A max LP has optimal tableau Assume columns labeled as 48 The original rhs was 20 8 however we have been asked to change ba to 25 Z £1 £2 23 81 [10 5 00 00 -2 0 1 00 -2 1 0 0 1 1.25 0 0 using the B-matrix method the new rhs for the constraints is 0 0 0 we can use that new equals y times new b hence new is 230 Now doing dual simplex on the updated tableau yields the new optimal tableau: DOO 82 10 280] 10 2 -8 24 2 -4 8 -0.5 1.5 2 0000 83 rhs 3000 OUUU 34 18 -.5 0000000 and the new rhs in row0
A max LP has optimal tableau Assume columns labeled as 48 The original rhs was 20 8 however we have been asked to change ba to 25 Z £1 £2 23 81 [10 5 00 00 -2 0 1 00 -2 1 0 0 1 1.25 0 0 using the B-matrix method the new rhs for the constraints is 0 0 0 we can use that new equals y times new b hence new is 230 Now doing dual simplex on the updated tableau yields the new optimal tableau: DOO 82 10 280] 10 2 -8 24 2 -4 8 -0.5 1.5 2 0000 83 rhs 3000 OUUU 34 18 -.5 0000000 and the new rhs in row0
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 3CEXP
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![A max LP has optimal tableau
Assume columns labeled as
[48]
The original rhs was 20
8
however we have been asked to change ba to 25
#1 2
3 81
[10 5 0 0
0 0 -2 0 1
00 -2 1 0
0 1 1.25 0 0
Z
using the B-matrix method the new rhs for the constraints is
0
0
82
10 280]
10
2 -8 24
2 -4
-0.5
-0.5 1.5 2
83 rhs
L
000
we can use that new equals y times new b hence new is 230
Now doing dual simplex on the updated tableau yields the new optimal tableau:
34
D☐☐☐☐☐☐
18
-.5
and the new rhs in row0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5dfcde0a-cc8e-4f7e-b5e2-9ec6ad36a3c9%2Faa7a7326-e51a-4df4-b90c-d89a84277471%2Fjab17wm_processed.png&w=3840&q=75)
Transcribed Image Text:A max LP has optimal tableau
Assume columns labeled as
[48]
The original rhs was 20
8
however we have been asked to change ba to 25
#1 2
3 81
[10 5 0 0
0 0 -2 0 1
00 -2 1 0
0 1 1.25 0 0
Z
using the B-matrix method the new rhs for the constraints is
0
0
82
10 280]
10
2 -8 24
2 -4
-0.5
-0.5 1.5 2
83 rhs
L
000
we can use that new equals y times new b hence new is 230
Now doing dual simplex on the updated tableau yields the new optimal tableau:
34
D☐☐☐☐☐☐
18
-.5
and the new rhs in row0
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