PROBLEM
It is a common practice to use the artificial variable when it is not possible to have a canonical form for a given L.P. problem. This usually happens with the constraints of types and =. By using the artificial variable (A), the problem that is not in standard canonical form will be made canonical.
When artificial variable is used, the usual methods of obtaining the Initial Basic Feasible Solution are:
(a) The Penalty or Big – M Method
(b) The Two-Phase Technique.
(a) Penalty or Big-M Method
In this method, the artificial variable that is added to the problem is assigned very large value per unit or cost M in the objective function.
This large value per unit cost M is negative for the Maximization problem and positive for the Minimization problem. The penalty M will make the basic solution infeasible.
The operation of simplex method will replace these artificial variables with decision variables in the basis with more realistic unit cost.
Succinctly, the summary of the procedural steps of this method is given below:
Step 1: Express the given L.P. problem in standard form.
Step 2: Add artificial variable (Ai, i = 1,2,…..) to the left hand side of all the constraints of type = or . Assign large penalty of negative M to the objective function if it is Maximization problem; while positive M for Minimization problem.
Step 3: Arrange the result of step 2 in tableau.
Step 4: Apply the regular steps of simplex method to what you have in step 3. The simplex iteration will drive out the artificial variables from the basis and allow for real decision variables to enter the basis. With further iteration, the optimal solution will be obtained.
ILLUSTRATION 18-9
Minimize Z = 3x1 + 2x2 Subject to: x1 + x2 5 -2x1 – x2 -6 x1 , x2 0
429 SUGGESTED SOLUTION 18-9
The given problem can be re-written as Min. Z = 3x1 + 2x2
Subject to: x1 + x2 5
2x1 + x2 6
x1 , x2 0
Step 1: To express the problem in standard form as follows:
Min. Z = 3x1 + 2x2 + OS1 + OS2 Subject to: x1 + x2 – S1 5
2x1 + x2 – S2 6
x1 , x2 , S1 , S2 0 Step 2: Artificial variables are added as follows:
Min. Z = 3x1 + 2x2 + OS1 + OS2 + MA1 +MA2 Subject to: x1 + x2 – S1 + OS2 + A1 + OA2 = 5 2x1 + x2 + OS1 – S2 + OA1 + A2 = 6 x1 , x2 , S1 , S2 , A1 , A2 0
Step 3: Table Formulation Table 1
Cj 3 2 0 0 M M
CB XB X1 X2 S1 S2 A1 A2 bi Ratio M
M A1 A2
1 1 -1 0 1 0 2 1 0 -1 0 1 5
6 5
3 To leave the basis
Zj 3M 2M -M -M M M Z=11M Z* =Cj-Zj 3-3M 2-2M M M 0 0
Most negative value of Z* allows X1 to enter into basis.
Based on the above ratios, variable A2 will leave the basis.
Table 2
Cj 3 2 0 0 M M
CB XB X1 X2 S1 S2 A1 A2 bi Ratio M 3 A1
X2
0 0.5 -1 0.5 1 -0.5
1 0.5 0 -0.5 0 0.5 2 3
2/0.5 = 4 To leave the Basis
3/0.5 = 6 Zj 3 (1.5+0.5M) -M (0.5M-1.5) M (1.5-0.5M) Z=9+2M
Z* =Cj-Zj
0 (0.5-0.5M) M (1.5-0.5M) 0 (0.5M-1.5)
Pivot Operations
New R2 = ẵ(old R2) New R1 = old R1 – New R
Table 3
Cj 3 2 0 0 M M
CB XB X1 X2 S1 S2 A1 A2 bi
2 3 X2
X1
0 1 -2 1 2 -1
1 0 1 -1 -1 1 5 6 Zj 3 2 -1 -1 1 1 Z=28
Z* =Cj-Zj 0 0 1 1 M-1 M-1 Pivot Operations
New R2 = 2(old R1)
New R2 = old R2 – ẵ (New R1)
Since All Z* values are positive, the problem is now optimal.
