Precedence diagram, positional weight and line efficiency
In one sentence
A precedence diagram establishes the required order of tasks; positional weight selects a candidate from the eligible tasks, and line efficiency and balance loss evaluate the chosen allocation on an explicit cycle time basis.
By the end of this lesson
- Build an acyclic precedence diagram from a task list and identify the tasks eligible at the start.
- Calculate each task's positional weight from its own time and the times of its unique successors.
- Generate a two-station candidate with the highest-weight tasks that are eligible and fit, at a target of 50 s/unit.
- Calculate line efficiency and balance loss for 80 s/unit of work and explain the limit to the optimality of the heuristic rule.
Why it matters
A precedence diagram is not a list that puts tasks in order of duration. Each node represents a concrete task, and each arrow represents the requirement that this task be completed before another. If several arrows reach a task, all of their conditions must be met. Branching in the diagram does not by itself prove that tasks can be done at the same time; it shows only that there are options with respect to the required order. To decide at which station tasks will be performed, these relationships are used together with an assignment rule. In this lesson, the relationships are built first, then a candidate is generated using positional weight. Equipment and competency checks are left to the acceptance process learned in the previous lesson; generating a candidate does not mean final approval.
Positional weight is the sum of a task's own time and the times of all unique tasks that must follow it. This value takes into account not only the long task but also how much work lies behind it. Reaching a successor by two different paths does not make it two separate tasks. The same time is counted once in the weight calculation. At the start, only tasks without predecessors can be selected. After each assignment, the eligible tasks are determined again. The highest weight does not override the eligibility condition. A task whose predecessor is not yet finished cannot be selected merely because its weight is large. Likewise, a task that does not fit the remaining time cannot be forced into the station.
A rule of this kind provides a systematic starting point among a large number of possible allocations. Because the same criterion is used at each selection, the decision is traceable; it can be explained why one task was selected before another. On the other hand, a heuristic ordering is not proof that it finds the best solution to every problem. Fewest stations, shortest highest workload and easiest implementation are different goals. An allocation may be good on one of these goals and weaker on another. The result is therefore written together with the objective used and the acceptance conditions. If it matches a lower bound, only what that criterion proves is stated; no general claim of optimality is made for all goals.
Line efficiency and balance loss summarize how the resulting allocation uses the total time capacity. Which cycle time is put in the denominator must be clear. Evaluation with the target cycle time and evaluation with a shorter ideal cycle time can give different results. Waiting is not added to work content to inflate the percentage; loss is not eliminated by reclassification. Assigning the same tasks to different stations can preserve total human work but change the highest workload. In this lesson the progression is from diagram to candidate, and from candidate to numerical comparison. At every stage it is kept in view that the final decision requires verifying resources and safe implementation as well as precedence.
Formula
Positional weight (s) = Task time + Times of all its unique successors
Lower bound on station count (stations) = Total work (s/unit) ÷ Target cycle time (s/unit), rounded up to a whole number
Remaining time (s/unit) = Target cycle time − Assigned workload
Line efficiency (%) = Total work ÷ (Number of stations × Selected cycle time) × 100
Balance loss (%) = 100 − Line efficiency
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- Step-by-step instructions
- Worked example with real numbers
- Where the method misleads
- Common mistakes
Lab instructions: Three-station line
- Measure 8 cycles at every station.
- Enter the elements with the names below and in this order. Each element starts at its start point and ends at its end point.
- Enter the operator's elements and waits on the Operator track and the machine's running on the Machine track; a machine row runs from the moment the machine starts until it stops.
- Enter a wait or a machine run as a row of its own every time you see it. Assign every row to its station.
Station 1
1.Pick up the body, clamp it in the fixture
Starts: right hand reaches for the bodies · Ends: clamp lever goes down, hand lets go
2.Fit the seat
Starts: left hand reaches for the yellow seat box on the top shelf · Ends: white seat sits in the body
3.Place the ball
Starts: left hand reaches for the ball tray · Ends: ball sits in the seat
4.Fit the cap by hand
Starts: right hand reaches for the cap box on the top shelf · Ends: cap is seated by hand, right hand lets go
5.Tighten the cap with the screwdriver
Starts: right hand reaches for the screwdriver on the bench · Ends: screwdriver is put down on the bench, hand lets go
6.Open the clamp, put the valve on the belt
Starts: right hand reaches for the clamp lever · Ends: valve is put down on the belt
Station 2
1.Take the valve from the belt, clamp it in the fixture
Starts: right hand reaches for the valve on the belt · Ends: valve is laid in the fixture, hand lets go
2.Fit the stem
Starts: left hand reaches for the stems in the yellow box · Ends: stem goes into the body
3.Fit the handle
Starts: right hand reaches for the handle box · Ends: handle sits on the stem
4.Fit the nut
Starts: left hand reaches for the nut tray · Ends: nut is seated by hand, left hand lets go
5.Tighten the nut
Starts: right hand reaches for the screwdriver · Ends: screwdriver is put down on the bench, hand lets go
6.Put the valve on the belt
Starts: left hand grips the finished valve · Ends: valve is put down on the belt
Station 3
1.Take the valve from the belt
Starts: right hand reaches for the valve on the belt · Ends: operator turns to the test chamber with the valve
2.Put the tested valve in the crate
Starts: left hand grips the valve in the test chamber · Ends: valve is put down in the crate
3.Put the new valve in the test chamber
Starts: right hand lowers the valve into the chamber · Ends: valve sits in the chamber, hand lets go
4.Start the test
Starts: both hands reach for the buttons at the front of the bench · Ends: hood comes down, both hands leave the buttons
5.Test (machine)Machine track
6.Waiting: test runningWait
7.Waiting: no valve on the beltWait