#5A
The Simplex method, PART 1
Example:
Beaver Creek Pottery Company
X_{1} = # of bowls to produce each day
X_{2} = # of mugs to produce each day
Z_{max } = $40X_{1} + $50X_{2 }+ 0S_{1} + 0S_{2}
The company needs to know how many bowls and mugs to produce each day in order to maximize profit.
_{ }Subject to the following constraints:
X_{1} + 2X_{2} + 1S_{1} + 0S_{2} = 40 labor, hr
4X_{1} + 3X_{2} + 0S_{1} + 1S_{2} = 120 clay, lb
X_{1}, X_{2}, S_{1}, S_{2} >= 0 nonnegativity
Initial Simplex Tableau for this Model


C_{j} 
Basic Variables 
Quantity (RHS) 










Z_{j} 

C_{j}  Z_{j} 
Step 1
(2^{nd} row) Record the model variables:
Step 2
Determine the basic feasible solution. A basic feasible solution:
Why is the following the basic feasible solution?
X_{1} = 0 and X_{2} = 0 and S1 = 40 and S2 = 120
Record the nonzero variables (basic variables) in this solution (2^{nd} column) and the quantity of those variables (in the 3^{rd} column).
Step 3
Record the Cj values (the objective function coefficients, representing the per unit contribution to profit or cost):
Step 4
(4^{th}7^{th} columns) Record the coefficients for the decision variables and the slack variables in the constraint equations.
Step 5
(5^{th} row) Compute the Z_{j} values (the gross profit). To do this, multiply each C_{j} value in the 1^{st} column by each column value under Quantity, the decision variables, and the slack variables; then sum each set of values.
Example:
Z_{j }for Quantity
0 * 40 = 0
0 * 120 = 0
0 + 0 = 0
Example:
Z_{j }for X_{1}
0 * 1 = 0
0 * 4 = 0
0 + 0 = 0
Step 6
(6^{th} row) Compute the Cj – Zj values (net increase in profit associated with one additional unit of each variable). To do this, subtract the Z_{j} row values from the C_{j} (top row) values.
Example:
C_{j} – Z_{j} for X_{1}
40 – 0 = 40
Step 7
Examine the profit represented by this solution. The profit (the Z value) is found in the Z_{j} row under the Quantity column. Recall that the solution is to produce no bowls and no mugs. Why is this solution not optimal?
Therefore, we want to move to a solution point that will give a better solution. We want to produce some bowls or some mugs; therefore one of these nonbasic variables (variables not in the present basic feasible solution) will enter the solution and become basic.
And one of the variables must leave the solution. Why?
Step 8
Which variable shall enter the solution?
Choose the highest positive value in C_{j} – Z_{j} row. It is 50, in this case, and belongs to X_{2}.
X_{2} becomes the pivot column.
Step 9
Which variable shall leave the solution?
This analysis is performed by dividing the Quantity values of the basic solution variables by the pivot column variables. The leaving variable is the one with the smallest positive quotient.
for S_{1}
40 / 2 = 20
for S_{2}
120 / 3 = 40
S_{1} becomes the pivot row.
The value 2 at the intersection of the pivot row and the pivot column is called the pivot number.
We are now ready to the second simplex tableau and a better solution.