# 计算机代写|组合优化代写Combinatorial optimization代考|CSC205

## 计算机代写|组合优化代写Combinatorial optimization代考|Bland’s Rule

The second method for dealing with degeneracy is to modify the simplex algorithm by Bland’s rule as follows:

• Choose the entering column index $j^$ satisfying $$j^=\min \left{j \in \bar{I} \mid c_j>0\right} .$$
• Choose the row index $i^$ which is the smallest one if there are more than one $i^$ satisfying
$$\frac{b_i{ }^}{a_{i^ j^}}=\min \left{\frac{b_i}{a_{i j^}} \mid a_{i j^*}>0\right} .$$
Theorem 6.3.1 With Bland’s rule, simplex algorithm will not run into a cycle, so that within finitely many iterations, the algorithm is able to determine whether the optimal value goes to infinity or not, and if the optimal value is finite, then the algorithm will obtain an optimal solution.

Proof It is sufficient to show that with Bland’s rule, simplex algorithm will not run into a cycle. For contradiction, suppose a cycle exists. For simplicity of discussion, we delete all constraints with row indices not selected in the cycle. Thus, for remaining row index $i, b_i=0$ since objective function value cannot be changed during computation of the cycle. In this cycle, there also exist some column indices entering the feasible basis and then leaving or vice versa. Let $t$ be the largest column index among them. For simplicity of discussion, we also delete all columns with index $j>t$ since we will always assign 0 to variable $x_j$ for $j>t$. Next, let us consider two moments in this cycle.

At the first moment, $t$ leaves the feasible basis. Assume column index $s$ enters the feasible hasis. Denote hy $a_{i j}$ and $c_j$ coefficients of constraints and cost, respectively, at this moment.

At the second moment, $t$ enters the feasible basis. Denote by $a_{i j}^{\prime}$ and $c_j^{\prime}$ coefficients of constraints and cost, respectively.

## 计算机代写|组合优化代写Combinatorial optimization代考|Initial Feasible Basis

How do we find the initial feasible basis? A popular way is to introduce artificial variables $y=\left(y_1, y_2, \ldots, y_m\right)^T$ and solve the following LP:
\begin{aligned} \max & w=-e y \ \text { subject to } & A x+I_m y=b \ & x \geq 0, y \geq 0, \end{aligned}
where $e=(1,1, \ldots, 1)$ and $I_m$ is the identity matrix of order $m$. In this I.P, those artificial variables form a feasible basis. There are three possible outcomes resulting from solving this LP.
(1) The cost function value $w$ is reduced to 0 and all artificial variables are removed from the feasible basis. In this case, the final feasible basis can be used as initial feasible basis in original LP.
(2) The cost function reaches a negative maximum value. In this case, the original LP has no feasible solution.
(3) The cost function value $w$ is reduced to 0 ; however, there is an artificial variable $y_i$ in the feasible basis. Let $b_i$ and $a_{i j}$ denote coefficients of constraints at the last moment. In this case, we must have $y_i=b_i=0$; otherwise, $w=e y>0$. Note that there exists a variable $x_j$ such that $a_{i j} \neq 0$ since $\operatorname{rank}(A)=m$. This means that we may take $a_{i j}$ as pivot element to move $y_i$ out from feasible basis and to move in $x_j$, preserving cost function value 0 . When all artificial variables are moved out from the feasible basis, this case is reduced to case (1).

# 组合优化代考

## 计算机代写|组合优化代写Combinatorial optimization代考|Bland’s Rule

• 选择输入列索引j^j^令人满意的j^=\min \left{j \in \bar{I} \mid c_j>0\right} 。j^=\min \left{j \in \bar{I} \mid c_j>0\right} 。
• 选择行索引我^我^如果有多个，哪个是最小的我^我^令人满意的
\frac{b_i{ }^}{a_{i^ j^}}=\min \left{\frac{b_i}{a_{i j^}} \mid a_{i j^*}>0\right} 。\frac{b_i{ }^}{a_{i^ j^}}=\min \left{\frac{b_i}{a_{i j^}} \mid a_{i j^*}>0\right} 。
定理 6.3.1 根据 Bland 规则，单纯形算法不会陷入循环，因此在有限多次迭代内，算法能够确定最优值是否趋于无穷大，如果最优值是有限的，则算法会得到一个最优解。

## 计算机代写|组合优化代写Combinatorial optimization代考|Initial Feasible Basis

(1)成本函数值在减少到 0 并且所有人工变量都从可行基础中删除。在这种情况下，最终可行基可以用作原始 LP 中的初始可行基。
(2)成本函数达到负最大值。在这种情况下，原来的 LP 没有可行的解决方案。
(3)成本函数值在减少到 0 ; 但是，有一个人为变量是一世在可行的基础上。让b一世和一个一世j表示最后时刻的约束系数。在这种情况下，我们必须有是一世=b一世=0; 否则，在=和是>0. 请注意，存在一个变量Xj这样一个一世j≠0自从秩⁡(一个)=米. 这意味着我们可以采取一个一世j作为枢轴元素移动是一世从可行的基础上移出并搬入Xj，保留成本函数值 0 。当所有人工变量都从可行基中移出时，这种情况就简化为情况（1）。

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