Let us take the LP max z = 9x1 + 15x₂ 8.t. *1 + 3x 4x1 +502 1, ₂0 which has BV equal {*, *1} Find B = -188--188---10 max 8.t. Hint: same B -1 €8₂ =; č0₂ = bi11 = bi1a = the range for cg so that this BV remains optimal is 0 the same BV is still a feasible solution and gives the tableau 5₁-5₂- = 30 100 and = z = 30 ≤100 and [00] but new cay then recalculate all new row0 entries. = 0; bha1 = 0; bíza · 81 82 rhs 1 0 0 C8₂ Cox Z 00 1 bi bi2 by 0 1 O biz biz b₂ 2
Let us take the LP max z = 9x1 + 15x₂ 8.t. *1 + 3x 4x1 +502 1, ₂0 which has BV equal {*, *1} Find B = -188--188---10 max 8.t. Hint: same B -1 €8₂ =; č0₂ = bi11 = bi1a = the range for cg so that this BV remains optimal is 0 the same BV is still a feasible solution and gives the tableau 5₁-5₂- = 30 100 and = z = 30 ≤100 and [00] but new cay then recalculate all new row0 entries. = 0; bha1 = 0; bíza · 81 82 rhs 1 0 0 C8₂ Cox Z 00 1 bi bi2 by 0 1 O biz biz b₂ 2
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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![Let us take the LP
max z = 9x1 + 15%₂
s.t.
*1 + 3x1
401 +502
1, ₂ > 0
which has BV equal (1,1)
Find B =
< 30
<100
the range for c so that this BV remains optimal is
<<
(i.e. allowable decrease/increase)
If we change c₂ to 11 we obtain the LP:
bi11
and B-1
=
max z = 9x1 + 11x₂
8.t.
*1 + 3x₂
4x1 + 5*
1, ₂0
the same BV is still a feasible solution and gives the tableau
=
Hint: same B-¹ but new cay then recalculate all new row0 entries.
€8₂ = ; ₂= |
Di tr
3₁ = ; 5₂ =
Z =
and CBy =
< 30
< 100
[00]
| bíz1 = bizz
Z DY
1
0
0
0
0 1
0
1
0
81
82
rhs
C8₂
Cox
Z
bi bi12 by
big big b](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc80b4d1f-cc91-4668-ace5-1850d4a765ec%2F81554e19-7ba0-4757-be8c-2e444d7ea9f3%2F745mj9c_processed.png&w=3840&q=75)
Transcribed Image Text:Let us take the LP
max z = 9x1 + 15%₂
s.t.
*1 + 3x1
401 +502
1, ₂ > 0
which has BV equal (1,1)
Find B =
< 30
<100
the range for c so that this BV remains optimal is
<<
(i.e. allowable decrease/increase)
If we change c₂ to 11 we obtain the LP:
bi11
and B-1
=
max z = 9x1 + 11x₂
8.t.
*1 + 3x₂
4x1 + 5*
1, ₂0
the same BV is still a feasible solution and gives the tableau
=
Hint: same B-¹ but new cay then recalculate all new row0 entries.
€8₂ = ; ₂= |
Di tr
3₁ = ; 5₂ =
Z =
and CBy =
< 30
< 100
[00]
| bíz1 = bizz
Z DY
1
0
0
0
0 1
0
1
0
81
82
rhs
C8₂
Cox
Z
bi bi12 by
big big b
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