91Ó°ÊÓ

Since we want to maximize p = x - y + z + w, we can rewrite the problem as minimizing the objective function -p = -(x - y + z + w). We can also write the inequality constraints as equalities by introducing slack variables s1, s2, s3, and s4: Minimize: \((-p = -x + y - z - w\)) Subject to: \(x + y + z + s1 = 3\) \(y + z + w + s2 = 3\) \(x + z + w + s3 = 4\) \(x + y + w + s4 = 4\) with all variables \(x\), \(y\), \(z\), \(w\), \(s1\), \(s2\), \(s3\), and \(s4\) non-negative.

We will now set up the initial simplex tableau by putting the objective function and constraints in a table: ``` p x y z w s1 s2 s3 s4 | RHS ---------------------------------------------------- 1 -1 1 -1 -1 0 0 0 0 | 0 0 1 1 1 0 1 0 0 0 | 3 0 0 1 1 1 0 1 0 0 | 3 0 1 0 1 1 0 0 1 0 | 4 0 1 1 0 1 0 0 0 1 | 4 ```

We will use the simplex method to pivot the tableau until all entries in the bottom row (except the RHS value) are non-negative. 1. Find the pivot column: The most negative entry in the bottom row is -1, which occurs twice. Choose the leftmost one, corresponding to the variable x (Column 1). 2. Find the pivot row: Divide the RHS of each row by the corresponding entry in the pivot column. Choose the row with the smallest non-negative ratio. Row 3 has the smallest ratio (4/1), so we pivot on the third row. 3. Update the tableau after pivoting: ``` p x y z w s1 s2 s3 s4 | RHS ---------------------------------------------------- 1 0 1 0 0 0 0 1 0 | 4 0 0 1 0 -1 1 0 -1 1 | 0 0 0 1 1 1 0 1 0 0 | 3 0 1 0 1 1 0 0 1 0 | 4 0 1 1 0 1 0 0 0 1 | 4 ``` 4. Continue applying the simplex method: - Step 1: Find the pivot column: The most negative entry in the bottom row is -1 (variable y, Column 2). - Step 2: Find the pivot row: Row 4 has the smallest ratio (3/1). - Step 3: Update the tableau after pivoting: ``` p x y z w s1 s2 s3 s4 | RHS ---------------------------------------------------- 1 0 0 -1 1 0 1 1 -1 | 1 0 0 1 0 -1 1 0 -1 1 | 0 0 0 0 -1 2 1 1 -1 -1 | 3 0 1 0 1 0 -1 0 2 -1 | 7 0 1 1 0 1 0 0 0 1 | 4 ``` All the entries in the bottom row, except the RHS value, are now non-negative. So we have found the optimal solution.

The optimal solution is obtained from the current tableau by reading the values of the basic variables (p, x, y, z, w, s1, s2, s3, and s4). The values are: \(p = 1\) \(x = 7\) \(y = 4\) \(z = 0\) \(w = 0\) The optimal solution is \(x = 7\), \(y = 4\), \(z = 0\), \(w = 0\), and the maximum value of p is \(p = 1\).