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value in the objective function row. For
example, suppose that the objective function
for the Acme Bicycle Company problem was
actually Z = 15x1 + 15x2 (i.e. Z – 15x1 – 15x2 =
0). What is the interpretation? This means
that both x1 and x2, if allowed to become basic
and increase in value, would both increase Z
at the same rate. See Figure 4.2 for an
example of a cornerpoint at which the rate of
increase would be the same for two entering
basic variables.
So how to handle this situation? Simple:
choose the entering basic variable arbitrarily
from among those tied at the most negative
value, because there is no good way to
determine in advance which selection will
reach the optimum solution in the smallest
number of pivots. Figure 4.3 shows an
example in which two variables are tied, but
for which many more pivots are required to
reach the optimum via one choice (clockwise)
than via the other choice (counterclockwise). Figure 4.2: A tie for the entering basic
variable. In fact, the choice of the “most negative
objective function coefficient” is only a
Figure 4.3: One entering basic variable
heuristic method (rule of thumb) for choosing
may be better than another.
the entering basic variable.
It usually
provides a good entering basic variable, but not always. On the other hand it is very
quick. In commercial solvers for very large LPs, the solver may not even calculate the
objective function coefficients for all of the variables. Instead it might only calculate
them for say 1000 of the variables in a 500,000 variable problem, and take the most
negative from among those 1000. Next time around it might calculate the objective
function coefficients for the next 1000 variables and choose the entering basic variable
only from among that 1000. The only time it might calculate the objective function
coefficients for all 500,000 variables is at the optimum, when it must check that there are
no negative coefficients left at all. Tie for the Leaving Basic Variable
Recall the mantra: “nonbasic, variable set to zero, corresponding constraint is active”.
This should remind you that the choosing the leaving basic variable via the minimum
ratio test is the same as finding out which constraint you first bump into as the entering Practical Optimization: a Gentle Introduction http://www.sce.carleton.ca/faculty/chinneck/po.html ©John W. Chinneck, 2000 7 basic variable increases in value. So
what does it mean when there is a tie
during the minimum ratio test? As
A
C
Figure 4.4 shows, it must mean that
there are two constraints (with their
corresponding basic variables) that are
current basic
B
feasible solution
tied as the first constraints you bump
into.
Choosing the basic variable
associated with constraint A or the basic
variable associated with constraint B as
the leaving basic variable will define the
same new cornerpoint.
Technically
Figure 4.4: A tie for the leaving basic variable.
speaking, the basic feasible solutions
will be different, because ea...
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This note was uploaded on 09/22/2013 for the course IEOR 4004 taught by Professor Sethuraman during the Fall '10 term at Columbia.
 Fall '10
 SETHURAMAN

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