Solving (cont)
z =
5M
+ (M
1)x
M e
(10)
a =
5
x +
e
(11)
Remember that M is a big number. We choose x as the entering variable,
and a as the leaving variable.
z =
5
e + (1
M )a
(12)
x =
5 + e
a
(13)
All the coefficients in the objective function are negative, so we have an
optimal solution. Notice that the value of a is 0, which means that the
original LP is feasible. The value of x is 5 and the objective function is
5.
Negating that we get that the optimal objective function value is 5, as we
expected.