We assign X1 = 6 , X2 = 5 and the optimal value is
Z = 3(6) + 2(5)
= 28.
ILLUSTRATION 18-10 Minimize Z = 5y1 + y2 Subject to: 2y2 10 2y1 6
10y1 + 4y2 40 y1 , y2 0 SUGGESTED SOLUTION 18-10
Step 1: The problem in standard form.
Minimize Z = 5y1 + y2 OS1 + OS2 + OS3
Subject to: Oy1 + 2y2 + S1 = 10
2y1 + Oy2 – S2 = 6
10y1 + 4y2 + S3 = 40
y1 , y2 , S1 , S2 , S3 0 Step 2: Artificial variable added to the result in step 1.
Minimize Z = 5y1 + y2 OS1 + OS2 + OS3 MA1
Subject to: Oy1 + 2y2 + S1 + OS2 + OS3 + OA1 = 10 2y1 + Oy2 + OS1 – S2 + OS3 + A1 = 6 10y1 + 4y2 + OS1 + OS2 + S3 + OA1 = 40
y1 , y2 , S1 , S2 , S3 , A1 0
431 Step 3: Table formulation.
Table 1
Cj 5 1 0 0 0 -M
CB XB y1 y2 S1 S2 S3 A1 bi Ratio 0 -M
0
S1
A1 S3
0 2 1 0 0 0 2 0 0 -1 0 1 10 4 0 0 1 0
10 6 40
Invalid 3 4 Z -2M 0 0 M 0 -M Z = -6M
Z* =Cj-Zj 5+2M 1 0 -M 0 0
The largest Z* is 5 + 2M. Therefore y1 goes into the basis, while A1 (with ratio 3) leaves the basis.
Table 2
Cj 5 1 0 0 0 -M
CB XB y1 y2 S1 S2 S3 A1 bi Ratio 0
5 0 S1
y1
S3
0 2 1 0 0 0 1 0 0 -0.5 0 0.5 0 0.4 0 1 0.1 -1
10 6 4
Not possible Not possible 4
Zj 5 0 0 -2.5 0 2.5 Z = 15 Z* =Cj - Zj 0 1 0 2.5 0 -2.5
Pivot Operations:
New Row 2 = ẵ (old R2) (R = Row) New Row 1 = old R1
New Row 3 = 1/10 (old R3) – New R3
S3 goes into the basis while S3 leaves the basis.
Table 3
Cj 5 1 0 0 0 -M
CB XB y1 y2 S1 S2 S3 A1 bi
0 5 0
S1 y1
S2
0 2 1 0 0 0 1 0.2 0 0 0.05 0 0 0.4 0 1 0.1 -1
10 5 4 Zj 5 1 0 0 0.25 0 Z = 25
Z* =Cj-Zj 0 0 0 0 -0.25 -M
Pivot Operations
New R3 = old R3
New R1 = old R1
New R2 = old R2 + ẵ (New R3)
Since all values of Z* 0, the problem is optimal. Therefore, y1 = 5, y2 = 0 with optimal value of Z = 5(5) + 0 = 25
(b) The Two-Phase Method
The method involved two phases of simplex optimization procedure. In the first phase, the sum of the artificial variables is minimized subject to the constraint of the given L.P. problem. This will help to obtain the initial basic feasible solution, which can be used in the next phase (second phase).
The second phase optimizes the original objective function starting with the basic feasible solution obtained at the end of the first phase.
The summary of procedural steps of the method is as follows:
In phase 1, these are the following steps:
Step 1: Express the given L.P. problem in standard form. Also ensure that all bi constant terms on the right hand side of constraints are non-negative and the constraints of type or = are assigned necessary artificial variables.
Step 2: Obtain a new objective function (w), which consists of the sum of artificial variables and the decision variables with assigned zero coefficients.
Step 3: Form the initial simplex table.
Step 4: Use the simplex method to minimize the function w subject to original constraints (as obtained in step 1) in order to have the optimal basic feasible solution.
If optimal basic feasible solution gives Min. w = 0, then continue with phase II of solving. Otherwise, if Min w 0, you stop because feasible solution does not exist.
