HOMEWORK 3 SOLUTIONS, MATH 340, FALL 2015
JOEL FRIEDMAN
Copyright: Copyright Joel Friedman 2015. Not to be copied, used, or revised without explicit written permission from the copyright owner.
(1) x
3 x
4 x
5 z =
= 1 − x
1
= − 1 − 2 x
1
= − 4 +3 x
1
3 x
1
− x
2
+ x
2
+2 x
2
+ x
2
This is not feasible; we use the two-phase method. Introduce x
0 imize w = − x
0
: and maxx x
3
4 x
5
= 1 − x
1
= − 1 − 2 x
1
= w =
− 4 +3 x
1
− x
2
+ x
2
+2 x
2
+ x
+ x
+ x
− x
0
0
0
0
The most negative variable is x
5
, so x
0 enters and x
5 leaves.
x
0 x
3
= 4 − 3 x
1
= 5 − 4 x
1 x
4
= 3 − 5 x
1 w = − 4 +3 x
1
− 2 x
2
− 3 x
2
− x
2
+2 x
2
+ x
+ x
+ x
− x
5
5
5
5 x
1 enters and x
4 leaves: x x
0
1 x
3
= 11 / 5 − (7 / 5) x
2
= 3 / 5 − (1 / 5) x
2
= 13 / 5 − (11 / 5) x
2 w = − 11 / 5 +(7 / 5) x
2
+(3 / 5) x
− (1 / 5) x
+(4 / 5) x
− (3 / 5) x
4
4
4
4
+(2 / 5) x
+(1 / 5) x
+(1 / 5) x
− (2 / 5) x
5
5
5
5 x
2 enters and x
3 leaves x x
0
1 x
2
= 6 / 11 +(7 / 11) x
3
= 4 / 11 +(1 / 11) x
3
= 13 / 11 − (5 / 11) x
3 w = − 6 / 11 − (7 / 11) x
3
+(1 / 11) x
4
− (3 / 11) x
4
+(4 / 11) x
4
− (1 / 11) x
4
+(3 / 11) x
5
+(2 / 11) x
5
+(1 / 11) x
5
− (3 / 11) x
5
The maximum value of w = − 6 / 11. Hence the minimum value of x
0
= 6 / 11.
Since this is non-zero, we conclude that the original LP problem is not feasible.
Research supported in part by an NSERC grant.
1

2 JOEL FRIEDMAN
(2) Again, we introduce slacks and then x
0 to the right-hand-sides.
x x
2
3 x
4
= 7 − x
1
= − 1 + x
1
= w =
− 4 + x
1
+ x
+ x
+ x
− x
0
0
0
0
(You could omit the x
0 different.) We pivot x
0 leaves.
x
2 from the in and x x
4
2 line; the dictionaries would be slightly
, which has the most negative constant, x
3 x
0
= 11 − 2 x
1
= 3
= 4 − x
1 w = − 4 + x
1
+
+
− x
+ x x x
4
4
4
4
So x
1 enters and x
0 leaves.
x
2 x
3
= 3 +2 x
= 3
0 x w
1
= 4 − x
0
= − x
0
− x
4
+ x
4
+ x
4
So x
0 is eliminated and we bring in the old objective: x
2 x
3
= 3 − x
4
= 3 + x
4 x
1
= 4 + x
4 z = 12 +3 x
4
So x
4 enters and x
2 leaves and we get: x x
4
3 x
1
= 3 − x
2
= 6 − x
2
= 7 − x
2 z = 21 − 3 x
2
This is a final dictionary, so we are done.
Had we pivoted x
3 obtained instead of x
0 into the first dictionary, we would have x x
2
0 x
4
= 8 − 2 x
1
= 1 − x
1
= − 3 z = − 1 + x
1
+
+
+ x x x
− x
3
3
3
3
In this case we get a negative constant in the x
4 row.
Department of Computer Science, University of British Columbia, Vancouver, BC
V6T 1Z4, CANADA, and Department of Mathematics, University of British Columbia,
Vancouver, BC V6T 1Z2, CANADA.
E-mail address : jf@cs.ubc.ca or jf@math.ubc.ca
URL : http://www.math.ubc.ca/~jf