Linear Programming Reading: CLRS, Ch. 29 or references

CSE 6331 Algorithms Steve Lai

1 2

1 2

1 2 1 1 2 2

Let , , , , be real numbers and let , , , be variables.

Linear function: ( , , , )

Linear equality

Linear functions and linear constraints

n

n

n n n

a a a bx x x

f x x x a x a x a x= + + +

•

•

•

1 2

1 2

1 2

: ( , , , ) (hyperplane)

Linear inequalities: ( , , , ) (half-space) ( , , , ) (half-space)

n

n

n

f x x x b

f x x x bf x x x b

=

≤≥

•

2

1 2

1 1 1 2

Given real numbers , , , where 1 , 1 ,

find real numbers , , , to

optimize a linear objective function

General Linear Programming Problems

i j ij

n

n n

b c a i m j nx x x

c x c x c x

≤ ≤ ≤ ≤

+ + +

•

1

subject to linear constrainlts

for 1 . n

ij j ij

a x b i m=

≤ = ≤ ≤ ≥

∑

3

1 2

1 1 2 2

1

find real rnu

1 , 111

, ,

Given r

, mbers

eal numbers

that

for 1

0 f

m

aximize

subject to

o

Standard Maximum Form

ij

i

j

n

n n

n

ij j ij

j

a i m j nb i mc j n

x x x

c x c x c x

a x b i m

x=

≤ ≤ ≤ ≤≤ ≤≤ ≤

+ + +

≤ ≤ ≤

≥

∑

r 1 j n≤ ≤4

11 12 1 1 1

21 22 2 2 2

1 1

and

maximiz

e

subject t

o

In Matrix NotationT

n

n

m m mn n n

a a a x ba a a x b

a a a x b

≤ ≥

≤

c x

Ax b x 0

5

( ) ( )

( ) ( )

1 2 1 2

1 2 1 2

11 12 1

21 22 2

1 1

-vectors:

-ve

, , , , , , ,

, , , , , , ,

cto

rs:

-matrix:

NotationT T

m m

T Tn n

n

n

m m mn

b b b y y y

x x x c c c

a a aa a a

a a

m

n

m n

a

= =

= =

=

×

b y

x c

A

6

( ) ( )

1 1 2 2

11 12 1

21 22 21 2 1 2

1 1

minimize

subject to

and

Standard Minimum Form (in Matrix Notation)T

m m

T T

n

nm m

m m mn

y b y b y b

a a aa a a

y y y c c c

a a a

= + + +

≥ ≥

≥

y b

y A c y 0

7

1 2

1 2

Variables: , , , Feasible solution: a vector ( , , , ) that

satisfies all the constraints. Feasible region : the set of all feasible solutions. Optimal solut

Terminology

n

n

x x xx x x

••

Ω••

ion: a solution that optimizes the objective function. Optimal objective value *: A linear program may be

or ( is nonempty or empty)

feasible infeasible bound or ed unbounded

f•

Ω•

( * is finite or infinite)f

8

1 2

1 2

1 2

1 2

1 2

maximize

subject to

2 4 4 2 12

1 , 0

objective function

main constraint

s

Examplex x

x xx xx x

x x

+

+ ≤ + ≤ − + ≤

≥ nonnegativity const ra ints

9

10 Source: Linear programming, Thomas S. Ferguson, UCLA

persons are available for jobs. If person works on job for one day, the value is . Choose an

assignment of persons to jobs to maximize th

Probl

e to l

em:

ta

Example: Optimal Assignment Problem

ij

I J ij a

1 1

1

1

Let

maximize

subjec

value. be the portion of time of person to work on job .

1 for 1,2, ,

1 for 1,2, ,

0

t to

ij

I J

ij iji j

I

ijiJ

ijj

ij

x i j

a x

x j J

x i I

x

= =

=

=

≤ =

≤ =

≥

∑∑

∑

∑

for 1,2, , and 1,2, ,i I j J= = 11

Two maximization LPs, and , are if for each feasible solution to with objective value there is a

equivalent

correspond feasible solution ting o with

Equivalence of Linear ProgramsL L

zLL

′

′ ′

•x

x

equivalent

objective value , and .

A maximization LP, , and a minimization LP, , are if for each feasible solution to with objective value there is a cor

vi

re

ce

sponding f

ve

ea

rsa

sibl

z

z

L LL′•

xe

solution to with objective value , and vice versa.

L z′ −′x

12

We can always convert an LP into an equivalent LP in standard (maximum or minimum) form.

