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Here is the original problem from McCarl and Spreen
Max
3x1 – 2x2 + X3
St
x1 + x2 + 2x3
= 20
2x1 +x2 +
x3
ge
10
X1 less than or equal to zero.
X2 nonnegative and X3
unrestricted.
x1
x2
x3
Used
SP
obj
3
2
1
10
c1
1
1
2
eq
20
20
0.5
c2
2
1
1
ge
10
10
0
answer
0
0
10
rc
2.5
2.5
0
The original primal problem, set up and solved in Excel with
appropriate restrictions on signs of X1 and X2. (X3 unrestricted).
You should be able to pop this open and see the constraints.
This dual can be taken directly, using our "rules for duals."
Min 20y1 +10 y2
s.t.
y1
 2y2
le 3
y1
+ y2
ge 2
2y1
+ y2
= 1
Y1 is unrestricted, y2 le 0.
Here is the solution of the direct dual.
Note that the SP of the dual
is equal to the solution to the primal and vice versa.
Also notice
that the reduced costs in the primal equal the surplus/slack in the
dual.
(The reduced costs in the dual are 0 because both constraints
in the primal are exactly satisfied)
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View Full DocumentY1
Y2
used
SP
min
20
10
10
1
2 le
3
0.5
0
1
1 ge
2
0.5
0
2
1 eq
1
1
10
answers
0.5
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 Fall '08
 Duffy,P

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