# 数学代写|线性规划作业代写Linear Programming代考|Initialization

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## 数学代写|线性规划作业代写Linear Programming代考|Initialization

In the previous section, we presented the simplex method. However, we only considered problems for which the right-hand sides were all nonnegative. This ensured that the initial dictionary was feasible. In this section, we shall discuss what one needs to do when this is not the case.
Given a linear programming 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}
the initial dictionary that we introduced in the preceding section was
\begin{aligned} \zeta & =\sum_{j=1}^n c_j x_j \ w_i & =b_i-\sum_{j=1}^n a_{i j} x_j \quad i=1,2, \ldots, m . \end{aligned}
The solution associated with this dictionary is obtained by setting each $x_j$ to zero and setting each $w_i$ equal to the corresponding $b_i$. This solution is feasible if and only if all the right-hand sides are nonnegative. But what if they are not? We handle this difficulty by introducing an auxiliary problem for which
(1) a feasible dictionary is easy to find and
(2) the optimal dictionary provides a feasible dictionary for the original problem.

## 数学代写|线性规划作业代写Linear Programming代考|Unboundedness

In this section, we shall discuss how to detect when the objective function value is unbounded.

Let us now take a closer look at the “leaving variable” computation: pick $l$ from $\left{i \in \mathcal{B}: \bar{a}_{i k} / \bar{b}_i\right.$ is maximal $}$. We avoided the issue before, but now we must face what to do if a denominator in one of these ratios vanishes. If the numerator is nonzero, then it is easy to see that the ratio should be interpreted as plus or minus infinity depending on the sign of the numerator. For the case of $0 / 0$, the correct convention (as we’ll see momentarily) is to take this as a zero.

What if all of the ratios, $\bar{a}_{i k} / \bar{b}_i$, are nonpositive? In that case, none of the basic variables will become zero as the entering variable increases. Hence, the entering variable can be increased indefinitely to produce an arbitrarily large objective value. In such situations, we say that the problem is unbounded. For example, consider the following dictionary:
$$\begin{array}{rrr} \zeta & =5+x_3-x_1 \ \hline x_2 & =5+2 x_3-3 x_1 \ x_4 & =7 & -4 x_1 \ x_5 & = & x_1 . \end{array}$$
The entering variable is $x_3$ and the ratios are
$$-2 / 5, \quad-0 / 7, \quad 0 / 0$$
Since none of these ratios is positive, the problem is unbounded.
In the next chapter, we will investigate what happens when some of these ratios take the value $+\infty$.

# 线性规划代写

## 数学代写|线性规划作业代写Linear Programming代考|Initialization

\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} \zeta & =\sum_{j=1}^n c_j x_j \ w_i & =b_i-\sum_{j=1}^n a_{i j} x_j \quad i=1,2, \ldots, m . \end{aligned}

(2)最优字典为原问题提供一个可行的字典。

## 数学代写|线性规划作业代写Linear Programming代考|Unboundedness

$$\begin{array}{rrr} \zeta & =5+x_3-x_1 \ \hline x_2 & =5+2 x_3-3 x_1 \ x_4 & =7 & -4 x_1 \ x_5 & = & x_1 . \end{array}$$

$$-2 / 5, \quad-0 / 7, \quad 0 / 0$$

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