Simplex

Part 3 of 4Professor Ed Dansereau

3. Test for Optimality

To check to determine if the current extreme point is the optimal solution is a simple interpretation of the Net Eval row.

●For a Maximization problem○ All numbers in the Net Eval row must be Greater or Equal than zero

●For a Minimization problem○ All numbers in the Net Eval row must be Less than or Equal to zero

3. Test for Optimality

Our example we want to maximize profit

For a Max problem the net eval row must be <= 0 for an optimal solution

We do not have an optimal solution.

Net Eval 250 400 0 0 0

4. Identify Entering & Leaving VariablesTo move to the next extreme point, we must identify which variable will move into the basis and which will leave. We start by identifying the Entering Variable by looking at the Net Evaluation row.

For a Maximization problem (Zmax)Choose the number that has the largest positive value, the farthest from zero.

For a Minimization problem (Zmin)Choose the number that has the largest negative value, the farthest from zero.

The column of the entering variable is referred to as the Pivot Column.

4. Identify Entering & Leaving Variables

Initial Tableau

250 400 0 0 0

Basis C X1 X2 S1 S2 S3 Q

S1 0 2.5 3 1 0 0 30

S2 0 8 4 0 1 0 80

S3 0 2 6 0 0 1 48

Z row 0 0 0 0 0 0

Net Eval 250 400 0 0 0

X2 is the entering variable because it is the most positive

4. Identify Entering & Leaving VariablesThe number of values in the basis can not change. To add the entering variable, we must take one away, the leaving variable. To find the leaving variable

For each variable in the basis, divide the Q by the sub.rate of the pivot column. Of the numbers generate select the smallest positive Q/(sub.rate). Negative and undefined numbers may not be selected.

An undefined number is any number divided by zero, 48/0 = undefined.

4. Identify Entering & Leaving Variables The Leaving Variable is S3 because it has the smallest positive Q/(sub.rate)

250 400 0 0 0

Basis C X1 X2 S1 S2 S3 Q Q/(Sub.rate)

S1 0 2.5 3 1 0 0 30 30/3 = 10

S2 0 8 4 0 1 0 80 80/4 = 20

S3 0 2 6 0 0 1 48 48/6 = 8

Z row 0 0 0 0 0 0

Net Eval 250 400 0 0 0

Leaving Variable