A linear program may not be in standard maximum form for any of thes

Converting to Standard Maximum Form•

•e reasons:

1. The objective funtion may be a minimization.

2. There may be " " (instaed of " ") inequality constraints.

3. There may be equality constraints.

4. There may be some without nonnegix

≥ ≤

ativity constraints.

13

1 1

1 1

1. Convert minimization to maximixation: Minimize ( , , ) maximize ( , , )

2. Convert a " " constraint to a " " constraint:

( )

3. Convert an equality c

n n

n n

ij j i ij j ij j

f x x f x x

a x b a x b= =

−

≥ ≤

≥ − ≤ −

⇒

⇒∑ ∑

1 1

1

1 1

onstraint into two inequalities:

( )

n n

ij j i ij j inj j

ij j i n nj

ij j i ij j ij j

a x b a x ba x b

a x b a x b

= =

=

= =

⇒ ⇒

≤ ≤

= ≥ − ≤ −

∑ ∑∑

∑ ∑

14

1

4. If is without nonnegativity constraint,

we replace with and add 0, 0.

Thus, becomes , and

becomes .

A feasible solution ,

j

j j j j j

j j j j j j

j j j j j j

xx x x x x

c x c x c xa x a x a x

x

′ ′′ ′ ′′− ≥ ≥

′ ′′−

′ ′′−

( )( )1

, , , , to the new LP

corresponds to a feasible solution , , , ,

to the original LP.

j j n

j n

x x x

x x x

′ ′′

15

1 2 1 2

1 21 2

1 21 2

1 21

1

2 3 2 3

subject to

min

subject

im

to

77

7 2 4

ize maximiz

2 4

0

e

0

Examplex x x x

x xx x

x xx x

x xx

x

− + −

+ ≤+ =

− − ≤ −− ≤

− ≤≥

≥

16

1 1

2 2 2

2

1 2 1 2 2

1 2 1 2

1

2

1 2

2maximize 2 maximize 2

subject to

Replace wit

subject t

h :

3 3 3

o

7 77

2 4

0

x x

x x x x xx x x xx x

x x

x x x

x

x

′ ′′+ ≤ + − ≤−

′ ′′−

′ ′′−

− ≤ −

−

−

≥

+

− +− ≤

2

1 2 2

1 2 2

72 2 4

, , 0

xx x xx x x

′′ ≤ −′ ′′− + ≤

′ ′′ ≥

17

1 2 3

1 2 3

1 2 3

1 2 3

2 3

3

2 2

1 2

maximize 2 3 3

subject to

77

Renaming ,

2 2

as , yie

4, , 0

lds:x

x x x

x x xx x xx x xx x x

x x x

− +

+

′

− ≤− − + ≤ −

− + ≤≥

′′

18

1 1 2 2

Each LP has an intimately related LP, called its dual. The original LP is called the pri

maximi

mal.

Primal (standar

ze

subjec

d maximum )

t

L :

P

Duality

Tn nc x c x c x

•

•

= + + +c x

1 1 2 2

to

minimiz

and

e

Dua

subject

l (stan

andto

dard minimum LP

):

Tm m

T T

y b y b y b

≤ ≥

= + + +

≥ ≥

•

Ax b x 0

y b

y A c y 0

19

1 2

1 11 12 1 1

2 21 22 2

The dual of the Standard Minimum LP is the Standard Maximum LP. The primal max LP and the dual min LP can be

simultaneously displayed as:

n

n

n

x x xy a a a by a a a

•

•

≤≤

2

1 2

1

m m m mn m

m n

b

y a a a bc c c

≤≥ ≥ ≥

20

1 2 1 2 3

1 2

1 2

1 2

maximize minimize 4 12

subject to

Primal problem: Dual

subject to

2 4

problem

4 2 12

:

Example

x x y y y

x xx xx x

+ + +

+ ≤+ ≤

− +

1 2 3

1 2 3

1 2 1 2 3

4 1 2 2 1

1 , 0 , , 0

y y yy y y

x x y y y

+ + ≥+ − ≥

≤≥ ≥

21

1 2

1

2

3

1 2 44 2 121 1 11 1

Examplex x

yyy

≤≤

− ≤≥ ≥

22

If is a feasible solution to the standard maximumLP ( , , ) and a feasible solution to its dual, thenTheorem.

Proof.

.

,

,

Weak Duality Theorem

T T

T T

T T T T

≤

≤ ≥ ⇒ ≤

≥ ≥ ⇒ ≥

xA b c y c x y b

Ax b x 0 y Ax y by A c y 0 y Ax c x

.

If a standard LP and its dual are both feasible,then they are bounded feasible.

If and are feasible solutions to the primal anddual L

Corollary

Ps, respe

.

