Line Balancing: How Many Workstations a Line Needs, Worked Through on Ten Tasks

What line balancing means

Line balancing is the job of splitting the tasks on a product across the stations of a line so that no station takes longer than takt time and as little time as possible goes on waiting. The order the product dictates still holds. You can't fit the shaft before the bearing bush is in.

The minimum number of workstations follows from two figures you already have:

Minimum number of workstations = total work content ÷ takt time, rounded up to the next whole number

If all the tasks together take 260 s and the customer needs one part every 60 s, the answer is 4.33, so the line needs at least 5 stations. Whether 5 is actually enough only shows once you assign the tasks, because tasks can't be split at will and many of them have to run in a fixed order.

What goes into the calculation

Tasks and their times. A task is the smallest piece of work that makes sense to keep at one station, such as "press in the bearing bush". The times come from a stopwatch or a time study, in seconds. The finer you cut the tasks, the more evenly you can spread them.

Precedence. For every task, you note which tasks have to be finished first. Drawn out, this is the precedence diagram. For ten tasks, a table with a "predecessors" column does the job.

Takt time. Available time divided by customer demand. How to calculate takt time covers what counts as available time and why it is a customer figure, not a shop-floor one.

One limit before you start. A task that takes longer than takt time won't fit any station, however you balance. You have to split it, shorten it, or run it on two parallel stations that take alternate parts.

The formulas

Let t be the time of one task, T the takt time and n the number of stations the line ends up with after balancing.

Minimum number of workstations = Σ t ÷ T, rounded up

Balance efficiency = Σ t ÷ (n × T)

Balance delay = 1 − balance efficiency

Idle time per cycle = n × T − Σ t

Balance efficiency is the share of paid station time that goes into the product. Balance delay is the rest, the share spent waiting. Idle time is the same loss in seconds, added up across all stations for one part.

Some textbooks and plants use the time of the slowest station in place of takt time. That gives a different number for the same line, which is mistake 4 below.

Worked example: a pump assembly line with ten tasks

The plant, the product and every time in this example are invented. They are chosen so you can check each figure by hand.

A line assembles a small circulator pump in a single variant. As in the takt time article's assembly cell, each shift has 440 minutes available, which is 26,400 s. The customer calls off 2,200 pumps a week over five shifts, so 440 pumps a shift.

Takt time = 26,400 s ÷ 440 pumps = 60 s

The routing lists ten tasks:

Most of the work sits on one long chain. After A, the tasks D, E, F, G, H, I and J have to follow one another. B and C only need A, so B can go anywhere before G and C anywhere before J.

Step 1: Minimum number of workstations

260 s ÷ 60 s = 4.33, rounded up to 5 stations

Four stations would give 4 × 60 s = 240 s per pump, 20 s short of the work. That is why you round up.

Step 2: Positional weights

To assign the tasks, this example uses the ranked positional weight method, which Helgeson and Birnie published in 1961. Each task gets a positional weight. That is its own time plus the times of every task that can only run after it, directly or further down the chain. A heavy task has a lot of work queued behind it, so it should go to a station early.

The table is already sorted by weight. That order is your list for the next step.

Step 3: Assign tasks to stations

Open a station and go down the list. Take the first task whose predecessors are already assigned and which still fits in the time left before 60 s. Repeat until nothing fits, then open the next station.

Station 1. A tops the list and has no predecessors. It takes 15 s and leaves 45 s. D comes next, A is assigned, and 35 s fits. That puts station 1 at 50 s with 10 s left. E would need 40 s, B and C 15 s each, and the rest are still waiting on predecessors. Station 1 is closed.

Station 2. E takes 40 s and leaves 20 s. F at 35 s no longer fits. B at 15 s does, which brings station 2 to 55 s. Nothing takes 5 s or less.

Station 3. F takes 35 s and leaves 25 s. G is now free, since B and F are both assigned, but it needs 30 s. C at 15 s fits and brings station 3 to 50 s.

Station 4. G at 30 s and H at 25 s make 55 s. I at 30 s doesn't fit.

Station 5. I at 30 s and J at 20 s make 50 s. All ten tasks are assigned.

Check the result against the predecessors column. Every task sits at the same station as its predecessors or a later one. The 5 stations from step 1 are enough.

Step 4: Balance efficiency and idle time

Balance efficiency = 260 s ÷ (5 × 60 s) = 260 s ÷ 300 s = 86.7%

Balance delay = 100% − 86.7% = 13.3%

Idle time per cycle = 300 s − 260 s = 40 s

Over a shift of 440 pumps that comes to 440 × 40 s = 17,600 s, about 293 minutes or 4.9 hours of waiting spread across five people.

That looks like a lot. Here it is the cost of rounding up. Four stations can't do the work, and five leave 40 s of slack on every pump. Even a perfect split would give each station 260 s ÷ 5 = 52 s of work and 8 s of idle time. Less idle time would need finer tasks or a different takt.

What the line can make

The two slowest stations need 55 s, and they set the output.

