5.2 Lecture_1g

8/9/2019 5.2 Lecture_1g

1/108

Lecture1g:

The Revised Simplex Method

Jeff Chak-Fu WONG

Department of Mathematics

Chinese University of Hong [email protected]

MAT581SS

Mathematics for Logistics

Produced by Jeff Chak-Fu WONG 1

8/9/2019 5.2 Lecture_1g

2/108

TABLE OF CONTENTS

1. Introduction

2. Revised Simplex Method For Standard Form I

3. Revised Simplex Method For Standard Form II

BLE OF CONTENTS 2

8/9/2019 5.2 Lecture_1g

3/108

Let us consider the LPP in the standard form, i.e.

max z=CT

Xsubject to

AX=b

and X 0,

(1)

with

1. b 0;

2. rankA=m(< n),

whereX Rn,C Rn,b Rm andA= [aij ]is an(m n)matrix.

BLE OF CONTENTS 3

8/9/2019 5.2 Lecture_1g

4/108

WriteA=

B R

and accordingly partition the vectorX as

X= XB

XR

. Then the system

AX=b

can be written as

B R XB

XR =b,BXB+RXR =b,

XB =B1b B1RXR, (2)

where the inverse ofB,B1, exists.

BLE OF CONTENTS 4

8/9/2019 5.2 Lecture_1g

5/108

And

z = CB CR

XBXR

=CBXB+ CRXR (3)

Substituting (3) into (2), we have

z =CBXB+ CRXR

=CB B1b B1RXR + CRXR=CBB1b CBB1R CRXR (4)

BLE OF CONTENTS 5

8/9/2019 5.2 Lecture_1g

6/108

For the sake of simplicity, we have

XB =B1b B1RXR

=B1b B1 am+1 am+2 an1 an

xm+1

xm+2...

xn1

xn

nm,1

=B1

b

nj=m+1

B1

ajxj

and

BLE OF CONTENTS 6

8/9/2019 5.2 Lecture_1g

7/108

z = CBB1b

CBB1 am+1 am+2 an cm+1 cm+2 cn

xm+1

xm+2

.

.

.

xn1

xn

= CBB1

b

n

j=m+1 CBB

1aj cj

xj

BLE OF CONTENTS 7

8/9/2019 5.2 Lecture_1g

8/108

max z=CBB1b

nj=m+1

CBB

1aj cj

xj

subject to

XB =B1

b B1

RXR =B1

b nj=m+1B1ajxjand XB, XR 0,

(5)

1. The condition of optimality

The values ofCBB

1aj cj

lead to the choice of a basic

variablexj

Therefore

j =CBB1aj cj

=CByj cj

=zj cj , j {m+ 1, m+ 2, , n}

BLE OF CONTENTS 8

8/9/2019 5.2 Lecture_1g

9/108

Put

=

m+1 m+2 n

T,

we have=CBB

1R CR.

Concerning the maximization LPP, we have

=CBB1

R CR 0

2. The choice of a basic variable

If

j = mini

CBB

1aj cj

, j {m+ 1, m+ 2, , n}

then the non-basic variablexj will become a basic variable.

BLE OF CONTENTS 9

8/9/2019 5.2 Lecture_1g

10/108

3. The choice of a non-basic variable

When obtaining the basic variable, we have

XB =B

1b

n

j=m+1 B

1ajxj =B

1b

B

1ajxj .

Recall thatB1b=XB Rm andB1aj =yj R

m

The so-called minimum ratio criteria is:

= mini

B1b

i

(B1aj)i

B1aj

i>0

= mini

(XB)i(yj)i

(yk)i >0=

B1b

r

(B1aj)r=

XBryrj

.

BLE OF CONTENTS 10

8/9/2019 5.2 Lecture_1g

11/108

Put the non-basic variable as

xBr = B1br(B1aj)r = xBryrj .

XB =

xB1

xB2...

xBr

.

..

xBm

.

BLE OF CONTENTS 11

8/9/2019 5.2 Lecture_1g

12/108

A USEFUL OBSERVATION

USEFUL OBSERVATION 12

8/9/2019 5.2 Lecture_1g

13/108

Theorems 7and11 (see Lecture Note 1e) could have also been

proved in an alternate way.

This is because the current basis matrix

B= b1 b2 br1 br br+1 bm1 bm

and the next basis matrix

B= b1 b2 br1 aj br+1 bm1 bm

differs only by one column.

Since in the current tableau the inverse of the current basis

matrix, namelyB1, is known, we could possibly use this to get

B1.


8/9/2019 5.2 Lecture_1g

14/108

8/9/2019 5.2 Lecture_1g

15/108

We shall discuss the revised simplex method in the following section

where a relationship betweenB1 and(B)1 will be utilized.


8/9/2019 5.2 Lecture_1g

16/108

RELATIONSHIP BETWEEN B1

AND(B)1

ELATIONSHIP BETWEEN B1

AN D(B)1 16

8/9/2019 5.2 Lecture_1g

17/108

Let

B=

b1 b2 br1 br br+1 bm1 bm

;

yj =B1aj ;

B = b1 b2 br1 aj br+1 bm1 bm

.


AN D(B)1 17

8/9/2019 5.2 Lecture_1g

18/108

Let

F = e1 e2 er1 yj er+1 em1 em

be anm mmatrix, where

e1 =

1

0

0

...

0

...

0

, e2 =

0

1

0

...

0

...

0

, , em =

0

0

0

...

0

...

1

, andyj =

y1j

y2j

y3j...

yrj

...

ymj

m1

.


AN D(B)1 18

8/9/2019 5.2 Lecture_1g

19/108

Then

BF = B e1 e2 er1 yj er+1 em1 em =

b1 b2 br1 Byj br+1 bm1 bm

=

b1 b2 br1 aj br+1 bm1 bm

=

B.

Hence

(B)1 = (BF)1 =F1B1 =EB1. (6)

But the matrixE(=F1)can be computed very easily provided

yrj = 0.


AN D(B)1 19

8/9/2019 5.2 Lecture_1g

20/108

1 0 y1j 0 1 0 0 0

0 1 y2j 0 0 1 0 0

......

......

......

......

. . ....

. . ....

0 0 yrj 0 0 0 1 0...

......

......

......

.... . .

.... . .

...

0 0 ymj 1 0 0 0 1

1 0 0 0 1 0 y1jyrj

0

0 1 0 0 0 1 y2jyrj

0

......

......

......

......

. . ....

. . ....

0 0 1 0 0 0 1yrj 0

......

......

......

......

. . ....

. . ....

0 0 0 1 0 0 ymj

yrj

1


AN D(B)1 20

8/9/2019 5.2 Lecture_1g

21/108

E.g.,

1 3 0 1 0 0

0 2 0 0 1 0

0 6 1 0 0 1

(1/2) R2

1 3 0 1 0 0

0 1 0 0 1/2 0

0 6 1 0 0 1

R1 3 R2

R3 6 R2

1 0 0 1 3/2 0

0 1 0 0 1/2 0

0 0 1 0 6/2 1

y1j

y2j

y3j

=

3

2

6

y1j/y2j

1/y2j

y3j/y2j

=

3/2

1/2

6/2


AN D(B)1 21

8/9/2019 5.2 Lecture_1g

22/108

It can be verified that

E= e1 e2 er1 er+1 em1 em

where

=

y1j

yrjy2jyrj

...yr1j

yrj

1yrj

yr+1jyrj

..

.ymjyrj

SoEis almost an identity matrix except that therth column is the

vector.


AN D(B)1 22

8/9/2019 5.2 Lecture_1g

23/108

Once(B)1 is known, we can compute all entries in the tableaueasily.

For example

XB = (B)1b

= EB1b=E(B1b) =EXB

=

e1 e2 er1 er+1 em1 em

XB,

=

1 0 y1jyrj

0

0 1 y2jyrj

0

......

. . ....

. . ....

0 0 1yrj

0

......

. . ....

. . ....

0 0 ymjyrj

1

xB1

xB2...

xBr...

xBm


AN D(B)1 23

8/9/2019 5.2 Lecture_1g

24/108

which, as before gives

xBi =

xBi xBryij

yrj, (i=r)

xBr

yrj, (i=r).

Of course XB so far is only a basic solution because we have only

enforced the conditionyrj = 0.


AN D(B)1 24

8/9/2019 5.2 Lecture_1g

25/108

To make XB a b.f.s and that too an improved one, we have to

again put conditions onr andj as inTheorem 7which will give the

usual rules for

a column to enter;

a column to leave.


AN D(B)1 25

8/9/2019 5.2 Lecture_1g

26/108

Let us illustrate the working of the revised simplex method.

Example 1 Solve the following LPP by the revised simplex method

max z= 2x1+x2

subject to

3x1+ 4x2 6

6x1+x2 3

x1, x2 0. (7)


AN D(B)1 26

8/9/2019 5.2 Lecture_1g

27/108

SolutionIntroducing the slack variablesx3 andx4 will give a(2 2)

identity matrix to start with. So we rewrite the problem as

max z= 2x1+x2+ 0x3+ 0x4

subject to

3x1 + 4x2 + x3 = 6

6x1 + x2 + x4 = 3

x1, x2, x3, x4 0


AN D(B)1 27

8/9/2019 5.2 Lecture_1g

28/108

FIRST ITERATION

1. Consider

XB =

x3

x4

, B=

1 0

0 1

=B1

CB =

0 0

.

2. Let

=CBB1 = 0 0 1 0

0 1

= 0 0

IRST ITERATION 28

8/9/2019 5.2 Lecture_1g

29/108

Calculate

z1 c1 =CBB1a1 c1

= a1 c1

=

0 0 3

6

c1= 0 2 =2

8/9/2019 5.2 Lecture_1g

30/108

Therefore,x1 becomes a basic variable since

mini

{zj cj}=z1 c1.

3. Take

y1 =B1

a1 = 1 00 1 3

6= 36

Consider

= mini

6

3,3

6= 3

6 =

1

2

Therefore,x4 becomes a non-basic variable.

IRST ITERATION 30

8/9/2019 5.2 Lecture_1g

31/108

SECOND ITERATION1. Consider

XB = x3

x1

, B = 1 30 6

CB =

0 2

.

B I2

=

1 3 1 0

0 6 0 1

I2 B

1

=

1 0

0 1

Thus,

B1 =

1 1/2

0 1/6

.

ECOND ITERATION 31

8/9/2019 5.2 Lecture_1g

32/108

y1 =B1a1 =

1 0

0 1

3

6

=

3

6

F I2

=

e1 y1 I2

I2 E

=

I2 e1

where

=

y11/y121/y12

= 3/61/6

= 1/21/6

.

(B)1 = 1 1/20 1/6

1 00 1

= 1 1/20 1/6

.

ECOND ITERATION 32

8/9/2019 5.2 Lecture_1g

33/108

B1 y1

=

1 0 3

0 1 6

B1 e2

=

0

1

x4 1 0 3

0 1 6

1 1/2 0

0 1/6 1

x4

ECOND ITERATION 33

8/9/2019 5.2 Lecture_1g

34/108

2. Let

=C B

B1 = 0 2 1 1/2

0 1/6 = 0 1/3 Calculate

z2 c2 =CBB1

a2 c2

= a2 c2

= 0 1/3 4

1 c2=

1

3 1 =

2

3

8/9/2019 5.2 Lecture_1g

35/108

8/9/2019 5.2 Lecture_1g

36/108

Consider

= mini

9/2

7/2

,1/2

1/6= 9

7Therefore,x3 becomes a non-basic variable.

ECOND ITERATION 36

8/9/2019 5.2 Lecture_1g

37/108

THIRD ITERATION

1. Consider

X B

= x2x1

, B = 4 31 6

C B

=

1 2

.

HIRD ITERATION 37

8/9/2019 5.2 Lecture_1g

38/108

8/9/2019 5.2 Lecture_1g

39/108

B1 y2 = 1 1/2 7/20 1/6 1/6

(B)1 e1 = 1 0

x3

1 1/2 7/20 1/6 1/6

27 R1 2/7 1/7 10 1/6 1/6

R2

16

R1

2/7 1/7 11/21 4/21 0

HIRD ITERATION 39

8/9/2019 5.2 Lecture_1g

40/108

2. Let

=C B

(B)1 = 1 2 2/7 1/71/21 4/21 = 4/21 5/21

Calculate

z3 c3 =C B(

B)

1

a3 c3= a3 c3

= 4/21 5/21

1

0

c3

= 4

21 0 =

4

21 >0

HIRD ITERATION 40

8/9/2019 5.2 Lecture_1g

41/108

z4 c4 =C B(

B)

1

a4 c4= a4 c4

= 4/21 5/21

0

1

c4

= 5

21 0 =

5

21 >0

3.

X B

=

x2

x1

= (B)1b=

2/7 1/7

1/21 4/21

6

3

=

9/7

2/7

andz= 2(2/7) + (9/7) = 13/7

Therefore, the optimal valuez of the given LPP is13/7and the

optimal solution is(x1 = 9/7, x

2 = 2/7).

HIRD ITERATION 41

8/9/2019 5.2 Lecture_1g

42/108

THE REVISED SIMPLEX METHOD IN A TABLEAU FORM

HE REVISED SIMPLEX METHODIN A TABLEAU FOR M 42

8/9/2019 5.2 Lecture_1g

43/108

INTRODUCTION

TRODUCTION 43

8/9/2019 5.2 Lecture_1g

44/108

In the last section, while deriving the updation formula

(B)1 =EB1,

we noted that we can implement the simplex method

somewhat differently where we store the elements of current

B1 (andnottheyj columns) in the tableau.

