optimization lecture 2

8/10/2019 optimization Lecture 2

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Lecture 2 Simplex methodsbasic idea of the is to confine the

search to corner oints of the feasible re ion

o which

The simplex method

) in a most intelligent wthere are only finitely many ay.

see corner points;

simplex algorithe can be used to solve anthe must be converted into a problem where all the

thm ,

Before LPLP

constraints are and all variables are nonnegatiequati veons .

in this f standardorm is said formto be in a .

An LP


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Standard FoConvert the rmto theLP LP

not in standard form isTheLP

in isstandard formTheLP

1 2

Subject to

1 2

Subject to

.

.

1 2x x 40

2x

x 60

.

.

11 2x x 40

2x x 60

s

s

,1 2 x x , , ,1 21 20 x x 0s s

u y, vit to an constraint by adding a slack variable to the thequality

,

i is .


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an has both and constraints apply the previous ,If LP

an below:

.Consider LP

standarNon d form

Standard form

1 2ax

Sub ect to

z x x

1 2x xax

Sub ect to

1 x 1 1100 x 100 s

2

1 2

50x 35x 6000

2

1 2

2

3

x

50x 35x 6000

s

s

1 2 20x 15x 2000

-1 2 420x e15x 2000

,1 2 , , , , ,1 2 1 2 3 4


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2. Preview of the Simplex Algorithm

an with m constraints and n variableshas been converted into the standard form

.

Suppose LP

from of such an is:The LPor 1 1 2 2 n nc xax n

Sub ec

c x c x

t

o

11 1 12 2 1n n 1 a x a x a x a x a x a x b

m m mn

, .,

n m

i x 0, i = 1 n


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we define

If

11 12 1n 1 1a a a x b

,, ,

21 22 2n 2 2a a a x bA x b

m1 m2 mn n ma a a x b

. Ax b

a basic solution to is obtained by settingvariables equal to and solving for the remaining

,Ifn m Ax bn m 0

var a es

assumes that setting the variables equal to

.

This

m

n 0myields a value for the remaining variables orequivalently the columns for the remaining variables

unique ,,

mm

are near y n epen ent.


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find a basic solution to we choose a set ofvariables and set each of ,

the ornonbasic variablesTo A

Nx b n m

BV

these variables equal to

we solve for the values of the ( ) variabl

.

Then

0

n m mn es

nonbasic variables basic solutions

( ) that satisfy

lead to different

the , or .

.

basic variables

Different

Ax bBV

basic sothe to the following equations:lutionConsider

1 2 xx 3

number of nonbasic variables .The

2 3 - x x -13 2 1-

for example { } then { }

can obtain the values for th basic variableese b settis n

, , , , .Setting

e

3 1 2xNB x xVV B

,3x 0 that is

.1 2x x 3

- x -0 1

basic solutionand is a, , .Thus 1 2 3x 2 x 1 x 0


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{ }and { } the becomesbasic solution, ,If 1 2 3NBV BVx x x

basic solutionan

{ }and { } the becomes , .

, ,If

1 2 3

2 1 3NBx x xx x xBVV

and

set

, .

Some

1 2 3x 3 x 0 x -1

s of variables do not ield a basic solution .m

the following linear system:Consider

1 2 3

1 2 3 2x x x4

.3

sol

would satisfut on yi

, ,3 1 2

1 2x 2x 1

t

.

Since

1 2 2x 4x 3

basihis c sos stem h lutionas no solution, there is no

corresponding to { }, .1 2xB xV


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Preview of the Simplex Algorithm

basic solution in which all variables are

is called abasic feasible soluti

nonnegativ

o

e

n

or .

Any

b s

the basic solutions in the previous exa empl , ,For 1x 2,2x x an are as c eas e

solutions the basic solution fails

, ,

, , ,but3 1 2 3

1 2 3

x x x

x 3 x 0 x -1

to e a basic feasiblfollowin show that a e

ecause .two tThe heorems

3s x

is ofsolution great importance in .LP


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:Theorem 1

eas e reg on or any near programm ng pro emis a convex set

.

an as an , t ere must e an

extreme point

o

of

pt m

the

a so ut on

feasible region optimalthat is

,

.

so

Theoremuni ue extremean there is a oint of the

: , s

For L PP L

2

each basic feasibfeasible region corresponding to solutionle .

point in the fe

,

asible region.


