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# 数学代写|运筹学代写Operations Research代考|Determining the Direction of Movement

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## 数学代写|运筹学代写Operations Research代考|Determining the Direction of Movement (Step 1 of an Iteration)

Increasing one nonbasic variable from zero (while adjusting the values of the basic variables to continue satisfying the system of equations) corresponds to moving along one edge emanating from the current $\mathrm{CPF}$ solution. Based on solution concepts 4 and 5 in Sec. 4.1 , the choice of which nonbasic variable to increase is made as follows:
$$Z=3 x_1+5 x_2$$
Increase $x_1$ ? Rate of improvement in $Z=3$.
Increase $x_2$ ? Rate of improvement in $Z=5$.
$5>3$, so choose $x_2$ to increase.
As indicated next, we call $x_2$ the entering basic variable for iteration 1 .
At any iteration of the simplex method, the purpose of step 1 is to choose one nonbasic variable to increase from zero (while the values of the basic variables are adjusted to continue satisfying the system of equations). Increasing this nonbasic variable from zero will convert it to a basic variable for the next BF solution. Therefore, this variable is called the entering basic variable for the current iteration (because it is entering the basis).

## 数学代写|运筹学代写Operations Research代考|Determining Where to Stop (Step 2 of an Iteration)

Step 2 addresses the question of how far to increase the entering basic variable $x_2$ before stopping. Increasing $x_2$ increases $Z$, so we want to go as far as possible without leaving the feasible region. The requirement to satisfy the functional constraints in augmented form (shown below) means that increasing $x_2$ (while keeping the nonbasic variable $x_1=0$ ) changes the values of some of the basic variables as shown on the right.
\begin{aligned} & \text { (1) } x_1+x_3=4 \ & x_1=0, \quad \text { so } \ & \text { (2) } 2 x_2+x_4=12 \ & x_3=4 \ & x_4=12-2 x_2 \ & \text { (3) } 3 x_1+2 x_2+x_5=18 \ & x_5=18-2 x_2 \text {. } \ & \end{aligned}
The other requirement for feasibility is that all the variables be nonnegative. The nonbasic variables (including the entering basic variable) are nonnegative, but we need to check how far $x_2$ can be increased without violating the nonnegativity constraints for the basic variables.
\begin{aligned} & \boldsymbol{x}_3=4 \geq 0 \quad \Rightarrow \text { no upper bound on } x_2 . \ & \boldsymbol{x}_4=12-2 x_2 \geq 0 \Rightarrow x_2 \leq \frac{12}{2}=6 \quad \leftarrow \text { minimum. } \ & \boldsymbol{x}_5=18-2 x_2 \geq 0 \Rightarrow x_2 \leq \frac{18}{2}=9 . \end{aligned}
Thus, $x_2$ can be increased just to 6 , at which point $\boldsymbol{x}_4$ has dropped to 0 . Increasing $x_2$ beyond 6 would cause $x_4$ to become negative, which would violate feasibility.

These calculations are referred to as the minimum ratio test. The objective of this test is to determine which basic variable drops to zero first as the entering basic variable is increased. We can immediately rule out the basic variable in any equation where the coefficient of the entering basic variable is zero or negative, since such a basic variable would not decrease as the entering basic variable is increased. [This is what happened with $x_3$ in Eq. (1) of the example.] However, for each equation where the coefficient of the entering basic variable is strictly positive $(>0)$, this test calculates the ratio of the right-hand side to the coefficient of the entering basic variable. The basic variable in the equation with the minimum ratio is the one that drops to zero first as the entering basic variable is increased.
At any iteration of the simplex method, step 2 uses the minimum ratio test to determine which basic variable drops to zero first as the entering basic variable is increased. Decreasing this basic variable to zero will convert it to a nonbasic variable for the next $\mathrm{BF}$ solution. Therefore, this variable is called the leaving basic variable for the current iteration (because it is leaving the basis).
Thus, $\boldsymbol{x}_4$ is the leaving basic variable for iteration 1 of the example.

## 数学代写|运筹学代写Operations Research代考|Determining the Direction of Movement (Step 1 of an Iteration)

$$Z=3 x_1+5 x_2$$

$5>3$，所以选择$x_2$增加。

## 数学代写|运筹学代写Operations Research代考|Determining Where to Stop (Step 2 of an Iteration)

\begin{aligned} & \text { (1) } x_1+x_3=4 \ & x_1=0, \quad \text { so } \ & \text { (2) } 2 x_2+x_4=12 \ & x_3=4 \ & x_4=12-2 x_2 \ & \text { (3) } 3 x_1+2 x_2+x_5=18 \ & x_5=18-2 x_2 \text {. } \ & \end{aligned}

\begin{aligned} & \boldsymbol{x}_3=4 \geq 0 \quad \Rightarrow \text { no upper bound on } x_2 . \ & \boldsymbol{x}_4=12-2 x_2 \geq 0 \Rightarrow x_2 \leq \frac{12}{2}=6 \quad \leftarrow \text { minimum. } \ & \boldsymbol{x}_5=18-2 x_2 \geq 0 \Rightarrow x_2 \leq \frac{18}{2}=9 . \end{aligned}

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