Apply Johnson's rule, then tabulate in/out-times.

Step 1 — Johnson's rule.

• Smallest overall $= 1$ (job $B$, $M_1$) $\to$ first: $(B,-,-,-)$.
• Next $= 2$ (job $A$, $M_2$) $\to$ last: $(B,-,-,A)$.
• Remaining $C, D$: smallest $= 3$ (job $D$, $M_1$) $\to$ earliest free slot; then $C$ takes the last free slot: $(B, D, C, A)$.

Optimal sequence: $B \to D \to C \to A$.

Step 2 — In/out-time table (on $M_2$, in-time $= \max(M_1$ out, previous $M_2$ out$)$):

Step 3 — Results. Total elapsed time $=$ last $M_2$ out-time $= \mathbf{24}$ minutes. …

A quick check: $M_2$ never has to wait once it starts except at the very beginning here, so elapsed time $\approx$ (time first job reaches $M_2$) …

Placing job $A$ early because its $M_1$ time ($5$) is small — Johnson's rule looks at the smallest across both machines; $A$'s $M_2$ time …