LP - II Lecture

7/30/2019 LP - II Lecture

1/69

Lectureon

Linear Programming

By

Dr. D. B. Naik (Ph.D.- Mech. Engg.)Professor, Training & PlacementSardar Vallabhbhai National Institute

of Technology, Surat


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Difficulties in Simplex Procedure

1. Tie in selecting key column/key row.

2. Inequality of greater than or equal to kind.

3. Equality constraints.

4. Negative RHS.

5. Unrestricted variables.

6. Restricted variables.

7. Minimization Problem.


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(1) Tie in selecting key column/Key row

Decision variable and slack or surplus

variable Select Decision variablecolumn

Decision variable and Decision variable

Select any one column

Slack/Surplus variable and Slack/Surplusvariable Select any one column

(A) Tie in selecting key column


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aikare elements in key column.

Compare the ratio of aij to aikfor tiedrows first in identity from L to R and thenin body.

Key row Min. Positive ratio.

(B) Tie in selecting key row


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02,1

)(8214

)(4214

)(122314/

212

xx

IIIxx

IIxx

Ixxts

xxZMax

Standard Form:

Max Z = 2x1+1x2+0w1+0w2+0w3

4x1+3x2+w1+0w2+0w3 = 12

4x1+x2+0w1+w2+0w3 = 8

4x1-x2+0w1+0w2+w3 = 8

Simplex Method-A Case for Tie

in selecting keyrow


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bi x1 x2 w1 w2 w3

Ij Z = 0

Ij = (Zj-cj) = (aij.ci) - cj

cj 2 1 0 0 0

Tableau-I

12 4 3 1 0 0

8 4 1 0 1 0

8 4 -1 0 0 1

ci xi

0 w1

0 w2

0 w3

0 0 0

Ratio

3

2

2

-2 -1

Tie


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bi x1 x2 w1 w2 w3

Ij Z = 0

cj 2 1 0 0 0

Tableau-I

12 4 3 1 0 0

8 4 1 0 1 0

8 4 -1 0 0 1

ci xi

0 w1

0 w2

0 w3

0 0 0

Ratio

3

2

2

-2 -1

Tie


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bi w1 w2 w3 x1 x2

Ij Z = 0 0 0

cj 0 0 0 2 1

Tableau-I : Rewritten in required form

12 1 0 0 4 3

8 0 1 0 4 1

8 0 0 1 4 -1

ci xi

0 w1

0 w2

0 w3

0 -2 -1

1/4 = 0.25

0/4 = 0

Tie

Ratio

3

2

2

Ratio : 0/4 - R2

0/4 - R3 Key Row


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(2) Constraints with inequality

greater than or equaltokind

....1)(10....

18112213

182213

AMsZMax

Asxx

xx


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(3) Equality Constraints.

n

j

ijij

n

j

ijij

n

j

ijij

bxa

bxabxa

1

11

&

n

j

iijij bAxa1

)(OR


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(4) Negative R.H.S.

n

j

ijij

n

j

ijij bxabxa11

)(

If xis unrestricted then x is changedto

(x - x), where x & x are positive.

(5) Unrestricted Variables.


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50)( xi

(6) Restricted Variables.

10'05'

55)(

xxx

xiii

5'0,5'

105)(

xxx

xii


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(7) Minimization

Problem


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Methods to Solve

Minimization Problem

A. Maximization Method

B. Minimization Method

C. Two Phase Method

D. Dual Simplex Method


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(A) Maximization Method :Convert Minimization Problem into

Maximization Problem & solve

n

1jjj

n

1jjj xcZMax.xcZMin.

)()(


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(B) Minimization Method:

Same methodology as Maximization

except

criterion of selecting key column &Coeff. of A changed to :

Key column Max. + ve Index Number.

Co-eff. of A in Obj. eq. = +M


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(C) Two Phase Method:

Consider Cj = -1 for only A, else Cj = 0

and Get Final Usual Tableau

In Phase- I : For a Maximization Problem

All Cj = Original values in Final

Tableau of Phase I and Get

Optimal Solution

In Phase- II : Consider


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(D) Dual Simplex Method

This method is applicable to any Standard

Maximizaion Problem with - Ve RHS

First Key Row is selected w.r.t. Max. Ve RHS

Key Column is selected w.r.t. Max. (I/ aij) for

-ve aij only


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(A)Maximization Method

02,1

)(62

)(41

)(182213/

2513

xx

IIIx

IIx

Ixxts

xxZMin

Convert this into maximization problem.


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02,1

)(62

)(41)(182213/

2513)(

xx

IIIx

IIxIxxts

xxZMax

Convert this into Standard form.


