There are more jobs than machines, so the problem is unbalanced; add a dummy machine with zero costs to square it, then apply the Hungarian method.

Step 0 — Balance the problem. There are $4$ jobs but only $3$ machines, so add a dummy machine $M_4$ with all costs $0$:

Step 1 — Row reduction. Row minima $2, 3, 4, 0$:

Step 2 — Column reduction. Column minima are $0, 0, 0, 0$ — no change.

Step 3 — Assign the zeros.

• $M_1$ single zero at $J_4$ $\to$ assign $M_1 \to J_4$.
• $M_2$ single zero at $J_2$ $\to$ assign $M_2 \to J_2$.
• $M_3$ single zero at $J_3$ $\to$ assign $M_3 \to J_3$.
• $M_4$ (dummy) takes the only column left, $J_1$ (a zero) $\to$ $M_4 \to J_1$.

Four independent zeros — optimal. …

Because the three cells that are each the minimum of their row happen to lie in different columns, the optimum can also be spotted directly — but the dummy-row …

Forgetting to add the dummy row (and trying to run the Hungarian method on a non-square matrix), or giving dummy cells a large cost instead of $0$. Also, failing to repor …