3. Solve the LP problem using the dual simplex method. minimize x₁ +45x2 + 3x3 x₁ +5x2-x3 subject to x1 + x2 + 2x3 -x1 + 3x2 + 3x3 -3x1 +8x25x3 X1, X2, X3 ≥ 0. ≥ 4 ≥ 2 ≥5 ≥ 3
3. Solve the LP problem using the dual simplex method. minimize x₁ +45x2 + 3x3 x₁ +5x2-x3 subject to x1 + x2 + 2x3 -x1 + 3x2 + 3x3 -3x1 +8x25x3 X1, X2, X3 ≥ 0. ≥ 4 ≥ 2 ≥5 ≥ 3
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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Question
Linear Programming.
Please follow the same way the example was done in 2nd pic
Transcribed Image Text:3. Solve the LP problem using the dual simplex method.
minimize
subject to
x1 +45x2 + 3x3
x1 +5x2-x3
x1 + x2 + 2x3
-x₁ + 3x₂ + 3x3
-3x1 + 8x2 - 5x3
X1, X2, X3 ≥ 0.
≥ 4
≥ 2
≥ 5
≥ 3
![8:34
< CamScanner 04-09-2023 20.33
Image
Word
solve the following LP problem, Using the am simples
min X₁ + x₂ + x3
st. X₁ + X₂ + X3 23.
4x₁ + x₂ + 2x3 25
X₁, X₂, X3 20
Solution
by inspection, we see that the problem has no bf.s. and it is not in
canonical form. Thus, proceed.
1. Change constraints inequality to "<" by multiplying across by -1.
add slack variables which then puts problem in canonical form.
"stmin x₁ + x₂ + x3
-X₁ X₂ X3 ≤-3
3-4х, - Х2 -2х3 5-5
X,, Х2, Х3 20
The Augmented Matrix Form.
br
X₁
X3
X4
XS
Py +R₂
-1
-4
I
**
6
So, (2,1)
by y
X-1
Xs
Add
X₂
-1
-1
1
X₁ 0
I
1
I
Since the basis matrix B= [94 95] therefore X= [X4 X5] = [-3
XB is a basic non-feasible solution (why?) Since both B₁ ²0₁ ; = 1, 2 and -
is more negative than-3, we pivot in row 2..
>
Now for the pivot column, compute,
-2
1
Corresponds to colurm:1
and the piviot is -4
th
X4
I
1
6
max
^{ 2²/1²: 0; 20} = max {- +₁+34 where
4443-4
I
-34-1/₂
1/4
min X₁ + X₂ + X3
Xs
Oz
Sign
entry = -4 is our pivot element
Y₂
-1 -1
X4
Xs
L
1
O
1/4 1/2
O
I
1/₂. O
-X₁ X2 X3 +₁ ≤-3
-4₁-Y₂-2x3 + £5
X₁, X₂ X3 X4 X5 20
-1/4
O
O
I -1/4
-1/4
b
-3
-5
-3
5/4
O
-7/4
5/4
Share
Now XB = [X₁ X4] [7]
since by <D, we pivo+ in row 1.
For pivot column, consider
max
агу соз
L
مات -
{aij
= max
Collage
I
3
More](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F974a9f57-bd63-4f80-bc4b-5de2cd412f05%2Fg8pah4d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:8:34
< CamScanner 04-09-2023 20.33
Image
Word
solve the following LP problem, Using the am simples
min X₁ + x₂ + x3
st. X₁ + X₂ + X3 23.
4x₁ + x₂ + 2x3 25
X₁, X₂, X3 20
Solution
by inspection, we see that the problem has no bf.s. and it is not in
canonical form. Thus, proceed.
1. Change constraints inequality to "<" by multiplying across by -1.
add slack variables which then puts problem in canonical form.
"stmin x₁ + x₂ + x3
-X₁ X₂ X3 ≤-3
3-4х, - Х2 -2х3 5-5
X,, Х2, Х3 20
The Augmented Matrix Form.
br
X₁
X3
X4
XS
Py +R₂
-1
-4
I
**
6
So, (2,1)
by y
X-1
Xs
Add
X₂
-1
-1
1
X₁ 0
I
1
I
Since the basis matrix B= [94 95] therefore X= [X4 X5] = [-3
XB is a basic non-feasible solution (why?) Since both B₁ ²0₁ ; = 1, 2 and -
is more negative than-3, we pivot in row 2..
>
Now for the pivot column, compute,
-2
1
Corresponds to colurm:1
and the piviot is -4
th
X4
I
1
6
max
^{ 2²/1²: 0; 20} = max {- +₁+34 where
4443-4
I
-34-1/₂
1/4
min X₁ + X₂ + X3
Xs
Oz
Sign
entry = -4 is our pivot element
Y₂
-1 -1
X4
Xs
L
1
O
1/4 1/2
O
I
1/₂. O
-X₁ X2 X3 +₁ ≤-3
-4₁-Y₂-2x3 + £5
X₁, X₂ X3 X4 X5 20
-1/4
O
O
I -1/4
-1/4
b
-3
-5
-3
5/4
O
-7/4
5/4
Share
Now XB = [X₁ X4] [7]
since by <D, we pivo+ in row 1.
For pivot column, consider
max
агу соз
L
مات -
{aij
= max
Collage
I
3
More
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