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1. Simplex (BigM) method example ( Enter your problem )

9. Unbounded solution example

Unbounded solution
In simplex table, if a variable should enter into the basis, but all the coefficients in that column are negative or zero. So this variable can not be entered into the basis, because for minimum ratio, negative value in denominator can not be considered and zero value in denominator would result oo.
Hence, the solution to the given problem is unbounded.
Example
Find solution using Simplex(BigM) method
MAX Z = 3x1 + 5x2
subject to
x1 - 2x2 <= 6
x1 <= 10
x2 >= 1
and x1,x2 >= 0

Solution:
Problem is
 Max Z =  3 x_1  +  5 x_2
subject to
   x_1  -  2 x_2 ≤ 6   x_1 ≤ 10   x_2 ≥ 1
and x_1,x_2 >= 0;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '<=' we should add slack variable S_1

2. As the constraint-2 is of type '<=' we should add slack variable S_2

3. As the constraint-3 is of type '>=' we should subtract surplus variable S_3 and add artificial variable A_1

After introducing slack,surplus,artificial variables
 Max Z =  3 x_1  +  5 x_2  +  0 S_1  +  0 S_2  +  0 S_3  -  M A_1
subject to
   x_1  -  2 x_2  +   S_1 = 6   x_1  +   S_2 = 10   x_2  -   S_3  +   A_1 = 1
and x_1,x_2,S_1,S_2,S_3,A_1 >= 0

 Iteration-1 C_j 3 5 0 0 0 -M B C_B X_B x_1 x_2 Entering variable S_1 S_2 S_3 A_1 MinRatio (X_B)/(x_2) S_1 0 6 1 -2 1 0 0 0 --- S_2 0 10 1 0 0 1 0 0 --- A_1 Leaving variable -M 1 0 (1)  (pivot element) 0 0 -1 1 (1)/(1)=1-> Z=0 0=Z_j=sum C_B X_B Z_j Z_j=sum C_B x_j 0 0=0xx1+0xx1+(-M)xx0Z_j=sum C_B x_1 -M -M=0xx(-2)+0xx0+(-M)xx1Z_j=sum C_B x_2 0 0=0xx1+0xx0+(-M)xx0Z_j=sum C_B S_1 0 0=0xx0+0xx1+(-M)xx0Z_j=sum C_B S_2 M M=0xx0+0xx0+(-M)xx(-1)Z_j=sum C_B S_3 -M -M=0xx0+0xx0+(-M)xx1Z_j=sum C_B A_1 C_j-Z_j 3 3=3-0 M+5 M+5=5-(-M)uarr 0 0=0-0 0 0=0-0 -M -M=0-M 0 0=(-M)-(-M)

Positive maximum C_j-Z_j is M+5 and its column index is 2. So, the entering variable is x_2.

Minimum ratio is 1 and its row index is 3. So, the leaving basis variable is A_1.

:. The pivot element is 1.

Entering =x_2, Departing =A_1, Key Element =1

R_3(new)= R_3(old)

R_1(new)= R_1(old)+ 2 R_3(new)

R_2(new)= R_2(old)

 Iteration-2 C_j 3 5 0 0 0 B C_B X_B x_1 x_2 S_1 S_2 S_3 Entering variable MinRatio (X_B)/(S_3) S_1 0 8 8=6+2xx1R_1(new)= R_1(old)+ 2 R_3(new) 1 1=1+2xx0R_1(new)= R_1(old)+ 2 R_3(new) 0 0=(-2)+2xx1R_1(new)= R_1(old)+ 2 R_3(new) 1 1=1+2xx0R_1(new)= R_1(old)+ 2 R_3(new) 0 0=0+2xx0R_1(new)= R_1(old)+ 2 R_3(new) (-2) -2=0+2xx(-1) (pivot element)R_1(new)= R_1(old)+ 2 R_3(new) --- S_2 0 10 10=10R_2(new)= R_2(old) 1 1=1R_2(new)= R_2(old) 0 0=0R_2(new)= R_2(old) 0 0=0R_2(new)= R_2(old) 1 1=1R_2(new)= R_2(old) 0 0=0R_2(new)= R_2(old) --- x_2 5 1 1=1R_3(new)= R_3(old) 0 0=0R_3(new)= R_3(old) 1 1=1R_3(new)= R_3(old) 0 0=0R_3(new)= R_3(old) 0 0=0R_3(new)= R_3(old) -1 -1=-1R_3(new)= R_3(old) --- Z=5 5=5xx1Z_j=sum C_B X_B Z_j Z_j=sum C_B x_j 0 0=0xx1+0xx1+5xx0Z_j=sum C_B x_1 5 5=0xx0+0xx0+5xx1Z_j=sum C_B x_2 0 0=0xx1+0xx0+5xx0Z_j=sum C_B S_1 0 0=0xx0+0xx1+5xx0Z_j=sum C_B S_2 -5 -5=0xx(-2)+0xx0+5xx(-1)Z_j=sum C_B S_3 C_j-Z_j 3 3=3-0 0 0=5-5 0 0=0-0 0 0=0-0 5 5=0-(-5)uarr

Variable S_3 should enter into the basis, but all the coefficients in the S_3 column are negative or zero. So S_3 can not be entered into the basis.

Hence, the solution to the given problem is unbounded.

This material is intended as a summary. Use your textbook for detail explanation.
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