Chapter4

# Second basic feasible solution or cornerpoint if

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Unformatted text preview: ally). basic variable Z x1 s2 s3 eqn. no. 0 1 2 3 Z x1 x2 s1 s2 s3 RHS MRT 1 0 0 0 0 1 0 0 -10 0 1 1 15 1 0 -1 0 0 1 0 0 0 0 1 30 2 3 2 never Tableau 4.3: The second basic feasible solution. As before, the values of the basic variables at the second feasible solution can be read directly from Tableau 4.3. For example, the value of x1 is now 2. Recalling that nonbasic variables are always zero, the new basic feasible solution is (x1,x2,s1,s2,s3) = (2,0,0,3,2). The value of Z can also be read directly from the tableau because it is the basic variable for equation 0: its value now is 30, which is an improvement over the value at the initial basic feasible solution. Practical Optimization: a Gentle Introduction http://www.sce.carleton.ca/faculty/chinneck/po.html ©John W. Chinneck, 2000 5 In putting the tableau into proper form, we have reached the second basic feasible solution. This process of moving from one basic feasible solution to the next is called pivoting, and so we have just completed one pivot. Completing the ABC solution We can see from Tableau 4.3 that have not yet reached the optimal solution because there are still negative coefficients available in the objective function row, equation 0. So the solution process continues for another iteration, as follows: • Step 2.2: x2 is the entering basic variable (see Tableau 4.4). • Step 2.3: the minimum ratio test shows that s3 is the leaving basic variable (see Tableau 4.4). • Step 2.4: Tableau 4.5 shows the revised tableau in proper form. basic variable Z x1 s2 s3 eqn. no. 0 1 2 3 Z x1 x2 s1 s2 s3 RHS MRT 1 0 0 0 0 1 0 0 -10 0 1 1 15 1 0 -1 0 0 1 0 0 0 0 1 30 2 3 2 never no limit 3/1 = 3 2/1 = 2 Tableau 4.4: The second iteration. basic variable Z x1 s2 x2 eqn. no. 0 1 2 3 Z x1 x2 s1 s2 s3 RHS MRT 1 0 0 0 0 1 0 0 0 0 0 1 5 1 1 -1 0 0 1 0 10 0 -1 1 50 2 1 2 never Tableau 4.5: The optimal tableau. Applying Step 2.1 shows that we have reached the optimum solution in Tableau 4.5: there are no negative variable coefficients in the objective function. This means that there is no improving direction in which to move. We can now read the optimum solution from the tableau (remembering that any variables not shown in the basic variables column must be nonbasic and hence zero): (x1,x2,s1,s2,s3) = (2,2,0,1,0). The optimum value of the objective function is the value of Z shown in the tableau: 50. The optimum solution shows that the Acme Bicycle Company should produce 2 mountain bikes per day and 2 racing bicycles per day for a total daily profit of \$50 per day. Special Cases in Tableau Manipulation There are several unusual cases that arise as you progress from tableau to tableau while carrying out the simplex method. Each of these special cases has an interesting interpretation. Practical Optimization: a Gentle Introduction http://www.sce.carleton.ca/faculty/chinneck/po.html ©John W. Chinneck, 2000 6 Tie for the Entering Basic Variable Suppose that there are two or more nonbasic variables that have the same most n...
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