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OPERATIONS RESEARCHChapter 04 - The Simplex Method Tableau Form
2
The Simplex MethodTableau Form
Start with an LPP in standard form: (Example 1)
.0,...,,x 7 43x 8 22:.
325 .max
521
5321
4321
5432 1
xxxxx
xxxxTSxxxxxZ
Constants5 2 3 -1 1
8
7
1 2 2 1 0
3 4 1 0 1
-1
1
jC
BasisBC
4x
5x
54321 x x x x x
3
The Simplex MethodTableau Form
Where,Basis: Basic variables in the current bfs.Constants: Values of the basic variables. : Coefficients of the variables in the objective function. : Coefficients of the basic variables in the objective function.From the above table, we have: Iteration #1
Relative profits ( ), where
= - {inner product of and the column corresponding to in the canonical system}. Thus,
jC
BC
17878
).1,1(
0,7,8 32154
Z
xxxxx
jC
jC jCBC jx
4)1(312
).1,1(3
02242
).1,1(2
32531
).1,1(5
3
2
1
C
C
C
4
The Simplex MethodTableau Form
Constants5 2 3 -1 1
8
7
1 2 2 1 0
3 4 1 0 1
-1
1
Z=-1 3 0 4 0 0row
BC
jC
Basis54321 x x x x x
4x
C
5x
5
The Simplex MethodTableau Form
Since there are positive values in row, the current solution is not optimal. is the entering variable, because it has the highest relative profit. Thus is the entering variable, and is the leaving variable.Iteration # 2Basic variables:Now, rewrite the system in canonical form wrt the new basic variables.
C
3x 4}7,4min{3 x
43 x 4x
15,0:,3,4 42153 Zxxxnonbasicxx
Constants 5 2 3 -1 1
43
½1 1 ½ 0 5/2 3 0- 1/2 1
31
Z=15 1- 4 0- 2 0row
BC
jC
Basis 54321 x x x x x
3x
5x
C
6
The Simplex MethodTableau Form
Thus,
And the relative profits are:
Iteration # 3So, is the new entering variable, , and is the leaving variable. Now, rewrite the system in canonical form wrt to the new basic variables.
1531234
).1,3(
Z
2112/1
2/1).1,3(1
46231
).1,3(2
1452/52/1
).1,3(5
4
2
1
C
C
C
1x 5/6}5/6,8min{1 x 5x.0,5/17,5/6 54231 xxxxx
7
The Simplex MethodTableau Form
The tableau now is:
Constants5 2 3 -1 1
17/56/5
0 2/5 1 3/5- 1/51 6/5 0- 1/5 2/5
35
Z=81/5 0- 26/5 0- 9/5- 2/5 row
BC
jC
Basis
54321 x x x x x
C
3x
1x
8
The Simplex MethodTableau Form
Therefore,
And the relative profits are:
Since all relative profits are negative, the current solution is optimal. That is
5/815/65/17
).5,3(
Z
05/225/315/25/1
).5,3(1
05/915/915/1
5/3).5,3(1
05/2665/625/65/2
).5,3(2
5
4
C
C
2C
.5/81,0,5/6,5/17 54213 Zxxxxx
9
Summary
1. Express the problem in standard form.2. Start with an initial bfs in canonical form, and set the initial tableau.3. Use the inner product rule to find the relative profits of the nonbasic variables.4. If all the relative profits are nonpositive, then the current solution is optimal. Otherwise, select the
entering variable as the nonbasic variable with the highest relative profit.5. Apply the minimum ratio rule to determine the leaving variable and the value of the entering
variable.6. Perform the pivot operation to get the new tableau and the bfs.( canonical system wrt the new basic
variables).7. Go to step 3.
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Example
0,..., x 3x x 14 x 23x 4 2:.
23 .
51
521
421
321
21
xxx
xxxTSxxZMax
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Important Remark
When the relative profits of all the nonbasic variables are all negative, the optimal solution is unique. And if one relative profit or more of the nonbasic variables is zero, the optimal solution is not unique. For example, if we carry the above example one more step, we get:
is the new entering variable, and is the leaving variable. And the tableau becomes:4/15}10,3/5,4/15min{5 x
5x 3x
constants 3 2 0 0 0
basis
15/413/45/2
0 0 5/8- 1/8 1 0 1 3/8 1/8 01 0- 1/4 ¼ 0
023
Z=14 0 0 0- 1 0
BC
jC
54321 x x x x x
5x
2x
1x
jC
12
Important Remark
Notice now that:
Thus, is the entering variable, andSo, we get the previous iteration. Which means that we have two optimal solutions, both give
Z=14.
14/34/1004/18/18/1
).3,2,0(0
04/34/3004/1
8/38/5
).3,2,0(0
4
3
C
C
3x variable.leaving theis xand ,6},,6min{ 53 x
13
Minimization LPP
A. First Approach1. A negative coefficient in the relative profits row indicates that the corresponding nonbasic
variable when increased will reduce the value of the objective function. Hence, in minimization problems, only those nonbasic variables with negative relative profits are eligible to enter the basis and improve the objective function.
2. The optimal solution is obtained when all coef’s in the rel. profits row are nonnegative. Thus, all previous steps in maximization LPP are the same, except the following modified step: If all the coef.’s in the rel. profits row are positive or zero, then the current basic feasible solution is optimal. Otherwise, select the nonbasic variable with the lowest (most negative) value in the rel. profits row to enter the basis.
B. Second ApproachExample: minimize becomes maximize , and then set the
required minimum value to be the negative of the solution you obtain. 21 3640 xxZ 21 3640 xxZ