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relationship between andextreme points basic feasibleThe

so u ons ou ne n s seen n( ) is:in the standard form

, . with slack variabl esThe

xamp eLP

eorem

X2

60 D

1 2Maximiz z 4x 3xe

50

Labor Constraint

.11 2x x 40s

3

4

0

Feasible Region

B

.

, , ,2

1

1

2

2

1 2

s

x x s 0

20

ELeather Constraintinequalities show the

.

Two shaded

area

10extrem of the

feasible region are

e points

, , ,

The

B C E .FX110 20 30 40 50F


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Table 1

, , , ( )

1 2 1 21 2 1 2s s s sx x 0 x x 20 E z 140

, , , , ( )1 2 2 11 2 2 1x x x 0 x 30 10 C z 120s s s s

, , ,1 22 1 11 2x x xs s s x -

-

,

, , , , (

2 2s s

s s s s s

s

b sx x x 0 x 60 20 0

no

no

, , , , ( )2 1 1 22 1 1 2s s s s x x x 0 x 40 20 B z 120

, , , , ( )1 2 1 21 2 1 2 s s x x x x 0 40 60 F z 0s s

to the and the extreme points of the feasible region .LP

in a natural fashion to the ex

sLP treme points.


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The Simplex Algorithm For MaxLPs

standathe to rd form

Convert LPStep 1

Deter

Step 3 whether the current is optimalmine bfst e current s not opt ma eterm ne w c

should be comnonbasic variable basice a andvariable ,step

to find a with a better objective function value

.

'

bfs

with a better objective function value b

. Go

ack to .Step 3

:s s mp ex a gor m e s n e orm

,n o ec ve unc no

1 1 2 2 n n

z c x c x c x 0.

.call this format the row version of the objective functionWe 0


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: 1 2 360x 30x 20xExamp maximizele 2

1 2 3

1 2 3

x x x4x 2x 1.5x 20

su ec o

1 2 3 2x 1.5x 0.

x

5 8

2x 5 , ,

Standar form1 2 3 x x x 0.

1 2 3maximi z 60x 30x 20x 8x 6x

zesub ect t

o

x 48s

1 2 23 4x 2x 1.5x 20s

1 2 3

2

3 . .

x

4 5s

, , , , , ,1 2 3 1 2 3 4

x x x s s s .s


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Solution

-

Example 2 - con

yields the equations and basic variables

.

1 2 3 z 60x 30x 20x 0 z 0

Row 0

1 2 3

1 2 3

1 1

2

x x xow s s

Ro 4x 2x 1.5xw 2 s

2s20 20

3 3

4

1

4

2 3

2

ow s sRow

x . x . xx 54 s 5s

we setIf 1 2 3x = x = x we can solve for the values

{ }and { }

, , , , .

, , , , , , .Thus

1 2 3 4

1 2 3 1 2 34

=

BV

0

=

s s s s

s s s NB = x x xVseach constraint is then in canonical form ( s have a coefficient

in one row and zeros in

Since = 1BV

right - hand side) r) with a nonnegative ( hsall other rows ,

.


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Solution -Example 2 - con Step 2

perform the simplex algorithm we need a basic ( , although .To

a n a as c eas e o u on

) variable for

a ears in w

not necessarily nonnegative .

Since

row 0

row 0z ith a coefficient of and does,1 znot appear in any other we use as thebasic variable, .row z

initial canonical form has:

= = ,

this initial

, , , , , .

, , , , , ,For 1 2 3 4z = 0 = 48 = 20 = 8 = 5bfs s s ss

this example indicates a can be used as aslack variable

.

,As

1 2 3

.


