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

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Lecture 5 Simplex Method II February 4, 2009
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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
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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
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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 0 , 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
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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 0 , 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
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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. Keep increasing the chosen nonbasic variable until one of the basic variable drops to 0. (We want to increase as much as possible, but we need to satisfy the nonnegativity con- straints.) The basic variable which drops to 0 actually leaves the basis.
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