Solve the LP problem using the dual simplex method. (Please follow the example structure below) minimize x1 + 45x2+ 3x3 subject to: x1 + 5x2 - x3 >= 4 x1 + x2 + 2x3 >= 2 -x1 + 3x2 + 3x3 >= 5 -3x1 + 8x2 - 5x3 >= 3 Х1, X2, X3 ≥ 0.
Solve the LP problem using the dual simplex method. (Please follow the example structure below) minimize x1 + 45x2+ 3x3 subject to: x1 + 5x2 - x3 >= 4 x1 + x2 + 2x3 >= 2 -x1 + 3x2 + 3x3 >= 5 -3x1 + 8x2 - 5x3 >= 3 Х1, X2, X3 ≥ 0.
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
3. Solve the LP problem using the dual simplex method. (Please follow the example structure below)
minimize x1 + 45x2+ 3x3
subject to:
x1 + 5x2 - x3 >= 4
x1 + x2 + 2x3 >= 2
-x1 + 3x2 + 3x3 >= 5
-3x1 + 8x2 - 5x3 >= 3
Х1, X2, X3 ≥ 0.
![ex.
Solve the following LP problem, using the dual simplex method
min X₁ + x₂ + x3
s.t.
Solution
by inspection, we see that the problem has no b.f.s. and it is not in
canonical form. Thus, proceed.
1. Change constraints inequality to "I" by multiplying across by
11 add slack variables which then puts problem in canonical form.
-1
1.
(st min x₁ + x₂ + x3
= x₁ - x₂-x3 = -
- 3
-4x₁ - x₂ - 2x3 25
X₁ X ₂ 1 X3 20
-4
1
The Augmented Matrix Form
br
X₁
X4
XS
X₁ + X₂ + X3 ≥ 3
4x₁ + x₂ + 2x3 25
X₁, X₂, X3 20
X₂
-1
-1
X₂
ī
-2
1
X4
-
0
(min X₁ + X₂ + X3
st. -X₁-x₂-x3 + x4 = -3
-4X₁-X2-2x3+X5 ≤5
11
X₁, X₂, X3, X4, X5 20
Xs
-
0
b
-3
-S
-
Since the basis matrix B= [94 95]
therefore X₂ = [X₁ Xs] = [-3
XB is a basic non feasible solution (why?) Since both B₁ ²0₁ ; = 1, 2 and - E
is more negative than-3, we pivot in row 2..](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F96783f09-bc48-4e35-9356-1a3f7492a61b%2F6sfmqr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:ex.
Solve the following LP problem, using the dual simplex method
min X₁ + x₂ + x3
s.t.
Solution
by inspection, we see that the problem has no b.f.s. and it is not in
canonical form. Thus, proceed.
1. Change constraints inequality to "I" by multiplying across by
11 add slack variables which then puts problem in canonical form.
-1
1.
(st min x₁ + x₂ + x3
= x₁ - x₂-x3 = -
- 3
-4x₁ - x₂ - 2x3 25
X₁ X ₂ 1 X3 20
-4
1
The Augmented Matrix Form
br
X₁
X4
XS
X₁ + X₂ + X3 ≥ 3
4x₁ + x₂ + 2x3 25
X₁, X₂, X3 20
X₂
-1
-1
X₂
ī
-2
1
X4
-
0
(min X₁ + X₂ + X3
st. -X₁-x₂-x3 + x4 = -3
-4X₁-X2-2x3+X5 ≤5
11
X₁, X₂, X3, X4, X5 20
Xs
-
0
b
-3
-S
-
Since the basis matrix B= [94 95]
therefore X₂ = [X₁ Xs] = [-3
XB is a basic non feasible solution (why?) Since both B₁ ²0₁ ; = 1, 2 and - E
is more negative than-3, we pivot in row 2..
![R₁ + R₂
R3-R2
max
Now for the pivot column, compute
So, (2, 1)
X₁
ла
X4 -1
1
Xs
'X
X4
X₁
X₂
Corresponds to column:I
and the piviot is -4
1
0
1
az;
{a: arj20 }
azj
O
1
O
1/4
-3/4 - 1/2
1/4 1/2
3/4
y = x₁
X2
1
O
O
entry = - 4 is our pivot element
X₂ X3
L
X4
Xs
-1
1
X3
X4
XSJ
1/2
11
N
00-8202
POL Aleny
112
O
= max
2/37
7/3
aintenance
O
1/2
2/3 -4/3
O
2
3 3
+-= {²+²+-}
O
-1/4
-1/4
-114
27868
1/4
-1/4
1/3
30084
-3
5/4
O
-7/4
5/4
-574
5/4
7/3
- 3
3 3
where
Now XB = [X₁ X4] = []
since b, 40, we pivo + in row 1.
For pivot column, consider
хош
{ax fix fid}
хош =
E S
[9₁, 9₂]
Since the basic matrix B
the basis vector X₂ = [X₁ X₂] = [ 1/²/3/3/3/3]
this is basic and feasible since all B₁ '0, the current basic
is optimal
2-4
-3
4
= max {-1,-1
TNTN
2
the corresponding optimal valve for the objective function is
[1,1 2/3
7/3
2
-11-15 T
2+/-/
M](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2F96783f09-bc48-4e35-9356-1a3f7492a61b%2Fjmcpva9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:R₁ + R₂
R3-R2
max
Now for the pivot column, compute
So, (2, 1)
X₁
ла
X4 -1
1
Xs
'X
X4
X₁
X₂
Corresponds to column:I
and the piviot is -4
1
0
1
az;
{a: arj20 }
azj
O
1
O
1/4
-3/4 - 1/2
1/4 1/2
3/4
y = x₁
X2
1
O
O
entry = - 4 is our pivot element
X₂ X3
L
X4
Xs
-1
1
X3
X4
XSJ
1/2
11
N
00-8202
POL Aleny
112
O
= max
2/37
7/3
aintenance
O
1/2
2/3 -4/3
O
2
3 3
+-= {²+²+-}
O
-1/4
-1/4
-114
27868
1/4
-1/4
1/3
30084
-3
5/4
O
-7/4
5/4
-574
5/4
7/3
- 3
3 3
where
Now XB = [X₁ X4] = []
since b, 40, we pivo + in row 1.
For pivot column, consider
хош
{ax fix fid}
хош =
E S
[9₁, 9₂]
Since the basic matrix B
the basis vector X₂ = [X₁ X₂] = [ 1/²/3/3/3/3]
this is basic and feasible since all B₁ '0, the current basic
is optimal
2-4
-3
4
= max {-1,-1
TNTN
2
the corresponding optimal valve for the objective function is
[1,1 2/3
7/3
2
-11-15 T
2+/-/
M
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