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we have obtained a we need to determine whether it is optimal:

, .Oncebfs

bfsDetermine if the Current is OptimalStep 3

do this we try to determine if there is any way can be increasedby inc ,To z

nonbasic variablereasing some from its current value of zeronon as c var aw e o ng a o er a e r curr es en va ues o

zero.

non

eachFor

1 2 3

basic variablenonbasic variab

we can use the equation above to determineif increasin a le

,but hold nonbasicin all other

nonbasic) will increase

any of the wvaria i lble lvari to .a l

b es

Increasing0 z

cause an increase in .z

increasing causes the greatest rate of increase in

increases from its current value of it will have to become a

.

,

However

If

1

1 zero

x

xas c var a

this reason is

e.

,For 1x called the entering variable.

as e mos nega ve coe c en n .serve 1 rowx


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:Step 4

choose the ( ) to thewith the most negative coefficient

entering variableinnonbasic variable

in a max problemWerow 0.

,current basi 1

c variables ( ) will change value, , , .1 2 3 4s s s s

., 1

RATIO

we have

, . ,

, . ,In Since

1 1

2

1 1

1 2 1

row 2 s s

-

= 20 - 4x 0 x 20 4 = 5

, . ,

,In

3 31 1 -

row 4

we have any will always be . , .For 414 5= xs s 0

means o eep a e as c var a es nonnega ve e argeswe can make is { } ,

, , .s

1 6 5 4 4m nx i


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The Ratio TeSolution - stExample 2 - con

entering a variable into the basis, compute the ratio:

of of enterin variable in row

When

rhs row coe icient

every constraint in which the enteringFor variable has a

constraint with the smallest ratio is called the winner

.

The

smallest ration is the large enterinst value of the g

.

The

nonnegative.

e en er ng var a e a as c var a e nsince this ( ) was the winner of the ratio test

constraint

a e 1x rowrow

.


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Solu Theti Ratio Teston -Example 2 - con

ma e a n we use e emen aryrow operations ( ) to make have a coefficient of

as c va er a ,s

o 11 1

rowxxero

row r

procedure is called pivoting on ; and

.

This

ows

row 3 row 3s ca e e

final result is that replaces as the basic variable

.

The 31

p vot row

sx

forterm in the that involves th

. The

row 3pivot row e entering basic

variable is called the .pivot term


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Gauss - Jordan method using and: sThe ero simplexStep 5

ero 1 z x1 x2 x3 s1 s2 s3 s4 rhs ero

0 1 -60 -30 -20

1 8 6 1 1 48

2 4 2 1.5 1 20

3 1 0.75 0.25 0.5 4 row 3 divided by 1/2

4 1 1 5ero z x x x s s s s r s

0 1 15 -5 30 240 60 times row 3 added to row 0

1 8 6 1 1 48

2 4 2 1.5 1 20

3 1 0.75 0.25 0.5 4

4 1 1 5

ero 3 z x1 x2 x3 s1 s2 s3 s4 rhs

-

1 -1 1 -4 16 - 8 times row 3 added to row 1

2 4 2 1.5 1 20

3 1 0.75 0.25 0.5 4


0 1 15 -5 30 240

1 -1 1 -4 16

2 -1 0.5 1 -2 4 - 4 times row 3 added to row 2

3 1 0.75 0.25 0.5 44 1 1 5


21/36

The ResultExample 2 - con

32 3

z 15x 5x 30 240 z 240Row 0 s

ow 1 3 1

2 3 2

3

2 3

x

x 0.5x 2 4 4

s s

Row 2 s s s

1 2 33 1 x 0.75x 0.25x 0.5 4 x 4

Row 3 s

Row 4 x

2 4 45s s 5

{ , , , , }and { }, , , .In 1 21 32 4 3BV NBVCanonical Form x x x1 s s s s

, , , ,

.1 2

3

1

2 3

4

x

0xs

called an iteration ( ) of the simplex algorithm

or sometime pivot .


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Example 2 - con.

is optimal

,.

2 3 3

z 240 15x 5

sx 30 current i os nThe bfs optimal because increasing to

( ) will increaset

while holding the other tononbasic variable3

1x0

the value ofeither or basic will caused the value of

. Making 2 3x zsto

- Reca

decre

ll

a

th

se.

