#5C

The Simplex method, PART 3

Example:

Beaver Creek Pottery Company

HOMEWORK EXERCISE

For the second iteration of the table (handout #5B) …

… Enter the values in the Zj row.

… Enter the values in the Cj – Zj row.

… What is the entering variable?

… What is the leaving variable?

Enter the above values in the tableau on handout #5B. Highlight the pivot column and the pivot row.

THIRD ITERATION OF THE TABLEAU

Simplex Tableau for this Model

3rd Iteration

 Cj Basic Variables Quantity (RHS) Zj Cj - Zj

New pivot row values

new pivot row values = old pivot row values / pivot number

Quantity:

X1:

X2:

S1:

S2:

Remaining rows

(In this case, there is only one remaining row.)

new row values = old row values -

(corresponding coefficients in pivot column * corresponding new tableau pivot row value)

Quantity:

X1:

X2:

S1:

S2:

Zj row

These values will be computed in the same way they were in the first and second iterations (see handout 5A, Step 5).

Quantity:

X1:

X2:

S1:

S2:

Cj – Zj row

These values will be computed in the same way they were in the first and second iteration (see handout 5A, Step 6).

This is the optimal solution! How do we know this?