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