Operations · Simulator 074
Parts move through a line of workstations, one after another. Change how long each station takes, how much those times vary, and how much space there is between stations. Find the bottleneck, and see why a line with random delays loses output unless it has buffers.
A serial line can never produce faster than its slowest station, the bottleneck. If station times are t₁, t₂, … then
throughput ≤ 1 / max(tᵢ)
With fixed (deterministic) times, the line reaches that limit exactly, whatever the buffers. Stations faster than the bottleneck spend part of their time blocked (finished, but no room downstream) or starved (waiting for a part from upstream).
When times vary, a slow part at one station can block the station before it and starve the one after it, so output falls below the limit. Buffers between stations absorb those swings. For two stations with exponential times of equal mean t and a buffer of B parts between them, the exact result is
throughput = (1/t) × (B + 2) / (B + 3)
so no buffer gives 2/3 of the limit, and each extra space recovers part of the loss. The simulator uses blocking after service: a finished part stays on its station until there is space downstream. Variability uses Erlang times (coefficient of variation 0, 0.5 or 1).
Station 1 never runs out of raw material, and finished parts always leave. No breakdowns, set-ups or rework. Measured throughput is an average since the last reset, so it needs a long run to settle when times vary.
A three-station line has fixed times of 2.0, 3.0 and 2.5 min. A manager halves the time of station 1. What happens to output? What change would raise output to 24 parts per hour, and which station limits the line after that?
Nothing changes: station 2 (3.0 min) is the bottleneck, so the line makes 60/3 = 20 parts/h either way; station 1 just spends more time blocked. To reach 24/h every station needs a time of 2.5 min or less, so station 2 must drop to 2.5 min. Then stations 2 and 3 tie as the bottleneck. Try it with times 1.0, 2.5, 2.5.