Corctiv

ollary.ely, and

T T

T T

⇒ ≤

=

c x y b

x yc x y b, then

and are optimal solutions to their respective problems.x y

23

A feasible solution to the primal linear program( , , ) is optimal iff there is a feasible solution to its duallinear program such

Theore

that . In this ca

m.

se, is

Strong Duality Theorem

T T=

xA b c y

c x y b y also optimal.

24

Primal

feasible bounded feasible unbounded infeasiblef.b. yes no no

Dual f.u. no no yes

i no yes yes

25

( ) ( )( ) ( )

1 2 1 2

1 2 1 2

11 1

to and

, , , , , , ,

Findmaximiz

,

where

esubject t

, , , , , ,

o

LP in equality form

T

T Tm m

T Tn m n m

b b b y y y

x x x c c c

a x a

+ +

= ≥

= =

= =

+

xc xAx b x 0

b y

x c

12 2 1, 1

21 1 22 2 2,

1 1 1 2 ,

n m n m

n m n m n

m m m n m n m m

x a x ba x a x a x b

a x a x a x b

+ +

+ +

+ +

+ + =+ + + =

+ + + =

26

Assume rank( ) . Let be a matrix formed by choosing columns out ofthe columns of . If B is a non-singular matrix and if all the variables not associated with the columns i

Basic SolutionsA m

B mm n A

n

=

+

basic solutio

n are set to zero, then thesolution to the resulting system of equations is called a

. The variables associated with thecolumns in

nbasic variables; are called the other va b ria

i

B

m xB

non-basic variablesbasic feasible soluti

les are . If a basic solution is feasible, it iscalld a . There are at most

( )! basic solutions

n

.! !

o

m nm

m nCm n

+ +=27

1 2

1 2

A set is said to be if for all pointsconvex

extreme poin

, we have (1 ) for all 0 1. A point in a convex set is called an

t

Basic feasible solutions are extreme pointsC x x C

x x Cx Cλ λ λ

∈+ − ∈

•≤ ≤

•

1 2 1 2

1 2

if there exist no , , where , and 0 1 such that (1 ) . Geometrically, extreme points are . The feasible region of an LP is a conv

corner points (verti

B

cesex set.

as i f

)

c

x x C x xx x x

λλ λ

∈

••

≠+

•

≤ ≤= −

of an LP (in the equality form) are of the feasible regions. If the LP is feasible bounded, t

easible solutionsextreme points

optimum soluthen at least one

occurs at an extreme poion int!•

28

Simplex Method

A system of linear inequalities define a polytope (or simplex) as a feasible region. The simplex algorithm begins at a starting vertex and moves along the edges of the polytope until it reaches the vertex of the optimum solution.

29 Source: Linear programming, Thomas S. Ferguson, UCLA

v̂

http://en.wikipedia.org/wiki/Vertex_(geometry)

1 2

Find maximize su

to bject to

Find

ma

Stand maximum form:

Equivalent sl

and

and ( ,

, , ) to

ack f r o m:

Converting standard form to slack form

T

Tms s s

•

•

≤ ≥

=

xc xAx b x 0

x s (or minimize ) and ,

may be written

ximize subje as

ct tr .

oo

T T−+ = ≥

• + = − − = −≥

=

c x c xAx s b x 0 s 0

Ax s b s b Ax s Ax b30

1 2

1 2

1 11 12 1 1

2 21 22 2 2

1

, , , are called .

The equality and the objective function can be displayed in a tableau:

1 1

slack variables

m

T

n

n

n

m m

s s s

x x xs a a a bs a a a

z

b

s a a

− = −

=

•

− −

• s Ax bc x

2

1 0m mn m

m n

a bz c c c− − −

31

{ }

{ }1 2

1 2

and , :

Basic variables: , , ,

Non-basic variab

minimiz

les: , , ,

The solution i

esubje

s a, b

ct

as

to

Basic and non-basic variablesT

m

n

B s s s

N x x x

•

•

−+ = ≥ ≥

=

=

=• =

c xs Ax b x 0 s

x 0 s b

0

ic solution.

basic feasible solut If , the solution is a .

If and , the solution is an

ion

optimum solution.

≥

≥ −

•

≥•

b 0

b 0 c 0

32

33

1

1 11 1 1 1

1

1

1

s.t. and , :

1 1

where 0 initially If 0, we can arp

Maximize

ounv i ot d

Pivot OperationT

n

j n

i in i

m m mj mn m

j n

i i

j

j

i

j

v

x xs a a a b

a a b

s a a a bz c c c v v

x

s

a

a

+ + = ≥ ≥

− −

− − − =

• ≠

c x s Ax b x 0 s 0

, switching the roles

of and ij

i j

as x

34

( )1 1 , 1 1 , 1 1

, 1 , 111 1 1

Dividing by the equation

,

we have an equivalent equation:

.

