#7B
The Simplex method, PART 2
Example:
Beaver Creek Pottery Company
Exercise
a) When one bowl is produced, how much labor slack is used?
b) When one bowl is produced, how much clay slack will be used?
c) When one mug is produced, how much labor slack is used?
d) When one mug is produced, how much clay slack is used?
e) Represent these values in the objective function.
Recall (from handout 7A):
The entering variable: We chose X_{2} because it provides more profit ($50). This is the highest positive value in the C_{j} – Z_{j} row.
The leaving variable: We choose S_{1} as the leaving variable.
We want to produce mugs (the entering variable).
a) How many mugs can be produced with our 40 hours of labor?
b) How many mugs can be produced with our 120 pounds of clay?
Simplex Tableau for this Model
2^{nd} Iteration


C_{j} 
Basic Variables 
Quantity (RHS) 










Z_{j} 

C_{j}  Z_{j} 
The various rows are computed using several simplex functions.
New pivot row values
new pivot row values = old pivot row values / pivot number
Quantity: 40 / 2 = 20
X_{1}: 1 / 2 = ½
X_{2}:
S_{1}:
S_{2}:
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: 120  (3 * 20) = 60
X_{1}: 4  (3 * ½) = 5/2 (or 2.5)
X_{2}:
S_{1}:
S_{2}:
Z_{j} row
These values will be computed in the same way they were in the first iteration (see handout 7A).
Quantity = (50 * 20) + (0 * 60) = 1000
X_{1}: = (50 * ½) + (0 * 5/2) = 25
X_{2}:
S_{1}:
S_{2}:
Cj – Zj row
These values will be computed in the same way they were in the first iteration (see handout 7A).
What is the entering variable?
What is the leaving variable?
Exercise:
Simplex Tableau for this Model
3rd Iteration


C_{j} 
Basic Variables 
Quantity (RHS) 










Z_{j} 

C_{j}  Z_{j} 