Optimization Using Simplex Method: Essay On Efficient Problem-solving.

Problem 1

Z= x₁-x₂

Our Constraint is

            x₁+x₂ ≤ - 1

It can be written as

x₁+x_{2 =} - 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

C_{i -} Z_i                                         1 -1 0

Z_{i =} 

For Minimization:

All C_{i -} Z_i         ≥ 0

Iteration 1:

Here we check the C_{i -} Z_i value as maximum that is 1 first column.  The corresponding column value is S_i = 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

C_{i -} Z_i                                          0 -2 -1

The above C_{i –} Z_i values does not satisfy minimization constraint.

All C_{i -} Z_i         ≥ 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+x_{2 =} - 1

x₁ & x₂ are non negative values.

Simplex Method:

Z₁= x₁-x₂ + OS,

x₁ + x₂ + s_{1 = -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

C_{i -} Z_i                                          1 -1 0 

For Minimization:

All C_{i -} Z_i         ≥ 0

Iteration 1:

Here we check the C_{i -} Z_i value as maximum that is 1 first column.  The corresponding column value is S_i = 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

C_{i -} Z_i                                             0 -2 -1

The above C_{i –} Z_i values does not satisfy minimization constraint.

All C_{i -} Z_i         ≥ 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+x_{2 =} - 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

 C_{i -} Z_i                                           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(x_1, 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