Line balancing is the work of making the slowest station faster or the fastest station smaller. Linear programming is one way to write that down so a solver can answer two questions precisely: with this sequence, what is the lowest cycle time I can reach, and how many stations does a target output actually require.
The model itself is small. What usually fails is the input: standard minutes that were never measured, a sequence nobody wrote down, and changeover time that gets dropped because it is inconvenient. This page builds the model from a five-operation example, shows where the numbers come from, and marks the point where engineering judgement still has to decide.
What decision the model supports
A line engineer usually faces one of two questions:
- Fixed headcount, best output. Four operators are on the line today. What is the lowest cycle time this sequence allows, and what daily output does that give?
- Fixed output, fewest operators. The plan needs 140 pieces a day. What is the smallest number of stations that can carry that?
Both questions become the same linear program with a different objective. Answer the question you actually have before writing the model; otherwise the solver optimises a number nobody is measured on.
Measure three things first
The model is only as good as these three inputs. They come from a time study, not from a machine catalogue.
- Standard minutes per operation (SAM). The time a trained operator needs at a defined pace, including allowances for handling and personal needs. For a sewing operation this is typically the garment-part handling time plus the machine run time at the set speed.
- Precedence. Which operations must finish before another can start. On a shirt front placket this is close to a straight chain; on a multi-panel garment it branches.
- Available minutes per shift. Working minutes minus planned breaks, minus changeover, minus expected downtime. A 480-minute shift with two 10-minute breaks and 40 minutes of changeover leaves 420 minutes, not 480.
Take the measurement on the machine the line will actually use. A SAM taken on a different machine class, at a different stitch density, is a different operation.
The five-operation example
Five operations in a chain, measured on the line:
| Operation | SAM (min) | Must follow |
|---|---|---|
| A | 1.8 | — |
| B | 2.4 | A |
| C | 1.2 | B |
| D | 2.6 | C |
| E | 1.4 | D |
| Total work content | 9.4 |
Total work content is 9.4 minutes per piece. That number does not change with balancing; it is the floor every arrangement has to carry.
Write the model
Let x[i][j] be 1 when operation i is assigned to station j, and let C be the cycle time.
“ minimise C subject to sum over j of x[i][j] = 1 for every operation i load[j] = sum over i of SAM[i] * x[i][j] <= C for every station j station(i) <= station(k) for every precedence i -> k x[i][j] in {0, 1} “
Three practical notes on reading this:
- The cycle time constraint does the balancing. Every station’s accumulated SAM must sit at or under C. Lowering C forces a finer split, which raises the station count.
- Precedence is what makes it hard. Without it, greedy packing by longest operation almost always works. With it, a large operation in the middle of the chain leaves idle time on both sides.
- Changeover enters as capacity, not as a variable. Subtract setup minutes from available time before you solve. Modelling changeover as a per-batch decision is a different (integer) model and usually more than the question needs.
Solve it by hand for a target output
Plan requirement: 140 pieces per shift, 420 available minutes.
“ required cycle time = 420 / 140 = 3.0 min per piece “
Every station must finish its share in 3.0 minutes or less. Now test the chain against that limit. Because operations sit in a strict sequence, a station can only take a consecutive run of them:
| Stations | Best split | Highest station load | Output per shift (420 min) | ||||
|---|---|---|---|---|---|---|---|
| 3 | A+B \ | C+D \ | E | 4.2 (A+B) | 100 | ||
| 4 | A \ | B+C \ | D \ | E | 3.6 (B+C) | 116 | |
| 5 | A \ | B \ | C \ | D \ | E | 2.6 (D) | 161 |
At 140 pieces the required cycle time is 3.0 minutes. No four-station split reaches it — the best four-station arrangement is limited by B+C at 3.6 minutes — so the answer is five stations, and once five stations are in place the line can actually run at 2.6 minutes, which is 161 pieces per shift.
That gap is the honest result and the reason to solve rather than guess: the requirement says 140, the sequence only offers 161 or 116, and nothing in between. A planner who assumes “4 stations should cover 140” will miss by 24 pieces a shift.
