Which OEE factor is costing you the most time?

When OEE (overall equipment effectiveness) is low, look first at the factor that loses the most minutes of planned production time, not the factor with the lowest percentage. Convert availability, performance and quality into lost minutes, check the inputs behind each one, then weigh the minutes by what they cost on that machine. The largest loss that survives those checks is where you start.

One week on a CNC mill runs through every step below. If you need the formulas first, How to calculate OEE works one shift by hand.

Why the lowest percentage misleads#

OEE multiplies three factors, and each one is a ratio of a different amount of time. Availability divides run time by planned production time. Performance divides the ideal time for the parts made by run time, and quality divides good parts by all parts.^[1]

Ideal cycle time is the best cycle the operation sustains in good conditions, timed at the machine rather than taken from a quote. Total count × ideal cycle time is the ideal time for the parts made: the least time they could have taken.

Because the base shrinks from one factor to the next, a percentage point is worth less time at each step. A point of availability is 1% of planned production time. A point of performance is 1% of run time, and a point of quality is 1% of the ideal time for the parts made.

For example, a mill with 2,250 planned minutes, 1,800 run minutes and 1,400 ideal minutes of output loses 22.5 minutes per point of availability, 18 per point of performance and 14 per point of quality. The glossary entry for performance gives the second factor’s full definition.

Convert each factor into lost minutes#

Each factor’s loss is the planned time it took away from good output. Three subtractions give the minutes.

The three add up to planned production time minus good count × ideal cycle time. That is the gap the short form of OEE measures: good count × ideal cycle time ÷ planned production time.^[1]

A team that chases the lowest percentage starts on the smaller loss here. Minutes put all three factors on one scale, so you can rank them.

Split the leading factors into their losses#

A factor tells you the kind of loss. The six big losses tell you the cause, and each one belongs to exactly one factor.^[2]

Minor stops last a minute or two and need no maintenance visit.^[2] Where they land depends on the cutoff your plant sets for logging a stop: shorter stops stay inside run time and lower performance. The post on micro-stops shows how that cutoff moves minutes between availability and performance while OEE stays the same.

Split the mill’s 920 minutes into buckets and rank them. You need a reason on each stop for availability, and a stop record that keeps short stops for performance.

Reduced speed leads at 250 minutes, ahead of changeovers at 210. Before you start a speed project, check the number that produced it. To rank stop reasons over a longer window, build a downtime Pareto on minutes, not counts.

Check the inputs before you trust the result#

Every factor rests on a number someone entered or a record someone kept. A wrong input moves minutes into the wrong factor, or invents them.

Performance depends on the ideal cycle time more than on anything the machine does. Suppose the mill’s 2.5-minute ideal came from the job quote. Timing 20 consecutive good cycles gives 2.8 minutes as the best the process sustains.

With 2.8 minutes, the 560 parts are worth 1,568 ideal minutes. Performance rises to 1,568 ÷ 1,800 = 87.1%, and its loss falls from 400 to 232 minutes. Of the 250 minutes booked as reduced speed, 168 were a gap between the quote and the machine, and 82 remain.

Changeovers now lead at 210 minutes, so measure changeover time before you plan that project. OEE becomes (532 × 2.8) ÷ 2,250 = 66.2%. Quality’s loss grows to 28 × 2.8 = 78.4 minutes, because each scrapped part carries more ideal time.

Keep 2.5 minutes only if a different program or tooling sustains it on this machine, since the ideal is the best cycle the operation sustains. Then the 168 minutes are a real performance loss with a known cause. Otherwise set the number from timed cycles, as the guide to setting the ideal cycle time describes.

Measured cycles also test the times you plan with. Jesel, a valvetrain manufacturer, compares measured cycle times against the times in its bills of material every day, as its customer story describes.

Quality counts only parts that pass the first time, so a reworked part is a reject.^[1] The guide to first pass yield explains why. A quality factor at 100% for weeks deserves a check against the scrap bin before anyone reads it as a result. Give each reject a reason when you log scrap, so quality minutes appear beside the stops.

Rank the losses by what they cost#

A lost minute isn’t worth the same everywhere. Two things set its price: what the loss consumes besides machine time, and whether the machine limits the plant’s output.

The bottleneck is the machine whose capacity limits what the whole line can ship. The Lean Enterprise Institute describes it as the weak link in the system, the one constraint that stops the next unit from being made with current resources.^[3]

A minute lost on the bottleneck is a minute of output lost for the line, while a minute lost on another machine is absorbed by its spare time. To prove which machine that is, see how to find the bottleneck.

Now suppose the mill has spare capacity and finishes the week’s work by Thursday. Its lost minutes cost little until demand rises, and the $1,120 of scrapped blanks is the only cash that leaves. The first project is the 18 startup rejects, and the stop and speed work moves to whichever machine is the bottleneck.

Read the pattern across shifts and machines#

One week on one machine gives you a ranking. The same ranking across shifts, part numbers and machines points at the cause before you walk to the floor.

Look at the factor before you look at the people. Low performance on one part number points at that operation’s cycle time on every shift that runs it, not at the operator who happened to run it last.

See each factor on its own#

IoTFlows reports OEE factor by factor, so the weakest factor is on screen without a calculation. A SenseAi on each machine decides running or stopped every second and counts parts from the machine’s own signal, with no PLC connection.

Each machine’s card shows Uptime, which is availability, and the parts gauge, which is performance against the ideal cycle time set on each operation. Quality reads on Shift Production as good parts over everything counted, from the scrap your team logs with a reason. The metrics reference defines each number.

For a month or a quarter, sort the Advanced Report by Worst Performing to put the machine with the lowest uptime at the top.

Open that machine, and its Downtime Classifications card ranks its stop reasons by hours lost.

Find your biggest loss this week#

1. Pick one machine and one week

   Choose the machine that limits output, or the one with the most scheduled hours.

2. Convert each factor to minutes

   Subtract run time from planned time and ideal output from run time, then multiply rejects by the ideal cycle time.

3. Check the inputs

   Compare the ideal cycle time with timed cycles, and the scrap log with the scrap bin.

4. Split the losses into buckets

   Rank stop reasons, short stops, slow cycles and rejects by minutes.

5. Price the top three losses

   Multiply minutes by what a minute of the machine earns, and add the material in each reject.

6. Fix one loss and remeasure

   Work on the largest bucket, then compare next week’s minutes with this week’s.