Line 20: Line 20:
 
Therefore the given problem can be converted into a linear programming problem:
 
Therefore the given problem can be converted into a linear programming problem:
  
<math>maximize -x_1^+ -x_1^- -x_2^+ -x_2^- -x_3^+ -x_3^- </math>
+
<math>maximize -x_1^+ -x_1^- -x_2^+ -x_2^- -x_3^+ -x_3^- \\
 +
subject\ to \begin{bmatrix}
 +
  1 & -1 & 1 & -1 & -1 & 1 \\
 +
  0 & 0  & -1& 1  & 0  & 0
 +
\end{bmatrix} \begin{bmatrix}
 +
  x_1^+  \\
 +
  x_1^- \\
 +
  x_2^+ \\
 +
  x_2^- \\
 +
  x_3^+ \\
 +
  x_3^-
 +
\end{bmatrix} = \begin{bmatrix}
 +
  2  \\
 +
  1
 +
\end{bmatrix}
  
subject to
+
</math>
 
+
<math> </math>
+
  
  

Revision as of 16:05, 23 January 2015


QE2013_AC-3_ECE580-3

Part 1,2,3,4,5

(i)
Solution:
To remove absolute values, each variable is separated into a positive component and a negative component. For $ x_1 $ for example:

$ x_1 = x_1^+ - x_1^- $

Then $ |x_1| = x_1^+ + x_1^- $

Therefore the given problem can be converted into a linear programming problem:

$ maximize -x_1^+ -x_1^- -x_2^+ -x_2^- -x_3^+ -x_3^- \\ subject\ to \begin{bmatrix} 1 & -1 & 1 & -1 & -1 & 1 \\ 0 & 0 & -1& 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1^+ \\ x_1^- \\ x_2^+ \\ x_2^- \\ x_3^+ \\ x_3^- \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} $



(ii)
Solution:

The schema will be destroyed if and only if the 2nd or 4th symbol change. Equivalently, the schema will not be destroyed if and only if both 2nd and the 4th symbols stay the same. As those events are independent:

$ P(Not\ destroyed) = P(2nd\ symbol\ does\ not\ change) \times P(4th\ symbol\ does\ not\ change) $

$ = (1-0.1)\times(1-0.1) = 0.81 $

Therefore

$ P(destroyed) = 1 - 0.81 = 0.19 $



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