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Standard matrix notation for the entries for the matrix B we have b11 = b21 = r2 = b31 = y2 = b12 = the missing values in rowo of the optimal tableau are the optimal solution for z, called z1 is b22= b32 = . y3 = b13 = b23 b33 = and

Standard matrix notation for the entries for the matrix B we have b11 = b21 = r2 = b31 = y2 = b12 = the missing values in rowo of the optimal tableau are the optimal solution for z, called z1 is b22= b32 = . y3 = b13 = b23 b33 = and

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Here is a Max LP (with some hidden values) for all answers enter answers in decimal format like these examples enter 3 as 3; enter 10 as 10 no decimal if not needed enter 2.3 just like that with no extra 0's and 0.6 as .6 max z = 60 x1 + 30 x2 + 20 x3 s.t. Z 1 0 0 0 8 x1 + 4 x1 + 2 x1 + r2 = And here is the optimal tableau with some values left as variables for you to determine X2 T3 $1 r2 0 0 0 1 1 0 1 17/16 0 0 x1 0 0 - 5/4 0 - 5/4 we have b11 = b21 = b31 = * x2 + x3 <= * x2 + 1.8 x3 <= - 5/2 9/8 Using B-matrix method (remember how to find them) answering the following and eventually fill in the boxes in the original LP Standard matrix notation for the entries for the matrix B * x2 + 5x3 <= 82 y2 5/4 5/4 - 5/16 y2 = $3 y3 - 13/2 the optimal solution for z, called z1 is b12 = b22= b32 = the missing values in rowo of the optimal tableau are rhs z1 204.5 52.5 28.875 " y3 = b13= b23 = b33 = and
Transcribed Image Text:Here is a Max LP (with some hidden values) for all answers enter answers in decimal format like these examples enter 3 as 3; enter 10 as 10 no decimal if not needed enter 2.3 just like that with no extra 0's and 0.6 as .6 max z = 60 x1 + 30 x2 + 20 x3 s.t. Z 1 0 0 0 8 x1 + 4 x1 + 2 x1 + r2 = And here is the optimal tableau with some values left as variables for you to determine X2 T3 $1 r2 0 0 0 1 1 0 1 17/16 0 0 x1 0 0 - 5/4 0 - 5/4 we have b11 = b21 = b31 = * x2 + x3 <= * x2 + 1.8 x3 <= - 5/2 9/8 Using B-matrix method (remember how to find them) answering the following and eventually fill in the boxes in the original LP Standard matrix notation for the entries for the matrix B * x2 + 5x3 <= 82 y2 5/4 5/4 - 5/16 y2 = $3 y3 - 13/2 the optimal solution for z, called z1 is b12 = b22= b32 = the missing values in rowo of the optimal tableau are rhs z1 204.5 52.5 28.875 " y3 = b13= b23 = b33 = and
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