You run one shift on a segment of a vehicle final-assembly line: five stations in series with small buffer lanes between them. Every stoppage ripples up and down the line, every defect is either fixed at the station or travels on, and the shift has a production plan to meet.
What you will learn
How OEE splits losses into availability, performance and quality — and why the bottleneck decides line output.
How finite buffers decouple stations (blocking and starving), and what that inventory costs in lead time (Little’s law).
Why stopping the line on a defect can beat running through it, and why running faster is not always more output.
Simulator
Time 0 min
▶Working
‖Starved (no unit)
⧗Blocked (no room downstream)
✕Broken down
⚑Andon stop
▪Unit in a buffer lane
Controls
Percent of rated speed. Faster cycles raise output — and the breakdown and defect rates.
Units each lane between stations can hold. More decouples stoppages, but adds work in process.
Each technician repairs one station; the bottleneck is served first, a second technician speeds up a repair.
Operators stop their station to fix a defect on the spot. It costs uptime — including stops for abnormalities that turn out to be nothing — but keeps defects from travelling on.
A team of three repairs units caught at end-of-line test for one hour, so they can ship.
Indicators
OEE
0.0%
critical
Output vs plan
100.0%
normal
Escaped defects
0
normal
Work in process
13
normal
Availability
100.0 %
Performance
0.0 %
Quality rate
100.0 %
Pace vs takt
100 %
Throughput
50.0 JPH
Units ahead (+) / behind (−) plan
0
Lead time through the line
16 min
Units waiting in the repair bay
0
Labour hours per unit
0.00 h
Stations down
0
Bottleneck idle (starved or blocked)
0.0 %
Trend
Crisis scenarios
Level 1 · Protect the bottleneck through an upstream stop
Plan: 50 units per hour for a four-hour half-shift. Station 3 is the line’s bottleneck and only one maintenance technician is on shift. Maintenance reports that a worn part on station 2 — right in front of the bottleneck — must be replaced at minute 35: a 40-minute job for one technician.
Output ≥ 101 % of plan at the end of the half-shift
Bottleneck idle ≤ 7 % of the time
Average work in process ≤ 18 units
Level 2 · Quality excursion at the trim station
Plan: 52 units per hour for a four-hour half-shift, and ten units from the last shift are still waiting in the repair bay. Twenty minutes in, a fastening problem appears at station 4: about one unit in eight leaves it with a defect. End-of-line test catches most of them — not all.
At most 1 defect escapes to the customer
Output ≥ 104 % of plan at the end (repaired backlog units count)
At most 3 units left in the repair bay
Average work in process ≤ 16 units
Level 3 · Demand surge
The line has been running at 80 % speed for a plan of 47 units per hour. Twenty minutes into the half-shift, sales raises the plan to 56 units per hour — takt time drops from about 1.28 to 1.07 minutes, and the bottleneck cannot keep up at 80 %. No overtime is allowed.
Output ≥ 102 % of plan after two hours
Output ≥ 103 % of plan at the end of the half-shift
At most 1 defect escapes to the customer
Average work in process ≤ 20 units
Basis — the model behind the numbers
Every relation the simulator uses, with its source. Constants marked as assumptions are illustrative calibrations.
OEE multiplies availability, performance and quality, measured at the bottleneck; it equals ideal cycle time × good units ÷ planned time.
OEE = A × P × Q; A = (T_plan − T_stop)/T_plan, P = CT_ideal · N_bn / (T_plan − T_stop), Q = N_first-pass / N_out; OEE = CT_ideal · N_good / T_plan when N_bn = N_out[1][2]The product equals ideal cycle time × good units ÷ planned time only while the bottleneck’s output equals the line’s output; units still between station 3 and the end of the line make a small difference.
A station passes its unit on only if the next buffer has room (else it is blocked) and starts only if the previous buffer has a unit (else it is starved).
station i: cycle = c_i / v; moves on iff buffer_i < K (else blocked); starts iff buffer_{i−1} > 0 (else starved)[3][4]
The line can never run faster than its slowest station; takt time is the available time divided by the demand.
r_line ≤ r_b = v / max_i c_i; takt = available time / demand = 60 / plan [min/unit][5][7]
Speed stress: losses driven by speed grow without bound as the line approaches the limit of its equipment and operators (105 % of rated speed).
σ_n(v) = ((1.05 − 0.85) / (1.05 − v))^n (σ = 1 at 85 %; σ_1: 2× at 95 %, 4× at 100 %; σ_2: 4× at 95 %, 16× at 100 %)Assumption: cycle times, failure and repair times, defect and false-stop rates, the speed-stress law and team sizes are illustrative values, not data from a real plant.
Stations break down only while working, more often the closer the line runs to its limit; repair time shrinks when a second technician joins.
P(fail in Δt) = 1 − exp(−λ_i · t_work), λ_i = σ_1(v) / MTBF_i; repair work ~ Erlang-3(mean MTTR_i), done at rate 1 + 0.6·(n−1), n ≤ 2[4][1]Assumption: cycle times, failure and repair times, defect and false-stop rates, the speed-stress law and team sizes are illustrative values, not data from a real plant.
Defects rise even more steeply with speed than breakdowns. With andon most defects are caught and fixed at the station, at the cost of short stops (some of them false alarms); otherwise end-of-line test catches some and the rest escape.
p_i = p0_i · σ_2(v) (+0.12 excursion at S4); andon: detect 90 %, stop 1.5 min, fix; false stops 0.3 % · σ_2(v) per unit and station; end-of-line test catches 70 %[7][1]Assumption: cycle times, failure and repair times, defect and false-stop rates, the speed-stress law and team sizes are illustrative values, not data from a real plant.
Little’s law: average time through the line equals work in process divided by throughput.
cycle c = 0.62/0.64/0.90/0.75/0.72 min at rated speed (S3 = bottleneck) · MTBF 120/100/200/110/150 min of work, MTTR 4/4/3/4/3 min at 85 % speed · S2 part replacement 40 technician-min · p0 = 0.15/0.15/0.2/0.2/0.05 % per unit · excursion contained after 3 andon pulls or 15 units caught at test · rework 4 min/unit, team of 3 for 1 h per call · 40 line operators · moving averages over 30 minAssumption: cycle times, failure and repair times, defect and false-stop rates, the speed-stress law and team sizes are illustrative values, not data from a real plant.
Randomness: a seeded mulberry32 generator; distributions used — uniform, exponential (inverse CDF), normal (Box–Muller), Poisson (Knuth). The seed is shown and shareable.
Sources
S. Nakajima — Introduction to TPM: Total Productive Maintenance (OEE = availability × performance × quality; six big losses) — Productivity Press, 1988
J. A. Buzacott, J. G. Shanthikumar — Stochastic Models of Manufacturing Systems (flow lines with finite buffers, blocking, starvation) — Prentice Hall, 1993