GE330_lect5 - Lecture 5 Simplex Method II February 4, 2009...

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Unformatted text preview: Lecture 5 Simplex Method II February 4, 2009 Moving From One BFS to Another Two adjacent vertices correspond to two BFSs whose basis have m- 1 common basic variables (only one different basic variable). In Example 1: A and B are adjacent vertices, the basis of the corresponding BFSs are ( s 1 ,s 2 ) and ( x 2 ,s 1 ), respectively. s 1 is the common one. When move from one BFS to another one, one variable leaves the basis and one enters the basic set. We need to determine: when to move: optimality conditions how to move: which variable enters the basis and which leaves the basis. 2 Checking Optimality At a BFS, increase a single nonbasic variable, leave other nonbasic variables unchanged, check if the objective value is improved. (Notice that we need to computer the cor- responding changes in basic variables to preserve equality constraints.) To check the improvement, we can solve the basic variables in terms of the nonbasic variables, and substitute into the ob- jective function. The coefficients of the nonbasic variables in the substituted objective function indicate the improve- ment. If a coefficient is negative (positive) then increasing the cor- responding nonbasic will decrease (increase) the objective value. These coefficients are called the reduced costs . 3 Checking Optimality: Examples Back to Example 1: max z = 2 x 1 + 3 x 2 2 x 1 + x 2 + s 1 = 4 x 1 + 2 x 2 + s 2 = 5 x 1 ,x 2 ,s 1 ,s 2 , If x 2 and s 1 are basic variables, we can get x 2 = 5 2- 1 2 x 1- 1 2 s 2 s 1 = 3 2- 3 2 x 1 + 1 2 s 2 Therefore z = 15 2 + 1 2 x 1- 3 2 s 2 . Now we know that if x 1 increases 1 unit, the objective value increases 1 2 unit. So the current solution is not optimal. Notice that, the objective value of the current solution is actually 15 2 . 4 Checking Optimality: Examples Still Example 1: max z = 2 x 1 + 3 x 2 2 x 1 + x 2 + s 1 = 4 x 1 + 2 x 2 + s 2 = 5 x 1 ,x 2 ,s 1 ,s 2 , If now x 1 and x 2 are basic variables, we can get x 1 = 1- 2 3 s 1 + 1 3 s 2 x 2 = 2 + 1 3 s 1- 2 3 s 2 Therefore z = 8- 1 3 s 1- 4 3 s 2 . Now the coefficients of both s 1 and s 2 are negative, therefore, increasing either of them can only decrease the objective value. The solution is optimal! The optimal value is 8. 5 The Entering and Leaving Variable Notice that the reduced costs not only tell us whether we should bring in a nonbasic variable but also tell us the rate of improvement for bringing in a nonbasic variable. We can choose to bring in the nonbasic with the best im- provement rate (break ties arbitrarily). This is the entering variable. To preserve the equality constraint, the values of some of the basic variables might have to decrease....
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GE330_lect5 - Lecture 5 Simplex Method II February 4, 2009...

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