In phase II, use the optimum basic feasible solution of the phase I as the starting solution for the original L.P problem. Here, make further iterations by the simplex method until the optimal basic feasible solution is obtained.
433 ILLUSTRATION 18-11
Use the two-phase method to solve the following:
Minimize Z = -3x1 + x2 + x3
Subject to: x1 – 2x2 + x3 11
-4x1 + x2 + 2x3 3
-2x1 + x3 = 1 x1 , x2 , x3 0 SUGGESTED SOLUTION 18-11
Step 1: To convert to standard form and add necessary artificial variables.
Minimize Z = -3x1 + x2 + x3
Subject to: x1 – 2x2 + x3 + S1 + 0S2 + 0S3 = 11 -4x1 + x2 + 2x3 + 0S1 – S2 + A1 = 3 -2x1 + 0x2 + x3 + 0S1 + 0S2 + A2 = 1 x1 , x2 , x3 , S1 , S2 , A1 , A2 0
Step 2: Objective function consisting of artificial variables alone:
Min. w = A1 + A2
Subject to: x1 – 2x2 + x3 + S1 + 0S2 + 0S3 = 11 -4x1 + x2 + 2x3 + 0S1 – S2 + A1 = 3 -2x1 + 0x2 + x3 + 0S1 + 0S2 + A2 = 1 x1 , x2 , x3 , S1 , S2 , A1 , A2 0
Step 3: Formulation of simplex table. Table 1
C 0 0 0 0 0 1 1
CB XB X1 X2 X3 S1 S2 A1 A2 bi Ratio 0 1
1 S1
A1 A2
1 -2 1 1 0 0 0 -4 1 2 0 -1 1 0 -2 0 1 0 0 0 0
11 3 1
1 1.5 1 Z -6 1 3 0 -1 1 1 w = 4 Z* =Cj-Zj 6 -1 -3 0 1 0 0
Table 2
Cj 0 0 0 0 0 1 1
CB XB X1 X2 X3 S1 S2 A1 A2 bi 0 1
1 S1
A1 X3
3 -2 0 1 0 0 -1 0 1 0 0 -1 1 -2 -2 0 1 0 0 0 1
10 1 1 Z 0 1 0 0 -1 0 -2 w = 1
Z* =Cj-Zj 0 -1 0 0 1 0 3 Table 3:
Cj 0 0 0 0 0 1 1 CB XB X1 X2 X3 S1 S2 A1 A2 bi
0 0 0
S1 X2
X3
3 0 0 1 -2 2 -5 0 1 0 0 -1 1 -2 -2 0 1 0 0 0 1
12 1 1 Z 0 0 0 0 0 0 0 w = 0
Since w = o, we move to phase II.
Phase II with objective function Min. Z = -3x1 + x2 + x3
Subject to the constraints in table 3.
Table 4
Cj -3 1 1 0 0 0 0
CB XB X1 X2 X3 S1 S2 A1 A2 bi Ratio 0 1
1
S1
X2 X3
3 0 0 1 -2 2 -5 0 1 0 0 -1 1 -2 -2 0 1 0 0 0 0
12 1 1
4 Not possible Not possible Z -2 1 1 0 -1 1 -1 w = 2
Z* =Cj-Zj -1 0 0 0 1 -1 1
We should not allow the artificial variable to go into Basis again. Hence, variable X1 enters the basis while S1 leaves the basis.
435 Table 5
Cj -3 1 1 0 0
CB XB X1 X2 X3 S1 S2 bi -3 1
1 X1
X2 X3
1 0 0 0.33 -0.67 0 1 0 0 -1 0 0 1 0.67 -1.33
4 1 9
Z -3 1 1 -0.33 -0.33 w = -2 Z* =Cj - Zj 0 0 0 0.33 0.33
The problem is optimal at X1 = 4 , X2 = 1 , X3 = 9 with the optimal value Z = -3(4) + 1(1) + 1(9)
= -12 + 1 + 9 = -2