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
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.
To solve this linear programming problem using the simplex method, we follow these steps:
Step 1: Set up the initial tableau
First, we need to convert the problem into canonical form:
Maximize Z = 2x1 + 4x2 + x3 + x4
Subject to:
x1 + 3x2 + x4 ≤ 4
2x1 + x2 ≤ 3
x2 + 4x3 + x4 ≤ 3
x1, x2, x3, x4 ≥ 0
We introduce slack variables (s1, s2, s3) to convert the inequalities into equalities:
x1 + 3x2 + x4 + s1 = 4
2x1 + x2 + s2 = 3
x2 + 4x3 + x4 + s3 = 3
We express the objective function in terms of these variables:
Z = 2x1 + 4x2 + x3 + x4 + 0s1 + 0s2 + 0s3
Now we can set up the initial tableau:
| x1 | x2 | x3 | x4 | s1 | s2 | s3 | RHS |
----------------------------------------------------------------------
Cj | 2 | 4 | 1 | 1 | 0 | 0 | 0 | 0 |
----------------------------------------------------------------------
4 | 1 | 3 | 0 | 1 | 1 | 0 | 0 | 4 |
----------------------------------------------------------------------
3 | 2 | 1 | 0 | 0 | 0 | 1 | 0 | 3 |
----------------------------------------------------------------------
3 | 0 | 1 | 4 | 1 | 0 | 0 | 1 | 3 |
----------------------------------------------------------------------
Zj-Cj| -2 | -4 | -1 | -1 | 0 | 0 | 0 | 0 |
-----------------------------------------------------------------------
Step 2: Check for optimality
Check if all the coefficients in the bottom row (Zj-Cj) are non-negative. If any of them are negative, the tableau is not optimal.
In this case, since there are negative coefficients (-2, -4, -1, -1), the current solution is not optimal.
Step 3: Determine the pivot column
Identify the most negative coefficient in the bottom row. The corresponding column will be the pivot column.
In this case, the most negative coefficient is -4, which corresponds to the x2 column. Therefore, the pivot column is x2.
Step 4: Determine the pivot row
To determine the pivot row, calculate the ratios of the RHS column to the positive elements in the pivot column. The smallest non-negative ratio corresponds to the pivot row.
In this case, the ratios are 4/3 and 3/1. The smallest non-negative ratio is 3/1, which corresponds to the first row. Therefore, the pivot row is 1.
Step 5: Pivot operation
Perform the pivot operation to make the pivot element (1 in the pivot row and pivot column) equal to 1 by dividing the pivot row by the pivot element.
New tableau:
| x1 | x2 | x3 | x4 | s1 | s2 | s3 | RHS |
----------------------------------------------------------------------
Cj | 0 | 0 | 1/3 | 1/3 | -8/3 | 0 | 0 | 4 |
----------------------------------------------------------------------
4 | 1 | 3 | 0 | 1 | 1 | 0 | 0 | 4 |
----------------------------------------------------------------------
3 | 2 | 1 | 0 | 0 | 0 | 1 | 0 | 3 |
----------------------------------------------------------------------
3 | 0 | 1 | 4 | 1 | 0 | 0 | 1 | 3 |
----------------------------------------------------------------------
Zj-Cj| -2 | 0 | -1 | -2 | 1 | 0 | 0 | -6 |
-----------------------------------------------------------------------
Step 6: Iterate until optimality is reached
Repeat steps 2-5 until all coefficients in the bottom row are non-negative.
Second tableau:
| x1 | x2 | x3 | x4 | s1 | s2 | s3 | RHS |
----------------------------------------------------------------------
Cj | 0 | 0 | 1/3 | 1/3 | -8/3 | 0 | 0 | 4 |
----------------------------------------------------------------------
1 | 1 | 3 | 0 | 1 | 1 | 0 | 0 | 4 |
----------------------------------------------------------------------
3/2| 0 | -1/2 | 0 | -1/2 | -1/2 | 1 | 0 | -3/2 |
----------------------------------------------------------------------
1 | 0 | 1 | 4 | 1 | 0 | 0 | 1 | 3 |
----------------------------------------------------------------------
Zj-Cj| 0 | 0 | 0 | 0 | -2 | -4 | -1 | -9 |
-----------------------------------------------------------------------
Since all coefficients in the bottom row (Zj-Cj) are non-negative, we have reached the optimal solution.
The optimal solution is:
x1 = 4
x2 = 0
x3 = 0
x4 = 3/2
The maximum value of Z is -9.