Ste 4 e rule for determinin the enterin variable is the coefficient with the greatest negative value

is the onl variable with a coefficientne ative

.

Sincex x

row 0

should be entered into eth basis.


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iteration ivotBe in 2

Example 2 - con.

the ratio test using as the entering variablePerforming 3x

for all vaules of

,From31

row 1 0 xs since1 3

s = 16 x

,

,

From if

From if

2 3row 2 s

row 3

0 x 4 0.5 = 8

x 0 x 4 0.25 = 16

for all vaules of, .From since34 4row 4 s 0 x 5=s

means o eep a e as c var a es nonnega ve el

,sargest we can make is { } , .3x 8 16 =min 8

pivotbecomes the

use to make a basic variable in

, .

- s, .

So

Now 3

row 2 ro

er

w

Step 5 rx ow 2o


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use to make a basic variable in- s, .Now 3eroStep 5 row 2x


0 1 15 -5 30 240

1 -1 1 -4 16

2 -2 1 2 -4 8 multiplied row 2 by 2

3 1 0.75 0.25 0.5 4

4 1 1 5


0 1 5 10 10 280 add 5 times row 2 to row 0

1 -1 1 -4 16

2 -2 1 2 -4 8

. . .

4 1 1 5ero 3 z x1 x2 x3 s1 s2 s3 s4 rhs

0 1 5 10 10 280

1 -2 1 2 -8 24 add row 2 to row 1

2 -2 1 2 -4 8

3 1 0.75 0.25 0.5 4

4 1 1 5


0 1 5 10 10 280

1 -2 1 2 -8 24

- -

3 1 1.25 -0.5 1.5 2 add -1/4 times row 2 to row 3

4 1 1 5


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The Result inExample 2 Canonical Form 2- con

Canonical Form 2z 5x 10 10 280 z 280

Basic VariableRow 0 s s

Row 1 1 2 3 12 2x 2 8 24 24s s s s

22 3 3

1

3

22 13

x 1.25x 0.5 1.5 2 x 2

Row 3 s s

ow

,In4 42

x s s

Canonical Form 2

{ , , , , }and { }

the and

, , .

Yieldin

1 4 3 23 21= z xBV NBV

b s

x = x

28

s s s s

s s0 24 x 8 x 2 5 .2 23s s x 0


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Example 2 - con.

for in yields:Solving z

row 0 z 280 5x 10 10

s

can see that increasing ( , , or while holding theWe 2 2 3s sx

solution at the end of is therefore optimal

.

.The iteration 2

following rule can be appliecanonical form d to determine whether aoptimals is : The bfs

optimalis ( for a max proAcanonical form ) if each

nonbasic variable has a nonnegative coefficient in the

blem

canonical form s . row 0


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The Big M Meth do

problems have found starting by using

.

Previous bfss ac var a ae as our

an have or constra

s c var a s

nts

e

i

es .

,If LP

however a starting,may no e rea y apparen

such a case the method may be used to solve

.

,In

s

Big M

t e 1 2

minimize z 2x 3x

.1 20.5x 0.25x 4subject to

..

2

1 2

x x 10

.,1 2 x x


28/36

LP standard formThe in

= 1 2 z 2x 3x 0.5x 0.25x 4MinimizeSub ect to

s

21 2 x 3x 20e

1 2

0, , , 1 21 2 x x s e

order to use the simplex method a is needed , . In sbf

variables will be labeled accord

ing t

, .

The

o the row in which they

.


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Example 3 - con. rtificial Variables

1 2

Row 0 z 2x 3x 0

1 2ow s . x . x

2

1

21 2

x 3x 20Row 2 e a

31 2 x x 10Row a3

b

e set equal to zero.

to the objective functio artificial variablesn for each, ,

.i

ia

function , i-

for each .ia

represents some very arge num er.


