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ESI 6314
Homework #2 Solutions
1. Problem 3, page 139
Basic Variables
Basic Feasible Solution
Corner Point
x
1
, x
2
x
1
=150 x
2
=100 s
1
=s
2
=0
(150, 100)
x
1
, s
1
x
1
=200, s
1
=150, x
2
=s
2
=0
(200, 0)
x
1
, s
2
x
1
=350, s
2
=300, x
2
=s
1
=0
Infeasible
x
2
, s
1
x
2
=400, s
1
=450, x
1
=s
2
=0
Infeasible
x
2
, s
2
x
2
=175, s
2
= 225, x
1
=s
1
=0
(0, 175)
s
1
, s
2
s
1
=350, s
2
=400, x
1
=x
2
=0
(0, 0)
2. Problem 5, page 139
+
332
122
26
10
144
0
14
0
will do the trick, although other correct representations are possible.
(the general procedure was discussed in class on Oct 11)
3. Problem 7, page 213
z
x
1
x
2
s
1
s
2
s
3
RHS
Basic Var.
______________________________________
1
4
1
0
0
0
0
z=0
______________________________________
0
2
3
1
0
0
4
s
1
=4
______________________________________
0
1
1
0
1
0
1
s
2
=1
______________________________________
0
4
1
0
0
1
2
s
3
=2
______________________________________
z
x
1
x
2
s
1
s
2
s
3
RHS
Basic Var.
______________________________________
1
0
0
0
0
1
2
z=2
______________________________________
0
0
5/2
1
0
1/2
3
s
1
=3
______________________________________
0
0
3/4
0
1
1/4
1/2
s
2
=1/2
______________________________________
0
1
1/4
0
0
1/4
1/2
x
1
=1/2
______________________________________
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View Full DocumentThis tableau yields the optimal solution z = 2, x
1
= 1/2, x
2
= 0. Pivoting in the nonbasic variable x
2
yields the alternative optimal solution z = 2, x
1
= 1/3, x
2
= 2/3. Averaging these two optimal solutions
yields a third optimal solution: z = 2, x
1
= 5/12, x
2
= 1/3. Obviously, this LP has an infinite number of
optimal solutions, so other correct answers for the 3
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 Fall '09
 VLADIMIRLBOGINSKI

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