In: Advanced Math
solve the following LP. Formulate and algebraically solve the problem.
what is the new optimal z value
show that the current basis is optimal
max z=65x1+25x2+20x3
8x1+6x2+x3<=48
4x1+2x2+1.5x3<=20
2x1+1.5x2+0.5x3<=8
x2<=5
x1,x2,x3>=0
max z = 65x1 + 25x2 + 20x3
8x1 + 6x2 + x3 <= 48
4x1 + 2x2 + 1.5x3 <= 20
2 x1 + 1.5x2 + 0.5x3 <= 8
x2 <= 5
x1,x2,x3,x4 >= 0
now we can add slack variables to get equations from inequalities.
Equations from given inequalities.
8x1 + 6x2 + x3 + s1 = 48
4x1 + 2x2 + 1.5x3 + s2 = 20
2 x1 + 1.5x2 + 0.5x3 + s3 = 8
x2 + s4 = 5
And - 65X1 - 25X2 - 20X3 + z = 0
Now we can matrix representation of initial tableau for above equations.
X1 X2 X3 S1 S2 S3 S4 p
8 6 1 1 0 0 0 0 48
4 2 3/2 0 1 0 0 0 20
2 3/2 1/2 0 0 1 0 0 8
0 1 0 0 0 0 1 0 5
-65 -25 -20 0 0 0 0 1 0
Here the most negative element in the bottom row will indicates the pivot element so here -65 ,so we have in column 1 so I am taking 1st column as a pivot column and for pivot row the least positive result when last column divided by pivot column will indicates so
i.e. +min (48/8 , 20/4 , 8/2) = 8/2 so 3rd row as a pivot row.
R3-> R3 (1/2)
X1 X2 X3 S1 S2 S3 S4 p
8 6 1 1 0 0 0 0 48
4 2 3/2 0 1 0 0 0 20
1 3/4 1/4 0 0 1/2 0 0 4
0 1 0 0 0 0 1 0 5
-65 -25 -20 0 0 0 0 1 0
R1-> R1 - 8R3 R2-> R2 - 4R3 R5-> R5 + 65R3
X1 X2 X3 S1 S2 S3 S4 p
0 0 -1 1 0 -4 0 0 16
0 -1 1/2 0 1 -2 0 0 4
1 3/4 1/4 0 0 1/2 0 0 4
0 1 0 0 0 0 1 0 5
0 95/4 -15/4 0 0 65/2 0 1 260
Here the most negative element in the bottom row will indicates the pivot element so here -15/4 ,so we have in column 3rd so I am taking 3rd column as a pivot column and for pivot row the least positive result when last column divided by pivot column will indicates so
i.e. +min (4/(1/2) , 4/(1/4)) = 4/(1/2) so 2nd row as a pivot row.
R2-> R2 (2)
X1 X2 X3 S1 S2 S3 S4 p
0 0 -1 1 0 -4 0 0 16
0 -2 1 0 2 -4 0 0 8
1 3/4 1/4 0 0 1/2 0 0 4
0 1 0 0 0 0 1 0 5
0 95/4 -15/4 0 0 65/2 0 1 260
R1-> R1 + R2 R3-> R3 - (1/4)R2 R5-> R5 + (15/4)R2
X1 X2 X3 S1 S2 S3 S4 p
0 -2 0 1 2 -8 0 0 24
0 -2 1 0 2 -4 0 0 8
1 5/4 0 0 -1/2 3/2 0 0 2
0 1 0 0 0 0 1 0 5
0 65/4 0 0 15/2 35/2 0 1 290
So now we did not have any negative elements in bottom row so we can stop the iterations. Now the optimum solution is Maximum Z = 290 At x1= 2 , x2 = 0 , x3 = 8