30/36

Example 3 - con.

modified in standard form then becomes

The1 2 2 3Row 0

LP z 2x 3x 0

- M - M a a

11 2Row 0.5x 0.25x 41 s

Row 2

e x 3x 20a

2 31Row x x3 1a 0

e o ec ve unc on s way ma esextremely costly for an a y

to be positivertificial variab sle .

optimal solution should force .The 2 3a a 0


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Initial TableauExample 3 - con.

Row z x1 x2 s1 e2 a2 a3 rhs

. - . - . - - .

1 0.50 0.25 1.00 4.00

. . . . .

3 1.00 1.00 1.00 10.00

2 3 in Row 0,To make = =0a a

Row 0+ M*(Row 2)+M*(Row 3)


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Pivot 1 z x1 x2 s1 e2 a2 a3 rhs ratio ero0 1.00 2M - 2 4M -3 -M 30M Row 0 + M(Row 2) + M(Row 3)

1 0.50 0.25 1.00 4.00 16

2 1.00 3 -1.00 1.00 20.00 6.67

3 1.00 1.00 1.00 10.00 10

ero 1 z x1 x2 s1 e2 a2 a3 rhs ero

0 1.00 2M-2 4M -3 -M 30M . . . .

2 0.33 1 -0.33 0.33 6.67 Row 2 divided by 3

3 1.00 1.00 1.00 10.00


0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3 Row 0 - (4M-3)*(Row 2)

1 0.50 0.25 1.00 4.00

2 0.33 1 -0.33 0.33 6.673 1.00 1.00 1.00 10.00


0 . - - - +

1 0.42 1.00 0.08 -0.08 2.33 Row 1 - 0.25*(Row 2)

2 0.33 1 -0.33 0.33 6.67

3 1.00 1.00 1.00 10.00

0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3

1 0.42 1.00 0.08 -0.08 2.33

2 0.33 1 -0.33 0.33 6.67

3 0.67 0.33 -0.33 1.00 3.33 Row 3 - Row 2


33/36

The Solution - con

Pivot 2 z x1 x2 s1 e2 a2 a3 rhs ratio

0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3

1 0.42 1.00 0.08 -0.08 2.33 5.60

2 0.33 1 -0.33 0.33 6.67 20.00

3 0.67 0.33 -0.33 1.00 3.33 5.00


0 1.00 (2M-3)/3 (M-3)/3 (3-4M)/3 (60+10M)/3-. . . . .

2 0.33 1 -0.33 0.33 6.67

3 1.00 0.50 -0.50 1.50 5.00 (Row 3)*(3/2)


0 1.00 -0.50 (1-2M)/2 (3-2M)/2 25.00 Row 0 + (3-2M)*(Row 3)/3

. . . - . .

2 0.33 1.00 -0.33 0.33 6.67

3 1.00 0.50 -0.50 1.50 5.00


0 1.00 -0.50 (1-2M)/2 (3-2M)/2 25.00

. - . . - . . ow - ow

2 0.33 1.00 -0.33 0.33 6.67

3 1.00 0.50 -0.50 1.50 5.00


0 1.00 -0.50 (1-2M)/2 (3-2M)/2 25.00 O timal Solution1 1.00 -0.13 0.13 -0.63 0.252 1.00 -0.50 0.50 -0.50 5.00 Row 2 -(1/3)*Row 3

3 1.00 0.50 -0.50 1.50 5.00


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The summary for the optimal solution

variable with the positive coefficient in rowshould enter the basis since this is a problem .

Themin

ratio test indicates that hould enter the sThe 2x

basis in which means the artificial variable,row 2will leave the basis.

2a

ratio test indicates that

.

should enter theThe 1

x

,

the basis.3

optimal solu is .tion ,The

e esu t

1 2z 25 x x 5


35/36

Tutorial

the simplex method to solve the following Using LP

1 2ax m ze z x x

.

.1

x

2x 12

.1 2 3x 2x 18

,1 2 x x .


36/36

LPsThe Simplex Algorithm for Min

the following :Consider LP

1 2 n m ze x x

.1 2 x x

.

,1 2

1 2

x x 0.

optimization lecture 2

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