Such an implementation of the simplex method is called therevised simplex methodin the tableau form.

Theoretically, the revised simplex method and the simplex

method are exactly same -

It is only the tableau which is different here and is of much smaller

size becauseB1 is a matrix of order(m m)only.

TRODUCTION 44

8/9/2019 5.2 Lecture_1g

45/108

The main advantage of the revised simplex method is that here

we handle a tableau ofmuch smaller size.

Also this is useful in solving large scale LPPs (wheren is verylarge) because

here the data (C,b, andA) is stored separately and

therefore doesnotchange during the implementation.

In contrast, for the simplex method, the initial data is stored in

the tableau which itself changes during the implementation

because we store the updatedyj columns in the tableau.

Here we may note that the simplex method requires allyj

columns to be evaluatedfirstso that(zj cj)can be computed.

TRODUCTION 45

8/9/2019 5.2 Lecture_1g

46/108

But in the revised simplex method, as currentB1 is known, we

first compute all(zj cj)(and this willnotrequire knowledge of

yj) and

then computeonlythatyj which is needed, i.e. the one which

corresponds tothe negative most valueof(zj cj).

Since the updation formula

(B)1 =EB1

is simple and efficient, the revised simplex methodimplementation is certainly attractive.

TRODUCTION 46

8/9/2019 5.2 Lecture_1g

47/108

8/9/2019 5.2 Lecture_1g

48/108

Here we discuss the implementation of the revised simplex

method for the case when artificial variables arenotrequired

andm slack variables themselves give the(m m)identity basis

matrix to start with. This case we term asstandard form I.

The case ofstandard form II, where some artificial variables are

required is illustrated by an example only.

TRODUCTION 48

8/9/2019 5.2 Lecture_1g

49/108

REVISED SIMPLEX METHOD FOR STANDARD FORM I

EVISED SIMPLEX METHODFOR STANDARDFOR M I 49

8/9/2019 5.2 Lecture_1g

50/108

In this form, them slack variables themselves give the(m m)

identity basis matrix to start with.

Hence the given LPP has the following form

max z=c1x1+c2x2+ +cnxn

subject to

a11x1+a12x2+ +a1nxn b1

a21x1+a22x2+ +a2nxn b2 (8)

...

am1x1+am2x2+ +amnxn bm

and x1 0, , xn0,

wherebi 0 (i= 1, , m).


8/9/2019 5.2 Lecture_1g

51/108

8/9/2019 5.2 Lecture_1g

52/108

Now them constraints of (9) constitute a system of(m+ 1)linearequations in(n+m+ 1)unknowns and this is expressed as

1 c1 c2 cn 0 0 0

0 a11 a12 a1n 1 0 0

0 a21 a22 a2n 0 1 0

......

. . ....

......

. . ....

...

0 am1 am2 amn 0 0 1

z

x1

x2...

xn+m

=

0

b1

b2...

bm

,

(10)

i.e. 1 CT0 A

m+1,n+m+1

zx

n+m+1,1

=

0b

m+1,1

(11)


8/9/2019 5.2 Lecture_1g

53/108

1 CT

0 A

z

x

=

0

b

where

CT =

c1 c2 cn 0 0

T,

and

A=

a11 a12 a1n 1 0 0

a21 a22 a2n 0 1 0

......

. . ....

......

. . ....

am1 am2 amn 0 0 1

.


8/9/2019 5.2 Lecture_1g

54/108

Let us define

aj,R=

cj

a1j

a2j

...

amj

= cj

aj (j= 1, , n+m)

and

bR =

0

b1

b2...

bm

=

0

b

,

where the subscript/superscriptR refers to the revised simplex

method.


8/9/2019 5.2 Lecture_1g

55/108

BR =

1 c1 c2 cm

0 a11 a12 a1m

0 a21 a22 a2m...

.... . .

......

0 am1 am2 amm

.

Now in (10),zhas to be a basic variable always because itrepresents the objective function which we wish to maximize and

therefore the first column of the basic matrixBR for (10) is always an

identity column.

1 CT

0 A

m+1,n+m+1

z

x

n+m+1,1=

0

b

m+1,1EVISED SIMPLEX METHODFOR STANDARDFOR M I 55

8/9/2019 5.2 Lecture_1g

56/108

BR =

1 c1 c2 cm

0 a11 a12 a1m

0 a21 a22 a2m...

.... . .

......

0 am1 am2 amm

.

Now in (10),zhas to be a basic variable always because itrepresents the objective function which we wish to maximize and

therefore the first column of the basic matrixBR for (10) is always an

identity column.

