nment Submission his assignment, you submit answers by question parts. The numbe

Question

nment Submission his assignment, you submit answers by question parts. The number of submissions remaining for euch question part only changes f y nment Scoring best submission for each question part is used for your score. remaining for each O -120 polnts TanFin11 4.1.024 Solve the linear programming problem by the simplex method. (There may be more than one correct answer.) Maximize P=2x+2y+3z subject toy + 2z s 10 3x 3y + 2z s 145 The maximum is P = 1 at (x, y, z)= Need Help? Resd Talk to a Tutor 120 points TanFin11 4.1.028 Solve the linear programming problem by the simplex method. Maximize x + 2y-z

Explanation / Answer

Max P = 2X + 2Y + 3Z

subject to   X + Y + 2Z 10

3X + 3Y + 2Z 145       

X,Y,Z >= 0

Given linear problem is a standard linear problem so we can add slack variables to get

Equations from given inequalities.

X + Y + 2Z + S1 = 10

3X + 3Y + 2Z + S2 = 145

And     -2X – 2Y - 3Z + P =0

Now we can matrix representation of initial tableau for above equations.

X      Y      Z      S1     S2     P            

1      1      2      1       0      0      10    

3      3      2      0      1      0     145   

-2     -2     -3     0      0        1         0     

   Here the most negative element in the bottom row will indicates the pivot column so we have maximum negative element is in 3rd   column and for For pivot row the least positive result when last column divided by pivot column will indicates

i.e. +min(10/2,145/2) = 10/2 so 1st    row is a pivoted row

R1-> R1 (1/2)

X Y Z      S1 S2      P            

1/2    1/2    1      1/2     0 0       5

3         3 2 0      1        0     145   

-2     -2     -3 0 0       1         0     

R2-> R2 – 2R1       R3 -> R3 + 3R1

X Y Z S1 S2      P            

1/2    1/2     1      1/2      0       0      5     

2 2       0      -1 1       0      135   

-1/2   -1/2    0      3/2    0       1      15    

we have maximum negative element is in 1st and 2nd   column are same so here I am taking 1st column is pivot column and for pivot row the least positive result when last column divided by pivot column will indicates

+min(5/(1/2),135/2) = 5/(1/2) so 1ST row is a pivot row.

R1-> R1 (2)

X Y Z      S1 S2      P            

1 1 2 1 0 0        10

2 2       0 -1 1       0      135   

-1/2   -1/2    0      3/2    0 1      15    

R2-> R2 – (2)R1   R3 -> R3 + (1/2)R1

X      Y Z     S1   S2      P            

1      1      2      1      0      0      10    

0      0     -4    -3 1      0      115   

0      0      1      2      0      1      20    

So now we did not have any negative elements in bottom row so we can stop the iterations.Now the optimum solution is Maximum P = 20   At    X = 10 , Y = 0 ,Z = 0.

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