Problem 1
Z= x1-x2
Our Constraint is
x1+x2 ≤ – 1
It can be written as
x1+x2 = – 1
x1 & x2 are non negative values.
Simplex Method:
Z1= x1-x2 + OS,
x1 + x2 + s1 = -1
s1 is the slack variable.
Initial Table:
|
CBi |
Ci |
1 -1 0 |
Solution |
Rate |
|
Basic Variable |
x1 x2 s1 |
|||
|
0 |
Si |
1 1 1 |
-1 |
-1/1 = -1 |
|
Zi |
0 0 0 |
0 |
Ci – Zi 1 -1 0
Zi =
For Minimization:
All Ci – Zi ≥ 0
Iteration 1:
Here we check the Ci – Zi value as maximum that is 1 first column. The corresponding column value is Si = 1. This is called key column.
|
CBi |
Ci |
1 -1 0 |
Solution |
Rate |
|
Basic Variable |
x1 x2 s1 |
|||
|
0 |
Si |
1 1 1 |
-1 |
|
|
Zi |
1 1 1 |
0 |
Ci – Zi 0 -2 -1
The above Ci – Zi values does not satisfy minimization constraint.
All Ci – Zi ≥ 0
So, we cannot find the solution for this problem.
Question – 2
Minimize:
Z= x1-x2
Our Constraint is
x1+x2 ≥ – 1
It can be written as
x1+x2 = – 1
x1 & x2 are non negative values.
Simplex Method:
Z1= x1-x2 + OS,
x1 + x2 + s1 = -1
s1 is the slack variable.
Initial Table:
|
CBi |
Ci |
1 -1 0 |
Solution |
Rate |
|
Basic Variable |
x1 x2 s1 |
|||
|
0 |
Si |
1 1 1 |
-1 |
|
|
Zi |
0 0 0 |
0 |
Ci – Zi 1 -1 0
For Minimization:
All Ci – Zi ≥ 0
Iteration 1:
Here we check the Ci – Zi value as maximum that is 1 first column. The corresponding column value is Si = 1. This is called key column.
|
CBi |
Ci |
1 -1 0 |
Solution |
Rate |
|
Basic Variable |
x1 x2 s1 |
|||
|
0 |
Si |
1 1 1 |
-1 |
|
|
Zi |
1 1 1 |
0 |
Ci – Zi 0 -2 -1
The above Ci – Zi values does not satisfy minimization constraint.
All Ci – Zi ≥ 0
So, we cannot find the solution for this problem.
Question – 3
Minimize:
Z= x1-x2
Our Constraint is
x1+x2 ≥ – 1
x1 & x2 are non negative values.
Simplex Method:
Z1= x1-x2 + 0
x1 + x2 + s1 = -1
s1 is the slack variable.
The Constraint can be written as,
x1+x2 = – 1
x1 + x2 + s1 = -1
The constraint confliction is >=. So, the independent variables should be in negative.
-x1 – x2 – s1 = 1
Initial Table:
|
CBi |
Ci |
1 -1 0 |
Solution |
Rate |
|
Basic Variable |
x1 x2 s1 |
|||
|
0 |
Si |
-1 -1 -1 |
-1 |
-1 |
|
Zi |
1 -1 0 |
0 |
Ci – Zi 0 0 0
The solution for this problem is unbounded.
Question – 4
Z= 10x1+x2
Our Constraint is
2x1+5x2 ≤ 11
It can be written as
2x1+5x2 = 11
Take x1=0,
2(0)+5x2 = 11
x2 = 11/5
x2 = 2.2
take x2 = 0
2x1+5(0) = 11
x1 = 11/2
x1 = 5.5
|
x1 |
0 |
2.2 |
|
X2 |
5.5 |
0 |
The shaded parts are the feasible solution.
(0, 2.2)
Z=10 x1 + x2
Z=10(0) +2.2
Z=2.2
(5.5, 0)
Z=10(5.5) + 0
Z=55
For maximization, Z=55 is the solution.
