Johnson's rule gives the sequence VII–I–IV–V–II–III–VI. Total elapsed time $= 91$; machine A idle $= 5$; machine B idle $= 13$.

Step 1 — Johnson's rule (order A then B). Repeatedly find the smallest time in the table: if it is under machine A, place that job as early as possible; if under machine B, place it as late as possible.

• Smallest $= 5$: at A(VII) $\Rightarrow$ VII first; at B(VI) $\Rightarrow$ VI last.
• Next $= 7$ at A(I) $\Rightarrow$ I next.
• Next $= 10$ at IV (both A and B) $\Rightarrow$ IV next.
• Next $= 14$ at A(V) $\Rightarrow$ V next; then B ties place II, III before VI.

Sequence: $\text{VII} \to \text{I} \to \text{IV} \to \text{V} \to \text{II} \to \text{III} \to \text{VI}$.

Step 2 — Elapsed and idle time table. Times (A, B): VII$(5,7)$, I$(7,12)$, IV$(10,10)$, V$(14,16)$, II$(16,14)$, III$(19,14)$, VI$(15,5)$.