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</math><br> | </math><br> | ||
Such that <math>x^T=[\dfrac{6}{5}, \dfrac{8}{5},\dfrac{1}{5}, 0, 0]</math> is a feasible solution. | Such that <math>x^T=[\dfrac{6}{5}, \dfrac{8}{5},\dfrac{1}{5}, 0, 0]</math> is a feasible solution. | ||
| + | |||
| + | ---- | ||
| + | ===Similar Problem=== | ||
| + | [https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob3] | ||
---- | ---- | ||
Revision as of 11:31, 25 February 2019
Automatic Control (AC)
Question 3: Optimization
August 2016 Problem 1
Solution
The problem equal to:
Minimize $ 2x_1+x_2 $
Subject to $ \begin{align*} &x_1+3x_2-x_3=6\\ &2x_1+x_2-x_4=4\\ &x_1+x_2+x_5=3\\ &x_1, x_2, x_3, x_4,x_5 >=0 \end{align*} $
such that $ A= \begin{bmatrix} 1 & 3 & -1 & 0 & 0 \\ 2 & 1 & 0 & -1 & 0 \\ 1 & 1 & 0 & 0 & 1 \end{bmatrix} $
we take $ B= \begin{bmatrix} 1 & 3 & 0 \\ 2 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \Rightarrow B\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =b \Rightarrow \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 0 \\ 2 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix}^{-1} \begin{bmatrix} 6\\ 4\\ 3 \end{bmatrix} = \begin{bmatrix} \dfrac{6}{5} \\ \dfrac{8}{5} \\ \dfrac{1}{5} \end{bmatrix} $
Such that $ x^T=[\dfrac{6}{5}, \dfrac{8}{5},\dfrac{1}{5}, 0, 0] $ is a feasible solution.
