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Pivot column corresponds to a negative value (normally the smallest, or
one of them if there are several smallest values) in the index row if
the problem is to maximize the object function, or to a positive value
(normally the greatest) in the index row if the problem is to minimize
(if all the values in the index row are non-negative/non-positive the
maximum/minimum is already attained). In the example given these values
are 0, 0, 2, -4, 0, 1, so when maximizing the pivot (leading) column
should be #4 (corresponding to value -4); when minimizing it is #3
(value 2 - the greatest positive) as shown.
2) The pivot column being determined, consider all POSITIVE components
in it (if there is no positive element in the pivot column, the object
function is unbounded - infinite optimum). Compare the ratios of each
free term and the corresponding positive element in the pivot column -
the least of them determines the pivot (leading) row. If there is more
than one least ratio, make an arbitrary choice between the corresponding
rows. In the example the pivot column is #3, both elements 2 and 4 are
positive, so we must compare 10/2 and 13/4 - the second one is the
smallest, so the pivot row is #2. If the pivot column was #4, the only
ratio is 13/3 in the 2nd row - it is again pivot row.
3) The common element of the pivot row and the pivot column is the pivot
element (it is always positive in the standard simplex-method according
the above). After the iteration the pivot column should become a unit
vector with 1 instead of the pivot element.
Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 16th, 2015
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