CS 600 Solutions to Test II
November 17, 1998



1 Introduce slack variables x3, x4, and x5to obtain



Max z = x1 + 3x2
Subject to
-x1 - x2 + x3 = -2
x1 - x2 + x4 = -1
2x1 + x2 + x5 = 4
$x_{i} \geq 0$, i = 1, ..., 5.. If we choose x1, and x2 as non-basic variables, then it is not feasible since x3, and x4 are negative.



Let us find an initial feasible solution by introducing a new variable x0, and solving the following LP problem:

Max z = -x0
Subject to
$-x_{1} - x_{2} - x_{0} \leq -2 $
$x_{1} - x_{2} - x_{0} \leq -1 $
$2x_{1} + x_{2} - x_{0} \leq 4 $
$x_{i} \geq 0$, i = 0, 1,2.. This can be solved starting with

Max z = -x0
Subject to
-x1 - x2 - x0 + x3 = -2
x1 - x2 - x0 + x4 = -1
2x1 + x2 - x0 + x5 = 4
$x_{i} \geq 0$, i = 0, ..., 5..

If we select x3, x4, and x5 as basic variables, it is not feasible as can be seen below.

x3 = -2 + x1 + x2 + x0
x4 = -1 - x1 + x2 + x0
x5= 4 - 2x1 - x2 + x0
z = -x0.
$x_{i} \geq 0$, i = 0, ..., 5..

Select 2 for the initial value for x0 while keeping x1 and x2 0. Then the initial dictionary is

x0 = 2 - x1 - x2 + x3
x4 = 1 - 2x1 + x3
x5= 6 - 3x1 - 2x2 + x3
z = -2 + x1 + x2 - x3.
$x_{i} \geq 0$, i = 0, ..., 5.. Interchanging x2 with x0 for a new dictionary, we obtain

x2 = 2 - x1 - x0 + x3
x4 = 1 - 2x1 + x3
x5= 2 - x1 + 2x0 - x3
z = -x0.

Thus the original problem is feasible, and an initial feasible solution for the original problem is

x0 = 0, x1 = 0, x2 = 2, x3= 0, x4 = 1, and x5 = 2.



2 a) The dual is



Max 5y1 + 8y2
Subject to
$y_{1} + 2y_{2} \leq 8 $
$-y_{1} + y_{2} \leq 3 $
$3y_{1} + y_{2} \leq 9 $
$2y_{1} -2 y_{2} \leq 3 $
$y_{1}, y_{2} \geq 0 $



b) Draw a graph of the feasible region and also two or three positions of z = c lines for the objective function. By sliding up the line z = c we can see that z is maximum at y1 = 2, y2 = 3, and z = 34.



3 a) Let f1(x1) = 4x1 - x12, and
f2(x2) = 4x2 - x23.

Then f's can be approximated by

f1(x11,x12,x13) = 3x11 + x12 - x13, and
f2(x21,x22,x23) = 3x21 -3x22 -15x23,

where $0 \leq x_{ij} \leq 1$ and xij = 0 if xij-1 < 1for all i and j, i = 1,2, and j = 1,2,3.

Hence the approximation problem is



Max z = 3x11 + x12 - x13 + 3x21 -3x22 -15x23
Subject to
$2(x_{11} + x_{12} + x_{13}) + 5(x_{21} + x_{22} + x_{23}) \leq 14$
$3(x_{11} + x_{12} + x_{13}) + (x_{21} + x_{22} + x_{23}) \leq 8$
$0 \leq x_{ij} \leq 1$ for all i and j, i = 1,2, and j = 1,2,3.



b) Let xij = 0 for all i and j, i = 1,2, and j = 1,2,3, and let all the slack variables a basic variable.

 
S. Toida
1999-08-30