Apply the Hungarian method: two reductions here already yield a complete zero assignment.

Row reduction then column reduction leaves zeros at $W_1J_1, W_2J_3, W_3J_2, W_4J_4$, which can all be boxed independently, so that is the optimal assignment.

Step 1 — Row reduction. Row minima: $10, 5, 12, 17$.

Step 2 — Column reduction. Column minima: $0, 8, 0, 3$.

Step 3 — Assign the zeros.

• $W_1$ has a single zero at $J_1$ $\to$ assign $W_1 \to J_1$; cross $W_4J_1$.
• $W_2$ has a single zero at $J_3$ $\to$ assign $W_2 \to J_3$; cross $W_3J_3$.
• $W_3$ now has a single zero at $J_2$ $\to$ assign $W_3 \to J_2$; cross $W_4J_2$.
• $W_4$ now has a single uncrossed zero at $J_4$ $\to$ assign $W_4 \to J_4$.

Four independent zeros are boxed — the assignment is optimal (a $4$-line cover exists).

Total cost (original matrix):

Dual-solve (independent check). Verify no swap improves it:

• Swap $W_3,W_4$ jobs: $W_3J_4 + W_4J_2 = 24 + 25 = 49$ vs current $W_3J_2 + W_4J_4 = 20 + 20 = 40$ — worse.
• Swap $W_1,W_4$ jobs: $W_1J_4 + W_4J_1 = 20 + 17 = 37$ vs current $W_1J_1 + W_4J_4 = 10 + 20 = 30$ — worse.
• Swap $W_1,W_2$ jobs: $W_1J_3 + W_2J_1 = 15 + 15 = 30$ vs current $W_1J_1 + W_2J_3 = 10 + 5 = 15$ — worse. No pairwise exchange reduces the total, confirming $55$ is optimal.

After the two reductions a complete set of independent zeros already existed, so no line-covering/revision step (Steps 4–5) was needed. When that happens, the two reductions alone have solved the problem.

Boxing zeros greedily and getting stuck (e.g. assigning $W_4J_1$ early blocks $W_1$). Always start with rows/columns having a single zero, and cross out the rest as you go.