Revision as of 13:46, 27 June 2012

ECE Ph.D. Qualifying Exam: Automatic Control (AC)- Question 3, August 2011

$\color{blue}\text{3. } \left( \text{20 pts} \right) \text{ Solve the following linear program, }$

$maximize - x 1 - 3 x 2 + 4 x 3$

subject to $x_{1}+2x_{2}-x_{3}=5$

$2x_{1}+3x_{2}-x_{3}=6$

$x_{1} \text{ free, } x_{2}\geq0, x_{3}\leq0.$

$\Rightarrow x_{1}=5-2x_{2}+x_{3}=3-\frac{3}{2}x_{2}+\frac{1}{2}x_{3}$

$\Rightarrow x_{2}-x_{3}=4$

@@ Line 7: / Line 7: @@
 &nbsp;<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>
-<math>\text{maximize }  -x_{1}-3x_{2}+4x_{3}</math><math>\text{maximize } -x_{1}-3x_{2}+4x_{3}</math><br>
+<span class="texhtml">maximize  − ''x''<sub>1</sub> − 3''x''<sub>2</sub> + 4''x''<sub>3</sub></span><br>
-===== <math>\color{blue}\text{Solution 1:}</math> =====
+<span class="texhtml"><sub></sub></span>subject to &nbsp;<math>x_{1}+2x_{2}-x_{3}=5</math>
+<math>2x_{1}+3x_{2}-x_{3}=6</math>
+<math>x_{1} \text{ free, } x_{2}\geq0, x_{3}\leq0.</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_{2}-x_{3}=4</math>