Applying the Hungarian method, the optimal assignment $P\to II,\;Q\to IV,\;R\to I,\;S\to III$ minimises the cost at $\text{₹}99$.

Step 1 — Row reduction (subtract the smallest entry of each row):

Row $P$ ($-25$): $6,\;0,\;8,\;4$

Row $Q$ ($-21$): $4,\;3,\;2,\;0$

Row $R$ ($-19$): $0,\;2,\;4,\;5$

Row $S$ ($-34$): $4,\;2,\;0,\;6$

Step 2 — Column reduction: each column already contains a $0$ (column minima are all $0$), so the matrix is unchanged:

$\begin{array}{c|cccc} & I & II & III & IV\\ \hline P & 6 & 0 & 8 & 4\\ Q & 4 & 3 & 2 & 0\\ R & 0 & 2 & 4 & 5\\ S & 4 & 2 & 0 & 6\end{array}$

Step 3 — Assign the zeros. Exactly one independent zero can be chosen in every row and column:

$R\to I$ (zero), $P\to II$ (zero), $S\to III$ (zero), $Q\to IV$ (zero).

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