Posted by Sharon on Saturday, November 10, 2007 at 6:57pm.
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a.
Let L = # of hours assigned to Lisa
D = # of hours assigned to David
S = # of hours assigned to Sarah
Max 30L + 25D + 18S
s.t.
L + D + S = 100
0.6L - 0.4D >= 0
-0.15L - 0.15D + 0.85S >= 0
-0.25L - 0.25D + S <= 0
L <= 50
a).
Let L = # of hours assigned to Lisa
D = # of hours assigned to David
S = # of hours assigned to Sarah
Max 30L + 25D + 18S
s.t.
L + D + S = 100
0.6L - 0.4D >= 0
-0.15L - 0.15D + 0.85S >= 0
-0.25L - 0.25D + S <= 0
L <= 50
L , D , S >= 0
b).
L = 48 Hours
D = 72 Hours
S = 30 Hours
Total Cost = $3780
c).
The Dual Price for Constraint 5 is 0. Therefore, additional hours for Lisa will not change the solution
d).
The dual price for constraint 3 is 0. Because there is No Lower Limit on the right and side range, the optimal solution will not change. Resolving the problem without this constraint will also show that the solution obtained in (b) does not change. Constraint 3, therefore, is really a redundant constraint.
You are supposed to minimize this problem, not maximize it.
L= 32
D= 48
S= 20
cost= $2,520
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