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[[Category:ECE]]
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[[Category:QE]]
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[[Category:CNSIP]]
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[[Category:problem solving]]
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[[Category:automatic control]]
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[[Category:optimization]]
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= [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]] in "Automatic Control" (AC)  =
 +
 
 +
=  [[ECE-QE_AC3-2011|Question 3, August 2011]], Part 2  =
  
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= [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]]: Automatic Control (AC)- Question 3, August 2011 =
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:[[ECE-QE AC3-2011 solusion-1|Part 1]],[[ECE-QE_AC3-2011_solusion-2|2]],[[ECE-QE AC3-2011 solusion-3|3]],[[ECE-QE AC3-2011 solusion-4|4]],[[ECE-QE AC3-2011 solusion-5|5]]
  
 
----
 
----
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&nbsp;<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>  
 
&nbsp;<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>  
  
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class="texhtml">maximize''x''<sub>1</sub> + ''x''<sub>2</sub></span>  
+
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<span class="texhtml">maximize &nbsp; &nbsp; &nbsp; &nbsp;''x''<sub>1</sub> + ''x''<sub>2</sub></span>  
  
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{subject to  }  x_{1}-x_{2}\leq2</math><font color="#ff0000" face="serif" size="4"><br></font>'''&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1}+x_{2}\leq6</math>'''
+
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\text{subject to  }  x_{1}-x_{2}\leq2</math><font color="#ff0000" face="serif" size="4"><br></font>'''&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1}+x_{2}\leq6</math>''' &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;  
−
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1},-x_{2}\geq0.</math>
+
  
−
===== <math>\color{blue}\text{Solution 1:}</math> =====
+
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1},x_{2}\geq0.</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>
+
'''Theorem:&nbsp;'''
  
−
<math>x_{1}+x_{2}+x_{4}=6</math>
+
A basic feasible solution is optimal if and only if the corresponding reduced cost coefficeints are all nonnegative.
  
−
<math>x_{i}\geq0    i=1,2,3,4</math>
+
'''Simplex Method:'''
  
 +
1. Transform the given problem into standard form by introducing slack variables&nbsp;<span class="texhtml">''x''<sub>3</sub></span>&nbsp;and&nbsp;<span class="texhtml">''x''<sub>4</sub></span>.<span class="texhtml"><sub></sub></span>
  
 +
2. Form a canonical augmented matrix corresponding to an initial basic feasible solution.
 +
 +
3. Calculate the reduced cost coefficients corresponding to the nonbasic variables.
 +
 +
4. If&nbsp;<math>r_{j}>\geq0</math>&nbsp;for all&nbsp;<span class="texhtml">''j''</span>, stop. -- the current basic feasible solution is optimal.
 +
 +
5. Select a&nbsp;<span class="texhtml">''q''</span>&nbsp;such that&nbsp;<span class="texhtml">''r''<sub>''q''</sub> &lt; 0</span>
 +
 +
6. If no&nbsp;<span class="texhtml">''y''<sub>''iq''</sub> &gt; 0</span>, stop. -- the problem is unbounded; else, calculate&nbsp;'''&nbsp;<math>p=argmin_{i}\left \{ y_{i0}/y_{iq}:y_{iq}>0 \right \}</math>.&nbsp; '''
 +
 +
7. Update the canonical augmented matrix by pivoting about the&nbsp;<span class="texhtml">(''p'',''q'')</span>&nbsp;th element.
 +
 +
8. Go to step 3.
 +
 +
----
 +
 +
<math>\color{blue}\text{Solution 1:}</math>
 +
 +
<span class="texhtml">&nbsp; &nbsp;min&nbsp;''&nbsp;&nbsp;'' − ''x''<sub>1</sub> − ''x''<sub>2</sub></span>&nbsp;<br> <span class="texhtml">&nbsp; &nbsp;subject to &nbsp; &nbsp;''x''<sub>1</sub> − ''x''<sub>2</sub> + ''x''<sub>3</sub> = 2</span>&nbsp;<br> <span class="texhtml">''&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; x''<sub>1</sub> + ''x''<sub>2</sub> + ''x''<sub>4</sub> = 6</span>&nbsp;
 +
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1},x_{2},x_{3},x_{4}\geq 0</math>
 +
 +
<math>\begin{matrix}
 +
1 & -1 & 1 & 0 & 2\\
 +
1 & 1 & 0 & 1 & 6 \\
 +
-1 & -1 & 0 & 0 & 0
 +
\end{matrix}
 +
\Rightarrow
 +
\begin{matrix}
 +
1 & -1 & 1 & 0 & 2\\
 +
0 & 2 & -1 & 1 & 4 \\
 +
0 & -2 & 1 & 0 & 2
 +
\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>\Rightarrow x_{1}=4, x_{2}=2, \text{the maximum value } x_{1}+x_{2}=6</math>
 +
 +
----
 +
 +
<math>\color{blue}\text{Solution 2:}</math>
 +
 +
<span class="texhtml">Get standard form for simplex method &nbsp; min&nbsp;''&nbsp;&nbsp;'' − ''x''<sub>1</sub> − ''x''<sub>2</sub></span>
 +
 +
<span class="texhtml">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;subject to &nbsp; &nbsp;''x''<sub>1</sub> − ''x''<sub>2</sub> + ''x''<sub>3</sub> = 2</span>
 +
 +
<span class="texhtml">''&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;x''<sub>1</sub> + ''x''<sub>2</sub> + ''x''<sub>4</sub> = 6</span>
 +
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>x_{i}\geq0,    i=1,2,3,4</math><br>
  