Substituting

i i j j i j j in n i

i j

ij

ij

ij ij i

i ji in ij j

j ij ijn

i j

j i

s a x a x a

a

a

a a a

x a x b

a aa a bx

x

x xsa

xa

x

− − + +

− +− +

•

•

= − + + + + + + +

= − + + + + + + +

11 1

this into every other equation yields:

k kj i kj ink k kn n

ij ij

kj ik i

ji

ij

j

j

a a a as a x a x

a a

a bb

a

xa

b

a −

+ − + + + + −

= −

35

1

1 11 11 1 1

1

1

1

1

After pivoting around , we have the equivalent LP:

1 1

ˆ

ˆ

ˆ ˆ ˆ ˆ

1

n

j in j in

ij ij

mj in mj im m m

i

j

ij

i in ij

i

n mi

j

ij

ij ij

mj

ij

j ij

j n

ij

x xa a a

saa

a a bx

bs a a b

a a

a a a bs a a b

a az c c c v

a

a

a

a a

a

a

− −

− −−

− −

−

−

− −

36

( )

( )

1

1. Pivot element: 1

2. Other elements in the pivot row:

3. Other elements in the pivot column:

4. All other elements

E xample:

:

3

Summary of Pivot Operation

r p r pc q c p q rc p

p p

r r p

c c p

q q rc

p

p

⇒− −

→

→

→−

−

•

→

0 5 3 2 9 2 7 23 1 1 2 3 2 1 2

0 4 3 0 4 3

−− ⇒ −2

1 1

Find minimize

to an

Stand min

d

and ( , , , ) to

imu

subject to

m form:

Equivalent slack

form:

Find

Converting minimum form to slack form

T

T

Tnu u u

≥ ≥

=

•

•

yy by A c y 0

y s minimize subject t

and , o

T

T= − ≥ ≥

y bs y A c y 0 s 0

37

{ }{ }

1 2

1 11 12 1 1

2 21 22 2 2 1 2

1 2

1 2

1

The equality and objective function can be displayed in a

tableau:

, , ,

, , ,

0

The s

n

n

n m

n

m m m mn m

n

T T

m

s s sy a a a by a a a b B s s s

N y y yy a a a b

c c c

==

− −

• = −

•

−

s y A c y b

, basic solution.

basic feasible solution

olution is a

If , the solution is a . If and , the solution is optimum soluan tion

=

− ≥≥ ≥•

=

•−

−

c 0b 0 c

y 0 s c

038

39

( )1

So we display both the primal and the dual in thesame tableau.

Pivoting for the dual follows the same rule:

r p r pc q c p q rc pp

⇒− −

1 2

1 11 12 1 1

2 21 22 2 2

1 2

1

0

n

n

n

m m m mn m

m n

x x xy a a a by a a a b

y a a a bz c c c− − −

40

1 2

1 11 12 1 1

2 21 22 2 2

1

Pivot madly until we suddently find that all entries in the lastrow and last column (exclusive of the corner) are nonnegative.

ˆˆ ˆ ˆˆˆ ˆ ˆ

ˆ ˆ

Pivot Madly Method

n

n

n

m m

r r r

t a a a b

t a a a b

t a

{ }{ }

{ }{ }

1

1

12

11

0 for , ,ˆ for ,

Solution for primal:

Soluti,

ˆ ˆ for , ,ˆ

on for dual:

ˆO

ˆ0 for , , 0

ˆ ˆ ˆ ˆptimal value

j j n

i i i m

j j n jm mn m

i i mm n

x x r r

x b x t t

y c r r ca a b

y y t tc c c v

v

= ∈

= ∈

= − = −

= ∈ =− −

=−

41

Suppose after pivoting for a while we have the tableau:

If , then is a feasible solution for the maximization problem. If , then is a f

,

, easible s

v

= =

−

≥

−

•

≥• = = −

r 0 t b

t 0

rt A b

c

r0 c

b 0

c olution for the minimization problem. If and , then we have an optimal solution

for both problems.≥ − ≥• b 0 c 0

42

The same as the pivot madly method except that we now choose the points to pivot on more systematically.

The goal is to make all entries in the last row and last column (ex

The Simplex Method•

•clusive of the corner) nonnegative.

v−

rt A b

c

43

0

0

0 0 0 0

0

, 0

, ,

Case 1: . Take any column with 0. Among

those with 0, choose the for which the ratio

is smallest. Pivot around . (I

Pivot rules for the simplex method (1)

j

i j

i i j i j

j c

i a i

b a a

≥ −

b 0

0

1 2 3 4 5 1 2 3

1 1

2 2

3 3

4 4

f no such ,

the max problem is feasible unbounded.)

