Posted on Categories:Operations Research, 数学代写, 运筹学

# 数学代写|运筹学代写Operations Research代考|KMA255 SIMPLEX METHOD SOLVES BOTH THE PRIMAL AND THE DUAL

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|运筹学代写Operations Research代考|SIMPLEX METHOD SOLVES BOTH THE PRIMAL AND THE DUAL

Let us explain this using the formulation in Illustration 1.1. The simplex table is shown in the usual notation in Table 3.3.

In the optimal tableau, let us observe the $C_j-Z_j$ values. These are $0,0,-3$ and $-1$ for variables $X_1, X_2, u_1$ and $u_2$. We also know (from complimentary slackness conditions) that there is a relationship between $X_j$ and $v_j$ and between $u_j$ and $Y_j$.

The values of $C_j-Z_j$ corresponding to $X_j$ are the negatives of the values of dual slack variables $v_j$ and the values of $C_j-Z_j$ corresponding to $u_j$ are the negatives of the values of dual decision variables $Y_j$.
(We will see the algebraic explanation of this subsequently in this chapter.)
Therefore, $Y_1^=3, Y_2^=1, v_1^=0, v_2^=0$. We also know from the complimentary slackness conditions that if $X_j$ is basic then $v_j=0$. This is also true because when $X_j$ is basic, $C_j-Z_j$ is zero. When $u_j$ is non-basic and has zero value, its $C_j-Z_j$ is negative at optimum indicating a nonnegative value of the basic variable $Y_j$.

The optimum solution to the dual can be read from the optimum tableau of the primal in the simplex algorithm. We need not solve the dual explicitly.

Let us look at an intermediate iteration (say with basic variable $u_1$ and $X_1$ ). The basic feasible solution is $u_1=1, X_1=4$ with non-basic variables $X_2=0$ and $u_2=0$. When we apply the above rule (and complimentary slackness conditions), we get the corresponding dual solution to be $Y_1=0, Y_2$ $=2, v_1=0, v_2=-1$ with $W=24$ (same value of $Z$ ).

This solution is infeasible to the dual because variable $v_2$ takes a negative value. This means that the second dual constraint $Y_1+2 Y_2 \geq 5$ is not feasible making $v_2=-1$. The value of $v_2$ is the extent of infeasibility of the dual, which is the rate at which the objective function can increase by entering the corresponding primal variable. A non-optimal basic feasible solution to the primal results in an infeasible dual when complimentary slackness conditions are applied. At the optimum, when complimentary slackness conditions are applied, the resultant solution is also feasible and hence optimal.

## 数学代写|运筹学代写Operations Research代考|THE DUAL SIMPLEX ALGORITHM

Consider the linear programming problem given by
ILLUSTRATION $3.4$
Minimize $Z=4 X_1+7 X_2$
Subject to
$$\begin{array}{r} 2 X_1+3 X_2 \geq 5 \ X_1+7 X_2 \geq 9 \ X_1, X_2 \geq 0 \end{array}$$
Normally, we would have added two artificial variables $a_1$ and $a_2$ to get an initial basic feasible solution. We do not add these now but write the constraints as equations with slack variables only. The equations are written with a negative RHS (something that the simplex method does not approve). We also convert it as a maximization problem by multiplying the objective function with $-1$.
The problem becomes
Maximize $Z=-4 X_1-7 X_2-0 X_3-0 X_4$
Subject to
$$\begin{array}{r} 2 X_1+3 X_2-X_3=5 \ X_1+7 X_2-X_4=9 \ X_1, X_2 \geq 0 \end{array}$$
We set up the simplex table as shown in Table $3.4$ with slack variables $X_3$ and $X_4$ as basic variables and with a negative RHS.