3. Solve the LP problem using the dual simplex method. minimize x1 +45x2 + 3x3 x1 +5x2-x3 x1 + x2 + 2x3 -x1 + 3x₂ + 3x3 -3x1 + 8x2 - 5x3 1, 2, 3 ≥ 0. subject to > 4 ≥ 2 ≥ 5 23
3. Solve the LP problem using the dual simplex method. minimize x1 +45x2 + 3x3 x1 +5x2-x3 x1 + x2 + 2x3 -x1 + 3x₂ + 3x3 -3x1 + 8x2 - 5x3 1, 2, 3 ≥ 0. subject to > 4 ≥ 2 ≥ 5 23
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
LP
Transcribed Image Text:3. Solve the LP problem using the dual simplex method.
minimize x1 +45x2 + 3x3
x1 + 5x2-x3
x1 + x2 + 2x3
-x1 + 3x2 + 3x3
-3x1 + 8x25x3
X1, X2, X3 ≥ 0.
subject to
> 4
≥2
> 5
≥ 3
![Ry +R₂
R3-R₂
Now for the pivot column, compute
C
азјсоз
max
So, (2,
byr
X₁
X+ -1
1
Xs
X₁
x₂
7
X4
X₁
Corresponds to colurn:1
and the piviot is -4
1
0
1
O
1
g:
Laz;
O
th
-1
1/4
I
entry = -4 is our pivot element
Хч
Xs
1
O
y = x₁
X2
1/4
1
-344-1/2
1/4
3/4
X3
X4
XsJ
X3
O
[0] 2/3
7/3
-1
112
I
1/2
ERS
2023-00
1/2
1/2
2/3
O
HOL Keny
= max
2/3
7/3
O
O
aintenance
4898 La
30084
{-4723-1
-1/4
O
1
O
O 1/4
O
-4/3
1
O
-1/4
-1/4
-3
5/4
-7/4
5/4
-514
5/4
7/3
1/3
0 -3
[a₁, 92]
Since the basic
matrix B=
the basis vector X₂ = [X₁ X₂] = [ ²/1/23/2]
this is basic and feasible since all B₁³0, the current basic
is optimal
where
[X, X4] = []
since b, 40, we pivo + in row 1.
For pivot column, consider
max
ату соз
3
도후, 흐,.
-3
32-4
= max {-1,-1+1}=
= 2
7
9
1440
3+3
3
Now XB =
sci
{aij
the corresponding optimal valve for the objective function is
=max](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2Febcd4e67-2c0e-4b22-b103-46f83ac72dd1%2Fawfh5l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Ry +R₂
R3-R₂
Now for the pivot column, compute
C
азјсоз
max
So, (2,
byr
X₁
X+ -1
1
Xs
X₁
x₂
7
X4
X₁
Corresponds to colurn:1
and the piviot is -4
1
0
1
O
1
g:
Laz;
O
th
-1
1/4
I
entry = -4 is our pivot element
Хч
Xs
1
O
y = x₁
X2
1/4
1
-344-1/2
1/4
3/4
X3
X4
XsJ
X3
O
[0] 2/3
7/3
-1
112
I
1/2
ERS
2023-00
1/2
1/2
2/3
O
HOL Keny
= max
2/3
7/3
O
O
aintenance
4898 La
30084
{-4723-1
-1/4
O
1
O
O 1/4
O
-4/3
1
O
-1/4
-1/4
-3
5/4
-7/4
5/4
-514
5/4
7/3
1/3
0 -3
[a₁, 92]
Since the basic
matrix B=
the basis vector X₂ = [X₁ X₂] = [ ²/1/23/2]
this is basic and feasible since all B₁³0, the current basic
is optimal
where
[X, X4] = []
since b, 40, we pivo + in row 1.
For pivot column, consider
max
ату соз
3
도후, 흐,.
-3
32-4
= max {-1,-1+1}=
= 2
7
9
1440
3+3
3
Now XB =
sci
{aij
the corresponding optimal valve for the objective function is
=max
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