Simplex two phase

transcript

Simplex MethodWhen decision variables are more than 2, we always use Simplex MethodSlack Variable: Variable added to a ≤ constraint to convert it to an equation (=).

A slack variable represents unused resourcesA slack variable contributes nothing to the objective function value.

Surplus Variable: Variable subtracted from a ≥ constraint to convert it to an equation (=).

A surplus variable represents an excess above a constraint requirement level.Surplus variables contribute nothing to the calculated value of the objective function.

Cont….

Basic Solution(BS) : This solution is obtained by setting any n variables (among m+n variables) equal to zero and solving for remaining m variables, provided the determinant of the coefficients of these variables is non-zero. Such m variables are called basic variables and remaining n zero valued variables are called non basic variables.

Basic Feasible Solution(BFS) : It is a basic solution which also satisfies the non negativity restrictions.

Cont…..

BFS are of two types:Degenerate BFS: If one or more basic variables are zero.

Non-Degenerate BFS: All basic variables are non-zero.

Optimal BFS: BFS which optimizes the objective function.

Example

Max. Z = 13x1+11x2

Subject to constraints:

4x1+5x2 << 1500 1500

5x5x11+3x+3x2 2 << 1575 1575

xx11+2x+2x2 2 << 420 420

xx11, x, x2 2 > 0

Solution :Step 1: Convert all the inequality constraints into equalities

by the use of slack variables.

Let S1, S2 , S3 be three slack variables.

Introducing these slack variables into the inequality constraints and rewriting the objective function such that all variables are on the left-hand side of the equation. Model can rewritten as:

Z - 13x1 -11x2 = 0

4x1+5x2 + S1 = 1500 1500

5x5x11+3x+3x2 2 +S+S22= 1575= 1575

xx11+2x+2x2 2 +S+S33 = 420 = 420

xx11, x, x22, S, S11, S, S22, S, S3 3 > 0

Cont…

Step II: Find the Initial BFS.

One Feasible solution that satisfies all the constraints is: x1= 0, x2= 0, S1= 1500,

S2= 1575, S3= 420 and Z=0.

Now, S1, S2, S3 are Basic variables.

Step III: Set up an initial table as:

Cont…Row NO.

Basic Variable

Coefficients of: Sol. RatioZ x1 x2 S1 S2 S3

A1 Z 1 -13 -11 0 0 0 0

B1 S1 0 4 5 1 0 0 1500 375

C1 S2 0 5 3 0 1 0 1575 315

D1 S3 0 1 2 0 0 1 420 420

Step IV: a) Choose the most negative number from row A1(i.e Z row). Therefore, x1 is a entering variable.

b) Calculate Ratio = Sol col. / x1 col. (x1 > 0) c) Choose minimum Ratio. That variable(i.e S2) is a departing

variable.

Cont….

Step V: x1 becomes basic variable and S2 becomes non basic

variable. New table is:

Row NO.

Basic Variable

Coefficients of: Sol. Ratio

Z x1 x2 S1 S2 S3

A1 Z 1 0 -16/5 0 13/5 0 4095

B1 S1 0 0 13/5 1 -4/5 0 240 92.3

C1 x1 0 1 3/5 0 1/5 0 315 525

D1 S3 0 0 7/5 0 -1/5 1 105 75

Cont…

Next Table is :

Row NO.

Basic Variable

Coefficients of: Sol.

Z x1 x2 S1 S2 S3

A1 Z 1 0 0 0 15/7 16/7 4335

B1 S1 0 0 0 1 -3/7 -13/7 45

C1 x1 0 1 0 0 2/7 -3/7 270

D1 x2 0 0 1 0 -1/7 5/7 75

Optimal Solution is : x1= 270, x2= 75, Z= 4335

Example

Max. Z = 3x1+5x2+4x3

2x1+3x2 << 8 8

2x2x22+5x+5x3 3 << 10 10

3x3x11+2x+2x22+4x+4x3 3 << 15 15

xx11, x, x22, x, x3 3 > 0

Cont…

Let S1, S2, S3 be the three slack variables.

