solve the simplex method LPP Z= 4X1=10X2 SUBJECT TO 2X1=X2<50, 2X1=5X2<100, 2X1=3X2<90 AND X1,X2>0.
Answers
Explanation:
solve the simplex method LPP Z= 4X1=10X2 SUBJECT TO 2X1=X2<50, 2X1=5X2<100, 2X1=3X2<90 AND X1,X2>0.
Answer:
Explanation:
To solve the linear programming problem using the simplex method, we need to convert the problem into standard form and then apply the simplex algorithm. Let's begin:
Convert the problem to standard form:
We have the objective function: Z = 4X1 + 10X2
Subject to the following constraints:
2X1 - X2 ≤ 50 (1)
2X1 - 5X2 ≤ 100 (2)
2X1 - 3X2 ≤ 90 (3)
X1, X2 ≥ 0
To convert the inequalities into equations, we introduce slack variables S1, S2, and S3:
2X1 - X2 + S1 = 50 (4)
2X1 - 5X2 + S2 = 100 (5)
2X1 - 3X2 + S3 = 90 (6)
X1, X2, S1, S2, S3 ≥ 0
Now, we can rewrite the objective function in terms of the decision variables and slack variables:
Z = 4X1 + 10X2 + 0S1 + 0S2 + 0S3
Create the initial tableau:
Construct a tableau with the coefficients of the variables in the objective function and the constraints:
Basic Variables X1 X2 S1 S2 S3 RHS
Z -4 -10 0 0 0 0
S1 2 -1 1 0 0 50
S2 2 -5 0 1 0 100
S3 2 -3 0 0 1 90
Perform the simplex algorithm:
a. Select the most negative coefficient in the Z row, which is -10. We choose X2 as the entering variable.
b. Calculate the minimum ratio test (RHS/entering variable coefficient) to determine the leaving variable. The minimum ratio should be positive.
For X2, the minimum ratio is min(50/-1, 100/-5, 90/-3), which is 10. We choose S2 as the leaving variable.
c. Perform row operations to make the entering variable coefficient 1 and other coefficients in its column 0.
Divide row 2 by -5:
Basic Variables X1 X2 S1 S2 S3 RHS
Z -4 -10 0 0 0 0
S1 2 -1 1 0 0 50
S2 -0.4 1 0 -0.2 0 -20
S3 2 -3 0 0 1 90
Multiply row 2 by -1/5:
Basic Variables X1 X2 S1 S2 S3 RHS
Z -4 -10 0 0 0 0
S1 2 -1 1 0 0 50
S2 0.08 -0.2 0 0.04 0 4
S3 2 -3 0 0 1 90
Perform row operations to eliminate the X2 column:
Multiply row 1 by -0.2:
Basic Variables X1 X2 S1 S2 S3 RHS
Z 0 0 -0.2 0 0 -10
S1 0.4 1 -0.2 0 0 -10
S2 0.08 0.2 0.04 1 0 4
S3 0.6 0 -0.6 0 1 -10
d. Repeat steps a, b, and c until all coefficients in the Z row are non-negative.
The final tableau is:
Basic Variables X1 X2 S1 S2 S3 RHS
Z 0 0 -0.2 -0.4 0 -10
X1 0.4 1 -0.2 0 0 -10
S2 0.08 0.2 0.04 1 0 4
S3 0.6 0 -0.6 0 1 -10
Read the solution:
The optimal solution is obtained when all coefficients in the Z row are non-negative.
X1 = -10
X2 = 0
Z = -10
However, since the solution for X1 is negative, the problem is infeasible. There is no feasible solution that satisfies all the given constraints.