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# 数学代写|线性规划代写Linear Programming代考|The Weak Duality Theorem

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## 数学代写|线性规划代写Linear Programming代考|The Weak Duality Theorem

As we saw in our example, the dual problem provides upper bounds for the primal objective function value. This result is true in general and is referred to as the Weak Duality Theorem:

THEOREM 5.1. If $\left(x_1, x_2, \ldots, x_n\right)$ is feasible for the primal and $\left(y_1, y_2, \ldots, y_m\right)$ is feasible for the dual, then
$$\sum_j c_j x_j \leq \sum_i b_i y_i .$$
Proof. The proof is a simple chain of obvious inequalities:
\begin{aligned} \sum_j c_j x_j & \leq \sum_j\left(\sum_i y_i a_{i j}\right) x_j \ & =\sum_{i j} y_i a_{i j} x_j \ & =\sum_i\left(\sum_j a_{i j} x_j\right) y_i \ & \leq \sum_i b_i y_i, \end{aligned}
where the first inequality follows from the fact that each $x_j$ is nonnegative and each $c_j$ is no larger than $\sum_i y_i a_{i j}$. The second inequality, of course, holds for similar reasons.

Consider the subset of the real line consisting of all possible values for the primal objective function, and consider the analogous subset associated with the dual problem. The weak duality theorem tells us that the set of primal values lies entirely to the left of the set of dual values. As we shall see shortly, these sets are both closed intervals (perhaps of infinite extent), and the right endpoint of the primal set butts up against the left endpoint of the dual set (see Figure 5.1). That is, there is no gap between the optimal objective function value for the primal and for the dual. The lack of a gap between primal and dual objective values provides a convenient tool for verifying optimality. Indeed, if we can exhibit a feasible primal solution $\left(x_1^, x_2^, \ldots, x_n^\right)$ and a feasible dual solution $\left(y_1^, y_2^, \ldots, y_m^\right)$ for which
$$\sum_j c_j x_j^=\sum_i b_i y_i^,$$
then we may conclude that each of these solutions is optimal for its respective problem. To see that the primal solution is optimal, consider any other feasible solution $\left(x_1, x_2, \ldots, x_n\right)$. By the weak duality theorem, we have that
$$\sum_j c_j x_j \leq \sum_i b_i y_i^=\sum_j c_j x_j^$$

## 数学代写|线性规划代写Linear Programming代考|The Strong Duality Theorem

The fact that for linear programming there is never a gap between the primal and the dual optimal objective values is usually referred to as the Strong Duality Theorem:
THEOREM 5.2. If the primal problem has an optimal solution,
$$x^=\left(x_1^, x_2^, \ldots, x_n^\right)$$
then the dual also has an optimal solution,
$$y^=\left(y_1^, y_2^, \ldots, y_m^\right)$$
such that
$$\sum_j c_j x_j^=\sum_i b_i y_i^ .$$
Carefully written proofs, while attractive for their tightness, sometimes obfuscate the main idea. In such cases, it is better to illustrate the idea with a simple example. Anyone who has taken a course in linear algebra probably already appreciates such a statement. In any case, it is true here as we explain the strong duality theorem.

The main idea that we wish to illustrate here is that, as the simplex method solves the primal problem, it also implicitly solves the dual problem, and it does so in such a way that $(5.2)$ holds.

To see what we mean, let us return to the example discussed in Section 5.1. We start by introducing variables $w_i, i=1,2$, for the primal slacks and $z_j, j=1,2,3$, for the dual slacks. Since the inequality constraints in the dual problem are greaterthan constraints, each dual slack is defined as a left-hand side minus the corresponding right-hand side. For example,
$$z_1=y_1+3 y_2-4$$

## 数学代写|线性规划代写Linear Programming代考|The Weak Duality Theorem

$$\sum_j c_j x_j \leq \sum_i b_i y_i .$$

\begin{aligned} \sum_j c_j x_j & \leq \sum_j\left(\sum_i y_i a_{i j}\right) x_j \ & =\sum_{i j} y_i a_{i j} x_j \ & =\sum_i\left(\sum_j a_{i j} x_j\right) y_i \ & \leq \sum_i b_i y_i, \end{aligned}

$$\sum_j c_j x_j^=\sum_i b_i y_i^,$$

$$\sum_j c_j x_j \leq \sum_i b_i y_i^=\sum_j c_j x_j^$$

## 数学代写|线性规划代写Linear Programming代考|The Strong Duality Theorem

$$x^=\left(x_1^, x_2^, \ldots, x_n^\right)$$

$$y^=\left(y_1^, y_2^, \ldots, y_m^\right)$$

$$\sum_j c_j x_j^=\sum_i b_i y_i^ .$$

$$z_1=y_1+3 y_2-4$$

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