# 数学代写|线性规划作业代写Linear Programming代考|Motivation-Finding Upper Bounds

#### Doug I. Jones

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## 数学代写|线性规划作业代写Linear Programming代考|Motivation-Finding Upper Bounds

We begin with an example:
\begin{aligned} \operatorname{maximize} & 4 x_1+x_2+3 x_3 \ \text { subject to } \quad x_1+4 x_2 & \leq 1 \ 3 x_1-x_2+x_3 & \leq 3 \ x_1, x_2, x_3 & \geq 0 . \end{aligned}
Our first observation is that every feasible solution provides a lower bound on the optimal objective function value, $\zeta^$. For example, the solution $\left(x_1, x_2, x_3\right)=(1,0,0)$ tells us that $\zeta^ \geq 4$. Using the feasible solution $\left(x_1, x_2, x_3\right)=(0,0,3)$, we see that $\zeta^* \geq 9$. But how good is this bound? Is it close to the optimal value? To answer, we need to give upper bounds, which we can find as follows. Let’s multiply the first constraint by 2 and add that to 3 times the second constraint:
\begin{aligned} 2\left(x_1+4 x_2 \quad\right) & \leq 2(1) \ +3\left(3 x_1-x_2+x_3\right) & \leq 3(3) \ \hline 11 x_1+5 x_2+3 x_3 & \leq 11 \end{aligned}
Now, since each variable is nonnegative, we can compare the sum against the objective function and notice that
$$4 x_1+x_2+3 x_3 \leq 11 x_1+5 x_2+3 x_3 \leq 11 .$$

## 数学代写|线性规划作业代写Linear Programming代考|The Dual Problem

Given a linear programming problem in standard form,
\begin{aligned} & \operatorname{maximize} \sum_{j=1}^n c_j x_j \ & \text { subject to } \sum_{j=1}^n a_{i j} x_j \leq b_i \quad i=1,2, \ldots, m \ & x_j \geq 0 \quad j=1,2, \ldots, n, \ & \end{aligned}
the associated dual linear program is given by
\begin{aligned} & \operatorname{minimize} \sum_{i=1}^m b_i y_i \ & \text { subject to } \sum_{i=1}^m y_i a_{i j} \geq c_j \quad j=1,2, \ldots, n \ & y_i \geq 0 \quad i=1,2, \ldots, m . \ & \end{aligned}
Since we started with (5.1), it is called the primal problem. Our first order of business is to show that taking the dual of the dual returns us to the primal. To see this, we first must write the dual problem in standard form. That is, we must change the minimization into a maximization and we must change the first set of greater-thanor-equal-to constraints into less-than-or-equal-to. Of course, we must effect these changes without altering the problem. To change a minimization into a maximization, we note that to minimize something it is equivalent to maximize its negative and then negate the answer:
$$\min \sum_{i=1}^m b_i y_i=-\max \left(-\sum_{i=1}^m b_i y_i\right)$$

# 线性规划代写

## 数学代写|线性规划作业代写Linear Programming代考|Motivation-Finding Upper Bounds

\begin{aligned} \operatorname{maximize} & 4 x_1+x_2+3 x_3 \ \text { subject to } \quad x_1+4 x_2 & \leq 1 \ 3 x_1-x_2+x_3 & \leq 3 \ x_1, x_2, x_3 & \geq 0 . \end{aligned}

\begin{aligned} 2\left(x_1+4 x_2 \quad\right) & \leq 2(1) \ +3\left(3 x_1-x_2+x_3\right) & \leq 3(3) \ \hline 11 x_1+5 x_2+3 x_3 & \leq 11 \end{aligned}

$$4 x_1+x_2+3 x_3 \leq 11 x_1+5 x_2+3 x_3 \leq 11 .$$

## 数学代写|线性规划作业代写Linear Programming代考|The Dual Problem

\begin{aligned} & \operatorname{maximize} \sum_{j=1}^n c_j x_j \ & \text { subject to } \sum_{j=1}^n a_{i j} x_j \leq b_i \quad i=1,2, \ldots, m \ & x_j \geq 0 \quad j=1,2, \ldots, n, \ & \end{aligned}

\begin{aligned} & \operatorname{minimize} \sum_{i=1}^m b_i y_i \ & \text { subject to } \sum_{i=1}^m y_i a_{i j} \geq c_j \quad j=1,2, \ldots, n \ & y_i \geq 0 \quad i=1,2, \ldots, m . \ & \end{aligned}

$$\min \sum_{i=1}^m b_i y_i=-\max \left(-\sum_{i=1}^m b_i y_i\right)$$

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