SIM12 · 约束理论 / 平衡产线的悖论SIM12 · Theory of Constraints / The balanced-line paradox
每站平均 3.4 的产线，输给了每站只有 3 的产线A line averaging 3.4 per station loses to a line fixed at 3
五个工序看起来产能配平，但每轮产能都随骰子波动，而且下游只能拿到轮初已有的在制品。上游这轮少做的，下游以后补不回来。固定 20×5 骰子表不重掷：三种模式面对的是同一组事实。Five stations look balanced, yet capacity changes with every roll and downstream can only use WIP available at the start of that round. Capacity lost upstream cannot be recovered later. The fixed 20-by-5 dice table is never rerolled, so every mode faces the same facts.
🎬 看骰子、缓冲堆和末端产出：波动会互相抵消吗？🎬 Watch the dice, buffers, and finished pile: does variation cancel out?
0/20
0
WIP0
机床=灰底Machines = neutral站间在制品Interstation WIP成品产出Finished output
💡 跑完看数据：平均产能更高，产出反而更低💡 Run it, then read the numbers: higher average capacity can still deliver less
固定骰子表 · 禁止重掷 · 数字由 stats() 计算Fixed dice table · no rerolls · values computed by stats()| 模式Mode | 20 轮产出Output / 20 rounds | 残余在制品Residual WIP |
|---|---|---|
| A · 每站固定 3A · Every station fixed at 3 | — | — |
| B · 固定骰子表B · Fixed dice table | — | — |
| C · 骰子表+每段 4 件C · Dice + 4 in each buffer | — | — |
每轮同步结算：每站转移量=min（本轮点数，轮初上游缓冲）。所有工序算完后再统一更新，绝不允许一件在同一轮穿过多个工序。Synchronous round rule: each transfer is min(this roll, upstream buffer at round start). Buffers update only after every station is calculated, so no part can cross multiple stations in one round.
🎯 波动规则🎯 Variation rule
波动加上工序依存，损失不会互相抵消，只会一路向下累积。Variation plus dependency does not average out; lost capacity compounds downstream.
⚠ 常见误区⚠ Common mistakes
✗ 把每站产能配平就是好产线✗ A balanced capacity plan guarantees a good line
✓ 平衡产线在波动下必然丢产出，保护瓶颈比追求配平有用。✓ A balanced line loses output under variation. Protecting the constraint matters more than nominal balance.
✗ 波动会自己平均掉✗ Variation averages itself out
✓ 上游丢掉的产能，下游永远补不回来。✓ Capacity lost upstream cannot be recovered downstream.