Modified form is:

Z - 3x1-5x2-4x3 =0

2x1+3x2 +S1= 8 8

2x2x22+5x+5x3 3 +S+S22= 10= 10

3x3x11+2x+2x22+4x+4x33+S+S33= 15= 15

xx11, x, x22, x, x33,, S1, SS22,, SS3 3 > 0

Initial BFS is : x1= 0, x2= 0, x3=0, S1= 8,

S2= 10, S3= 15 and Z=0.

Cont…

Basic Variable

Z x1 x2 x3 S1 S2 S3

Z 1 -3 -5 -4 0 0 0 0

S1 0 2 3 0 1 0 0 8 8/3

S2 0 0 2 5 0 1 0 10 5

S3 0 3 2 4 0 0 1 15 15/2

Therefore, x2 is the entering variable and S1 is the departing variable.

Cont…

Basic Variable

Z x1 x2 x3 S1 S2 S3

Z 1 1/3 0 -4 5/3 0 0 40/3

x2 0 2/3 1 0 1/3 0 0 8/3 -

S2 0 -4/3 0 5 -2/3 1 0 14/3 14/15

S3 0 5/3 0 4 -2/3 0 1 29/3 29/12

Cont…

Basic Variable

Z x1 x2 x3 S1 S2 S3

Z 1 -11/15 0 0 17/15 4/5 0 256/15

x2 0 2/3 1 0 1/3 0 0 8/3 4

x3 0 -4/15 0 1 -2/15 1/5 0 14/15 -

S3 0 41/15 0 0 2/15 -4/5 1 89/15 89/41

Cont…

Basic Variable

Z x1 x2 x3 S1 S2 S3

Z 1 0 0 0 45/41 24/41 11/41 765/41

x2 0 0 1 0 15/41 8/41 -10/41 50/41

x3 0 0 0 1 -6/41 5/41 4/41 62/41

x1 0 1 0 0 -2/41 -12/41 15/41 89/41

Optimal Solution is : x1= 89/41, x2= 50/41, x3=62/41, Z= 765/41

Example

Min.. Z = x1 - 3x2 + 2x3

3x1 - x2 + 3x3 << 7

-2x-2x1 1 + 4x+ 4x2 2 << 12 12

-4x-4x11 + 3x + 3x22 + 8x + 8x33 < < 1010

xx11, x, x22, x, x3 3 > 0

Cont…

Convert the problem into maximization problem

Max.. Z’ = -x1 + 3x2 - 2x3 where Z’= -Z

3x1 - x2 + 3x3 << 7

-2x-2x1 1 + 4x+ 4x2 2 << 12 12

-4x-4x11 + 3x + 3x22 + 8x + 8x33 < < 1010

xx11, x, x22, x, x3 3 > 0

Cont…

Let S1, S2 and S3 be three slack variables.

Modified form is:

Z’ + x1 - 3x2 + 2x3 = 0

3x1 - x2 + 3x3 +S1 = 7

-2x-2x1 1 + 4x+ 4x2 2 + S + S22 = 12 = 12

-4x-4x11 + 3x + 3x22 + 8x + 8x33 +S +S33 = = 1010

xx11, x, x22, x, x3 3 > 0Initial BFS is : x1= 0, x2= 0, x3=0, S1= 7, S2= 12, S3 = 10

and Z=0.

Cont…

Basic Variable

Z’ x1 x2 x3 S1 S2 S3

Z’ 1 1 -3 2 0 0 0 0

S1 0 3 -1 3 1 0 0 7 -

S2 0 -2 4 0 0 1 0 12 3

S3 0 -4 3 8 0 0 1 10 10/3

Cont…

Basic Variable

Z’ x1 x2 x3 S1 S2 S3

Z’ 1 -1/2 0 2 0 3/4 0 9

S1 0 5/2 0 3 1 1/4 0 10 4

x2 0 -1/2 1 0 0 1/4 0 3 -

S3 0 -5/2 0 8 0 -3/4 1 1 -

Cont…

Basic Variable

Z’ x1 x2 x3 S1 S2 S3

Z’ 1 0 0 13/5 1/5 8/10 0 11

x1 0 1 0 6/5 2/5 1/10 0 4

x2 0 0 1 3/5 1/5 3/10 0 5

S3 0 0 0 11 1 -1/2 1 11

Optimal Solution is : x1= 4, x2= 5, x3= 0,

Z’ = 11 Z = -11

Example

Max.. Z = 3x1 + 4x2

x1 - x2 << 1

-x-x1 1 + x+ x2 2 << 2 2

xx11, x, x2 2 > 0

Cont…Let S1 and S2 be two slack variables

Modified form is:

