Difference between revisions of "2016AC-3-1" - Rhea

Latest revision as of 11:37, 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.

@@ Line 1: / Line 1: @@
 [[Category:ECE]]
 [[Category:QE]]
-[[Category:AC]]
 [[Category:problem solving]]
@@ Line 22: / Line 21: @@
 Minimize <math>2x_1+x_2</math><br>
 Subject to <math>\begin{align*}
-x - 5y &=  8 \\
+&x_1+3x_2-x_3=6\\
-x + 9y &=  -12
+&2x_1+x_2-x_4=4\\
-\end{align*}</math>
+&x_1+x_2+x_5=3\\
+&x_1, x_2, x_3, x_4,x_5 >=0
-===Solution 2===
+\end{align*}</math><br>
-Let <math>Z=X+Y</math>,
+such that <math>A=
+\begin{bmatrix}
-<math>P_Z(k)=P_Z(z=k)=P_Z(x+y=k)\\
+& 3 & -1 & 0 & 0 \\
-=\sum_{i=0}^{k}P(x=i)(y=k-i)=\sum_{i=0}^{k}\frac{\lambda_1^i e^{\lambda_1}}{i!}\cdot\frac{\lambda_2^{k-i}e^{-\lambda_2}}{(k-i)!}
+& 1 & 0 & -1 & 0 \\
-=e^{-\lambda_1-\lambda_2}\cdot \frac{(\lambda_1+\lambda_2)^k}{k!}
+& 1 & 0 & 0 & 1
-</math>
+\end{bmatrix}
+</math><br>
-Using <math>(a+b)^k=\sum_{i=0}^{k}a^ib^{(k-i)}\cdot \frac{k!}{i!(k-i)!} </math>
+we take <math>B=
+\begin{bmatrix}
-Therefore,
+& 3 & 0 \\
+& 1 & 0 \\
-<math>
+& 1 & 1
-P_{X|Z}(x=k|z=n)=\frac{P(x=k,z=n)}{P(z=n)}=\frac{P(x=k,x+y=n)}{P(z=n)}=\frac{P(x=k,y=n-k)}{P(z=n)}\\
+\end{bmatrix} \Rightarrow
-=\frac{\lambda_1^k e^{-\lambda_1}}{k!}\frac{\lambda_2^{n-k} e^{-\lambda_2}}{(n-k)!}\frac{n!}{e^{-\lambda_1}e^{-\lambda_2}(\lambda_1+\lambda_2)^n}
+B\begin{bmatrix}
-= \frac{\lambda_1^k \lambda_2^{n-k} }{(\lambda_1+\lambda_2)^n}\frac{n!}{k!(n-k)!}
+x_1 \\
-</math>
+x_2 \\
+x_3
-===Solution 3===
+\end{bmatrix}
+=b \Rightarrow \begin{bmatrix}
-We will view this problem through the lens of Bayes' Theorem. As such, we can write the conditional distribution as
+x_1 \\
+x_2 \\
-<math>
+x_3
-P(X = x | X+Y = n) = \frac{P(X = x, X+Y = n)}{P(X+Y = n)} = \frac{P(X = x, Y = n - X)}{\sum^n_{k = 0}P(X = k, Y = n - k)}
+\end{bmatrix}
-</math>.
+=
+\begin{bmatrix}
-Since <math>X</math> and <math>Y</math> are independent, we can further write
+& 3 & 0 \\
+& 1 & 0 \\
-<math>
+& 1 & 1
-P(X = x | X+Y = n) = \frac{P(X = x)P(Y = n - X)}{\sum^n_{k = 0}(P(X=k)P(Y = n-k))}
+\end{bmatrix}^{-1}
-</math>.
+\begin{bmatrix}
+\\
-Now let us separate the above expression into numerator and denominator. Recalling that <math>X</math> and <math>Y</math> are independent Poisson r.v.s, the numerator is given by
+\\
-<math>
+\end{bmatrix}
-P(X = x)P(Y = n - X) = \frac{e^{-\lambda_1}\lambda_1^x}{x!}\cdot\frac{e^{-\lambda_2}\lambda_2^{n-x}}{(n-x)!}
+=
-</math>.
+\begin{bmatrix}
+\dfrac{6}{5} \\
-Multiplying the above by <math>\frac{n!}{n!}</math> gives
+\dfrac{8}{5} \\
+\dfrac{1}{5}
-<math>
+\end{bmatrix}
-P(X = x)P(Y = n - X) = \frac{e^{-\lambda_1 + \lambda_2}}{n!}{n\choose x}\lambda_1^x\lambda_2^{n-x}
+</math><br>
-</math>.
+Such that <math>x^T=[\dfrac{6}{5}, \dfrac{8}{5},\dfrac{1}{5}, 0, 0]</math> is a feasible solution.
-Now let us examine the denominator. Again, we make use of the fact that <math>X</math> and <math>Y</math> are independent Poisson r.v.s to write
-<math>
-\sum^n_{k = 0}(P(X=k)P(Y = n-k)) = \sum^n_{k = 0}\left(\frac{e^{-\lambda_1}\lambda_1^k}{k!}\frac{e^{-\lambda_2}\lambda_2^{n-k}}{(n-k)!}\right)
-</math>.
-Again, we multiply by <math>\frac{n!}{n!}</math> to obtain
-<math>
-\sum^n_{k = 0}(P(X=k)P(Y = n-k)) = \frac{e^{-\lambda_1 + \lambda_2}}{n!}\sum^n_{k = 0}{n\choose k}\lambda_1^k\lambda_2^{n-k}
-</math>.
-We can make use of the binomial formula to simplify this expression. Recall that the binomial formula is given by
-<math>
-(a+b)^n = \sum^n_{k = 0}{n\choose k}a^k b^{n - k}
-</math>.
-We use this to write
-<math>.
-\sum^n_{k = 0}(P(X=k)P(Y = n-k)) = \frac{e^{-\lambda_1 + \lambda_2}}{n!}\cdot(\lambda_1 + \lambda_2)^n
-</math>
-Putting this all together, we can finally write
-<math>
-P(X = x | X+Y = n) = \frac{\frac{e^{-\lambda_1 + \lambda_2}}{n!}{n\choose x}\lambda_1^x\lambda_2^{n-x}}{\frac{e^{-\lambda_1 + \lambda_2}}{n!}\cdot(\lambda_1 + \lambda_2)^n} = {n\choose x}\frac{\lambda_1^x\lambda_2^{n-x}}{(\lambda_1 + \lambda_2)^n}
-</math>
-and we are done.
+----
 ===Similar Problem===
+[https://engineering.purdue.edu/ECE/Academics/Graduates/Archived_QE_August_2015/AC-3?dl=1 2015 QE AC3 Prob1]<br>
-If <math>X</math> and <math>Y</math> are independent binomial random variables with success probabilities <math>p</math> and <math>q</math> respectively, find the probability mass function of <math>X</math> when <math>X + Y = k</math>. In addition, investigate what happens to this p.m.f. when <math>p = q</math>.
+[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>
 ----
-[[ECE-QE_CS1-2015|Back to QE CS question 1, August 2015]]
+[[QE2016_AC-3_ECE580|Back to QE AC question 3, August 2016]]
 [[ECE_PhD_Qualifying_Exams|Back to ECE Qualifying Exams (QE) page]]

Difference between revisions of "2016AC-3-1" - Rhea

Latest revision as of 11:37, 25 February 2019

Solution

Similar Problem

Alumni Liaison