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Measure quantities in thousands of copies. Use five supply rows: initial inventory, regular production in month 1, overtime in month 1, regular production in month 2, overtime in month 2. Their supplies/capacities are $(10,40,150,40,150)$. The three demand columns are the two due dates and a dummy column for unused capacity, with demands $(40,60,290)$, balancing the total supply $390$.

The transportation problem has cost matrixAn infinite entry prohibits delivering a later month's production to an earlier deadline. The extra $20$ in the second column of the early rows accounts for holding stock through the first month. For production rows, the dummy column represents capacity not used, not copies actually printed. Initial inventory assigned to the dummy column would instead remain in stock through both months, costing $40$; this distinguishes physical inventory from optional production capacity. No gratuitous disposal is needed in the optimum.

Let $x_{ij}\ge 0$ have the stated row and column sums, with the forbidden entries zero, and minimize $\sum c_{ij}{x}_{ij}$. A feasible solution isThus regular production is $40$ in each month; overtime is zero then $10$; first-month closing inventory is $10$, and final inventory is zero. Its total production-plus-holding cost isTo prove optimality, use transportation dual potentials $u=(-430,-30,0,-50,0)$ for rows and $v=(430,450,0)$ for columns. Every allowed edge satisfies $u_i+v_j\le c_{ij}$. Multiplying by any feasible shipment and summing gives the lower boundThe proposed shipment attains it, so the schedule is optimal. The dual calculation certifies the minimum without depending on an unshown transportation-simplex iteration.

A balanced transportation problem chooses nonnegative shipments $x_{ij}$ minimizing ${\sum }_{ij}{c}_{ij}{x}_{ij}$, subject to ${\sum }_j{x}_{ij}=s_i$ and ${\sum }_i{x}_{ij}=d_j$, where the total supply equals total demand. For multipliers $u_i,v_j$, its Lagrangian isTaking its infimum over $x_{ij}\ge 0$ produces the linear programming duality problem of maximizing ${\sum }_i{u}_i{s}_i+{\sum }_j{v}_j{d}_j$ subject to $u_i+v_j\le c_{ij}$. These are the transportation dual potentials.

The transportation simplex algorithm starts with a feasible transportation spanning tree of $m+n-1$ basic entries, obtainable by filling the north-west available cell with the smaller remaining supply or demand. Set $u_i+v_j=c_{ij}$ on basic cells; this determines all potentials up to adding a constant to all $u_i$ and subtracting it from all $v_j$. If all nonbasic reduced costs $r_{ij}=c_{ij}-u_i-v_j$ are nonnegative, feasibility and the dual bound prove optimality. Otherwise let a negative-cost cell enter. It closes a unique cycle in the basic tree. Alternate additions and subtractions around that cycle and choose the largest amount preserving nonnegativity, namely the minimum shipment on the subtraction cells. A cell that reaches zero leaves the basis. The cost changes by the entering reduced cost times the pivot amount. Zero basic cells and an anti-cycling rule can handle degeneracy; all pivots below are nondegenerate.

The north-west initial shipment matrix isWith $u_1=0$, its transportation dual potentials are $u=(0,1,0)$ and $v=(5,9,5)$. The most negative reduced cost is $r_{32}=-7$. The cycle is $+(3,2),-(3,3),+(2,3),-(2,2)$; the pivot amount is $\min (40,46)=40$. HenceNow $u=(0,1,-7)$ and $v=(5,9,5)$. Enter $(1,3)$, of reduced cost $-4$, on cycle $+(1,3),-(1,2),+(2,2),-(2,3)$. The amount is $\min (22,78)=22$, givingThe new potentials are $u=(0,5,-3)$ and $v=(5,5,1)$. Enter $(2,1)$, of reduced cost $-7$, on cycle $+(2,1),-(1,1),+(1,3),-(2,3)$. The amount is $\min (14,56)=14$. ThusTo certify optimality in the original transportation problem, take $u=(0,5,-3)$ and $v=(-2,5,1)$. Their entire reduced-cost matrix isThey satisfy dual feasibility, and their value is $5\cdot 84-3\cdot 40-2\cdot 14+5\cdot 68+78=690$, equal to the shipment cost. The minimum transportation cost is therefore $690$, with no unproved optimality assertion from the iterations alone.

Associate free real dual variables $u_i$ and $v_j$ with the row and column equality constraints. The Lagrangian for the transportation problem isIts infimum over $x_{ij}\ge 0$ is finite exactly when $u_i+v_j\le c_{ij}$ for every pair. The linear programming duality therefore givesThe dual variables have no sign restrictions because the primal constraints are equalities.

A feasible shipment $x$ is optimal if and only if there exist feasible transportation dual potentials satisfying complementary slackness:Thus every occupied route must have $u_i+v_j=c_{ij}$. To see sufficiency explicitly, the primal minus dual objective isand complementary slackness makes it zero. This is the weak duality certificate. Necessity follows from strong duality for the feasible, bounded linear program. If the total supply $M$ is positive, $x_{ij}=a_i{b}_j/M$ gives feasibility; if $M=0$, $x=0$ does. The feasible shipments form a closed bounded set, so the primal optimum is attained. This also covers zero supplies and demands. The one redundant equality merely leaves the dual gauge freedom $(u,v)\mapsto (u+t,v-t)$.

Use the transportation simplex algorithm, with transportation dual potentials $u_i,v_j$ and reduced costs $r_{ij}=c_{ij}-u_i-v_j$. On a basic cell choose $u_i+v_j=c_{ij}$. The initial shipment matrix, read from the circled entries in the PDF, isIts six positive cells form a transportation spanning tree. Setting $u_1=0$ givesEnter cell $(1,1)$, whose reduced cost is $-6$. The cycle pivot has signsThe maximum feasible increment is $\theta =\min (5,4)=4$, so cell $(4,1)$ leaves the basis. The result isFor the new transportation spanning tree,Enter cell $(2,3)$ with reduced cost $-4$. Its cycle pivot isNow $\theta =\min (5,1)=1$, and cell $(1,3)$ leaves. We obtainIts row sums are $(12,5,7,12)$ and its column sums $(9,14,13)$, as required. The final transportation dual potentials areAll reduced costs are nonnegative, proving optimality. More explicitly, for any feasible shipment matrix $X$,Our shipments use only zero-reduced-cost cells and attain this lower bound. This supplies an independent weak duality certificate for the transportation optimum.