26,400 s ÷ 55 s = 480 pumps a shift

3,600 s ÷ 55 s = 65.5 pumps an hour, rounded

The customer needs 440 a shift, which at 60 s takt is 60 an hour. Stations 2 and 4 run 5 s under takt and the other three 10 s under. The takt time article shows how to work out units per hour and per minute, including on a conveyor.

Four line balancing mistakes

1. Rounding down

Rounding 4.33 down to 4 means fitting 260 s of work into 240 s of station time. Even split perfectly, each station would need 65 s. At 65 s the line makes 406 whole pumps in 26,400 s, 34 short of the order. Always round the minimum up, even at 4.05.

2. Filling stations in routing order

The obvious way is to work down the routing from A to J and open a new station as soon as the next task won't fit.

That gives six stations instead of five and a balance efficiency of 260 s ÷ 360 s = 72.2%. One more operator for the same 440 pumps. The positional weight method skips a task that doesn't fit and keeps looking further down the list. That is how C ended up filling the gap at station 3.

3. Loading stations right up to takt

The rule in step 3 allows a station up to 60 s. In the example none gets there, the highest sit at 55 s. A station at exactly 60 s has no room at all, and every stoppage or scrapped part shows up as a pump missing at the end of the shift.

Balance against a planned cycle time below takt instead. Mistake 5 in the takt time article shows where that figure comes from, 112 s against a 120 s takt in that case. An OEE calculation tells you how much time stoppages and scrap cost on your own line. Suppose the pump line plans with a 55 s cycle time. Then 260 s ÷ 55 s = 4.73, which still rounds up to 5 stations, and the balance above already stays within 55 s.

4. Comparing efficiencies with different bases

Measured against the 60 s takt, the line runs at 86.7%. Measured against its slowest station, 55 s, it is 260 s ÷ (5 × 55 s) = 260 s ÷ 275 s = 94.5%. Same line, same people.

Both versions are in common use. The first tells you how well the line fits the customer. The second tells you how evenly the work is spread. Say which one you mean, and only compare figures calculated on the same basis.

When to re-balance

A balance holds for one takt time and one set of work content. When either changes, run the numbers again.

Demand goes up. If the customer calls off 2,640 pumps a week instead of 2,200, that is 528 a shift and a takt of 26,400 s ÷ 528 = 50 s. Five stations would have 250 s between them for 260 s of work. The line needs 260 s ÷ 50 s = 5.2, rounded up to 6 stations. Stations 2 and 4, at 55 s, would already be 5 s over the new takt.

Demand goes down. At 1,650 pumps a week it is 330 a shift and a takt of 26,400 s ÷ 330 = 80 s. Then 260 s ÷ 80 s = 3.25 rounds up to 4 stations. Stay at five and balance efficiency drops to 260 s ÷ (5 × 80 s) = 65%.

The work changes. A new fixture cuts the press-in time, a variant adds a task, or a task moves out to a sub-assembly area. Each of these shifts the total and often the positional weights too.

Available time changes. If the line changes over between variants, the changeover comes out of available time and takt gets shorter. SMED: cutting changeover time shows how to win some of it back.

A mixed-model line, with variants of different work content, is usually balanced on the volume-weighted average. Sequence then matters as well. Two long variants back to back push a station over takt even when the average fits.

Balancing sets how often the line releases a part. How long a single part spends on the line depends on the inventory between stations, which how to calculate lead time works through. To find which line in a plant runs above takt in the first place, start with a value stream map.

FAQ

What is the difference between takt time and line balancing?

Takt time is a number. It says how often the customer needs one part. Line balancing is the work that follows from it. It assigns the tasks to stations so that no station takes longer than that interval.

How do you calculate the number of workstations?

Divide the total work content by takt time and round up. With 260 s of work and a 60 s takt that is 4.33, so 5 stations. This is a lower bound. Whether it holds only shows once you assign the tasks in an order that respects precedence.

What is a good balance efficiency?

There is no general target. A line only reaches 100% if total work content is an exact multiple of takt and the tasks happen to split evenly. Compare your figure with what the minimum number of stations allows and with your last balance. In the example the ceiling is 86.7%, because five stations is the minimum.

Does the ranked positional weight method always find the best answer?

No. It is a rule of thumb that gets you a workable balance quickly. With many tasks, a different assignment may need fewer stations, and exact optimisation methods exist for that, run on a computer. When the rule of thumb already hits the minimum number of stations, as in the example, no answer with fewer stations exists.

What if one task takes longer than takt time?

Then it fits no station. You can split it, shorten it with a fixture or a different method, or double it up. With two parallel stations, each takes every second part and has two takt cycles to finish it.

The next step

Pick a line where one station keeps falling behind. Write down the tasks with stopwatch times and the predecessors of each. Work out the minimum number of stations and compare it with how many you staff today. For ten to twenty tasks, paper and half an hour are enough.

If you want to know whether your line is held back by how the work is split or by stoppages and inventory, that is the scope of our process analysis. Fixed scope, fixed duration, fixed price.