4 lp simplex big m

7/29/2019 4 LP Simplex Big M

1/4

1

Linear Programming

Simplex: Big-M Method

Learning Outcomes

Students should be able to:

Solve LP problems involving=or constraints Solve LP problems using the Big M Method

The Big M Method

Recall that the simplex algorithm requires a starting basic feasible solution (bfs). So far,the problems we solved using the simplex method involved constraints, in which we

added slack variables to convert the constraint into equality constraint. For problems

involving equality(=) or greater-than-or-equal () constraints, the bfs may be hard tofind. The Big M method can be used to cater to these situations.

Case 1: Equality (=) Constraints[refer to Hillier & Lieberman textbook, Chp. 4, sec 4.6]

Some LP problems may involve equality constraints.

For equality constraint a11X1 +a12X2 +..+ a1nX3 = k, a nonnegative artificial variable A1 isadded to the LHS of the equation as follows:

a11X1 +a12X2 +..+ a1nX3 + A1= k

The artificial variable A1 is just a dummy variable. It has no physical meaning. However,It is needed to set up the initial simplex tableau. Hence, it will become a basic variablefor the initial solution.

In the objective function, the artificial variable will be given an arbitrary coefficient value M if the problem is a maximization problem, and a coefficient value of+M if the problem is a minimization problem.

M symbolically represents a very large positive number. (M=mammoth?). The objectivefunction of the original LP must be modified to ensure that the artificial variables are all

equal to 0. Hence, iterations of the simplex method will force the artificial variable out oftableau. In the optimal tableau the artificial variable will have zero value.

Example: Wyndor Glass Co.Suppose Plant 3 is required to operate at full capacity. Hence the third constraint

3X1 +2X218 is rewritten as 3X1 2X2 =18. The new model for the real problem, andthe artificial problem model in the standard form are shown below

ECS716: Operations ResearchPn Paezah Hamzah

Max Z = 3X1 + 5X2 +0S1 + 0S2 MA1s.t.

X1 + S1 = 42X2 + S2 = 12

3X1 + 2X2 +A1 = 18

X1, X2, S1, S2, A1 0

Max Z = 3X1 + 5X2s.t.

X1 42X2 12

3X1 +2X2 = 18

X1, X2 0

Artificialvariable

technique

The Real Problem The Artificial Problem


2/4

2

The graphical solution to Wyndor Glass Co is as follows:

Converting the Objective Function to Proper Form

To prepare for the simplex iterations, the standard form is augmented as follows:

In the above model, the objective function (z-row of simplex tableau) contains the

artificial variable. This artificial variable will be in the starting basis for the initial simplextableau. All artificial variables must be algebraically eliminated from the z-row beforebeginning the simplex. This is to ensure that we begin with a canonical1 form.

To eliminate the artificial variable from the z-row or equation (1), form the followingsystem of linear equations.

Z - 3X1 - 5X2 + MA1 = 0 (1)3X1 + 2X2 + A1 = 18 (4)

Multiply both sides of equation (3) by M we get:

3MX1 + 2MX2 + MA1 = 18M (4)

Next, subtract equation (4) from equation (1):

Z - 3X1 - 5X2 + MA1 = 0 (1)- 3MX1 + 2MX2 + MA1 = 18M (4)

Z- 3X1 -3MX1 - 5X2 - 2MX2 = -18M

Corner Point Feasible (CPF) solution

CPF Z=3X1 + 5X2

(2,6) 36

(4,3) 27

Optimal solution:Produce 2 batches of doors and 6

batches of windows to attain a maximumprofit of $36,000

(0,6)

(4,0) 6X1

9

2X2 =12

X1 =4

3X1 +2X2 =18

(2, 6)

4 3

(0,0)

Max Z = 3X1 + 5X2s.t.

X1 42X2 12

3X1 +2X2 = 18X1, X2 0

[z-row] Z - 3X1 - 5X2 +MA1 = 0 (1)

X1 + S1 = 4 (2)2X2 + S2 = 12 (3)

3X1 + 2X2 + A1 = 18 (4)

Canonical form: a system of linear equation in which each equation has variable with

coefficient 1 in that equation, and coefficient 0 in all other equations.


3/4

3

The new equation gives the objective function in terms of just the nonbasic variables

X1 and X2.

Z - 3X1 -3MX1 5X2 2MX2 = -18M, or

Z - (3 +3M)X1 (5+ 2M)X2 = -18M

The new standard model (in a canonical form) is as follows:

Writing the z-row at the bottom gives:

The initial simplex tableau (with z-row at the bottom) is shown below.

Basic Var. X1 X2 S1 S2 A1 RHS

S1 1 0 1 0 0 4

S2 0 2 0 1 0 12

S3 3 2 0 0 1 18

Z -3M-3 -2M-5 0 0 0 -18M

Continue with the remaining iterations.

Iteration 2


X1 1 0 1 0 0 4

S2 0 2 0 1 0 12

S3 0 2 -3 0 1 6

Z 0 -2M-5 3M+3 0 0 -6M+12

Iteration 3


X1 1 0 1 0 0 4

S2 0 0 3 1 -1 6

X2 0 1 -3/2 0 1/2 3

Z 0 0 -9/2 0 M+5/2 27

Z - (3 +3M)X1 (5+ 2M)X2 = -18M

X1 + S1 = 42X2 + S2 = 12

3X1 + 2X2 + A1 = 18

X1 + S1 = 4

2X2 + S2 = 123X1 + 2X2 + A1 = 18

Z - (3 +3M)X1 (5+ 2M)X2 = -18M


4/4

4

Iteration 4

Basic Var. Z X1 X2 S1 S2 A1 RHS

X1 0 1 0 0 -1/3 1/3 2

S1 0 0 0 1 1/3 -1/3 2

X2 0 0 1 0 1/2 0 6

Z 1 0 0 0 3/2 M+1 36

All z-row coefficients 0. Hence, the optimal solution is reached.

Optimal solution:

X1= 2, X2 = 6, S1 = 2,S

2=0, S

3=0, A

1=0

Optimal value=36

Case 2: Constraints of the form

Refer to the text book [Radio Therapy Example]

[Demonstrated in class]

Exercise:

Solve the following LPs using the Big M method.

1.

Min Z = 0.4X1 + 0.5X2s.t.

0.3X1 + 0.1X2 2.7

0.5X1 + 0.5X2 = 60.6X1 + 0.4X2 6

X1, X2 0

Min Z = 4X1 + 4X2 + X3s.t.

X1 + X2 + X3 22X1 + X2 3

2X1 + X2 + 3X3 3X1, X2,X3 0

2. Max Z = 3X1 + X2

s.t.

X1 + X2 32X1 + X2 4X1 + X2 = 3

X1, X2 0

3. Min Z = 3X1 + X2s.t.

2X1 + X2 63X1 + 2X2 = 4

X1, X2 0

4 lp simplex big m

Documents