W chinneck 2000 6 tie for the entering basic variable

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Unformatted text preview: egative 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.

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