(2 intermediate revisions by the same user not shown)
Line 71: Line 71:
 
----
 
----
 
===Similar Problem===
 
===Similar Problem===
[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob3]
+
[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob1]<br>
 
+
[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob3]<br>
 +
[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2014/AC-3.pdf?dl=1 2014 QE AC3 Prob2]<br>
 
----
 
----
 
[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]]
 
[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]]
  
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]
 
[[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Latest revision as of 10:37, 25 February 2019


ECE Ph.D. Qualifying Exam

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.


Similar Problem

2015 QE AC3 Prob1
2015 QE AC3 Prob3
2014 QE AC3 Prob2


Back to QE AC question 3, August 2016

Back to ECE Qualifying Exams (QE) page

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett