Simplex, Example of Simplex, How to Simplex,

1. Simplex
Part 2 of 4
Professor Ed Dansereau

2. 2. Initial Feasible Solution
We take the standard form constraints and objective function and fill in the Tableau
(table).
Basis C Object Function in Std Form Q
Basis Variables Subrate RHS
Z row Z
Net Eval

3. 2. Initial Feasible Solution - O.F.
Begin to fill out table by putting Objective Function in Standard Form across top.
Initial Tableau
250 400 0 0 0
Basis C X1 X2 S1 S2 S3 Q
Basis Variables Subrate RHS
Z row Z
Net Eval

4. 2. Initial Feasible Solution - Basis
Next Fill in Basis
Three rules for basic variables within Basis
1. One variable per constraint in Basis
2. For constraints that have slack, use them - less than or equal to
3. If a constraint has an Artificial variable, use it - greater than or equal to and
equal to
The coefficient of the variable comes from the Objective Function

5. 2. Initial Feasible Solution - Basis
In our example we have three constraints, all constraints are “<=” so we use slack,
and the OF coefficient is zero (slack has zero profit).
Initial
Tableau
250 400 0 0 0
Basis C X1 X2 S1 S2 S3 Q
S1 0
Subrate RHS
S2 0
S3 0
Z row Z
Net Eval

6. 2. Initial Feasible Solution - Subrate
Put the coefficient of left hand side of the constraints (everything to the left of the
<= sign) into the subrate. Align with the variables from the Objective function -
make sure the coefficient for X1 is under the X1 variable.
Initial Tableau
250 400 0 0 0
Basis C X1 X2 S1 S2 S3 Q
S1 0 2.5 3 1 0 0
RHS
S2 0 8 4 0 1 0
S3 0 2 6 0 0 1
Z row Z
Net Eval

7. 2. Initial Feasible Solution - Subrate
Notice that the columns under the slack variables (S1, S2, and S3). The are unit
vector columns. A Unit Vector is a column of zeros and only a single 1 which occurs
at the intersection of that variable’s row (in basis) and column. The zeros in the
substitution rate indicates of each slack variables unit vector indicate that the
constraints are independent of each other. In other words, a change in the one
resource (constraint) has no effect on the other resources.
0
S1
1
0
0

8. 2. Initial Feasible Solution - RHS
In the RHS section, fill the the Right Hand Side of each constraint, everything to the
right of the “<=” sign.
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 Z
Net Eval

9. 2. Initial Feasible Solution - Z row
The values in the z row are calculated by multiplying the c (coefficient) values of the
basis by the corresponding value in the sub.rate column and then summing the
results of each row. Z row = ∑(c x sub. rate)
Let us look at just the X1 variable
● (0 * 2.5) + (0 * 8) + (0 * 2) = 0
● All three coefficients for S1, S2, & S3 are zero
● 2.5, 8, & 2 are the corresponding X1 values
Starting at the origin makes this calculation easy.
250
Basis C X1
S1 0 2.5
S2 0 8
S3 0 2
Z row 0

10. 2. Initial Feasible Solution - Z row
Repeat the procedure for all the column variables
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 Z
Net Eval

11. 2. Initial Feasible Solution - Z
Z represents the profit for a maximum or cost for a minimum at the extreme point.
The formula is Z = ∑(c x Q). Notice we use the Q column instead of sub.rate when
calculating Z.
Z = (0*30) + (0*80) + (0*48) = 0
This makes sense. We are starting at
the origin (0,0) and not producing any products. So we are not making any money.
Basis C Q
S1 0 30
S2 0 80
S3 0 48
0

12. 2. Initial Feasible Solution - Net Evaluation
Finally we calculate the Net Evaluation row subtracting the objective function
coefficient for each variable minus the value in the Z row. Net Eval = C - Zrow
● For X1 the OF coefficient is 250
● The Z row is 0
● Net Eval = 250 - 0 = 2503.
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