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<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{3. } \left( \text{20 pts} \right) \text{ Solve the following linear program, }</math></span></font> | <font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}\text{3. } \left( \text{20 pts} \right) \text{ Solve the following linear program, }</math></span></font> | ||
| − | <span class="texhtml">maximize | + | <span class="texhtml">maximize − ''x''<sub>1</sub> − 3''x''<sub>2</sub> + 4''x''<sub>3</sub></span><br> |
| − | <span class="texhtml"><sub></sub></span>subject to < | + | <span class="texhtml"><sub></sub></span>subject to <span class="texhtml">''x''<sub>1</sub> + 2''x''<sub>2</sub> − ''x''<sub>3</sub> = 5</span> |
| − | < | + | <span class="texhtml"> 2''x''<sub>1</sub> + 3''x''<sub>2</sub> − ''x''<sub>3</sub> = 6</span> |
| − | <math>x_{1} \text{ free, } x_{2}\geq0, x_{3}\leq0.</math> | + | <math>x_{1} \text{ free, } x_{2}\geq0, x_{3}\leq0.</math> |
===== <math>\color{blue}\text{Solution 1:}</math> ===== | ===== <math>\color{blue}\text{Solution 1:}</math> ===== | ||
| − | <math>\Rightarrow x_{1}=5-2x_{2}+x_{3}=3-\frac{3}{2}x_{2}+\frac{1}{2}x_{3}</math> | + | <math>\Rightarrow x_{1}=5-2x_{2}+x_{3}=3-\frac{3}{2}x_{2}+\frac{1}{2}x_{3}</math> |
| − | <math>\Rightarrow x_{2}-x_{3}=4</math> | + | <math>\Rightarrow x_{2}-x_{3}=4</math> |
| + | |||
| + | It is equivalent to <math>min x_{1}+3x_{2}-4x_{3} | ||
| + | =5-2x_{2}+x_{3}+3x_{2}-4x_{3} = x_{2}-3x_{3}+5, x_{2}\geq0, x_{3}\leq0</math> | ||
| + | |||
| + | <math>x_{2}-3x_{3}+5 = x_{2}-x_{3}-2x_{3}+5=9-2x_{3}\geq9</math> | ||
| + | |||
| + | Equicalently, <math>-x_{1}-3x_{2}+4x_{3}\leq-9</math> | ||
| + | |||
| + | Equality is satisfied when <math>x_{3}=0, x_{2} =4, x_{1}=5-2\times4=-3</math> | ||
| + | |||
| + | <math>\Rightarrow \left\{\begin{matrix} | ||
| + | x_{1}=-3\\ | ||
| + | x_{2}=4\\ | ||
| + | x_{3}=0 | ||
| + | \end{matrix}\right.</math> | ||
| + | |||
| + | |||
| + | <br> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | [[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]] | ||
| + | |||
| + | [[Category:ECE]] [[Category:QE]] [[Category:Automatic_Control]] [[Category:Problem_solving]] | ||
Revision as of 13:17, 27 June 2012
ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 3, August 2011
$ \color{blue}\text{3. } \left( \text{20 pts} \right) \text{ Solve the following linear program, } $
maximize − x1 − 3x2 + 4x3
subject to x1 + 2x2 − x3 = 5
2x1 + 3x2 − x3 = 6
$ x_{1} \text{ free, } x_{2}\geq0, x_{3}\leq0. $
$ \color{blue}\text{Solution 1:} $
$ \Rightarrow x_{1}=5-2x_{2}+x_{3}=3-\frac{3}{2}x_{2}+\frac{1}{2}x_{3} $
$ \Rightarrow x_{2}-x_{3}=4 $
It is equivalent to $ min x_{1}+3x_{2}-4x_{3} =5-2x_{2}+x_{3}+3x_{2}-4x_{3} = x_{2}-3x_{3}+5, x_{2}\geq0, x_{3}\leq0 $
$ x_{2}-3x_{3}+5 = x_{2}-x_{3}-2x_{3}+5=9-2x_{3}\geq9 $
Equicalently, $ -x_{1}-3x_{2}+4x_{3}\leq-9 $
Equality is satisfied when $ x_{3}=0, x_{2} =4, x_{1}=5-2\times4=-3 $
$ \Rightarrow \left\{\begin{matrix} x_{1}=-3\\ x_{2}=4\\ x_{3}=0 \end{matrix}\right. $
