let x1 denote number of calculus textbooks
x2 denote number of history textbooks
x3 denote number of marketing textbooks
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subject to | ||||||||||||||||||||||
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and x1,x2,x3≥0; |
After introducing slack variables
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subject to | ||||||||||||||||||||||||||||||||||
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and x1,x2,x3,S1,S2≥0 |
Iteration-1 | Cj | 10 | 4 | 8 | 0 | 0 | ||
B | CB | XB | x1 | x2 | x3 | S1 | S2 | MinRatio XBx1 |
S1 | 0 | 1100 | 1 | 1 | 1 | 1 | 0 | 1100/1=1100 |
S2 | 0 | 1900 | (2) | 1 | 2 | 0 | 1 | 1900/2=950→ |
Z=0 | Zj | 0 | 0 | 0 | 0 | 0 | ||
Zj-Cj | -10↑ | -4 | -8 | 0 | 0 |
Entering =x1, Departing =S2, Key Element
=2
Iteration-2 | Cj | 10 | 4 | 8 | 0 | 0 | ||
B | CB | XB | x1 | x2 | x3 | S1 | S2 | MinRatio |
S1 | 0 | 150 | 0 | 12 | 0 | 1 | -1/2 | |
x1 | 10 | 950 | 1 | 12 | 1 | 0 | 1/2 | |
Z=9500 | Zj | 10 | 5 | 10 | 0 | 5 | ||
Zj-Cj | 0 | 1 | 2 | 0 | 5 |
Hence, optimal solution is arrived with value of variables as
:
x1 = 950, x2 = 0, x3 = 0
Max Z = 9500
so maximum profit is $9500 and number of calculus texts is 950 , history and marketing text is 0
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