1 CT

0 A

m+1,n+m+1

z

x

n+m+1,1=

0

b

m+1,1EVISED SIMPLEX METHODFOR STANDARDFOR M I 56

8/9/2019 5.2 Lecture_1g

57/108

BR =

1 c1 c2 cm

0 a11 a12 a1m

0 a21 a22 a2m...

.... . .

......

0 am1 am2 amm

= e1 a1,R a2,R am,R m+1,m+1 .

The remainingm columns ofBR (we note thatBR is an

(m + 1) (m + 1)basis matrix for the system (10)) have to come from

columnsaj,R depending upon whichmcolumns ofA currently

constitute the basis matrixB for the systemAX=b.

1 CT

0 A

m+1,n+m+1

z

x

n+m+1,1

=

0

b

m+1,1


8/9/2019 5.2 Lecture_1g

58/108

8/9/2019 5.2 Lecture_1g

59/108

Now using the partition method of finding inverse, we get

B1R =

1 CTBB

1

0 B1

, (13)

where

(m+ 1)columns ofB1R are denoted ase1, 1, 2, , m;

e1 being the identity column, i.e.e1 =

10

...

0

.

Thus

B1R = e1 1 2 m

.


8/9/2019 5.2 Lecture_1g

60/108

8/9/2019 5.2 Lecture_1g

61/108

yj,R= (zj cj)

yj . By equation (14) we observe that

the first component ofyj,Rgives the value of(zj cj)and

remainingm components give the usualyj column which iscomputed in the simplex method.

The main point to be noted here is that the value of(zj cj)is

computed in a way which doesnotneedyj

, where as in the

simplex method wefirstcomputeyj and then use them to

compute(zj cj).


8/9/2019 5.2 Lecture_1g

62/108

yj,R= (zj cj)

yj . By equation (14) we observe that

the first component ofyj,Rgives the value of(zj cj)and

remainingm components give the usualyj column which iscomputed in the simplex method.

The main point to be noted here is that the value of(zj cj)is

computed in a way which doesnotneedyj

, where as in the

simplex method we first computeyj and then use them to

compute(zj cj).


8/9/2019 5.2 Lecture_1g

63/108

8/9/2019 5.2 Lecture_1g

64/108

In practice to compute(zj cj), equation (14) tells us that we have

to take the first row of the matrixB1R and then take its dot product

with the vectoraj,R, i.e.

(zj cj) = (First row ofB1R ) (columnaj,R). (17)

Also

XRB =B1R bR = 1 CTBB1

0 B1 0

b

= z(XB)XB

; (18)i.e.

the first component of the vectorXRB gives the current value ofthe objective function, and

the remainingm components ofXRB give the current b.f.sXB .


8/9/2019 5.2 Lecture_1g

65/108

As for the standard form - I, initiallyB =In inverse of the initial basis

matrix for the revised simplex method is

B1

R =

1 0 0 0

0 1 0 0

0 0 1 0......

. . ....

0 0 0 1

.

Also the initial revised simplex tableau looks like


8/9/2019 5.2 Lecture_1g

66/108

Variable in the basis e1 1 2 m XRB

(z) (xn+1) (xn+2) (xn+m)

z 1 0 0 0 0

xn+1 0 1 0 0 b1

xn+2 0 0 1 0 b2...

.

.

....

. . ....

xn+1 0 0 0 0 1 bm

The dataa

j,R(j= 1, , n)and bR isstored separatelyand

access to a particular column is made as and when it is

needed.

Now we compute(zj cj)for non-basic columnaj,R as per

equation (17), i.e.,

(zj cj) = (First row ofB1R ) (columnaj,R),

and findthe negative most valueof(zj cj), say(zk ck).


8/9/2019 5.2 Lecture_1g

67/108

Variable in the basis e1 1 2 m XRB

(z) (xn+1) (xn+2) (xn+m)

z 1 0 0 0 0

xn+1 0 1 0 0 b1

xn+2 0 0 1 0 b2...

.

.

....

.

.

.

xn+1 0 0 0 0 1 bm

So we next computeyk only by computing the full vectoryk,R,

no otheryj,R is computed as no otheryj is needed.

Onceyk is known, we find a variable to leave the basis as per

the usual minimum ratio criteria.

Now we need to update the current basis inverse(B1R )to the

new basis inverse, namely(BR)1.


8/9/2019 5.2 Lecture_1g

68/108

For this we use the relationship

(BR)1 =EB1R

as derived in the last section, where

E= e1 e2 er1 er+1 em

,

and

= y1kyrk

y2kyrk

yr1k

yrk

1yrk

yr+1kyrk

ymkyrk

T.