Question – 5
Maximize:
Z= 3x1+4x2
The constraints are,
2x1+x2 ≤ 6
2x1+3x2 ≤ 9
We can written as,
2x1+x2 = 6
2x1+3x2 = 9
If we take x1 = 0
2 (0) +x2 = 6
x2 = 6
If we take x2 = 0
2x1+0 = 6
x1 = 6/3
x1 =2
|
x1 |
0 |
3 |
|
X2 |
6 |
0 |
x1 = 0,
2x1+3x2 = 9
2(0) + 3x2 = 9
3x2 = 9
x2 = 9/3
x2 = 3
x2 = 0,
2x1+3x2 = 9
2 x1+ 3(0) = 9
2 x1 = 9
x1= 9/2
x1 = 4.5
|
x1 |
0 |
4.5 |
|
X2 |
3 |
0 |
point is (2.25, 1.5)
For maximization,
(0, 6)
Z = 3x1+4x2
Z = 3(0) + 4(6)
Z = 24
(4.5, 0)
Z = 3x1+4x2
Z = 3(4.5) + 4(0)
Z = 13.5
(2.25, 1.5)
Z = 3x1+4x2
Z = 3(2.25) + 4(1.5)
Z = 12.75
The solution is (0,6).
The shaded parts of the graph is feasible solution.
Question 6
Maximize: z = x (5π – x) on [0, 20]
The value of π is 3.14.
Z = x (5*3.14 – x)
Z = x (15.7 – x)
Z = 15.7x – x2
In [0, 20], the value of x is 0,
Z = 15.7 (0) – 0
Z = 0
Use the quadratic formula,
The equation is,
Z = 15.7x – x2
The solution is x = 16 and x = 16
X=0 X=0.064
V=0 V=0
Question 7
Maximize: z = |x2 – 8| on [-4, 4]
In [-4, 4], the value of x is -4,
Z = |(-42)-8|
Z = |16-8|
Z = 8
Use the quadratic formula,
The equation is,
Z = x2 – 8 X = -2.825 X = 0 X = 2.825
Z = 0 Z = 8
Question 8
Maximize: z = x1(x2-1) + x3(x32 – 3)
If we take x1 = 0, x2 = 1, x3 = -1
Z = 0(1-1) + (-1)((-12)-3)
Z = 2
If we take x1 = 1, x2 = -1, x3 = 0
Z = 1(-1-1) + 0(0-3)
Z = -2
If we take x1 = -1, x2 = 0, x3 = 1
Z = -1(0-1) + 1(1-3)
Z = 1 – 2
Z = -1
|
X1 |
X2 |
X3 |
Z |
|
0 |
1 |
-1 |
2 |
|
1 |
-1 |
0 |
-2 |
|
-1 |
0 |
1 |
-1 |
Question 9
Minimize: f(x1, x2) = (x1 – 4)2 +(x2 – 4)2
The constraints are,
x1+x2 ≤ 4
x1+3x2 ≤ 9
We can written as,
x1+x2 = 4
x1+3x2 = 9
If we take x1 = 0
(0) +x2 = 4
x2 = 4
If we take x2 = 0
x1+0 = 4
x1 = 4
x1 =4
|
X1 |
0 |
4 |
|
X2 |
4 |
0 |
If we take x1 = 0
(0) +3x2 = 9
x2 = 9/3
x2 = 3
If we take x2 = 0
x1+0 = 9
x1 = 9
x1 =9
|
X1 |
0 |
3 |
|
X2 |
9 |
0 |
(0,0) (4,0) (9,0)
We have,
|
L(x1, x2) |
= |
(x1 − 4)2 + (x2 − 4)2 − λ1(x1 + x2 − 4) − λ2(x1 + 3x2 − 9) |
The Kuhn-Tucker conditions are,
−2(x1 − 4) − λ1 − λ2 = 0
−2(x2 − 4) − λ1 − 3λ2 = 0
x1 + x2 ≤ 4, λ1 ≥ 0, and λ1(x1 + x2 − 4) = 0
x1 + 3x2 ≤ 9, λ2 ≥ 0, and λ2(x1 + 3x2 − 9) = 0
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