Maximize the following objective function using simplex method:
Z=10 X_(1 )+8 X_2
Subject To:
3 X_1+1X_2 ≤4500
2 X_1+2X_2 ≤4000
X_1+3X_2 ≤4500
With
X_1 ≥0
X_2 ≥0
To solve this linear programming problem using the Simplex method, we will proceed with the following steps:
Step 1: Set up the initial tableau
Convert the given linear programming problem into standard form. Introduce slack variables to convert the inequality constraints into equality constraints. The initial tableau will look like this:
```
-----------------------------------------------------------
| | X1 | X2 | S1 | S2 | S3 | RHS |
-----------------------------------------------------------
| Z | 10 | 8 | 0 | 0 | 0 | 0 |
-----------------------------------------------------------
| S1 | 3 | 1 | 1 | 0 | 0 | 4500 |
-----------------------------------------------------------
| S2 | 2 | 2 | 0 | 1 | 0 | 4000 |
-----------------------------------------------------------
| S3 | 1 | 3 | 0 | 0 | 1 | 4500 |
-----------------------------------------------------------
```
Step 2: Identify the pivot column
Look at the coefficients in the bottom row (Z-row). The most negative coefficient indicates the pivot column. In this case, the most negative coefficient is -10 in column X1.
Step 3: Identify the pivot row
To identify the pivot row, divide the right-hand side (RHS) column by the pivot column values. Find the smallest positive ratio. In this case, the smallest positive ratio is 1500 when dividing 4500 by 3. This means that row S1 is the pivot row.
Step 4: Perform row operations
Perform row operations to make the pivot element (in this case, 3) equal to 1. Divide the entire pivot row by the pivot element.
```
-----------------------------------------------------------
| | X1 | X2 | S1 | S2 | S3 | RHS |
-----------------------------------------------------------
| Z | 10 | 8 | 0 | 0 | 0 | 0 |
-----------------------------------------------------------
| S1 | 1 | 1/3 | 1/3 | 0 | 0 | 1500 |
-----------------------------------------------------------
| S2 | 2 | 2 | 0 | 1 | 0 | 4000 |
-----------------------------------------------------------
| S3 | 1 | 3 | 0 | 0 | 1 | 4500 |
-----------------------------------------------------------
```
Perform row operations to make all other elements in the pivot column equal to zero. Multiply row S1 by -2 and add it to row S2. Multiply row S1 by -1 and add it to row S3.
```
-----------------------------------------------------------
| | X1 | X2 | S1 | S2 | S3 | RHS |
-----------------------------------------------------------
| Z | 10 | 8 | 0 | 0 | 0 | 0 |
-----------------------------------------------------------
| S1 | 1 | 1/3 | 1/3 | 0 | 0 | 1500 |
-----------------------------------------------------------
| S2 | 0 | 4/3 | -2/3 | 1 | 0 | 1000 |
-----------------------------------------------------------
| S3 | 0 | 8/3 | -1/3 | 0 | 1 | 3000 |
-----------------------------------------------------------
```
Step 5: Repeat steps 2-4
Repeat steps 2-4 until all coefficients in the Z-row are non-negative, indicating the optimal solution has been found. The coefficients in the Z-row are currently negative, so we continue with the next iteration.
Step 6: Identify the new pivot column and pivot row
The most negative coefficient in the Z-row is -8 in the X2 column. To identify the pivot row, divide the RHS column by the pivot column values. The smallest positive ratio is 750 when dividing 3000 by 4/3. This means that row S3 is the pivot row.
Step 7: Perform row operations
Perform row operations to make the pivot element (in this case, 4/3) equal to 1. Divide the entire pivot row by the pivot element.
```
-----------------------------------------------------------
| | X1 | X2 | S1 | S2 | S3 | RHS |
-----------------------------------------------------------
| Z | 10 | 0 | 2/3 | -3/4 | -1/4 | -2250 |
-----------------------------------------------------------
| S1 | 1 | 0 | 1/4 | -1/4 | 1/12 | 250 |
-----------------------------------------------------------
| S2 | 0 | 4/3 | -2/3 | 1 | 0 | 1000 |
-----------------------------------------------------------
| S3 | 0 | 1 | -1/6 | 0 | 1/4 | 750 |
-----------------------------------------------------------
```
Perform row operations to make all other elements in the pivot column equal to zero. Multiply row S3 by -2/3 and add it to row S2. Multiply row S3 by -1/6 and add it to row S1. Multiply row S3 by 2/3 and add it to row Z.