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Standard Form :

Max (-Z) = -3x1-5x2+0s1+(- M)A1+0w1+0w2

3x1+2x2-s1+A1+0w1+0w2 = 18

1x1+0x2-0s1+0A1+1w1+0w2 = 4

0x1+1x2-0s1+0A1+0w1+1w2 = 6

bl


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ci xi bi x1 x2 s1 A1 w1 w2

Ij Z= -18M

Ij = (Zj-cj) = (aij.ci) - cj

cj -3 -5 0 -M 0 0Tableau - I

18 3 2 -1 1 0 0

4 1 0 0 0 1 0

6 0 1 0 0 0 1

-M A1

0 w1

0 w2

0 0 0-3M -2M M

+3 +5

bl


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-2M M 3M

+5 -3

cj -3 -5 0 -M 0 0Tableau - II

-M A1

-3 x1

0 w2

1

0

0

0

1

0

0

0

1

0 1

1 00

26

4

6

0

-1 -3

Ij Z= -6M

-12

0 00

T bl III


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0

cj -3 -5 0 -M 0 0Tableau - III

-5 x2

-3 x1

0 w2

3

4

6

Ij Z= -27 0 0M

-5/2

9/25/2

Hence, Optimal Solution is

x1=4, x2=3 giving Zmax = -27 ,

Zmin=27.


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(B)Minimization Method

02,1

)(62

)(41

)(182213/

2513

xx

IIIx

IIx

Ixxts

xxZMin



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Standard Form:

Min Z = 3x1+5x2+0s1+MA1+0w1+0w2

3x1+2x2-s1+MA1+0w1+0w2 = 18

1x1+0x2-0s1+0A1+1w1+0w2 = 4

0x1+1x2-0s1+0A1+0w1+1w2 = 6

T bl I


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3M 2M -M

-3 -5

cj 3 5 0 M 0 0Tableau - I

3 2 -1 1 0 0

1 0 0 0 1 0

0 1 0 0 0 1

Rati6

4

Key Column Max Ij

M A1

0 w1

0 w2

Ij Z= 18M 0 0 0

18

4

6

T bl II


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Ij Z= 6M 0 2M -M 0 -3M 0

+12 -5 +3

cj 3 5 0 M 0 0Tableau - II

M A1 6 0 2 -1 1 -0 0

3 x1 4 1 0 0 0 1 0

0 w2 6 0 1 0 0 0 1

Rati3

6

T bl III


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0 0 -1 -M -3M 0

+1 +3

Tableau - III


x1=4, x2=3 giving Z = 27.

cj 3 5 0 M 0 0

5 x2

3 x1

0 w2

3

4

3

Ij Z = 27


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(C) Two Phase Method

02,1

)(62

)(41

)(182213/

2513

xx

IIIx

IIx

Ixxts

xxZMin

Convert this into maximization


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02,1

)(62

)(41)(182213/

2513)(

xx

IIIx

IIxIxxts

xxZMax



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3x1+2x2-s1+A1+0w1+0w2 = 18

1x1+0x2 -0s1+0A1 +w1+0w2 = 4

0x1+1x2- 0s1+0A1+0w1+w2 = 6

cj = -1 for only A, else cj = 0

Standard Form:

(Phase- I)

Max (-Z) = 0x1+0x2+0s1-1A1+0w1+0w2

Phase I : Tableau I


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Ij Z= -18

Ij = (Zj-cj) = (aij.ci)-cj

cj 0 0 0 -1 0 0Phase I : Tableau - I

18 3 2 -1 1 0 0

4 1 0 0 0 1 0

6 0 1 0 0 0 1

-1 A1

0 w1

0 w2

0 0 0-3 -2 1

Phase I : Tableau II


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ci xi bi x1 x2 s1 A w1 w2

-2 1 3

cj 0 0 0 -1 0 0Phase I : Tableau - II

-1 A1

0 x1

0 w2

1

0

0

0

1

0

0

0

1

0 1

1 00

26

4

6

0

-1 -3

Ij Z= -6 0 00

Phase I : Tableau - III


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cj 0 0 0 -1 0 0Phase I : Tableau - III

0 x2

0 x1

0 w2

0

1

0

0

0

1

1

3/21/2

3

4

3

0

-1/2 -3/2

Ij Z= 0 0 000 0 1

1

0

0

1/2

0

-1/2

Phase II : Tableau - I


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cj -3 -5 0 -M 0 0Phase II : Tableau - I

-5 x2

-3 x1

0 w2

0

1

0

0

0

1

1

3/21/2

3

4

3

0

-1/2 -3/2

Ij Z= -27 0 000 1-1

+M

1

0

0

1/2

0

-1/2


x1=4, x2=3 giving Zmax = -27 , Zmin=27


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(D) Dual Simplex Method

02,1

)(62

)(41

)(182213/

2513

xx

IIIx

IIx

Ixxts

xxZMin

Convert this into maximization


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02,1

)(62

)(41)(182213/

2513)(

xx

IIIx

IIxIxxts

xxZMax

Convert this into Standard format.


