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An example of LP problem solved by the Simplex Method
Linear Optimization 2016
Fabio D'Andreagiovanni
Exercise 1
Solve the following Linear Programming problem through the Simplex Method.
max s.t
3 x 1 2 x 1 x 1 2 x 1 x 1
x 2 x 2 2 x 2 2 x 2 x 2
3 x 3 x 3 3 x 3 x 3 x 3
Solution
The rst step is to rewrite the problem in standard form as follows:
min s.t
− 3 x 1 2 x 1 x 1 2 x 1 x 1
x 2 x 2 2 x 2 2 x 2 x 2
3 x 3 x 3 3 x 3 x 3 x 3
x 4
x 4
x 5
x 5
x 6 x 6
Having added the slack variables x 4 , x 5 , x 6 , it is easy to nd the following initial basis:
B = [A 4 A 5 A 6 ] =
and thus to split the decision variables in the following way:
xB =
x 4 x 5 x 6
(^) xN =
x 1 x 2 x 3
The solution associated with the basis B is x = (0, 0 , 0 , 2 , 5 , 6) with value z = 0.
We can then de ne the following simplex tableau:
The rst thing to do is checking the value of the reduced costs in the 0th-row: if all reduced costs are non-negative, then we have a su cient condition of optimality and the solution associated with the current basis is optimal. In our case, three variables, namely x 1 , x 2 , x 3 are associated with the negative reduced costs (-3, -1, -3). The su cient condition is thus not satis ed and we thus proceed to operate a change of basis. Following Bland's rule, we choose as variable entering the basis that with the smallest subscript: we then choose xj with j = min{k : ¯ck < 0 } = min{ 1 , 2 , 3 } = 1. Therefore, x 1 enters the basis and column 1 of the tableau is the pivot column.
Simplex Method - Exercises
Looking at the entries of the pivot column, we can then derive the value θ∗^ considering the values associated with the basic variables So we have:
θ = min k=1, 2 ,3:uk > 0
xk uk
= min
So the minimum is attained for variable x 4 and x 4 exits the basis. The pivot row is thus the row 1 of the tableau and the pivot element is that at the intersection of row 1 and column 1.
In order to get the new tableau corresponding to the new basis:
B = [A 1 A 5 A 6 ] =
we operate the following row operations, aimed at transforming the rst column (2 1 2)T^ of the tableau into the column (1 0 0)T^ using the entries of the pivot row (all entries but the pivot element must become null, while the pivot element must become equal to 1):
- R 0 ←− R 0 + 32 R 1
- R 1 ←− 12 R 1
- R 2 ←− R 2 − 12 R 1
- R 3 ←− R 3 − R 1
The new tableau that we obtain is:
3 0 1/2 -3/2 3/2 0 0 1 1 1/2 1/2 1/2 0 0 4 0 3/2 5/2 -1/2 1 0 4 0 1 0 -1 0 1
associated with the solution (1, 0 , 0 , 0 , 4 , 4) of value z = − 3.
Again, we look at the 0-th row to check the presence of negative reduced costs. We have a single variable associated with negative reduced cost, namely x 3. Thus x 3 enters the basis and the third column of the tableau becomes the pivot column. We derive again the value θ considering the values associated with the basic variables So we have:
θ = min k=1, 2 ,3:uk > 0
xk uk
= min
1 2
5 2
The minimum is then attained for variable x 5 and x 5 exits the basis. The pivot row is thus the row 1 of the tableau and the pivot element is that at the intersection of row 1 and column 3.
In order to get the new tableau corresponding to the new basis:
B = [A 1 A 3 A 6 ] =
we operate the following row operations, aimed at transforming the third column (1/2 5/2 0)T^ of the tableau into a column (1 0 0)T^ using the entries of the pivot row (all entries but the pivot element must become null, while the pivot element must become equal to 1):
- R 0 ←− R 0 + 35 R 2
- R 1 ←− R 1 − 15 R 2
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