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

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## 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.

## 数学代写|运筹学代写运筹学代考|SIMPLEX METHOD既解决了原始的，也解决了对子的

$X_j$对应的$C_j-Z_j$的值是双松弛变量$v_j$的负数，$u_j$对应的$C_j-Z_j$的值是双决策变量$Y_j$的负数
(我们将在本章后面看到对此的代数解释)

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