Once(BR)1 is known the new revised simplex tableau is known

and then we continue till all(zj cj)0or

there is an indication of unbounded solution.


8/9/2019 5.2 Lecture_1g

69/108

8/9/2019 5.2 Lecture_1g

70/108

Solution

Problem (19) is clearly in the standard form - I because the slack

variablesx3

andx4

will give a(2 2)identity matrix to start with.

So we rewrite the problem as

Max z

subject to

z 2x1 x2 0x3 0x4 = 0

3x1+ 4x2+x3 = 6

6x1+x2+x4 = 3x1, x2, x3, x4 0. (20)


8/9/2019 5.2 Lecture_1g

71/108

8/9/2019 5.2 Lecture_1g

72/108

FIRST ITERATION

Step 1

Given the initial simplex tableau,

we first compute(zj cj)for non-basic columns and if all of

these are nonnegative then the current solutionXRB is optimal,

otherwise we find the negative most(zj cj)to identify(zk ck).

IRST ITERATION 72

8/9/2019 5.2 Lecture_1g

73/108

1 2 1

0 3 4

0 6 1

=

e1 a1,R a2,R

.

B1R =

1 CTBB

1

0 B1

=

1 012

0 B1

.

For our example

(z1 c1) = (first row ofB1R )a1,R

=

1 0 0 23

6

=2

IRST ITERATION 73

8/9/2019 5.2 Lecture_1g

74/108

1 2 1

0 3 4

0 6 1

= e1 a1,R a2,R

.

B1R =

1 CTBB1

0 B1

=

1 012

0 B1

.

(z2 c2) = (first row ofB1R )a2,R

= 1 0 0

1

4

1

=1,

giving the negative most value as2fork= 1. Thus the variablex1

becomes a basic variable.

IRST ITERATION 74

8/9/2019 5.2 Lecture_1g

75/108

1 2 1

0 3 4

0 6 1

= e1 a1,R a2,R

.

Step 2

Once negative most value(zk ck)is identified, we compute the

vector yk,R only, by using the relation

yk,R =B1R ak,R.

For our examplek = 1and so we computey1,R only to get

y1,R =B1R a1,R =

1 0 0

0 1 0

0 0 1

2

3

6

= 2

3

6

,

which we augment with the current tableau as shown.

IRST ITERATION 75

8/9/2019 5.2 Lecture_1g

76/108

Variable in the basis e1 1 2 XRB y1,R

(z) (x3) (x4)

z 1 0 0 0 -2

x3 0 1 0 6 3

x4 0 0 1 3 6

IRST ITERATION 76

8/9/2019 5.2 Lecture_1g

77/108

Step 3

Next we find a variable to leave the basis by employing the

usual minimum ratio criteria.

In this context we note thatz will never be considered to leave

the basis as it represents the objective function which we wish

to maximize.

In our example we evaluate

min

6

3,3

6

=

3

6

x4which corresponds to the variablex4 and therefore the variablex4

leaves the basis.

IRST ITERATION 77

8/9/2019 5.2 Lecture_1g

78/108


(z) (x3) (x4)

z 1 0 0 0 -2

x3 0 1 0 6 3

x4 0 0 1 3 6

IRST ITERATION 78

8/9/2019 5.2 Lecture_1g

79/108


(z) (x3) (x4)

z 1 0 0 0 -2

x3 0 1 0 6 3

x4 0 0 1 3 6

IRST ITERATION 79

Step 4

8/9/2019 5.2 Lecture_1g

80/108

Now we update the current inverseB1R to get the new inverse

(BR)1 which is given byEB1R .

In our example, as

x1 is becoming a basic variable and

x4 is becoming non-basic,

we have

E= e1 e2

, where=

1/31/21/6

.Thus

E=

1 0 1/3

0 1 1/2

0 0 1/6

.

IRST ITERATION 80

8/9/2019 5.2 Lecture_1g

81/108

1 0 2 1 0 0

0 1 3 0 1 0

0 0 6 0 0 1

1 0 0 1 0 1/3

0 1 0 0 1 1/2

0 0 1 0 0 1/6

and

=

y11/y13

y12/y13

1/y13

=

(2/6)

3/6

1/6

=

1/3

1/2

1/6

IRST ITERATION 81

8/9/2019 5.2 Lecture_1g

82/108

and

(BR)1 =EB1R =

1 0 1/3

0 1 1/2

0 0 1/6

1 0 0

0 1 0

0 0 1

=

1 0 1/3

0 1 1/2

0 0 1/6

.

