This question's explanation and your answers so far.
ある部品の生産ラインは二つの工程 A，B の順で構成されており，各工程の機械台数，部品 1 個の生産に要する作業時間，不良率は表のとおりである。1 日の稼働時間を 10 時間とするとき，この生産ラインの 1 日の生産能力（良品が生産される数）は何個か。ここで，工程 A での不良品は工程 B には送らないものとする。また，機械の故障時間や段取り時間，工程間の仕掛品在庫は考えないものとし，仕掛中のものは終了時間が来ても最後まで仕上げるものとする。
| 工程 | 機械台数 | 作業時間／個 | 不良率 |
|---|---|---|---|
| A | 1 台 | 3 分 | 5％ |
| B | 1 台 | 2 分 | 10％ |
正解 The correct answer is ア “171”. Work through the calculation step by step. First, convert the daily operating time of 10 hours into minutes: 10×60 = 600 minutes. Process A has 1 machine and takes 3 minutes per unit, so in one day it can handle 600÷3 = 200 units. However, the defect rate of process A is 5%, and the question states that “defective products from process A are not sent to process B,” so only the good units, 200×(1−0.05) = 190, are passed on to process B. Next, look at process B: 1 machine takes 2 minutes per unit, so in 600 minutes it can handle up to 600÷2 = 300 units. In other words, B’s capacity of 300 units is larger than the 190 units that actually arrive, so B is not the constraint (bottleneck), and all 190 units are processed by B (190×2 = 380 minutes is enough, and the question says that a part being worked on is finished to completion even if the end time has passed). Finally, subtracting the 10% defect rate of process B, the final number of good units is 190×(1−0.10) = 171. Therefore the daily production capacity of this line is 171 units, and ア is correct. There are two key points: the process with the smaller capacity, process A (3 min/unit), determines the flow of the whole line; and the defect rates must be applied in two stages, A and B, multiplied in turn.
180 units comes from 200×(1−0.10) = 180, that is, applying only the 10% defect rate of process B to process A’s output of 200 units while overlooking the 5% defect rate of process A. In this question, 5% defects also occur at process A, so by the time the units reach B the count has already dropped from 200 to 190. Defects occur at each process, so both A’s 5% and B’s 10% must be multiplied in turn (200×0.95×0.9 = 171). This is a wrong answer that misses the point that when processes are connected in series, the yield is the product of the good-unit rates of the individual processes.
200 units comes from 600÷3 = 200, which is simply the number of units process A can take in and work on in one day, with no defect rate taken into account at all. What the question asks for is “the number of non-defective units produced,” so A’s 5% defects and B’s 10% defects must be subtracted. 200 units is nothing more than the processing capacity of process A (output measured on an input basis); it is not the number of good units that come out at the end of the line.
257 units comes from taking 600÷2 = 300 units as the starting point and computing 300×0.95×0.9 = 256.5 ≈ 257; it is a wrong answer that mistakes which process is the bottleneck. Process B, at 2 minutes per unit, is indeed faster, but only parts that have passed through A flow into B, so what determines the flow of the whole line is the slower side, namely process A, which takes 3 minutes per unit. The production capacity of processes in series is determined by the process with the smallest capacity (the bottleneck), and the capacity of a fast process cannot be treated as the capacity of the whole line.