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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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 Fall '10
 SETHURAMAN

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