Question

In: Advanced Math

solve the following LP. Formulate and algebraically solve the problem. what is the new optimal z...

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

Solutions

Expert Solution

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


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