Here is the optimal tableau for a Max LP 2X1 2 38 81 83 rhs 1 0 5 0 0 10 280 0 0 0 1 5/2 -8 24 0 0 -2 1 0 3/2 -4 8 0 1 1.25 0 0 -0.5 1.5 2 1. For the B-matrix method what is the first row of B-¹ -2 row3 = column1 = 82 10 enter fractions like so 7/2 with no extra spaces 2. After calculating B (i.e. the inverse of B-¹) the original column has values: row1 = b1 = column2 = ; b2 = row2 = 3. Using B-matrix method, the original rhs vector b had these coordinate values: ; and b3 = column3 = Hint: remember new rhs equals B-¹ times original rhs, so...?
Here is the optimal tableau for a Max LP 2X1 2 38 81 83 rhs 1 0 5 0 0 10 280 0 0 0 1 5/2 -8 24 0 0 -2 1 0 3/2 -4 8 0 1 1.25 0 0 -0.5 1.5 2 1. For the B-matrix method what is the first row of B-¹ -2 row3 = column1 = 82 10 enter fractions like so 7/2 with no extra spaces 2. After calculating B (i.e. the inverse of B-¹) the original column has values: row1 = b1 = column2 = ; b2 = row2 = 3. Using B-matrix method, the original rhs vector b had these coordinate values: ; and b3 = column3 = Hint: remember new rhs equals B-¹ times original rhs, so...?
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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Transcribed Image Text:Here is the optimal tableau for a Max LP
2 X1
3 81
82 63
rhs
1
0
0 0
10
10 280
0
0
0
1
5/2
-8 24
0
0
-2
1 0
3/2
-4 8
0
1 1.25 0 0
-0.5 1.5 2
1. For the B-matrix method what is the first row of B-¹
X2
5
row3 =
-2
column1 =
enter fractions like so 7/2 with no extra spaces
2. After calculating B (i.e. the inverse of B-¹) the original
column has values: row1 =
b1 =
column2 =
; b2 =
row2 =
3. Using B-matrix method, the original rhs vector bhad these coordinate values:
; and b3 =
Hint: remember new rhs equals B-¹ times original rhs, so ...?
column3 =
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