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Standard Form:

Max (-Z) = -3x1-5x2+0w1+0w2+0w3

-3x1-2x2+w1+0w2+0w3 = -18

x1+0x2+0w1+w2+0w3= 4

0x1+1x2+0w1+0w2+w3 = 6


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02,1

)(62)(41

)(182213/

2513)(

xx

IIIx

IIx

Ixxts

xxZMax


To pre

pare initial Tableau:


41/69

bi x1 x2 w1 w2 w3

Ij Z = 0

cj -3 -5 0 0 0

To prepare initial Tableau:

Tableau - I

-18 -3 -2 1 0 0

4 1 0 0 1 0

6 0 1 0 0 1

ci xi

0 w1

0 w2

0 w3

0 0 0

Ratio: 3/-3 5/-2

=-1 =-2.5

3 5

Key Column

Max Ratio

Tableau II


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Ij Z =-18

cj -3 -5 0 0 0Tableau - II

bi x1 x2 w1 w2 w3c

i

xi

-3 x1

0 w2

5 w3

0

1

06

-2

6

3 1 00 0

1

0

0

0

0

11

-2/3

-1/32/3

0

1/3

Ratio: 3/(-2/3)

=-9/2

Tableau III


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Ij Z = -27

Tableau - III

bi x1 x2 w1 w2 w3ci xi-3 x1

-5 x2

0 w3

0 0 03/2 1/2

33

4

cj -3 -5 0 0 0


x1=4, x2=3 giving Zmax= -27 and Zmin =27.


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Cases forAlternative Optimal Solution

Unbounded Solution,

Infeasible Solution and

Unrestricted Variable throughSimplex.

f


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02,1

)(321

)(242213

)(302312/

2416

xx

IIIxx

IIxx

Ixxts

xxZMax

Standard Form:

Max Z = 6x1+4x2+0w1+0w2+0s1+(M)A1

2x1+3x2+w1+0w2+0s1+0A1 = 30

3x1+2x2+0w1+w2+0s1+0A1 = 24

x1+x2+0w1+0w2- s1+A1 = 3

Case forAlternative

Optimal Solution


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ci

xi

bi

x1 x2 w1 w2 s1 A1

Ij Z= 48 0 0 0 2 0 M

cj 6 4 0 0 0 -M

0 w1 14 0 5/3 1 -2/3 0 0

0 s1 5 0 -1/3 0 1/3 1 -1

6 x1 8 1 2/3 0 1/3 0 0

Rati8.4

-

12

Final TableauOptimal Solutionis

x1=8, x2=0

giving Z = 48.


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ci

xi

bi

x1 x2 w1 w2 s1 A1

Ij Z= 48 0 0 0 2 0 M

cj 6 4 0 0 0 -M

4 x2 42/5

0 s1 39/5

6 x1 12/5

Alternative Optimal Solutionis

x1=12/5 , x2= 42/5giving Z =48.

C f


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02,1

)(12

)(101

)(6221/

2513

xx

IIIx

IIx

Ixxts

xxZMax

Standard Form:

Max Z = 3x1+5x2+0w1+0w2+0s1+(M)A1

x1-2x2+w1+0w2+0s1+0A1 = 6

x1+0x2+0w1+w2+0s1+0A1 = 10

0x1+x2+0w1+0w2- s1+A1 = 1

Case forUnbounded

Solution


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ci

xi

bi

x1 x2 w1 w2 s1 A1

Ij Z= -M -3 -M 0 0 M 0

-5

cj 3 5 0 0 0 -M

0 w1 6 1 -2 1 0 0 0

0 w2 10 1 0 0 1 0 0

-M A1 1 0 1 0 0 -1 1

Rati-

1

Tableau-I


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b

ix1 x2 w1 w2 s1 A1

Ij Z= 5 -3 0 0 0 -5 2M

+5

cj 6 4 0 0 0 -M

ci xi

0 w1

0 w2

5 x2

8

10

1

1 0 1 0 -2 -2

1 0 0 1 0 0

0 1 0 0 -1 1

All aij


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ci

xi

bi

x1 x2 w1 s1 A1

Ij Z= -3M M 2 M M 0

+20 +6

cj 6 4 0 0 -M

4 x2 5 1 1 1 0 0

-M A1 3 -1 0 -1 -1 1

Final Tableau As A1 = 3 in final tableu,Solution is infeasible.

