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<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{2. } \left( \text{20 pts} \right) \text{ Use the simplex method to solve the problem, }</math></span></font> | <font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{2. } \left( \text{20 pts} \right) \text{ Use the simplex method to solve the problem, }</math></span></font> | ||
| − | < | + | <span class="texhtml">maximize''x''<sub>1</sub> + ''x''<sub>2</sub></span> |
| − | <math>\text{subject to } x_{1}-x_{2}\leq2</math><font color="#ff0000" face="serif" size="4"><br></font> | + | <math>\text{subject to } x_{1}-x_{2}\leq2</math><font color="#ff0000" face="serif" size="4"><br></font>''' <math>x_{1}+x_{2}\leq6</math>''' |
| + | <math>x_{1},-x_{2}\geq0.</math> | ||
| − | + | ===== <math>\color{blue}\text{Solution 1:}</math> ===== | |
| − | === | + | <math>\text{Get standard form for simplex method } min -x_{1}-x_{2}</math> |
| + | |||
| + | <math>\text{subject to } x_{1}-x_{2}+x_{3}=2</math> | ||
| + | |||
| + | <math>x_{1}+x_{2}+x_{4}=6</math> | ||
| + | |||
| + | <math>x_{i}\geq0 i=1,2,3,4</math> | ||
| + | |||
| + | |||
| + | |||
| + | <math>\begin{matrix} | ||
| + | & a_{1} & a_{2} & a_{3} & a_{4} & b\\ | ||
| + | & 1 & -1 & 1 & 0 & 2\\ | ||
| + | & 1 & 1 & 0 & 1 & 6 \\ | ||
| + | c^{T} & -1 & -1 & 0 & 0 & 0 | ||
| + | \end{matrix} | ||
| + | \Rightarrow | ||
| + | \begin{matrix} | ||
| + | 1 & -1 & 1 & 0 & 2\\ | ||
| + | 1 & 1 & 0 & 1 & 6 \\ | ||
| + | 0 & 0 & 0 & 1 & 6 | ||
| + | \end{matrix}</math> | ||
| + | |||
| + | <math>\Rightarrow | ||
| + | \begin{matrix} | ||
| + | 1 & -1 & 1 & 0 & 2\\ | ||
| + | 0 & 2 & -1 & 1 & 4 \\ | ||
| + | 0 & 0 & 0 & 1 & 6 | ||
| + | \end{matrix} | ||
| + | \Rightarrow | ||
| + | \begin{matrix} | ||
| + | 1 & 0 & \frac{1}{2} & \frac{1}{2} & 4\\ | ||
| + | 0 & 1 & -\frac{1}{2} & \frac{1}{2} & 2 \\ | ||
| + | 0 & 0 & 0 & 1 & 6 | ||
| + | \end{matrix}</math> | ||
| + | |||
| + | |||
| + | |||
| + | <math>\therefore \text{the optimal solution to the original problem is } x^{*}=\begin{bmatrix} | ||
| + | 4\\ 2 | ||
| + | \end{bmatrix}}</math> | ||
| + | |||
| + | <math>\text{The maximum value for } x_{1}+x-{2} \text{ is } 6</math> | ||
Revision as of 11:43, 27 June 2012
ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 3, August 2011
$ \color{blue}\text{2. } \left( \text{20 pts} \right) \text{ Use the simplex method to solve the problem, } $
maximizex1 + x2
$ \text{subject to } x_{1}-x_{2}\leq2 $
$ x_{1}+x_{2}\leq6 $
$ x_{1},-x_{2}\geq0. $
$ \color{blue}\text{Solution 1:} $
$ \text{Get standard form for simplex method } min -x_{1}-x_{2} $
$ \text{subject to } x_{1}-x_{2}+x_{3}=2 $
$ x_{1}+x_{2}+x_{4}=6 $
$ x_{i}\geq0 i=1,2,3,4 $
$ \begin{matrix} & a_{1} & a_{2} & a_{3} & a_{4} & b\\ & 1 & -1 & 1 & 0 & 2\\ & 1 & 1 & 0 & 1 & 6 \\ c^{T} & -1 & -1 & 0 & 0 & 0 \end{matrix} \Rightarrow \begin{matrix} 1 & -1 & 1 & 0 & 2\\ 1 & 1 & 0 & 1 & 6 \\ 0 & 0 & 0 & 1 & 6 \end{matrix} $
$ \Rightarrow \begin{matrix} 1 & -1 & 1 & 0 & 2\\ 0 & 2 & -1 & 1 & 4 \\ 0 & 0 & 0 & 1 & 6 \end{matrix} \Rightarrow \begin{matrix} 1 & 0 & \frac{1}{2} & \frac{1}{2} & 4\\ 0 & 1 & -\frac{1}{2} & \frac{1}{2} & 2 \\ 0 & 0 & 0 & 1 & 6 \end{matrix} $
$ \therefore \text{the optimal solution to the original problem is } x^{*}=\begin{bmatrix} 4\\ 2 \end{bmatrix}} $
$ \text{The maximum value for } x_{1}+x-{2} \text{ is } 6 $
