 h | 82.9% |
| M3 | 67 h | 75 h | 89.3% |
| Work centre | 190 h | 220 h | 86.4% |
The work centre figure comes from the summed hours, not from averaging the three percentages. A plain average would give 86.3%. The two figures only differ when the machines have different available hours and different percentages. The further apart both are, the larger the gap can get.
Step 5: Effective capacity
Suppose planning allowed 22 hours for changeovers and 15 hours for breakdowns this week, and that is exactly what happened.
Effective capacity = 220 h − 22 h − 15 h = 183 h
Machining time ÷ effective capacity = 168 h ÷ 183 h = 91.8%
The 15 hours missing to reach 183 h are the no-order hours. The work centre could have milled 15 hours more this week without any change to changeovers or breakdowns.
The cross-check
The 220 available hours split into 168 h machining, 22 h changeover, 15 h breakdown and 15 h with no order, which is 220 h again. Thirty hours were free. Half of them were lost to breakdowns and half because there was no work. Those are two separate problems with separate causes, and the percentage on its own can't tell them apart.
OEE would take the 15 no-order hours out of its planned production time, leaving 205 h. Availability in the OEE sense would be 168 h ÷ 205 h = 82.0%. The full OEE needs part counts and cycle times, which this example doesn't have.
Five ways to count it wrong
1. Switching the denominator
Against theoretical capacity, the work centre ran at 190 h ÷ 504 h = 37.7%. Against available capacity it ran at 86.4%. Same week, same three machines.
Both numbers are right. The first tells you how much of the whole week the machines worked, and so how much room a third shift or the weekend would give you. The second tells you how full the two scheduled shifts were. Put the denominator in every report, and only compare machines or plants that use the same one.
2. Using order time instead of occupied time
If utilization comes from shop floor bookings and operators only book the start and end of each job, everything in between counts as used. That includes a breakdown halfway through. In the example, the 15 breakdown hours land in the numerator:
(190 h + 15 h) ÷ 220 h = 93.2%
Utilization goes up by almost seven points and the breakdowns vanish from the figure. If your bookings don't record breakdowns separately, say so next to the number.
3. Counting changeovers on some machines and not on others
Leave changeovers out and the same week comes to 168 h ÷ 220 h = 76.4% instead of 86.4%. Ten points from the counting rule alone.
Which rule you choose matters less than using it on every machine. At the bottleneck, report changeover hours separately as well. Every changeover hour there is an hour in which the whole plant makes nothing. The SMED article shows how to cut them.
4. Adding overtime to the numerator only
Suppose that in another week M3 also runs a Saturday shift with 6 available hours and spends all of them machining. Its occupied time rises from 67 h to 73 h. Leave the denominator at the 75 scheduled hours and you get:
73 h ÷ 75 h = 97.3%
The correct figure is 73 h ÷ (75 h + 6 h) = 90.1%. Keep adding extra shifts to the top line only and sooner or later you report more than 100%. A utilization above 100% nearly always means the numerator and the denominator were counted on different rules.
5. Confusing planned load with measured utilization
Capacity planning in the ERP adds up the standard times of every operation scheduled on a work centre. Say the operations scheduled on milling for the example week carried 200 standard hours. The ERP then shows 200 h ÷ 220 h = 90.9%. The machine data shows 190 hours, or 86.4%.
People call both of these utilization. One is a plan, the other a measurement, and the 10-hour gap has a cause worth finding. Either the standard times are generous, or scheduled jobs didn't run. In the example the second is the likelier one, since there were 15 hours with no order even though more work was scheduled than got done. Compare plan and actual on the same list of orders.
What to do with the result
Find the bottleneck
M3 has the highest utilization at 89.3% and the fewest no-order hours at 3. That makes it the first suspect for the bottleneck. It doesn't prove it. A machine can be highly utilized because it is slow or changes over a lot, with nobody downstream waiting on it.
A bottleneck shows itself in inventory. Jobs pile up in front of it and the next station waits behind it. Count the jobs waiting in front of each of the three machines. The article on measuring lead time shows how waiting jobs turn into lead time, and a value stream map shows where the bottleneck sits across the whole plant.
Decide in hours, not percentages
When sales asks whether a rush job with 20 hours of milling fits into the week, the percentage won't answer it. The hours will. In the example there were 15 no-order hours, spread across three machines and five days. Whether the 20 hours fit depends on which machine can run the job and when its material arrives.
Why 100% is not the target
A machine with no free hour at all can't absorb a job that arrives late or catch up after a breakdown. Every disturbance pushes the following jobs back.
Queueing theory puts numbers on this for one simple model, the M/M/1 queue. Jobs arrive at random at a single machine, and processing times vary at random too. The average number of jobs in the system is then ρ ÷ (1 − ρ), where ρ is utilization. At 80% that is 4 jobs, at 90% it is 9, and at 95% it is 19. Going from 90% to 95% more than doubles the average number of jobs at the machine. At 100% the model has no steady state at all.
Your milling shop isn't an M/M/1 queue, and its numbers won't carry over one to one. The model still shows why the last few points are so expensive. More jobs in the queue means a longer wait for each of them.
Frequently asked questions
How do you calculate machine utilization?
Divide the machine's occupied time, machining plus changeovers, by its available time from the shift plan, both for the same period. With 65 occupied hours out of 75 available, that is 86.7%. Take planned breaks and maintenance out of the available time first.
What is the difference between utilization and OEE?
Utilization tells you whether the machine had work. OEE tells you how well it ran while it had work. A machine with few orders can have low utilization and high OEE at the same time.
Can capacity utilization be over 100%?
Measured against available capacity, only when the numerator and denominator don't match. The usual cause is overtime or Saturday shifts counted in the hours used but not in the hours available. Planned load in capacity planning can exceed 100%. That means more work is scheduled than the machine can do.
What is a good capacity utilization rate?
There is no figure that holds for every plant. At the bottleneck utilization should be high, because every hour lost there is lost to the whole plant. The machines around it need free time so the bottleneck never waits for work. Compare your figure with last week's on the same counting rules, and watch whether the queues in front of the machines grow.
Does changeover time count towards utilization?
That is your call. This article counts changeovers as occupied time, because the machine is tied to an order while it changes over. What matters is that every machine is counted the same way and that the report shows changeover hours separately.
Try it on one machine
Pick a machine that everyone calls fully loaded. Pull one week of hours from the machine data or the shop floor bookings and split them as in the table above, into machining, changeover, breakdown and no order. Calculate utilization against the available time and compare it with the figure in today's report.
If you want to know whether your bottleneck sits at the machine or in the planning in front of it, that is the scope of our process analysis.
