1

2

In our presentation of the Simplex method we have used the slack variables as the

starting solution. These were coming from the standardized form of constraints that

are type of “”

However, if the original constraint is a “≥” or “=” type of constraint, we no longer

have an easy starting solution.

Therefore, Artificial Variables are used in such cases. An artificial variable is a

variable introduced into each equation that has a surplus variable.

To ensure that we consider only basic feasible solutions, an artificial variable is

required to satisfy the nonnegative constraint.

The two method used are:

• The M- method

• The Two-phase method

The M-Method

The M-method starts with the LP in the standard form

For any equation (i) that does not have slack, we augment an

artificial variable Ri

Given M is sufficiently large positive value, The variable Ri is

penalized in objective function using (-M Ri ) in case of maximization

and (+ M Ri ) in case of minimization. ( Penalty Role)

The M-Method ( Example)

Minimized 214 xxz

Subject to:

0,

42

634

33

21

21

21

21

xx

xx

xx

xx

The M-Method ( Solution)

Minimized 214 xxz

Subject to:

0,,,

42

634

33

4321

421

321

21

xxxx

xxx

xxx

xx

By subtracting surplus x3 in second constraint and adding slack x4 in third constraint, thus we get:

By using artificial variables in equations that haven’t slack variables and penalized them in objective function, we got:

Minimized 21214 MRMRxxz Subject to:

0,,,,,

42

634

33

214321

421

2321

121

RRxxxx

xxx

Rxxx

Rxx

Then, we can use R1 , R2 and x4 as the starting basic feasible solution.

Basicx1x2x3R1R2x4Solution

z-4-10-M-M00

R13101003

R243-10106

x41200014

New z-row = Old z-row + M* R1-row + M*R2 -row

Basicx1x2x3R1R2x4Solution

z-4+7M-1+4M-M0009M

R13101003

R243-10106

x41200014

Artificial variables become zero

Basicx1x2x3R1R2x4Solution

z0(1+5M/)3-M(4-7M/)3004+2M

x111/301/3001

R205/3-1-4/3102

x405/30-1/3013

Thus, the entering value is (-4+7M) because it the most positive coefficient in the z-row.

The leaving variable will be R1 by using the ratios of the feasibility condition

After determining the entering and leaving variable, the new tableau can be computed by Gauss-Jordan operations as follow:

The last tableau shows that x2 is the entering variable and R2 is the leaving variable. The simplex computation must thus continued for two more iteration to satisfy the optimally condition.

The results for optimality are:

5

17,1,

5

9,5

2321 zandxxx

Observation regarding the Use of M-method:

1.The use of penalty M may not force the artificial variable to zero level in the final simplex iteration. Then the final simplex iteration include at least one artificial variable at positive level. This indication that the problem has no feasible condition.

2.( M )should be large enough to act as penalty, but it should not be too large to impair the accuracy of the simplex computations.

Example:

Maximize 21 5.02.0 xxz

Subject to :

0,

42

623

21

21

21

xx

xx

xx

Using Computer solution, apply the simplex method M=10, and repeat it using M=999.999. the first M yields the correct solution x1 =1 and x2 =1.5, whereas the second gives the incorrect solution x1 =4 and x2 =0

Multiplying the objective function by 1000 to get z= 200x1 + 500x2 and solve the problem using M=10 and M=999.999 and observe the second value is the one that yields the correct solution in this case

The conclusion from two experiments is that the correct choice of the value of M is data dependent .

When a basic feasible solution is not readily available, the two-phase simplex method may be used as an alternative to the big M method.

In the two-phase simplex method, we add artificial variables to the same constraints as we did in big M method. Then we find a basic feasible solution to the original LP by solving the Phase I LP.

In the Phase I LP, the objective function is to minimize the sum of all artificial variables.

At the completion of Phase I, we use Phase II and reintroduce the original LP’s objective function and determine the optimal solution to the original LP.

Two-phase method

In Phase I, If the optimal value of sum of the artificial variables are

greater than zero, the original LP has no feasible solution which

ends the solution process. Other wise, We move to Phase II

Note:

Example

Minimized 214 xxz Subject to:

0,

42

634

33

21

21

21

21

xx

xx

xx

xx

Solution:

Phase I :

Minimize :21 RRr

Subject to:

0,,,,,

42

634

33

214321

421

2321

121

RRxxxx

xxx

Rxxx

Rxx

Basicx1x2x3R1R2x4Solution

r000-1-100

R13101003

R243-10106

x41200014

New r-row = Old r-row + 1* R1-row + R2 -row

Basicx1x2x3R1R2x4Solution

r74-10009

R13101003

R243-10106

x41200014

Basicx1x2x3R1R2x4Solution

r000-1-100

x1101/53/5-1/503/5

x201-3/5-4/53/506/5

x40011-111

By using new r-row, we solve Phase I of the problem which yields the following optimum tableau

Because minimum r=0, Phase I produces the basic feasible solution:

1,5

6, 425

31 xandxx

Phase II

After eliminating artificial variables column, the original problem can be written as:

Minimize :

214 xxz

Subject to:

0,,,

15

6

5

32

5

3

5

14

4321

43

32

31

xxxx

xx

xx

xx

Basicx1x2x3x4Solution

z-4-1000

x1101/503/5

x201-3/506/5

x400111

Again, because basic variables x1 and x2 have nonzero coefficient in he z row, they must be substituted out, using the following computation:

New z-row = Old z-row + 4* x1-row + 1*x2 -row

Basicx1x2x3x4Solution

z001/5018/5

x1101/503/5

x201-3/506/5

x400111

The initial tableau of Phase II is as the following:

The removal of artificial variables and their column at the end of Phase I can take place only when they are all nonbasic. If one or more artificial variables are basic ( at zero level) at the end of Phase I, then the following additional steps must be under taken to remove them prior to start Phase II

Step 1. Select a zero artificial variable to leave the basic solution and designate its row as pivot row. The entering variable can be any nonbasic (nonartificial) variable with nonzero (positive or negative) coefficient in the pivot row. Perform the associated simplex iteration.

Step 2. Remove the column of the (Just-leaving) artificial from the tableau. If all the zero artificial variables have been removed , go to Phase II. Otherwise, go back to Step I.

Remarks:

21