To find all optimal solutions for the given LP, we need to solve the LP problem using the simplex method.

The initial LP problem is as follows: Maximize z = x1 + x2 Subject to: x1 + x2 + x3 ≤ 1 x1 + 2*x3 ≤ 1 xi ≥ 0

We convert the inequalities into equations by introducing slack variables: x1 + x2 + x3 + s1 = 1 x1 + 2x3 + s2 = 1

The initial tableau is:

Basic Variables  |  x1   x2   x3   s1   s2   z   |
-----------------------------------------------
        -z       |   -1   -1    0    0    0    0  |  0
-----------------|-------------------------------
        s1       |    1    1    1    1    0    0  |  1
-----------------|-------------------------------
        s2       |    1    0    2    0    1    0  |  1

The pivot column is x2 (the most negative coefficient in the -z row). The pivot row is s1 (the minimum positive ratio of the right-hand side divided by the pivot column coefficient).

Dividing the right-hand side of each row by the pivot column coefficient: s1: 1/(-1) = -1 s2: 1/(-1) = -1

The next tableau is:

Basic Variables  |  x1   x2   x3   s1   s2   z   |
-----------------------------------------------
        -z       |   -1    0    1    1    1    0  |  1
-----------------|-------------------------------
        x2       |    1    1    1    1    0    0  |  1
-----------------|-------------------------------
        s2       |    0    0    1    1    1    0  |  0

The pivot column is x3 (the most negative coefficient in the -z row). The pivot row is s2 (the minimum positive ratio of the right-hand side divided by the pivot column coefficient).

Dividing the right-hand side of each row by the pivot column coefficient: s2: 1/(-1) = -1

The next tableau is:

Basic Variables  |  x1   x2   x3   s1   s2   z   |
-----------------------------------------------
        -z       |   -1    0    0    2    1    0  |  1
-----------------|-------------------------------
        x2       |    1    1    0    1    1    0  |  1
-----------------|-------------------------------
        x3       |    0    0    1    1    1    0  |  0

Since there are no negative coefficients in the -z row, we have reached an optimal solution. The optimal solution is: x1 = 0 x2 = 1 x3 = 0 z = 1

Therefore, the only optimal solution for the given LP problem is x1 = 0, x2 = 1, x3 = 0, z = 1.


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