Balance efficiency is a separate number
Cycle time answers the output question. Efficiency answers the utilisation question:
“ efficiency = total work content / (stations x cycle time) “
For the three arrangements above: 74.6% at three stations, 65.3% at four, 72.3% at five. Notice that adding stations lowers efficiency in the middle of that range — the fourth station absorbs idle time rather than useful work. Fewer stations at a slower cycle is often the better labour decision, which is why the objective has to match the real question.
Changeover, and when the model stops being linear
A style change resets the line: retime the operations, re-sequence the bundle flow, re-check the machine settings. If changeover costs 60 minutes and the plan runs four styles a shift, 240 of the 420 available minutes are gone and the required cycle time doubles.
Two honest options:
- Include it as capacity. Subtract changeover from available minutes, as above. Simple, and correct when changeover is planned.
- Model batches. If batch size is a decision, the model needs a setup cost per style per station, which makes it a mixed-integer program. Run it only when the batch decision is genuinely open; otherwise the extra complexity buys nothing.
The point where the linear model stops being trustworthy is when operation times depend on what was sewn before it — needle heat, thread tension drift, or a style that changes the machine setup mid-batch. Those effects belong in the time study, not in the constraint matrix.
If you have no solver
Most shops do not run an LP solver for a five-operation line, and for a short chain you do not need one. The usable hand method:
- Compute required cycle time from the plan.
- Take operations in precedence order and add them to the current station while the load stays at or under the cycle time.
- Start a new station when the next operation does not fit.
- Count the stations and check the highest load against the cycle time.
This is the largest-candidate rule and it lands on the optimum for a straight chain like the example. It stops being reliable when precedence branches, because a greedy choice on one branch can leave a large operation stranded on another. When the line has branches and the answer matters, use the solver.
Common modelling errors
- Using average cycle time as the bottleneck. The bottleneck is the highest station load, not the mean. A line balanced at 3.4 average with one station at 4.2 runs at 4.2.
- Leaving rework out of the SAM. If one piece in twenty is re-sewn, the effective SAM is higher and the cycle time is optimistic.
- Quoting machine speed instead of operation time. Stitch speed sets the run time portion only; handling usually dominates on short seams.
- Balancing to the plan instead of the takt. The plan changes weekly; the sequence does not. Solve for a cycle time range, then pick the station count that survives the next plan.
- Forgetting the walking and bundling time. Moving bundles between stations is work, and on a wide line it is measurable.
What to verify before the change goes live
- Retime the two heaviest operations on the actual machine, at the actual stitch density.
- Run the new arrangement on one bundle lot and record the highest station load, not the average.
- Compare the predicted output with the observed output for a full shift; a gap above 10% usually means the SAMs or the available minutes are wrong, not the balance.
- Check the machine settings for each station in the new split. A station that gains a heavier seam may need a different needle, feed setting, or thread path.
The engineering judgement this model cannot make: whether the operators on the line can actually hold the pace, and whether the machine can hold the stitch quality at that pace. Measure both before committing to a cycle time.
Related guides
- Shirt production line optimization: speed settings and stitch patterns — the machine-side view of the same line.
- Shirt production line troubleshooting: common jams and thread breaks — what breaks when the pace is pushed.
- Shirt production line sewing machine configurations — which machine sits at each station.
- Underwear production line optimization — the same balancing logic on a shorter cycle.
- Underwear production sewing equipment configuration — equipment by operation.
- Automatic thread trimmer guide — trimming time inside the operation SAM.
- Automatic tension system guide — why tension drift changes the effective operation time.
- Programmable stitch patterns — pattern changes as a setup cost.
- Sewing machine automation guide — which operations are worth automating first.
- Sewing machine repair cost guide — downtime cost as an input to the available-minutes figure.
Limits of this guide
The numbers here are a worked example, not a benchmark from any factory. Standard minutes vary by machine class, garment construction, operator skill, and pace definition, and the same operation on two lines can differ by 20%. Measure your own line, then run the model with your figures. Nothing here replaces a time study or the judgement of the line supervisor who has to hold the pace.