We apply the Hungarian method: row-reduce, column-reduce, then box independent zeros.

Row and column reduction produces enough zeros to box one in each row and column at $O_1T_3, O_2T_1, O_3T_2$; adding the original times gives $54$ minutes.

Step 1 — Row reduction. Row minima are $20, 15, 17$. Subtract from each row:

Step 2 — Column reduction. Column minima are $0, 0, 2$. Subtract from each column:

Step 3 — Assign the zeros.

• Row $O_2$ has a single zero at $T_1$ $\to$ assign $O_2 \to T_1$; cross the other zero in column $T_1$ (namely $O_1T_1$).
• Row $O_3$ has a single zero at $T_2$ $\to$ assign $O_3 \to T_2$.
• Row $O_1$ now has one uncrossed zero, at $T_3$ $\to$ assign $O_1 \to T_3$.

Three zeros are boxed, one in each row and column, so the assignment is optimal.

Total time (from the original matrix):

Dual-solve (independent check). Compare all $3! = 6$ assignments:

The least is $54$, confirming the Hungarian result.

Direct enumeration is feasible for a $3\times3$ matrix: list all $6$ assignments and pick the smallest total. It agrees with the Hungarian method ($54$ minutes) and is a good check, but it does not scale to larger matrices.

Reading the final total from the reduced matrix (which would give $0$) instead of the original times. Always add the original $c_{ij}$ values at the boxed cells.