This is a maximisation problem, so convert it to minimisation by subtracting every entry from the largest ($16$), then apply the Hungarian method.

Subtracting each profit from $16$ gives a loss matrix; the Hungarian method on it yields the assignment $S_1T_1, S_2T_3, S_3T_2$, whose original profits sum to $46$ (thousand).

Step 0 — Convert to minimisation. Largest profit $M = 16$. Replace each entry by $16 - p_{ij}$ (the opportunity-loss matrix):

Step 1 — Row reduction. Row minima $0, 1, 1$:

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

Step 3 — Assign the zeros.

• $S_1$ single zero at $T_1$ $\to$ assign $S_1 \to T_1$; cross $S_3T_1$.
• $S_2$ single zero at $T_3$ $\to$ assign $S_2 \to T_3$.
• $S_3$ now single zero at $T_2$ $\to$ assign $S_3 \to T_2$.

Three independent zeros — optimal.

Maximum profit (from the original profit matrix):

Dual-solve (independent check). All $3! = 6$ one-to-one assignments' profits:

The largest total is $46$ (from $S_1T_1, S_2T_3, S_3T_2$), confirming the Hungarian result.

Instead of subtract-from-largest you may negate the matrix (use $-p_{ij}$) and minimise; the two routes give the same assignment. Whichever conversion is used, always read the final profit from the original matrix.

Running the Hungarian method directly on the profit matrix without converting — that minimises profit, giving the worst assignment. Also, quoting the loss-matrix total as the answer instead of the real profit.