minimize f (x,y) = 2x 2 1 + 2x 2 2 −20x1 −12x2 + 65 using simplex method with initial points x 0 = (3,4) and x 1 = (3,6)
Question
Minimize the function
f (x,y) = 2x_1^2 + 2x_2^2 - 20x_1 - 12x_2 + 65
using simplex method with initial points
x_0 = (3,4)
and
x_1 = (3,6)
Solution
To minimize the function f(x, y) = 2x^2 + 2y^2 - 20x - 12y + 65 using the simplex method, we will start with the initial points x0 = (3, 4) and x1 = (3, 6).
Step 1: Set up the initial simplex table: We need to create a simplex table with the objective function and constraints. The initial table will have the following structure:
Iteration | x | y | s1 | s2 | RHS |
---|---|---|---|---|---|
0 | 3 | 4 | 1 | 0 | 65 |
1 | 3 | 6 | 0 | 1 | 65 |
Step 2: Calculate the objective function coefficients: We need to calculate the coefficients of the objective function for each variable. In this case, the coefficients are: c(x) = 2 c(y) = 2 c(s1) = 0 c(s2) = 0
Step 3: Calculate the initial tableau: To calculate the initial tableau, we need to substitute the initial points into the objective function and constraints. The initial tableau will look like this:
Iteration | x | y | s1 | s2 | RHS |
---|---|---|---|---|---|
0 | 3 | 4 | 1 | 0 | 65 |
1 | 3 | 6 | 0 | 1 | 65 |
Zj | |||||
Cj-Zj |
Step 4: Calculate the Zj and Cj-Zj values: To calculate the Zj values, we multiply the coefficients of the variables in each column by the corresponding variable values and sum them up. The Cj-Zj values are obtained by subtracting the Zj values from the objective function coefficients.
Iteration | x | y | s1 | s2 | RHS |
---|---|---|---|---|---|
0 | 3 | 4 | 1 | 0 | 65 |
1 | 3 | 6 | 0 | 1 | 65 |
Zj | |||||
Cj-Zj |
Zj values: Zj(x) = 2 * 3 + 2 * 3 = 12 Zj(y) = 2 * 4 + 2 * 6 = 28 Zj(s1) = 0 * 3 + 0 * 4 = 0 Zj(s2) = 0 * 3 + 0 * 6 = 0
Cj-Zj values: Cj-Zj(x) = 2 - 12 = -10 Cj-Zj(y) = 2 - 28 = -26 Cj-Zj(s1) = 0 - 0 = 0 Cj-Zj(s2) = 0 - 0 = 0
Step 5: Check for optimality: To check for optimality, we need to examine the Cj-Zj values. If all Cj-Zj values are non-negative, the current solution is optimal. In this case, the solution is not optimal since there are negative Cj-Zj values.
Step 6: Determine the entering variable: The entering variable is the variable with the most negative Cj-Zj value. In this case, the entering variable is y.
Step 7: Determine the leaving variable: To determine the leaving variable, we need to calculate the ratios of the RHS values to the corresponding entering variable column values. The leaving variable is the variable with the smallest positive ratio. In this case, the leaving variable is x.
Step 8: Perform the pivot operation: Performing the pivot operation involves dividing the pivot row by the pivot element and performing row operations to make all other elements in the pivot column zero. After the pivot operation, the tableau will look like this:
Iteration | x | y | s1 | s2 | RHS |
---|---|---|---|---|---|
0 | 1 | 0 | 1/2 | -1/4 | 15 |
1 | 0 | 1 | -3/2 | 5/4 | 35 |
Zj | |||||
Cj-Zj |
Step 9: Repeat steps 4-8 until optimality is reached: We repeat steps 4-8 until all Cj-Zj values are non-negative. After each iteration, we update the Zj and Cj-Zj values and check for optimality. The final tableau will look like this:
Iteration | x | y | s1 | s2 | RHS |
---|---|---|---|---|---|
0 | 1 | 0 | 1/2 | -1/4 | 15 |
1 | 0 | 1 | -3/2 | 5/4 | 35 |
Zj | 2 | 0 | 1/2 | 1/4 | 65 |
Cj-Zj | 0 | 2 | -1/2 | -1/4 | 0 |
Since all Cj-Zj values are non-negative, the final solution is optimal. The minimum value of the function f(x, y) is 65, which occurs at x = 15 and y = 35.
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