Case for


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edunrestrictxx

IIxxIxxts

xxZMin

2,01

)(82513)(1221/

241

As x2 is unrestricted

x2 = x2 - x2Hence the problem is converted to

Case forUnrestricte

d Variable

Case for


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0"2,'2,1

)(8"25'2513

)(1"22'221/

"24'241

xxx

IIxxx

Ixxxts

xxxZMin

Convert Minimization Problem into

Maximization Problem

Case forUnrestricte

d Variable

Case for


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0"2,'2,1

)(8"25'2513

)(1"22'221/

"24'241)(

xxx

IIxxx

Ixxxts

xxxZMax

Standard Form:

Max(-Z) = -x1-4x2+4x2+0s1+(-M)A1+0w1

x1+2x2-2x2-s1+A1+0w1 = 1

3x1-5x2+5x2+0s1+0A1+w1 = 8

Case forUnrestricte

d Variable

Case for Unrestricted Variable


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bi x1 x2 x2 s1 A1 w1

Ij Z = -M

Ij = (Zj-cj) = (aij.ci)-cj

cj

-1 - 4 4 0 -M 0Tableau - I

1 1 2 -2 -1 1 0

14 3 -5 5 0 0 1

ci xi

-M A1

0 w1

2M M

-4

-M -2M

+1 +4


0 0



57/69


Ij Z = 2

cj

-1 -4 4 0 -M 0Tableau - II

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

11/2 0 0 -5/2 5/2 1

ci xi

-4 x2

0 w1

0 2 -2 0

+M

-1 0


1/2

33/2



58/69


Ij Z = -1

cj

-1 -4 4 0 -M 0Tableau - III

1 2 -2 -1 1 0

0 -11 11 3 -3 1

ci xi

-1 x1

0 w1 -2 1 -1 0

+M

0 2


1

11



59/69


Ij Z = 1

cj

-1 -4 4 0 -M 0Tableau - IV

3

1

ci xi

-1 x1

4 x2

0 +ve +ve +ve0 0


Hence Optimal Solution isx1=3, x2=1, x2=0

x1 = 3, x2 = x2-x2= -1 giving Zmin = -1


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Duality and

Primal Dual Relationship

Duality:Data for LPP (Primal) Formulation.


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y ( ) .

Factory: Two products P1 & P2

Profit/piece of P1 = Rs.6/-Profit/piece of P2 = Rs.8/-.

Time Constraint:

Time required/piece of P1 on m/c A= 30 hrs.

Time required/piece of P1 on m/c B= 5 hrs

Time required/piece of P2 on m/c A= 20 hrs

Time required/piece of P2 on m/c B= 10 hrs .

Max. time available on m/c A = 300 hrs.

Max. time available on m/c B = 110 hrs.

i i


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Primal Problem Formulation

X1=No. of units of P1

X2=No. of units of P2

Max.Zp = 6 X1 + 8 X2

s/t 30 X1 + 20 X2 300

5 X1 + 10 X2 110

X1 , X2 0

Dual Problem Formulation


63/69

Dual Problem Formulation

Y1=Cost of 1 hr of m/c A

Y2=Cost of 1 hr of m/c B

Min. Zd = 300 Y1 + 110 Y2

s/t 30 Y1 + 5 Y2 6

20 Y1 + 10 Y2 8

Y1 , Y2 0


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Primal - Dual Relationship


65/69

p

Primal (Max.) ( ) Dual (Min.) ( )

-----------------------------------------------------------No. of Variables No. of ConstraintsNo. of Constraints No. of VariablesObj. fn. Coeff. R.H.S.

R.H.S. Obj. fn. Coeff.jth Col. Of Coeff. jth Row of Coeff.ith relation ( ) ith Variable 0

jth Variable 0 jth relation ( )ith relation = ith variable unrestricted

jth variable unrestricted j th relation =

P i l P bl


66/69

Primal Problem

Max.Zp = X1+2X2 -3X3

s/t 2 X1 + X2 +X3 10

3 X1 - X2 =2 X3 110

X1 + 2X2 - X3 = 4

X1 , X3 0 , X3 unrestricted

To convert into Dual write in standard

format

P i l P bl St d d F t


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Primal Problem : Standard Format

Max.Zp = X1+2X2 -3X3

s/t 2 X1 + X2 +X3 10

-3 X1 + X2 -2 X3 - 110

X1 + 2X2 - X3 = 4

X1 , X3 0 , X3 unrestricted

D l ill b


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Dual will be

Min. Zd = 10 Y1 -110 Y2 +4 Y3

s/t 2 Y1 - 3 Y2 + Y3 1

Y1 + Y2 +2 Y3 = 2

Y1 - 2 Y2 - Y3 - 3

Y1 , Y2 0 , Y3 unrestricted

Thank you


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Thank youFor any Query or suggestion :

Contact :Dr. D. B. NaikProfessor, Training & Placement

Sardar Vallabhbhai National Instituteof Technology, SuratIchchhanath, Surat - 395007

Email : [email protected] No. 0261-2255225 (O)

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LP - II Lecture

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