Consider the LP: Trivially is optimal. How about ?
It is also easy to see that since In general, given , , and the LP constraints , we can assign each constraint a coefficient as follows: If and for all , then is an upper bound, which further implies that We want the upper bound as small as possible. Therefore, we would like to solve the following LP: which is termed as the dual linear program.

We note that the primal problem may have different forms. If the primal has a constraint , then we have the corresponding variable in the dual. If the primal has a constraint , then we have the corresponding variable unconstrained in sign in the dual.

Now we use the form , subject to , . By the discussion above, we know that the optimal value of the dual problem gives an upper bound of the primal problem. This property is called the weak duality.

However, if the primal problem is infeasible, we cannot conclude that the dual problem is unbounded below. It is possible that both primal and dual problems are infeasible.

We’ve already known that any feasible solution of the dual gives an upper to the primal. The question is that, is there any gap between the optimal value of the primal and the optimal value of the dual?

Now we can complete the following table.

The proof of the strong duality is an application of the Farkas’ lemma, which we have introduced in Lecture 4.

To apply this lemma, we consider the following corollary.

Let be the cone of column vectors of . The corollary tells us that either intersects the region , or there exists a hyperplane strictly separating and , where and .

We are now ready to prove the strong duality theorem.

In the proof of the corollary of the Farkas’ lemma, we employ the matrix , which looks similar to the standard form of the linear program. In fact, if we consider the standard form, the proof of the strong duality can apply the Farkas’ lemma directly, instead of using the corollary.

We now consider an application of the strong duality in linear programming.

Given an undirected graph , a matching is a set of edges such that no edge in the set is a loop and no two edges share common vertices, and a vertex cover is a set of vertices that includes at least one endpoint of every edge. Here are some examples of matchings and vertex covers.

Clearly the set is a matching and is a vertex cover. So we consider the problem of maximum matching and minimum vertex cover.

We can write an integer programming formulation of the maximum matching problem. Any matching can be represented by variables such that if the edge and otherwise. Conversely, any assignment to can represent a matching if and for all . So the problem can be formulated as If we relax the constraints to be , it can be reformulated as a linear program The relaxed problem is called the maximum fractional matching problem.

Analogously, for each vertex we can assign a variable to represent whether is in the vertex cover. We relax the constraints again. Then we obtain the problem of maximum fractional vertex cover as follows It is easy to verify that the fractional minimum vertex cover problem is the dual of the fractional maximum matching problem. So we know that for any graph, the size of the maximum fractional matching equals the size of the minimum fractional vertex cover.

In a bipartite graph , we can further show that the size of the maximum fractional matching equals the size of the maximum matching. Given a fractional matching , consider the subgraph consisting of fractional edges where .

• Case 1. There exists a cycle . Note that is an even number since is bipartite. Let . Then add to for all odd , and subtract from for all even (we assume that ). The resulting satisfy all constraints and the size of the fractional matching remains the same.
• Case 2. There is no cycles. Then choose any path . Note that all edges incident to has except . So if belongs to but . Similarly, if belongs to but . Again, let . Then subtract from for all odd , and add to for all even (we assume that ). Now the resulting satisfy all constraints and the size of the fractional matching is nondecreasing.
  Each operation decrease the number of fractional edges. So there is no fractional edges after finite many operations. Consequently, any fractional matching can be converted into an integral matching that is not worse, which implies that the size of the maximum fractional matching equals the size of the maximum matching

We can also show that the size of the minimum fractional vertex cover equals the size of the minimum vertex cover in any bipartite graph . Suppose where and . We construct a random vertex cover as follows.
Uniformly choose a real number . For every , let if , and otherwise. For every , let if , and otherwise. For every and , if , then . So at least one of is in , which gives that is a vertex cover.
Now we calculate the expected size of . For any , . Thus, by the linearity of expectation, , which is the size of the minimum fractional vertex cover. In addition, there exists such that the vertex cover constructed by has the size .

Overall, we conclude that in any bipartite graph, the size of the maximum matching equals the size of the minimum vertex cover.