Takt Time: What the Customer Needs and What the Cell Can Make
T. Krause
What takt time tells you
Takt time is the interval at which the customer takes one part, on average. You calculate it. You don't time it with a stopwatch. The formula holds what the customer needs and says nothing about what your own shop can do.
Takt time = available production time ÷ customer demand in the same period
The Lean Enterprise Institute's lexicon defines it the same way, as it stood in October 2026: available production time divided by customer demand. Takt is the German word for a beat in music, and the name fits. It is the beat the customer sets.
What counts as available time
Available time is the time the line or cell is scheduled to produce. Take out every scheduled block in which nobody makes a part. That covers breaks during which the cell stops, the shift meeting, cleaning at the end of the shift and planned maintenance. If operators relieve each other and the cell runs through the break, the break stays in.
Plants disagree about breakdowns, changeovers and scrap. Some subtract an allowance for breakdowns from available time and raise demand by the scrap. Others leave takt time as a pure customer figure and carry the losses in a second number, the planned cycle time. You can defend either, as long as the rule is written next to the figure.
This article follows the second rule. Takt time here contains no breakdowns and no scrap. What the two of them cost is in mistake 5.
The example: an early shift in an assembly cell
The plant, the customer and every figure are made up for this example. They are chosen so that you can check each line with a pocket calculator.
An assembly cell builds one hydraulic valve in a single variant, so there is no changeover during the shift. Three workstations stand in a row, each with one operator: pre-assembly, final assembly and leak test. The customer calls off 1,100 valves a week. The cell works five days a week on one shift.
| Input | Value |
|---|---|
| Shift length | 480 min |
| Breaks, cell stopped | 30 min |
| Shift meeting | 5 min |
| Cleaning at the end of the shift | 5 min |
| Customer call-off per week | 1,100 valves |
| Shifts per week | 5 |
Where a line doesn't divide evenly, the result is rounded.
Step 1: Available time
Available time = shift length − breaks − shift meeting − cleaning
480 min − 30 min − 5 min − 5 min = 440 min
440 min × 60 = 26,400 s
Step 2: Demand in the same period
The 440 minutes cover one shift, so demand has to be per shift as well.
1,100 valves ÷ 5 shifts = 220 valves per shift
Step 3: Takt time
Takt time = available time ÷ demand
26,400 s ÷ 220 valves = 120 s per valve = 2 min per valve
One tested valve has to leave the cell every two minutes. That is 30 an hour.
The cross-check
220 valves × 120 s = 26,400 s = 440 min
The sum has to land back on the available time. The week gives the same answer, because 5 shifts × 26,400 s is 132,000 s, and 132,000 s ÷ 1,100 valves is 120 s again.
Takt time vs cycle time
Takt time comes from the customer. Cycle time comes from the shop floor. It is how often a workstation hands on a finished part, and you measure it with a stopwatch.
In the example cell, the cycle time at all three stations is manual work.
| Workstation | Cycle time | Against takt | Possible per shift |
|---|---|---|---|
| Pre-assembly | 88 s | 32 s under | 300 |
| Final assembly | 100 s | 20 s under | 264 |
| Leak test | 132 s | 12 s over | 200 |
Possible per shift is 26,400 s divided by the cycle time.
Cycle time above takt
The leak test needs 132 s and sits 12 s above takt. It gets through 200 valves a shift and the customer needs 220. That leaves the cell 20 short per shift and 100 short per week. Every valve passes through the test, so the cell as a whole releases one valve every 132 s.
The only way to deliver is overtime.
220 valves × 132 s = 29,040 s = 484 min
That is 44 minutes more than the 440 available, on every shift. Over the week it comes to 220 minutes.
Cycle time below takt
Pre-assembly needs 88 s. If it works straight through, it builds 300 subassemblies a shift, 80 more than the customer takes, and they stack up in front of final assembly. If it works to takt, it waits 32 s on every valve. Over the shift that is 220 × 32 s = 7,040 s, which rounds to 117.3 minutes.
The time is lost either way. You can see the stack. You can't see the waiting.
Are three operators enough?
The three cycle times together are the work content of one valve.
88 s + 100 s + 132 s = 320 s
320 s ÷ 120 s = 2.67
On paper the cell needs 2.67 operators. Three are enough, then, and the work is just unevenly spread. An even split would be 320 s ÷ 3 = 106.7 s per station.
Suppose one 22-second task can move from the leak test to pre-assembly, fitting the nameplate for instance.
| Workstation | Before | After |
|---|---|---|
| Pre-assembly | 88 s | 110 s |
| Final assembly | 100 s | 100 s |
| Leak test | 132 s | 110 s |
| Total | 320 s | 320 s |
The slowest station now needs 110 s and sits 10 s below takt. The cell could make 26,400 s ÷ 110 s = 240 valves a shift, 20 more than it has to. Nobody works any faster than before.
Takt time vs lead time
Takt time is an interval between two parts. Lead time is how long a single part spends in the cell, from the first touch to a passed test. One valve every two minutes does not mean a valve takes two minutes.
The work content hardly matters here. 320 s rounds to 5.3 minutes. Lead time depends on how many valves are in the cell. Suppose a count finds 45 valves in the cell, at the stations and in the buffers between them. A valve that joins the back is finished roughly when 45 valves have left.