Z -3x1 - 4x2 = 0

x1 - x2 +S1 = 1

-x-x1 1 + x+ x2 2 +S+S2 2 = 2= 2

xx11, x, x22, S, S11, S, S2 2 > 0Initial BFS is : x1= 0, x2= 0, S1= 1, S2= 2

and Z=0.

Cont…

Basic Variable

Z x1 x2 S1 S2

Z 1 -3 -4 0 0 0

S1 0 1 -1 1 0 1 -

S2 0 -1 1 0 1 2 2

Cont…

Basic Variable

Z x1 x2 S1 S2

Z 1 -7 0 0 4 8

S1 0 0 0 1 1 3 -

x2 0 -1 1 0 1 2 -

x1 is the entering variable, but as in x1 column every no. is less than equal to zero, ratio cannot be calculated. Therefore given problem is having a unbounded solution.

Introduction

LPP, in which constraints may also have > and = signs, we introduce a new type of variable , called the artificial variable. These variables are fictitious and cannot have any physical meaning. Two Phase Simplex Method is used to solve a problem in which some artificial variables are involved. The solution is obtained in two phases.

Example

Min.. Z = 15/2 x1 - 3x2

3x1 - x2 - x3 > 3 3

xx11 - x - x22 + x + x33 > 2

xx11, x, x22, x, x3 3 > 0

Cont…

Convert the objective function into the maximization form

Max. Z’ = -15/2 x1 + 3x2 where Z’= -Z

3x1 - x2 - x3 > 3 3

xx11 - x - x22 + x + x33 > 2

xx11, x, x22, x, x3 3 > 0

Cont…

Modified form is :

Introduce surplus variables S1 and S2, and artificial variables a1 and a2

Z’ + 15/2 x1 - 3x2 = 0

3x1 - x2 - x3 –S1 + a1 = 3 3

xx11 - x - x22 + x + x33 –S –S22 + a + a22 = = 2

xx11, x, x22, x, x3 3 , , S1, SS22, , a1, aa22 > 0

Cont…

Phase I : Simplex method is applied to a specially constructed Auxiliary LPP leading to a final simplex table containing a BFS to the original problem.

•Step 1: Assign a cost –1 to each artificial variable and a cost 0 to all other variables in the objective function.•Step 2: Construct the auxiliary LPP in which the new objective function Z* is to be maximized subject to the given set of constraints.

Cont…

Max. Z* = -a1 –a2

Z* + a1 + a2 = 0

3x1 - x2 - x3 –S1 + a1 = 3 3

xx11 - x - x22 + x + x33 –S –S22 + a + a22 = = 2

xx11, x, x22, x, x3 3 , , S1, SS22, , a1, aa22 > 0

Auxiliary LPP is:

Initial solution is a1 = 3, a2 = 2 and Z* = 0

Cont…•Step 3: Solve the auxiliary problem by simplex method until either of the following three possibilities arise:

•Max Z* < 0 and at least one artificial variable appear in the optimum basis at a positive level. In this case given problem does not have any feasible solution.

•Max Z* = 0 and at least one artificial variable appear in the optimum basis at a zero level. In this case proceed to Phase II.•Max Z* = 0 and no artificial variable appear in the optimum basis. In this case also proceed to Phase II.

Cont…

Basic Variable

Z* x1 x2 x3 S1 S2 a1 a2

Z* 1 0 0 0 0 0 1 1 0

a1 0 3 -1 -1 -1 0 1 0 3

a2 0 1 -1 1 0 -1 0 1 2

This table is not feasible as a1 and a2 has non zero coefficients in Z* row. Therefore next step is to make the table feasible.

