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Write the primal in the form $\min c^{T}x$ with $Ax\ge b$ and $x\ge 0$. Its dual linear program is $\max b^T\lambda$ with $A^T\lambda \le c$ and $\lambda \ge 0$. Indeed, nonnegative combinations of the primal inequalities give $c^{T}x\ge {\lambda }^{T}Ax\ge {\lambda }^{T}b$, which is weak duality. The specific dual isThe four dual inequalities correspond, in order, to the four primal variables.

Introduce nonnegative slack variables $s_1,s_2,s_3$ and start the simplex algorithm at $x=y=z=0$, with slacks $9,8,2$. The objective's largest positive entering coefficient belongs to $y$. Its ratio test is $9/3=3$, $8/2=4$, $2/1=2$, so $s_3$ leaves and $y$ enters.

Solve that pivot row for $y$ and substitute into the other rows and the objective $v$:This is a feasible simplex dictionary at nonbasic variables $x=z=s_3=0$. All their objective coefficients are negative, so no feasible increase can improve the objective. ThusThe negative reduced coefficients also prove uniqueness of this optimum.

The dual linear program, with nonnegative variables $u_1,u_2,u_3$, isThe feasible choice $\boxed{(u_1,u_2,u_3)=(0,0,5)}$ has objective $10$. Weak duality proves that it and the primal point are optimal. Complementary slackness also forces $u_1=u_2=0$ because the first two primal slacks are positive, and then the positive $y$ forces $u_3=5$.

Use a directed bipartite graph with one vertex for each job and one for each agent. Job vertices have supply $1$ and agent vertices demand $1$. An arc from job $i$ to agent $j$ has cost $c_{ij}$ and nonnegative flow $x_{ij}$; an upper bound $1$ may be added, but is already implied by the balances. The minimum-cost flow formulation of the assignment problem isInteger supplies and demands give an integral optimum by integrality of the transportation problem. Its $n$ positive arcs form a perfect matching, hence give one job per agent.

A network simplex algorithm basis for this connected $2n$-vertex network has $2n-1$ tree arcs, whereas an integral assignment has only $n$ positive flows. For $n>1$, at least $n-1$ basic tree arcs therefore carry zero flow. This unavoidable degeneracy in linear programming can cause zero-step pivots, extra bookkeeping and cycling unless an appropriate pivot rule is used. The Hungarian algorithm exploits the assignment structure directly rather than maintaining many zero-flow basic arcs.

Suppose the assignment dual potentials satisfy ${\lambda }_i-{\mu }_j\le c_{ij}$. For every feasible assignment $y$,If the selected assignment $x$ uses only arcs with ${\lambda }_i-{\mu }_j=c_{ij}$, its cost equals this lower bound. Weak duality and complementary slackness certify optimality. The PDF correctly denotes the feasible assignment by $x_{ij}$; the TeX aid mistakenly replaces it with ${\lambda }_{ij}$.

The Hungarian algorithm maintains feasible assignment dual potentials and a matching in a graph of zero reduced cost edges, where $r_{ij}=c_{ij}-{\lambda }_i+{\mu }_j\ge 0$. Search for an augmenting path in a matching in this equality graph. If there is none, let $S$ be the reachable jobs and $T$ the reachable agents in the alternating search from unmatched jobs. Every agent in $T$ is matched to a job in $S$. SetReduced costs from $S$ to its complement decrease by $\Delta$, those from outside $S$ into $T$ increase, and the remaining ones do not change. Thus feasibility and current matching equalities are preserved, while at least one new equality edge appears. Continue until an augmenting path in a matching increases the matching size. A perfect matching in the equality graph then has the same value as the dual linear program and is optimal.

For the example, start with the row minima $\lambda =(10,50,50)$ and $\mu =(0,0,0)$. The reduced cost matrix isMatch job $1$ to agent $1$. Searching from both unmatched jobs gives $S=\{1,2,3\}$ and $T=\{1\}$; the minimum cost to an agent outside $T$ is $\Delta =10$. Update to $\lambda =(20,60,60)$ and $\mu =(10,0,0)$:The new equality edge $(2,3)$ augments the matching. From unmatched job $3$, the next alternating search reaches agent $1$ and its matched job $1$, giving $S=\{1,3\}$ and $T=\{1\}$. Now $\Delta =15$. The updated feasible assignment dual potentials and reduced costs areThe augmenting path in a matching $J_3\to A_1\to J_1\to A_2$ uses job $3$ to agent $1$, the reversed matched edge to job $1$, then the new edge to agent $2$. It gives the optimal assignmentThe dual linear program value is $35+60+75-25=145$, an independent optimality certificate.

Substitute the dual linear program from the previous part and optimize jointly over $x,t,s$. The result isThe objective is linear, and the constraints are equalities or inequalities between affine functions, so this is a linear program. For each fixed feasible $x$, minimizing over $t,s$ gives exactly $f_k(x)$ by the threshold formula for the sum of the largest components; hence the reformulation preserves the optimal value and optimal $x$ whenever an optimum exists.

Introduce slack variables $s_1,s_2,s_3$. In the initial simplex algorithm dictionary, $x_2$ has the largest positive objective coefficient. The ratio test givesso $x_2$ enters and $s_3$ leaves. Solving the third constraint for $x_2$ givesThe objective becomesEvery reduced cost is now nonpositive, so the simplex optimality criterion givesThe other two slacks both equal one.

The dual linear program isThe vectoris dual feasible and has objective value $12$. By weak duality, it and the displayed primal point are optimal; they also satisfy complementary slackness.