#7C
The Simplex method, PART 3
Example:
Beaver Creek Pottery Company
EXERCISE
For the second iteration of the table (handout #7B) …
… 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 #7B. Highlight the pivot column and the pivot row.
THIRD ITERATION OF THE TABLEAU
Simplex Tableau for this Model
3rd Iteration
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Cj |
Basic Variables |
Quantity (RHS) |
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Zj |
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Cj - Zj |
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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 7A, 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 7A, Step 6).
This is the optimal solution! How do we know this?