 
<math>\begin{matrix}
 
<math>\begin{matrix}
−
  & a_{1} & a_{2} & a_{3} & a_{4} & b\\  
+
  & x_{1} & x_{2} & x_{3} & x_{4} & b\\  
 
  & 1 & -1 & 1 & 0 & 2\\  
 
  & 1 & -1 & 1 & 0 & 2\\  
 
  & 1 & 1 & 0 & 1 & 6 \\  
 
  & 1 & 1 & 0 & 1 & 6 \\  
 
c^{T} & -1 & -1 & 0 & 0 & 0
 
c^{T} & -1 & -1 & 0 & 0 & 0
−
\end{matrix}
+
\end{matrix}</math>&nbsp; &nbsp; &nbsp; <math>\Rightarrow
−
\Rightarrow
+
 
\begin{matrix}
 
\begin{matrix}
 
  1 & -1 & 1 & 0 & 2\\  
 
  1 & -1 & 1 & 0 & 2\\  
 
  1 & 1 & 0 & 1 & 6 \\  
 
  1 & 1 & 0 & 1 & 6 \\  
 
  0 & 0 & 0 & 1 & 6
 
  0 & 0 & 0 & 1 & 6
−
\end{matrix}</math>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;
+
\end{matrix}
−
 
+
\Rightarrow
−
<math>\Rightarrow
+
 
\begin{matrix}
 
\begin{matrix}
 
  1 & -1 & 1 & 0 & 2\\  
 
  1 & -1 & 1 & 0 & 2\\  
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  0 & 1 & -\frac{1}{2} & \frac{1}{2} & 2 \\  
 
  0 & 1 & -\frac{1}{2} & \frac{1}{2} & 2 \\  
 
  0 & 0 & 0 & 1 & 6
 
  0 & 0 & 0 & 1 & 6
−
\end{matrix}</math>
+
\end{matrix}</math><font color="#ff0000"><br></font>  
  
 +
<math>\therefore \text{the optimal solution to the original problem is } x^{*}=  \begin{bmatrix} 4\\ 2 \end{bmatrix}</math><font face="serif" color="#ff0000" style="font-size: 17px;">'''<br>'''</font>
  
 +
The maximum value for &nbsp;<font face="serif"><span class="texhtml">&nbsp;''x''<sub>1</sub> + ''x''<sub>2</sub> is 6.</span><br></font>
  
−
<math>\therefore \text{the optimal solution to the original problem is } x^{*}=\begin{bmatrix}
+
----
−
4\\ 2
+
----
−
\end{bmatrix}}</math>
+
<math>\color{blue}\text{Related Problem: Solve the following problem using simplex method}</math>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; max &nbsp;<span class="texhtml">2''x''<sub>1</sub> + 3''x''<sub>2</sub></span>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; subject to &nbsp;<math>2x_{1}+x_{2}\leq4</math>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1}+x_{2}\leq3</math>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{1},x_{2}\geq0.</math>
 +
 
