Johnson's rule gives the optimal sequence $3-5-2-4-1$; total elapsed time $=41$ h, idle time $M_1=11$ h, $M_2=5$ h.

Step 1 — Apply Johnson's rule. The processing times are:

$\begin{array}{c|ccccc} \text{Job} & 1 & 2 & 3 & 4 & 5\\ \hline M_1 & 6 & 8 & 4 & 5 & 7\\ M_2 & 3 & 7 & 6 & 4 & 16\end{array}$

Repeatedly pick the smallest time: if it is on $M_1$ place the job as early as possible, if on $M_2$ place it as late as possible.

• Smallest is $3$ (Job 1 on $M_2$) $\Rightarrow$ Job 1 goes last.
• Next smallest is $4$: Job 3 on $M_1$ $\Rightarrow$ first; Job 4 on $M_2$ $\Rightarrow$ next-to-last.
• Next is $7$: Job 5 on $M_1$ $\Rightarrow$ next available first; Job 2 on $M_2$ $\Rightarrow$ next available last.

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

Step 2 — Compute the elapsed time table (In/Out on each machine):

$\begin{array}{c|cc|cc} \text{Job} & M_1\text{ in} & M_1\text{ out} & M_2\text{ in} & M_2\text{ out}\\ \hline 3 & 0 & 4 & 4 & 10\\ 5 & 4 & 11 & 11 & 27\\ 2 & 11 & 19 & 27 & 34\\ 4 & 19 & 24 & 34 & 38\\ 1 & 24 & 30 & 38 & 41\end{array}$

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