Also

XRB = (BR)

1bR =

1 0 1/30 1 1/20 0 1/6

06

3

= 19/2

1/2

.

We shall follow the convention that current basis inverse inB1R and

the next basis inverse in(BR)1 irrespective of what is the number of

current iteration.

IRST ITERATION 82

8/9/2019 5.2 Lecture_1g

83/108

We shall always writeXRB (XRB)no matter at what iteration we are


(z) (x3) (x1)

z 1 0 1/3 1 -2/3

x3 0 1 -1/2 9/2 7/2

x1 0 0 1/6 1/2 1/6

IRST ITERATION 83

8/9/2019 5.2 Lecture_1g

84/108

Also we shall always writeXRB no matter at what iteration we are


(z) (x3) (x1)

z 1 0 1/3 1 -2/3

x3 0 1 -1/2 9/2 7/2

x1 0 0 1/6 1/2 1/6

IRST ITERATION 84

8/9/2019 5.2 Lecture_1g

85/108

SECOND ITERATION

Step 1 We have

(z4 c4) =

1 0 1/3

0

0

1

= 1/3,

(z2 c2) =

1 0 1/3

1

4

1

=2/3

8/9/2019 5.2 Lecture_1g

86/108

Step 2 We find

y2,R= B1R a2,R =

1 0 1/30 1 1/20 0 1/6

14

1

= 2/37/2

1/6

.

Step 3 As

min

9/2

7/2,1/2

1/6

= min

9

7, 3

occurs for the variablex3 it becomes a non-basic variable.

ECOND ITERATION 86

8/9/2019 5.2 Lecture_1g

87/108


(z) (x3) (x1)

z 1 0 1/3 1 -2/3

x3 0 1 -1/2 9/2 7/2

x1 0 0 1/6 1/2 1/6

ECOND ITERATION 87

8/9/2019 5.2 Lecture_1g

88/108


(z) (x3) (x1)

z 1 0 1/3 1 -2/3

x3 0 1 -1/2 9/2 7/2x1 0 0 1/6 1/2 1/6

ECOND ITERATION 88

Step 4 No

8/9/2019 5.2 Lecture_1g

89/108

Step 4 NowE=

e1 e3

,

where

=

y21/y22

1/y22

y23/y22

== (2/3)

7/21

7/2

(1/6)7/2

= 4/21

2/7

1/21

,

(BR)1 =EB1R =

1 4/21 0

0 2/7 0

0 1/21 1

1 0 1/3

0 1 1/2

0 0 1/6

=

1 4/21 5/21

0 2/7 1/7

0 1/21 4/21

ECOND ITERATION 89

8/9/2019 5.2 Lecture_1g

90/108

X

R

B = (BR)

1

b

R

= 1 4/21 5/21

0 2/7 1/70 1/21 4/21

0

63

= 13/7

9/72/7

.Therefore the next revised simplex tableau is

Variable in the basis e1 1 2 XRB

(z) (x2) (x1)

z 1 4/21 5/21 13/7

x2 0 2/7 -1/7 9/7

x1 0 -1/21 4/21 2/7

ECOND ITERATION 90

THIRD ITERATION

8/9/2019 5.2 Lecture_1g

91/108

Step 1 We have

(z3 c3) = 1 4/21 5/21 0

1

0

= 4/21and

(z4 c4) =

1 4/21 5/21 00

1

= 5/21.

As both of these are non-negative, the current solution is optimal.

Therefore, the optimal valuez of the given LPP is13/7and the

optimal solution is(x1 = 9/7, x

2 = 2/7).

HIRD ITERATION 91

8/9/2019 5.2 Lecture_1g

92/108

THE REVISED SIMPLEX METHOD FOR THE STANDARD FORM-II

HE REVISED SIMPLEX METHODFOR THE STANDARDFOR M-II 92

8/9/2019 5.2 Lecture_1g

93/108

We next consider the standard form-II, where the artificial

variables are required.

There could be many versions for the implementation of the

revised simplex method for this case but the easiest and mostnatural seems to be to use the BigMmethod and solve the

same by the revised simplex method.