 +
<math>\color{blue}\text{Solution:}</math>
 +
 
 +
Transform to standard form: &nbsp;min &nbsp;<span class="texhtml"> − 2''x''<sub>1</sub> − 3''x''<sub>2</sub></span>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;subject to &nbsp;&nbsp;<span class="texhtml">2''x''<sub>1</sub> + ''x''<sub>2</sub> + ''x''<sub>3</sub> = 4</span>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<span class="texhtml">''x''<sub>1</sub> + ''x''<sub>2</sub> + ''x''<sub>4</sub> = 3</span>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>x_{i}\geq0,    i=1,2,3,4</math>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;<math>\begin{matrix}
 +
& x_{1} & x_{2} & x_{3} & x_{4} & b\\
 +
& 2 & 1 & 1 & 0 & 4\\
 +
& 1 & 1 & 0 & 1 & 3 \\
 +
c^{T} & -2 & -3 & 0 & 0 & 0
 +
\end{matrix}</math>
 +
 
 +
We have&nbsp;<span class="texhtml">''r''<sub>1</sub> =  − 2 &lt; 0</span>&nbsp; and &nbsp;<span class="texhtml">''r''<sub>2</sub> =  − 3 &lt; 0</span>. &nbsp;We introduce&nbsp;<span class="texhtml">''a''<sub>2</sub></span>&nbsp;into the new basis and pivot&nbsp;<span class="texhtml">''y''<sub>22</sub></span>, by calculating the ratios&nbsp;<span class="texhtml">''y''<sub>''i''0</sub> / ''y''<sub>''i''2</sub>,''y''<sub>''i''2</sub> &gt; 0</span>.<sub></sub>
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{matrix}
 +
& x_{1} & x_{2} & x_{3} & x_{4} & b\\
 +
& 2 & 1 & 1 & 0 & 4\\
 +
& 1 & 1 & 0 & 1 & 3 \\
 +
c^{T} & 1 & 0 & 0 & 3 & 9
 +
\end{matrix} \Rightarrow
 +
\begin{matrix}
 +
& x_{1} & x_{2} & x_{3} & x_{4} & b\\
 +
& 1 & 0 & 1 & -1 & 1\\
 +
& 1 & 1 & 0 & 1 & 3 \\
 +
c^{T} & 1 & 0 & 0 & 3 & 9
 +
\end{matrix}</math>
 +
 
 +
All the reduced cost coefficients are positive, hence the optimal solution to the problem in standard form is  
 +
 
 +
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;<math>x^{*}=\begin{bmatrix}
 +
0 &
 +
3 &
 +
1 &
 +
0
 +
\end{bmatrix}^{T}.</math>
 +
 
 +
The optimal solution to the original problem is&nbsp;<math>x^{*}=\begin{bmatrix}
 +
0 &
 +
3
 +
\end{bmatrix}^{T}.</math>&nbsp;and the optimal objective value is 9.<br>
 +
 
 +
<br>
 +
 
 +
----
 +
 
 +
Automatic Control (AC)- Question 3, August 2011
 +
 
 +
Go to
 +
 
 +
*Part 1: [[ECE-QE AC3-2011 solusion-1|solutions and discussions]]
 +
*Part 2: [[ECE-QE_AC3-2011_solusion-2|solutions and discussions]]
 +
*Part 3: [[ECE-QE AC3-2011 solusion-3|solutions and discussions]]
 +
*Part 4: [[ECE-QE AC3-2011 solusion-4|solutions and discussions]]
 +
*Part 5: [[ECE-QE AC3-2011 solusion-5|solutions and discussions]]
 +
 
 +
----
  
−
<math>\text{The maximum value for } x_{1}+x-{2} \text{ is } 6</math>
+
[[ECE PhD Qualifying Exams|Back to ECE Qualifying Exams (QE) page]]

Latest revision as of 09:10, 13 September 2013


ECE Ph.D. Qualifying Exam in "Automatic Control" (AC)

Question 3, August 2011, Part 2

Part 1,2,3,4,5

 $ \color{blue}\text{2. } \left( \text{20 pts} \right) \text{ Use the simplex method to solve the problem, } $

               maximize        x1 + x2

               $ \text{subject to } x_{1}-x_{2}\leq2 $
                                        $ x_{1}+x_{2}\leq6 $                                         

                                        $ x_{1},x_{2}\geq0. $

Theorem: 

A basic feasible solution is optimal if and only if the corresponding reduced cost coefficeints are all nonnegative.

