Check the conversion condition, form fictitious machines $G = M_1+M_2$ and $H = M_2+M_3$, apply Johnson's rule, then tabulate across all three real machines.

Step 1 — Test the conversion condition.

Since $\min M_1 = 6 \ge \max M_2 = 6$, the condition $\min A_i \ge \max B_i$ is satisfied, so the problem may be converted to a two-machine problem.

Step 2 — Form the fictitious machines $G_i = A_i + B_i$ and $H_i = B_i + C_i$:

Step 3 — Johnson's rule on $(G, H)$.

• Smallest $= 8$ (job $3$, on $G$) $\to$ first: $(3,-,-,-,-)$.
• Next $= 10$ (job $4$, on $G$) $\to$ next earliest: $(3,4,-,-,-)$.
• Remaining $1,2,5$: smallest $= 13$ — job $1$ on $G$ ($\to$ earliest free) and job $5$ on $H$ ($\to$ latest free): $(3,4,1,-,5)$.
• Last job $2$ takes the middle: $(3,4,1,2,5)$.

Optimal sequence: $3 \to 4 \to 1 \to 2 \to 5$.

Step 4 — In/out-time table on the real machines (in-time $= \max(\text{out-time on previous machine, previous job's out-time on this machine})$):

Step 5 — Results. Total elapsed time $=$ last $M_3$ out-time $= \mathbf{55}$ hours. …

The alternative form of the condition, $\min C_i \ge \max B_i$, would serve equally; here $\min M_3 = 8 \ge \max M_2 = 6$ also holds, so either route just …

Computing the elapsed time from the fictitious $G, H$ machines — they are only for finding the order. Also, skipping the condition check: if neither $\min A \ge \max B$ …