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SPECIAL CASES 247

This then gives an initial tableau:

x 1 x 2 x 3 s 1 s 2 s 3 s 4 s 5 a 5 Value
Basis c B 1 1.2 2 0 0 0 0 0 M
0 1 2 0 1 0 0 0 0 0 150
s 1
0 1 0 2 0 1 0 0 0 0 150
s 2
0 2 1 0 0 0 1 0 0 0 80
s 3
0 2 3 1 0 0 0 1 0 0 225
s 4
a 5 M 1 0 0 0 0 0 0 1 1 25
z j M 0 0 0 0 0 0 M M 25M
1þM 1.2 2 0 0 0 0 M 0
c j – z j
On inspection of the c j – z j row, we see that three non-basic variables would improve the current basic solution: x 1 ,
x 2 and x 3 . However, because of the M value, we identify x 1 as the pivot column. Calculating the appropriate ratios
(not shown here) indicates row a 5 as the pivot row and therefore the pivot element is 1 (the intersection of column
x 1 and row a 5 ). So, a 5 is set to leave the solution and x 1 to enter. In the problem context this makes sense. We must
produce at least 25 litres of x 1 no matter what else we do, so the Simplex method forces this solution as a first step.
Completing the necessary arithmetic transformation on the initial table we have our next basic solution as:

Value
x 1 x 2 x 3 s 1 s 2 s 3 s 4 s 5 a 5
Basis c B 1 1.2 2 0 0 0 0 0 M
0 0 2 0 1 0 0 0 1 1 125
s 1
0 0 0 2 0 1 0 0 1 1 125
s 2
0 0 1 0 0 0 1 0 2 2 0
s 3
0 0 3 1 0 0 0 1 2 2 175
s 4
1 1 0 0 0 0 0 0 1 1 25
x 1
1 0 0 0 0 0 0 1 1 25
z j
0 1.2 2 0 0 0 0 0 M 1
c j – z j
For FJC this means they are producing 25 litres of Sweet Grape and nothing else to give a profit contribution of 25R.
They still have 125 kilos each of Grade A and Grade B grapes unused, have 30 kilos of natural flavourings left and 175
hours of labour remain unused. Is there an even better solution? We note from the c j – z j row that there are still positive
values in this row so further improvement is possible. We also note that the artificial variable has been removed from
thebasis sophaseIof thesolutioniscomplete. Wecanomitthe a 5 column from further iterations.
The pivot column is x 3 , we determine from the ratio calculations that the pivot row is s 2 so the pivot element
is 2. Using this to transform the tableau in the usual way we have:

Value
x 1 x 2 x 3 s 1 s 2 s 3 s 4 s 5
Basis c B 1 1.2 2 0 0 0 0 0

s 1 0 0 2 0 1 0 0 0 1 125
x 3 2 0 0 1 0 0.5 0 0 0.5 62.5
s 3 0 0 1 0 0 0 1 0 2 30
s 4 0 0 3 0 0 0.5 0 1 1.5 112.5
x 1 1 1 0 0 0 0 0 0 1 25
1 0 2 0 1 0 0 0 150
z j
0 1.2 0 0 1 0 0 0
c j – z j

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