Lead time ≈ inventory in the cell × interval at which parts leave the cell
Today a valve leaves the cell every 132 s.
45 valves × 132 s = 5,940 s = 99 min
Somebody works on the valve for 320 s of that. 320 s ÷ 5,940 s rounds to 5.4%.
After the task has moved, the cell builds to takt. With the same inventory that gives 45 valves × 120 s = 5,400 s = 90 min. Cap the inventory at 15 valves as well and you are left with 15 valves × 120 s = 1,800 s = 30 min. Takt time is 120 s in both cases. It tells you nothing about lead time until you know the inventory.
This is an approximation, and it holds while the inventory stays about level. The value stream map of a bearing flange uses the same one in days, dividing each stock by daily demand. On a whole order, from receipt to dispatch, setup, queue and move time come on top. How to calculate lead time works that through on six stations.
The three figures side by side:
| Figure | Question | Where it comes from | In the example |
|---|---|---|---|
| Takt time | How often does the customer need a part? | calculated from time and demand | 120 s |
| Cycle time | How often does a workstation hand on a part? | timed | 88 s, 100 s, 132 s |
| Lead time | How long is one part in the cell? | measured, or estimated from inventory | 99 min |
Five ways to get takt time wrong
Each mistake comes with the wrong and the right number from the example.
1. Using the shift length
Put in the full 480 minutes and you get 28,800 s ÷ 220 valves = 130.9 s. The right figure is 120 s.
At 130.9 s the leak test looks almost fine, only 1.1 s over takt. The real gap is 12 s. The break still happens. A cell laid out for 130.9 s gets 201 whole valves out of the 26,400 seconds it has. That is 19 short.
2. Taking demand and time from different periods
The 120 s belong to a week with five shifts. Put a public holiday in the week, leave the call-off at 1,100 valves, and only 4 shifts × 440 min = 1,760 min are available. That is 105,600 s.
105,600 s ÷ 1,100 valves = 96 s
Keep planning with 120 s that week and you expect 220 valves a shift. The customer needs 275. Even the rebalanced cell at 110 s makes only 240. It is 35 short per shift and 140 short over the week. At 96 s the cell would need 320 s ÷ 96 s = 3.33 operators, so three no longer cover it.
Takt time is not a fixed property of the cell. Recalculate it whenever the call-off or the working time changes, and always take time and demand from the same period.
3. Calling cycle time takt time
"The leak test has a takt time of 132 seconds." That sentence describes a cycle time. You get the same number by feeding the formula with what the cell last delivered.
26,400 s ÷ 200 valves = 132 s
The right figure is 120 s, calculated with the 220 valves the customer needs.
Use your own output and the formula will always say the cell is on takt. The 20 valves missing every shift never appear in that sum. Say takt time when you mean the 120 s and cycle time when you mean the 132 s.
4. Averaging the cycle times
The mean of the three cycle times is 320 s ÷ 3 = 106.7 s. That is under 120 s, so the cell appears to hold takt. It doesn't. The slowest station decides the count, and it needs 132 s. That makes 200 valves instead of 220.
Compare each workstation with takt on its own. The mean answers a different question. If it is below takt, you have enough people and rebalancing is worth the effort. That is how the 132 s at the leak test became 110 s above.
5. Planning right up to takt
The 120 s assume a shift with no breakdowns and no scrap. Suppose the shift log shows 20 minutes of breakdowns per shift on average, and 5 valves fail the leak test and are scrapped. Then 440 min − 20 min = 420 min remain, which is 25,200 s. To ship 220 good valves the cell has to build 225.
Planned cycle time = 25,200 s ÷ 225 valves = 112 s
A leak test running exactly at the 120 s takt gets through 210 valves in 25,200 s. Of those, 205 are good, which is 15 too few. The cell needs 8 s of margin below takt. At 110 s the rebalanced cell has it. It tests 229 whole valves, 224 of them good, 4 more than ordered.
This is where plants calculate differently. Plants on the first rule take the 20 minutes off available time, plan for 225 valves rather than 220, and call what comes out takt time. Here that is 25,200 s ÷ 225 valves = 112 s, the planned cycle time under a different label. This article keeps the figures apart. The customer sets the 120 s, and the cell has to be capable of the 112 s. The 8 s margin stays visible that way, and it shrinks as breakdowns and scrap come down.
An OEE calculation shows what the 20 minutes are made of. How to calculate OEE splits one shift into availability, performance and quality.
Where takt time is little help
The calculation assumes that demand arrives fairly evenly over the period and that the parts carry a similar amount of work. If a cell builds variants with very different work content, takt time is an average that no single part matches. You then need the work content per variant and a sequence that mixes long and short parts.
In one-off production the figure does little. A shop that builds different parts to customer drawings every week is better off planning with lead time.
Takt time is also not a target to beat. Building faster than the customer takes parts means inventory. You can improve a cycle time. A takt time you can only meet.
The next step
Pick one line or cell and the current week. Write down the shift length, every scheduled stop and the customer's call-off, then calculate takt time. Time a few cycles at each workstation and put the cycle time next to takt. Paper and a stopwatch are enough.
If you want to know whether your delivery dates hang on a workstation above takt or on the inventory between stations, that is the scope of our process analysis. Fixed scope, fixed duration, fixed price.