use simplex method solbe LPP
maximize Z=2x1+4x2+x3+x4
subject to x1+3x2+x4<4
2x1+x2<3
x2+4x3+x4<3
x1,x2,x3,x4>0
4 answers
Simplex method
Pls
No way there is 2000+ views but no answer wth bruh
Sorry for the delay in response. Here is the solution using the simplex method:
First, we convert the inequality constraints into equality constraints by adding slack variables:
x1 + 3x2 + x4 + x5 = 4
2x1 + x2 + x6 = 3
x2 + 4x3 + x4 + x7 = 3
We can now write the problem in standard form:
Maximize Z = 2x1 + 4x2 + x3 + x4
subject to:
x1 + 3x2 + x4 + x5 = 4
2x1 + x2 + x6 = 3
x2 + 4x3 + x4 + x7 = 3
x1, x2, x3, x4, x5, x6, x7 ≥ 0
We start with the initial feasible solution:
x1 = x2 = x3 = x4 = 0
x5 = 4
x6 = 3
x7 = 3
The pivot column is chosen as the one with the most negative coefficient in the objective function, which is x2. The pivot row is chosen as the one with the smallest non-negative ratio of the right-hand side to the pivot column coefficient, which is the second constraint:
x1 + 3x2 + x4 + x5 = 4
2x1 + x2 + x6 = 3
x2 + 4x3 + x4 + x7 = 3
The ratio for the second constraint is 3/1 = 3, so x2 enters the basis and x6 leaves the basis. Using row operations, we get:
x1 + 3x2 + x4 + x5 = 4
(2/3)x2 - (1/3)x1 + (1/3)x6 = 1
x4 + (1/2)x3 - (1/2)x2 + (1/2)x7 = 3/2
The next pivot column is x1, and the pivot row is the first constraint:
(4/3)x1 + x2 - (1/3)x5 - (1/3)x4 = 4/3
(2/3)x2 - (1/3)x1 + (1/3)x6 = 1
x4 + (1/2)x3 - (1/2)x2 + (1/2)x7 = 3/2
The ratio for the first constraint is 4/3, so x1 enters the basis and x5 leaves the basis. Using row operations, we get:
x1 + 3x2 + x4 + x8 = 4
(2/3)x2 - (1/3)x1 + (1/3)x6 = 1
x4 + (1/2)x3 - (1/2)x2 + (1/2)x7 = 3/2
The optimal solution is:
Z = 2x1 + 4x2 + x3 + x4 = 2(4/3) + 4(1/3) + 0 + 0 = 10/3
x1 = 4/3
x2 = 1
x3 = 0
x4 = 0
x5 = 0
x6 = 1
x7 = 3/2
x8 = 0
Therefore, the optimal value of Z is 10/3, which occurs when x1 = 4/3, x2 = 1, x3 = 0, and x4 = 0.
First, we convert the inequality constraints into equality constraints by adding slack variables:
x1 + 3x2 + x4 + x5 = 4
2x1 + x2 + x6 = 3
x2 + 4x3 + x4 + x7 = 3
We can now write the problem in standard form:
Maximize Z = 2x1 + 4x2 + x3 + x4
subject to:
x1 + 3x2 + x4 + x5 = 4
2x1 + x2 + x6 = 3
x2 + 4x3 + x4 + x7 = 3
x1, x2, x3, x4, x5, x6, x7 ≥ 0
We start with the initial feasible solution:
x1 = x2 = x3 = x4 = 0
x5 = 4
x6 = 3
x7 = 3
The pivot column is chosen as the one with the most negative coefficient in the objective function, which is x2. The pivot row is chosen as the one with the smallest non-negative ratio of the right-hand side to the pivot column coefficient, which is the second constraint:
x1 + 3x2 + x4 + x5 = 4
2x1 + x2 + x6 = 3
x2 + 4x3 + x4 + x7 = 3
The ratio for the second constraint is 3/1 = 3, so x2 enters the basis and x6 leaves the basis. Using row operations, we get:
x1 + 3x2 + x4 + x5 = 4
(2/3)x2 - (1/3)x1 + (1/3)x6 = 1
x4 + (1/2)x3 - (1/2)x2 + (1/2)x7 = 3/2
The next pivot column is x1, and the pivot row is the first constraint:
(4/3)x1 + x2 - (1/3)x5 - (1/3)x4 = 4/3
(2/3)x2 - (1/3)x1 + (1/3)x6 = 1
x4 + (1/2)x3 - (1/2)x2 + (1/2)x7 = 3/2
The ratio for the first constraint is 4/3, so x1 enters the basis and x5 leaves the basis. Using row operations, we get:
x1 + 3x2 + x4 + x8 = 4
(2/3)x2 - (1/3)x1 + (1/3)x6 = 1
x4 + (1/2)x3 - (1/2)x2 + (1/2)x7 = 3/2
The optimal solution is:
Z = 2x1 + 4x2 + x3 + x4 = 2(4/3) + 4(1/3) + 0 + 0 = 10/3
x1 = 4/3
x2 = 1
x3 = 0
x4 = 0
x5 = 0
x6 = 1
x7 = 3/2
x8 = 0
Therefore, the optimal value of Z is 10/3, which occurs when x1 = 4/3, x2 = 1, x3 = 0, and x4 = 0.
1 answer