```
-----------------------------------------------------------
| | X1 | X2 | S1 | S2 | S3 | RHS |
-----------------------------------------------------------
| Z | 10 | 0 | 1/6 | -3/8 | 0 | 0 |
-----------------------------------------------------------
| S1 | 1 | 0 | 1/4 | -1/4 | 1/12 | 250 |
-----------------------------------------------------------
| S2 | 0 | 0 | -4/3 | 5/3 | 1/6 | 250 |
-----------------------------------------------------------
| S3 | 0 | 1 | -1/6 | 0 | 1/4 | 750 |
-----------------------------------------------------------
```
Step 8: Repeat steps 5-7
Repeat steps 5-7 until all coefficients in the Z-row are non-negative. The coefficients in the Z-row are now non-negative, indicating the optimal solution has been found.
Step 9: Interpret the final solution
The optimal values of the decision variables X1 and X2 can be found in the last column of the tableau.
X1 = 250
X2 = 750
The maximum value of the objective function Z is equal to 2250.
Therefore, the optimal solution to the given linear programming problem is X1 = 250, X2 = 750, and Z = 2250.
To maximize the objective function Z = 10X1 + 8X2 using the simplex method, we need to set up the initial Simplex tableau and then perform iterations. Here are the steps:
Step 1: Set up the initial Simplex tableau
-------------------------------------------
Basic Variables | X1 | X2 | S1 | S2 | S3 | RHS
----------------------------------------------------------------------
Z (Coefficient) | 10 | 8 | 0 | 0 | 0 | 0
----------------------------------------------------------------------
S1 (Constraint 1) | 3 | 1 | 1 | 0 | 0 | 4500
----------------------------------------------------------------------
S2 (Constraint 2) | 2 | 2 | 0 | 1 | 0 | 4000
----------------------------------------------------------------------
S3 (Constraint 3) | 1 | 3 | 0 | 0 | 1 | 4500
----------------------------------------------------------------------
Step 2: Determine the pivot column
---------------------------------
In the bottom row (Z row), find the most negative coefficient. In this case, the most negative coefficient is -10 in the X1 column.
Thus, the pivot column is X1.
Step 3: Determine the pivot row
------------------------------
Divide the right-hand side (RHS) column by the corresponding coefficients in the pivot column. Take the smallest positive value. In this case, the pivot row is S2 with a ratio of 2000 (4000/2).
Step 4: Perform the pivot operation
----------------------------------
Perform the pivot operation to make the pivot element (element in the pivot row and pivot column) equal to 1.
Basic Variables | X1 | X2 | S1 | S2 | S3 | RHS
----------------------------------------------------------------------
Z (Coefficient) | 0 | 4 | 0 | 10 | -30 | 20000
----------------------------------------------------------------------
S1 (Constraint 1) | 0 | 0.5 | 1 | -0.5| 0 | 3500
----------------------------------------------------------------------
S2 (Constraint 2) | 1 | 1 | 0 | 0.5 | 0 | 2000
----------------------------------------------------------------------
S3 (Constraint 3) | 0 | 2.5 | 0 | -0.5| 1 | 2500
----------------------------------------------------------------------
Step 5: Repeat Steps 2-4 until the optimal solution is reached
Perform Steps 2-4 on the current tableau until there are no negative coefficients in the Z row:
Step 2: The most negative coefficient in the Z row is -30 in the S3 column.
So, the pivot column is S3.
Step 3: The pivot row is S1 with a ratio of 14 (3500/0.5).
Step 4: Perform the pivot operation to make the pivot element equal to 1.
Basic Variables | X1 | X2 | S1 | S2 | S3 | RHS
----------------------------------------------------------------------
Z (Coefficient) | 0 | 3 | 0 | 20 | 0 | 30500
----------------------------------------------------------------------
S1 (Constraint 1) | 0 | 0 | 1 | -1 | 0.5 | 500
----------------------------------------------------------------------
S2 (Constraint 2) | 1 | 0 | 0 | 1 | -0.25| 1500
----------------------------------------------------------------------
X1 (Constraint 3) | 0 | 1 | 0 | -0.2 | 0.4 | 1000
----------------------------------------------------------------------
Step 2: All coefficients in the Z row are non-negative. The optimal solution is reached.
The optimal solution is Z = 30500, X1 = 500, X2 = 1500.