Cont…

Basic Variable

Z* x1 x2 x3 S1 S2 a1 a2

Z* 1 -4 2 0 1 1 0 0 -5

a1 0 3 -1 -1 -1 0 1 0 3 1

a2 0 1 -1 1 0 -1 0 1 2 2

Therefore, x1 is the entering variable and a1 is the departing variable.

Cont…

Basic Variable

Z* x1 x2 x3 S1 S2 a2

Z* 1 0 2/3 -4/3 -1/3 1 0 -1

x1 0 1 -1/3 -1/3 -1/3 0 0 1 -

a2 0 0 -2/3 4/3 1/3 -1 1 1 3/4

Cont…

Basic Variable

Z* x1 x2 x3 S1 S2

Z* 1 0 0 0 0 0 0

x1 0 1 -1/2 0 -1/4 -1/4 5/4

x3 0 0 -1/2 1 1/4 -3/4 3/4

As there is no variable to be entered in the basis, this table is optimum for Phase I. In this table Max. Z* = 0 and no artificial variable appears in the optimum basis, therefore we can proceed to Phase II.

Cont…Phase II: The artificial variables which are non basic at the end of Phase I are removed from the table and as well as from the objective function and constraints. Now assign the actual costs to the variables in the Objective function. That is, Simplex method is applied to the modified simplex table obtained at the Phase I.

Basic Variable

Z’ x1 x2 x3 S1 S2

Z’ 1 15/2 -3 0 0 0 0

x1 0 1 -1/2 0 -1/4 -1/4 5/4

x3 0 0 -1/2 1 1/4 -3/4 3/4

Again this table is not feasible as basic variable x1 has a non zero coefficient in Z’ row. So make the table feasible.

Cont…

Basic Variable

Z’ x1 x2 x3 S1 S2

Z’ 1 0 3/4 0 15/8 15/8 -75/8

x1 0 1 -1/2 0 -1/4 -1/4 5/4

x3 0 0 -1/2 1 1/4 -3/4 3/4

Optimal Solution is : x1= 5/4, x2= 0, x3= 3/4,

Z’ = -75/8 Z = 75/8

Example

Min.. Z = x1 - 2x2 –3x3

-2x1 + x2 + 3x3 = 2 2

2x2x11 + 3x + 3x22 + 4x + 4x33 = = 1

xx11, x, x22, x, x3 3 > 0

Cont…Phase I:

Introducing artificial variables a1 and a2

Auxiliary LPP is: Max. Z* = -a1 - a2

Z* + a1 + a2 = 0

-2x1 + x2 + 3x3 + a1 = 2 2

2x2x11 + 3x + 3x22 + 4x + 4x33 + a + a2 2 == 1

xx11, x, x22, x, x3 3 > 0

Cont…

Basic Variable

Z* x1 x2 x3 a1 a2

Z* 1 0 0 0 1 1 0

a1 0 -2 1 3 1 0 2

a2 0 2 3 4 0 1 1

Cont…

Basic Variable

Z* x1 x2 x3 a1 a2

Z* 1 0 -4 -7 0 0 -3

a1 0 -2 1 3 1 0 2 2/3

a2 0 2 3 4 0 1 1 1/4

Cont…

Basic Variable

Z* x1 x2 x3 a1

Z* 1 7/4 5/4 0 0 -5/4

a1 0 -7/2 -5/4 0 1 5/4

x3 0 1/2 3/4 1 0 1/4

As there is no variable to be entered in the basis, therefore Phase I ends here. But one artificial variable is present in the basis and Z* < 0. Therefore we cannot proceed to Phase II.

Given problem is having a non-feasible solution.

Example

Min.. Z = 4x1 + x2

3x1 + x2 = 3 3

4x4x11 + 3x + 3x22 > 6

x1 +2x2 < < 44

xx11, x, x22 > 0

Cont…Phase I:

Introducing artificial variable a1 and a2, surplus variable S1 and slack variable S2

