Maximize P= 8x1+12x2
Subject to
2x1+4x2<=24
3x1+3x2<=21
4x1+2x2<=20
X1, x2 => 0
Using the SIMPLEX method. Clearly state the row operations used.
For each tableau used, complete the leftmost column by providing
the appropriate basic variables.
a visit to
http://www.zweigmedia.com/RealWorld/simplex.html
will show the tableaux and solution so you can verify your work.
To maximize the objective function P = 8x1 + 12x2, subject to the given constraints, we can use the simplex method. This method involves creating a tableau and performing row operations to iteratively improve the objective function value.
Step 1: Convert the given problem into a standard form by introducing slack variables for each constraint.
The problem becomes:
Maximize P = 8x1 + 12x2
Subject to:
2x1 + 4x2 + s1 = 24 (constraint 1)
3x1 + 3x2 + s2 = 21 (constraint 2)
4x1 + 2x2 + s3 = 20 (constraint 3)
x1, x2, s1, s2, s3 ≥ 0
The initial tableau is as follows:
| Basis | x1 | x2 | s1 | s2 | s3 | RHS |
|-------|----|----|-----|-----|-----|-----|
| s1 | 2 | 4 | 1 | 0 | 0 | 24 |
| s2 | 3 | 3 | 0 | 1 | 0 | 21 |
| s3 | 4 | 2 | 0 | 0 | 1 | 20 |
| P | -8 | -12| 0 | 0 | 0 | 0 |
Step 2: Find the pivot element. The pivot element is the most negative entry in the bottom row (excluding the rightmost column).
In this case, the pivot element is -12.
Step 3: Perform row operations to make all other entries in the pivot column zero. Divide each entry in the pivot row by the pivot element (-12) to make the pivot element equal to 1.
The row operations for this step are as follows:
R1 → R1/(-12)
R2 → R2 - (R1 * 3)
R3 → R3 - (R1 * 4)
P → P - (R1 * (-12))
The updated tableau is as follows:
| Basis | x1 | x2 | s1 | s2 | s3 | RHS |
|-------|------|------|------|------|------|-----|
| x2 | 1/3 | 2/3 | 1/12| 0 | 0 | 8 |
| s2 | 7/3 | -1/3 | -1/4 | 1 | 0 | 45 |
| s3 | 10/3 | -2/3 | -1/4 | 0 | 1 | 40 |
| P | 8/3 | 0 | -1/4 | 0 | 0 | 96 |
Step 4: Repeat steps 2 and 3 until all coefficients in the bottom row (excluding the rightmost column) are non-negative.
In the updated tableau, the most negative entry in the bottom row is -1/4. Therefore, we need to continue with the row operations.
The pivot element is -1/4.
Step 3: Perform row operations to make all other entries in the pivot column zero. Divide each entry in the pivot row by the pivot element (-1/4) to make the pivot element equal to 1.
The row operations for this step are as follows:
R1 → R1 - (R2 * (1/3))
R3 → R3 - (R2 * (10/3))
P → P - (R2 * (8/3))
The updated tableau is as follows:
| Basis | x1 | x2 | s1 | s2 | s3 | RHS |
|-------|----|----|-------|------|------|-----|
| x2 | 0 | 1 | -1/12 | 2/3 | -1/12| 11 |
| x1 | 1 | 0 | 1/12 | -1/3 | 1/12| 13 |
| s3 | 0 | 0 | 1/3 | 4/3 | 1/3 | 45 |
| P | 0 | 0 | 5/6 | 8/3 | 2/3 | 48 |
Step 4: Repeat steps 2 and 3 until all coefficients in the bottom row (excluding the rightmost column) are non-negative.
In the updated tableau, all coefficients are non-negative.
Therefore, the optimal solution is x1 = 13, x2 = 11, with a maximum objective function value of P = 48.
The row operations used were:
1. R1 → R1/(-12)
2. R2 → R2 - (R1 * 3)
3. R3 → R3 - (R1 * 4)
4. P → P - (R1 * (-12))
5. R1 → R1 - (R2 * (1/3))
6. R3 → R3 - (R2 * (10/3))
7. P → P - (R2 * (8/3))
I hope this helps! Let me know if you have any further questions.