A three-machine ($M_1 \to M_2 \to M_3$) sequencing problem can be converted to a two-machine problem using Johnson's rule if: (a) $\min A_i \le \max B_i$ (b) $\min A_i \ge \max B_i$ or $\min C_i \ge \max B_i$ (c) $\max A_i \ge \min B_i$ (d) all processing times are equal

A three-machine (M₁ → M₂ → M₃) sequencing problem can be converted to a two-machine problem using Johnson's ru

Exercises · Q12

M_1 \to M_2 \to M_3

(a)

\min A_i \le \max B_i

(b)

\min A_i \ge \max B_i

\min C_i \ge \max B_i

(c)

\max A_i \ge \min B_i

(d) all processing times are equal

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Rule. For $n$ jobs on three machines $M_1 \to M_2 \to M_3$ with times $A_i, B_i, C_i$ , Johnson's two-machine method may be applied (via $G_i = A_i + B_i$ , $H_i = B_i + C_i$ ) only if at least one of

\min_i A_i \ \ge\ \max_i B_i \qquad \text{or} \qquad \min_i C_i \ \ge\ \max_i B_i

holds. This ensures the middle machine $M_2$ is never the bottleneck, so the two-machine optimum is also optimal for three machines.

Options.

(a) $\min A_i \le \max B_i$ — this is (almost always) true and guarantees nothing; wrong.
(b) matches the rule exactly — correct. …

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Q.A three-machine ( $M_1 \to M_2 \to M_3$ ) sequencing problem can be converted to a two-machine problem using Johnson's rule if:

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Q.A three-machine (M1→M2→M3M_1 \to M_2 \to M_3M1​→M2​→M3​) sequencing problem can be converted to a two-machine problem using Johnson's rule if:

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Q.A three-machine ( $M_1 \to M_2 \to M_3$ ) sequencing problem can be converted to a two-machine problem using Johnson's rule if:

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