| Line 13: | Line 13: | ||
<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>\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}, | + | <math>x_{1},x_{2}\geq0.</math> |
---- | ---- | ||
| Line 33: | Line 33: | ||
5. Select a <span class="texhtml">''q''</span> such that <span class="texhtml">''r''<sub>''q''</sub> < 0</span> | 5. Select a <span class="texhtml">''q''</span> such that <span class="texhtml">''r''<sub>''q''</sub> < 0</span> | ||
| − | 6. If no <span class="texhtml">''y''<sub>'' | + | 6. If no <span class="texhtml">''y''<sub>''iq''</sub> > 0</span>, stop. -- the problem is unbounded; else, calculate ''' <math>p=argmin_{i}\left \{ y_{i0}/y_{iq}:y_{iq}>0 \right \}</math>. ''' |
7. Update the canonical augmented matrix by pivoting about the <span class="texhtml">(''p'',''q'')</span> th element. | 7. Update the canonical augmented matrix by pivoting about the <span class="texhtml">(''p'',''q'')</span> th element. | ||
| Line 80: | Line 80: | ||
<math>\begin{matrix} | <math>\begin{matrix} | ||
| − | & | + | & 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 \\ | ||
| Line 106: | Line 106: | ||
The maximum value for <font face="serif"><span class="texhtml"> ''x''<sub>1</sub> + ''x''<sub>2</sub> is 6.</span><br></font> | The maximum value for <font face="serif"><span class="texhtml"> ''x''<sub>1</sub> + ''x''<sub>2</sub> is 6.</span><br></font> | ||
| + | |||
| + | ---- | ||
| + | |||
| + | <font face="serif"><span class="texhtml" /></font><math>\color{blue}\text{Related Problem: Solve the following problem using simplex method}</math> | ||
| + | |||
| + | min <math>2x_{1}+3x_{2}</math> | ||
| + | |||
| + | subject to <math>2x_{1}+x_{2}\leq4</math> | ||
| + | |||
| + | <math>x_{1}+x_{2}\leq3</math> | ||
| + | |||
| + | <math>x_{1},x_{2}\geq0.</math> | ||
| + | |||
| + | <math>\color{blue}\text{Solution:}</math> | ||
| + | |||
| + | Transform to standard form: min <math>-2x_{1}-3x_{2}</math> | ||
| + | |||
| + | subject to <math>2x_{1}+x_{2}+x_{3}=4</math> | ||
| + | |||
| + | <math>x_{1}+x_{2}+x_{4}=3</math> | ||
| + | |||
| + | <math>x_{i}\geq0, i=1,2,3,4</math> | ||
| + | |||
| + | <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 <math>r_{1}=-2<0</math> and <math>r_{2}=-3<0</math>. We introduce <math>a_{2}</math> into the new basis and pivot <math>y_{22}</math>, by calculating the ratios <math>y_{i0}/y_{i2},y_{i2}>0</math>.<sub></sub> | ||
| + | |||
| + | <math>\begin{matrix} | ||
| + | x_{1} & x_{2} & x_{3} & x_{4} & b\\ | ||
| + | 2 & 1 & 1 & 0 & 4\\ | ||
| + | 1 & 1 & 0 & 1 & 3 \\ | ||
| + | 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 | ||
| + | |||
| + | <math>x^{*}=\begin{bmatrix} | ||
| + | 0 & | ||
| + | 3 & | ||
| + | 1 & | ||
| + | 0 | ||
| + | \end{bmatrix}^{T}.</math> | ||
| + | |||
| + | The optimal solution to the original problem is <math>x^{*}=\begin{bmatrix} | ||
| + | 0 & | ||
| + | 3 | ||
| + | \end{bmatrix}^{T}.</math> and the optimal objective value is 9. | ||
| + | |||
| + | |||
---- | ---- | ||
| Line 113: | Line 167: | ||
Go to | Go to | ||
| − | *Part 1: [[ECE-QE AC3-2011 solusion-1|solutions and discussions]] | + | *Part 1: [[ECE-QE AC3-2011 solusion-1|solutions and discussions]] |
| − | *Part 2: [[ECE- | + | *Part 2: [[ECE-QE_AC3-2011_solusion-2|solutions and discussions]] |
*Part 3: [[ECE-QE AC3-2011 solusion-3|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 4: [[ECE-QE AC3-2011 solusion-4|solutions and discussions]] | ||
Revision as of 20:23, 28 June 2012
Contents
ECE Ph.D. Qualifying Exam in "Automatic Control" (AC)
Question 3, Part 2, August 2011
$ \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.
<span class="texhtml" />$ \color{blue}\text{Related Problem: Solve the following problem using simplex method} $
min $ 2x_{1}+3x_{2} $
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 $ -2x_{1}-3x_{2} $
subject to $ 2x_{1}+x_{2}+x_{3}=4 $
$ x_{1}+x_{2}+x_{4}=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 $ r_{1}=-2<0 $ and $ r_{2}=-3<0 $. We introduce $ a_{2} $ into the new basis and pivot $ y_{22} $, by calculating the ratios $ y_{i0}/y_{i2},y_{i2}>0 $.
$ \begin{matrix} x_{1} & x_{2} & x_{3} & x_{4} & b\\ 2 & 1 & 1 & 0 & 4\\ 1 & 1 & 0 & 1 & 3 \\ 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
Go to
- Part 1: solutions and discussions
- Part 2: solutions and discussions
- Part 3: solutions and discussions
- Part 4: solutions and discussions
- Part 5: solutions and discussions
