Presets
Controls
Live readouts
In-page checks
What to try
- Raise arrivals above 55 kits/h: the kiln (M04) saturates and its queue fills.
- Give M04 a second server and watch the bottleneck move to M20 (83.7 kits/h).
- Shrink buffers to 0.25x: parts are rejected at full buffers and kits are lost.
The science
Each kit is a fork/AND-join job: after the crosscut saw (M07) it splits into parallel part streams (wheels, axles, chassis, cab, chimney, carton) that must all finish before the assembly stations start.
Capacity of a station with s parallel servers and mean cycle t: C = s·3600/t kits/h. The line capacity is mini Ci (the bottleneck), reduced by breakdowns: the mean service time becomes t + t/(MTBF)·MTTR.
Queueing check: for a single exponential station (M/M/c) with ρ = λ/(cμ) the Erlang-C formula gives Pw = T/(S+T) with S = ∑n<c an/n!, T = (ac/c!)/(1-ρ), then Lq = Pwρ/(1-ρ) and W = Lq/λ + 1/μ. Little's law: L = λW holds in any stable system.
Simplifications: cycle times are uniform ±16% around the mean (not fitted to a real plant), failures are exponential, a part arriving at a full buffer is rejected (the kit is then lost, where a real plant would block upstream), transport is a fixed delay per edge with no conveyor accumulation. This is a teaching model of a fictional plant, not a prediction for a real one.