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Let $x_{ij}\ge 0$ be shipments, with column sums eight and row sums at most the four capacities. Total demand is $32$, below total capacity $36$, so unused capacity is allowed. This is a transportation problem with capacity inequalities; equivalently add a dummy shop demanding four units with zero shipping cost.

Using the original PDF's cost $c_{12}=10$, take the feasible solutionIts row totals are $(7,8,8,9)$ and all column totals are eight. Four units of capacity remain unused at the second factory. Its shipping cost is $35+32+28+2+7+16=\boxed{120}$.

To prove that the cost is a global minimum, use the transportation dual certificate with capacity inequalities. If row potentials satisfy $u_i\le 0$ and $u_i+v_j\le c_{ij}$, every feasible solution satisfieswhere $s=(7,12,8,9)$. The second inequality uses the sign of $u_i$: multiplying a capacity upper bound by a nonpositive number reverses it. Choose $u=(0,0,-3,0)$ and $v=(4,7,2,5)$. The reduced costs areso every entry is nonnegative. The lower bound is $-24+8(4+7+2+5)=120$, equal to the displayed shipping cost. The minimum shipping cost is $120$.

For the production costs in a transportation problem, add the production cost of each factory to every real shipment from that factory. Unused capacity incurs no production cost, so the dummy-shop costs remain zero. The resulting real-shop cost matrix isA feasible solution isIts row totals are $(7,12,8,5)$ and column totals remain eight, leaving four units unused at the fourth factory. Its combined cost is $18+32+48+40+32+24+35=229$.

For a matching transportation dual certificate with capacity inequalities, take $u^{\prime }=(-1,-1,-3,0)$ and $v^{\prime }=(7,11,7,9)$. Their reduced costs areall nonnegative. The lower bound is $-7-12-24+8(7+11+7+9)=229$. Thus the minimum combined production and shipping cost is $229$, and the second allocation attains it.