Auxiliary LPP is:Max. Z* = -a1 - a2

Z* + a1 + a2 = 0

3x1 + x2 +a1 = 3 3

4x4x11 + 3x + 3x22 –S –S11 + a + a22 = 6

x1 +2x2 +S2 = 44

xx11, x, x22 > 0

Cont…

Basic Variable

Z* x1 x2 S1 S2 a1 a2

Z* 1 0 0 0 0 1 1 0

a1 0 3 1 0 0 1 0 3

a2 0 4 3 -1 0 0 1 6

S2 0 1 2 0 1 0 0 4

Cont…

Basic Variable

Z* x1 x2 S1 S2 a1 a2

Z* 1 -7 -4 1 0 0 0 -9

a1 0 3 1 0 0 1 0 3 1

a2 0 4 3 -1 0 0 1 6 3/2

S2 0 1 2 0 1 0 0 4 4

Cont…

Basic Variable

Z* x1 x2 S1 S2 a2

Z* 1 0 -5/3 1 0 0 -2

x1 0 1 1/3 0 0 0 1 3

a2 0 0 5/3 -1 0 1 2 6/5

S2 0 0 5/3 0 1 0 3 9/5

Cont…

Basic Variable

Z* x1 x2 S1 S2

Z* 1 0 0 0 0 0

x1 0 1 0 1/5 0 3/5

x2 0 0 1 -3/5 0 6/5

S2 0 0 0 1 1 1

As there is no variable to be entered in the basis, this table is optimum for Phase I. In this table Max. Z* = 0 and no artificial variable appears in the optimum basis, therefore we can proceed to Phase II.

Cont…Phase II:

Original Objective function is:Min.. Z = 4x1 + x2

Convert the objective function into the maximization form

Max. Z’ = -4 x1 - x2 where Z’= -Z

Basic Variable

Z* x1 x2 S1 S2

Z* 1 4 1 0 0 0

x1 0 1 0 1/5 0 3/5

x2 0 0 1 -3/5 0 6/5

S2 0 0 0 1 1 1

Again this table is not feasible as basic variable x1 and x2 has a non zero coefficient in Z’ row. So make the table feasible.

Cont…

Basic Variable

Z’ x1 x2 S1 S2

Z’ 1 0 0 -1/5 0 -18/5

x1 0 1 0 1/5 0 3/5 3

x2 0 0 1 -3/5 0 6/5 -

S2 0 0 0 1 1 1 1

Therefore, S1 is the entering variable and S2 is the departing variable.

Cont…

Basic Variable

Z’ x1 x2 S1 S2

Z’ 1 0 0 0 1/5 -17/5

x1 0 1 0 0 -1/5 2/5

x2 0 0 1 0 3/5 9/5

S2 0 0 0 1 1 1

Optimal Solution is : x1= 2/5, x2= 9/5,

Z’ = -17/5 Z = 17/5

Introduction

At the stage of improving the solution during Simplex procedure, if a tie for the minimum ratio occurs at least one basic variable becomes equal to zero in the next iteration and the new solution is said to be Degenerate.

Example

Max.. Z = 3x1 + 9x2

x1 + 4x2 << 8 8

xx11 + 2x + 2x22 << 4

xx11, x, x22 > 0

Cont…

Let S1 and S2 be two slack variables.

Modified form is:

Z - 3x1 - 9x2 = 0

x1 + 4x2 + S1 = 8 8

xx11 + 2x + 2x2 2 +S+S22 = = 4

xx11, x, x22, S, S11, S, S22 > 0

Initial BFS is : x1= 0, x2= 0, S1= 8, S2= 4 and Z=0.

Cont…

Basic Variable

Z x1 x2 S1 S2

Z 1 -3 -9 0 0 0

S1 0 1 4 1 0 8 2

S2 0 1 2 0 1 4 2

In this table S1 and S2 tie for the leaving variable. So any one can be considered as leaving variable. Therefore, x2 is the entering variable and S1 is the departing variable.

Cont…

Basic Variable

Z x1 x2 S1 S2

Z 1 -3/4 0 9/4 0 18

x2 0 1/4 1 1/4 0 2 8

S2 0 1/2 0 -1/2 1 0 0

Cont…

Basic Variable

Z x1 x2 S1 S2

Z 1 0 0 3/2 3/2 18

x2 0 0 1 1/2 -1/2 2

x1 0 1 0 -1 2 0

Optimal Solution is : x1= 0, x2= 2

Z = 18

It results in a Degenerate Basic Solution.

Simplex two phase

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