8/9/2019 5.2 Lecture_1g

94/108

Example 3 Consider the following LPP

max z= 4x1+ 3x2

subject to

x1+x2 8

2x1+x2 10

x1, x2 0. (21)

Solution

Let problem (21) be solved by the BigMmethod but rather than

employing the simplex method we wish to employ the revised


simplex method. We have

max z = 4x1 + 3x + 0x + 0x4 M A1

8/9/2019 5.2 Lecture_1g

95/108

max z= 4x1+ 3x2+ 0x3+ 0x4 M A1

subject to

x1+x2+x3 = 8

2x1+x2 x4+A1 = 10

x1, x2, x3, x4, A1 0 (22)

which in the revised simplex format is expressed as

max z

subject to

z 4x1 3x2 0x3 0x4+M A1 = 0

x1+x2+x3 = 8

2x1+x2 x4+A1 = 10

zis unrestricted, x1, x2, x3, x4, A1 0. (23)


8/9/2019 5.2 Lecture_1g

96/108

8/9/2019 5.2 Lecture_1g

97/108

because

X

R

B =B

1

R b

R

= 1 0 M

0 1 00 0 1

0

810

= 10M

810

.


FIRST ITERATION

Step 1 We have

8/9/2019 5.2 Lecture_1g

98/108

Step 1 We have

(z1 c1) = 1 0 M 4

12

= 42M

(z2 c2) = 1 0 M

3

1

1

= 3M

(z4 c4) =

1 0 M

0

0

1

=M.

So the negative most value of (zj cj), i.e. (zk ck) = 42M, i.e. k = 1

andx1 becomes a basic variable.

IRST ITERATION 98

8/9/2019 5.2 Lecture_1g

99/108

Step 2 Next we computey1,R, i.e.

y1,R =

1 0 M

0 1 0

0 0 1

4

1

2

=

42M

1

2

,

and append the same to the initial tableau as shown here.

IRST ITERATION 99

8/9/2019 5.2 Lecture_1g

100/108

Step 3

The minimum ratio criteria gives, namely

min8

1

,10

2 =

10

2

= 5

is attained for the variable A1 soA1 leaves the basis as it becomes a

non-basic variable.

IRST ITERATION 100

Step 4

8/9/2019 5.2 Lecture_1g

101/108

Now

E= e1 e2

,

where

=

y11/y13

y12/y13

1/y13

=

(42M)2

1/2

1/2

,

(BR)1 =EB1R =

1 0 2 +M

0 1 1/2

0 0 1/2

1 0 M

0 1 0

0 0 1

=

1 0 2

0 1 1/2

0 0 1/2

IRST ITERATION 101

8/9/2019 5.2 Lecture_1g

102/108

XRB = (BR)

1b1 =

1 0 2

0 1 1/2

0 0 1/2

0

8

10

=

20

3

5

.

Therefore the next revised simplex tableau is


(z) (x3) (x1)

z 1 0 2 20 -2

x3 0 1 -1/2 3 1/2

x1 0 0 1/2 5 -1/2

IRST ITERATION 102

SECOND ITERATION

8/9/2019 5.2 Lecture_1g

103/108

Step 1

We have

(z2 c2) =

1 0 2

4

1

2

= 0

(z4 c4) =

1 0 2

0

0

1

=2.

Thereforex4 becomes a basic variable.

ECOND ITERATION 103

Step 2

8/9/2019 5.2 Lecture_1g

104/108

We evaluatey4,R =

1 0 2

0 1 1/2

0 0 1/2

0

0

1

=

2

1/2

1/2

Step 3

The minimum ratio occurs for the variablex3 sox3 becomes a

nonbasic variable.


y4,R

(z) (x3) (x1)

z 1 0 2 20 -2

x3 0 1 -1/2 3 1/2

x1 0 0 1/2 5 -1/2

ECOND ITERATION 104

Step 4

8/9/2019 5.2 Lecture_1g

105/108

Step 4

Now

E= e1 e3 ,where

= y41/y42

1/y42

y43/y42

= (2)1/2

2(1/2)

2

= 4

21

,

(BR)1 =

1 4 0

0 2 0

0 1 1

1 0 2

0 1 1/2

0 0 1/2

= 1 4 0

0 2 1

0 1 0

ECOND ITERATION 105

8/9/2019 5.2 Lecture_1g

106/108

XRB =

1 4 0

0 2 1

0 1 0

0

8

10

=

32

6

8

Therefore the next revised simplex tableau is


(z) (x4) (x1)

z 1 4 0 32

x4 0 2 -1 6

x1 0 1 0 8

ECOND ITERATION 106

THIRD ITERATION

Step 1

8/9/2019 5.2 Lecture_1g

107/108

We have

(z2 c2) =

1 4 0

4

1

0

= 0

(z3 c3) =

1 4 0

3

1

1

= 1

(z4 c4) =

1 4 1

M

0

1

=M 1.

HIRD ITERATION 107

8/9/2019 5.2 Lecture_1g

108/108

As all(zj cj) 0, the current solution namely,(x

1 = 8, x

2 = 0)is

optimal and the optimal valuez = 32.

HIRD ITERATION 108

Documents

5.2 Lecture_1g