Simplex Method:

1. Transform the given problem into standard form by introducing slack variables x3 and x4.

2. Form a canonical augmented matrix corresponding to an initial basic feasible solution.

3. Calculate the reduced cost coefficients corresponding to the nonbasic variables.

4. If $ r_{j}>\geq0 $ for all j, stop. -- the current basic feasible solution is optimal.

5. Select a q such that rq < 0

6. If no yiq > 0, stop. -- the problem is unbounded; else, calculate  $ p=argmin_{i}\left \{ y_{i0}/y_{iq}:y_{iq}>0 \right \} $. 

7. Update the canonical augmented matrix by pivoting about the (p,q) th element.

8. Go to step 3.

$ \color{blue}\text{Solution 1:} $

   min    − x1 − x2 
   subject to    x1 − x2 + x3 = 2 
                      x1 + x2 + x4 = 6 

                      $ x_{1},x_{2},x_{3},x_{4}\geq 0 $

$ \begin{matrix} 1 & -1 & 1 & 0 & 2\\ 1 & 1 & 0 & 1 & 6 \\ -1 & -1 & 0 & 0 & 0 \end{matrix} \Rightarrow \begin{matrix} 1 & -1 & 1 & 0 & 2\\ 0 & 2 & -1 & 1 & 4 \\ 0 & -2 & 1 & 0 & 2 \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} $

$ \Rightarrow x_{1}=4, x_{2}=2, \text{the maximum value } x_{1}+x_{2}=6 $

$ \color{blue}\text{Solution 2:} $

Get standard form for simplex method   min    − x1 − x2

                                                           subject to    x1 − x2 + x3 = 2

                                                                             x1 + x2 + x4 = 6

                                                                             $ x_{i}\geq0, i=1,2,3,4 $

$ \begin{matrix} & x_{1} & x_{2} & x_{3} & x_{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} $

The maximum value for   x1 + x2 is 6.

$ \color{blue}\text{Related Problem: Solve the following problem using simplex method} $

                      max  2x1 + 3x2

            subject to  $ 2x_{1}+x_{2}\leq4 $

                              $ x_{1}+x_{2}\leq3 $

                              $ x_{1},x_{2}\geq0. $

$ \color{blue}\text{Solution:} $

Transform to standard form:  min   − 2x1 − 3x2

                                       subject to   2x1 + x2 + x3 = 4

                                                          x1 + x2 + x4 = 3

                                                          $ x_{i}\geq0, i=1,2,3,4 $

              $ \begin{matrix} & x_{1} & x_{2} & x_{3} & x_{4} & b\\ & 2 & 1 & 1 & 0 & 4\\ & 1 & 1 & 0 & 1 & 3 \\ c^{T} & -2 & -3 & 0 & 0 & 0 \end{matrix} $

We have r1 = − 2 < 0  and  r2 = − 3 < 0.  We introduce a2 into the new basis and pivot y22, by calculating the ratios yi0 / yi2,yi2 > 0.

               $ \begin{matrix} & x_{1} & x_{2} & x_{3} & x_{4} & b\\ & 2 & 1 & 1 & 0 & 4\\ & 1 & 1 & 0 & 1 & 3 \\ c^{T} & 1 & 0 & 0 & 3 & 9 \end{matrix} \Rightarrow \begin{matrix} & x_{1} & x_{2} & x_{3} & x_{4} & b\\ & 1 & 0 & 1 & -1 & 1\\ & 1 & 1 & 0 & 1 & 3 \\ c^{T} & 1 & 0 & 0 & 3 & 9 \end{matrix} $

All the reduced cost coefficients are positive, hence the optimal solution to the problem in standard form is

               $ x^{*}=\begin{bmatrix} 0 & 3 & 1 & 0 \end{bmatrix}^{T}. $

The optimal solution to the original problem is $ x^{*}=\begin{bmatrix} 0 & 3 \end{bmatrix}^{T}. $ and the optimal objective value is 9.


Automatic Control (AC)- Question 3, August 2011

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