max z = 3x + 2x2 12 s1 82 rhs 2x1 + 5x2 < 8 3x1 + 7x2 < 10 *1, 12 2 1 0 0 0 1/3 1 -2/3 4/3 0 1 7/3 0 s.t. 1 10 1/3 10/3 (a) The range of values of the right hand side bị of the first constraint for which the current basis remains optimal is from to (b) The range of values of the right hand side b2 of the second constraint for which the current basis remains optimal is from to (c) If the right hand side of the first constraint is changed to bị = 7, the new optimal solution will be #1= 22=O (d) If the right hand side of the second constraint is changed to b=5, the new optimal solution will be x1=, r2:

Practical Management Science
6th Edition
ISBN:9781337406659
Author:WINSTON, Wayne L.
Publisher:WINSTON, Wayne L.
Chapter2: Introduction To Spreadsheet Modeling
Section: Chapter Questions
Problem 20P: Julie James is opening a lemonade stand. She believes the fixed cost per week of running the stand...
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rhs
max z = 3x1 + 2x2
2x1 + 5x2 <
3x1 + 7x2 < 10
S1
82
s.t.
8
1
1
10
0 1/3 1 -2/3 4/3
7/3 0
1/3
X1, X2 > 0
0 1
10/3
(a) The range of values of the right hand side b1 of the first constraint for which
the current basis remains optimal is from to
(b) The range of values of the right hand side b2 of the second constraint for which
the current basis remains optimal is from
to
(c) If the right hand side of the first constraint is changed to bi
optimal solution will be r1=O, x2
7, the new
(d) If the right hand side of the second constraint is changed to b2=5, the new
optimal solution will be x1:
Transcribed Image Text:rhs max z = 3x1 + 2x2 2x1 + 5x2 < 3x1 + 7x2 < 10 S1 82 s.t. 8 1 1 10 0 1/3 1 -2/3 4/3 7/3 0 1/3 X1, X2 > 0 0 1 10/3 (a) The range of values of the right hand side b1 of the first constraint for which the current basis remains optimal is from to (b) The range of values of the right hand side b2 of the second constraint for which the current basis remains optimal is from to (c) If the right hand side of the first constraint is changed to bi optimal solution will be r1=O, x2 7, the new (d) If the right hand side of the second constraint is changed to b2=5, the new optimal solution will be x1:
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