1 1 1 6 6 1 3 10 4 4 4 0 2 4

0 3 1 7 7 2 3 15 1 4 2

33 2

23

31212

0 5 4 25 2 4 2 1 0 5 4 0

i

r r r r r r r rt tt tt tt t

−

− − − −− − − − −

−−−−−

44

45

( )

0

0

0

0

0

0

0

,

,

After the pivot, stays nonnegative, the entry

becomes positive, and the value never gets less ( gets greater if 0).

1

Properties of rule 1:

j

i

i

i j i

j

i j

c

vv b

r p r pc q c p q rc p

b

a b

c

a

v

p

−

≠

⇒− −

−

• b

( )( )

0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

, ,

, , , ,

, ,

1

i j i i j

i j i j i i j i i j

j i j j i i j

a b a

a a b a b a

c a v c b a

⇒ − −

+

0, 0

Case 2: some are negative. Take the first 0. Find any negative entry in row , say 0. (If there is no such ,

the max problem is infeasible.)

Pivot rules for the simplex method (2)

i k

k j

b bk a j

<<

•

[ ] [ ][ ]

[ ]

0 0

0 0 0 0 0

0 0

, ,

, 0 ,

0 ,

1 2 3 4

1

2

3

4

Compare with

for which 0, 0 and choose such that

is smallest ( may be .) Pivot around .

1 3 11 2 4 4

2 1 3 1 21 4 2 0

5 2 4 2 0

1 24

5

k k j i i j

i i j i i j

i j

b a b a

b a i b a

i k a

r r r rtttt

≥ >

− − −−

− − −− −

−

−

1 2 3

1

2

3

4

0 1 2 41 3 3 1

2 2 3 15 1 4 0

r r rtttt

− −− −

−−

46

47

( )

0

The objective in Case 2 is to get to Case 1. With the pivot, the nonnegative stay nonnegative,

and becomes no smaller (it gets larger if 0).

1

Properties of rule 2:

i

k i

bb b

r p r pc q c cp

p q r p

≠

⇒− −

••

( )( )

0 0 0 0 00

0 0 0 0 0 0 0 0

00 0 0 0 0 0 0

0 0, ,

, , , ,

, ,

,

,

1

i j i i ji

i j i i j i j i i j i i j

k j k k j i j

i j

k k j i i j

a b ab

a b a a b a b a

a

a

b a a b a b a

⇒ − −

− −

0

0 0

0

, 0 ,

Case 3: . Treat it as the dual problem. Take any column such that 0. Among those with

0, choose the for which the ratio is

Pivot rules for the simplex method (3)

i

i j j i j

i b j

a j c a

≥<

< −

• − c 0

0 0, 0

1 2 3 4

1

2

3

closest to zero. Pivot around . (If there is no such ,

the dual problem is feasible unbounded.)

1 1 3 1 60 3 2 4 4 2 0 3 1 25 1 4 2 3

i ja j

r r r rttt

− −

−

48

1 2 3

2 3

1 3

1 2 3

1 2 3

2 3

1

Primal: maximize 2 subject to all 0 and

2 3 3 2

2 1

Dual: minimize 3 2 subject to all 0 and

2 1

Example

i

i

x x x x

x xx xx x x

y y y y

y yy y

+ + ≥

+ ≤− + + ≤

+ + ≤

+ + ≥

−

•

≥

•

++ + 3

1 2 3

12 3 2y y y

≥+ + ≥

49

[ ][ ]

1 2 3 1 3 3

1 1

2 2

3 2

1 3 2

1 1

3

2

0 1 2 3 2 1 1 21 0 3 2 1 0 2

2 1 1 2 1 1 11 1 2 0 1 1 1 1

5 3 1 1 3 4 3 1 3 0 1 3 2 3 7 3 1 1 3 1 3

Maximum solution for

2 3 1 1 3 5 3

primal:

31

x x x x y xy yy yy x

x y yy xxx

→

− −− −→

− − − −

− − −−

−

2 3

1 2 3

0, 1 3, 2 3l

0, Minimum solution for dua

Optimal value for bot1 3

h:, 1

5 3

x x

y y y

= = =

= = =

50

The simplex rules as stated may lead to cycling, even though it is rare in practical problems. There are strategies (modified simplex methods) for

avoiding cycling. Smallest-subscript r

Cycling•

•

• ule: If there is a choice of pivot rows or a choice of pivot columns, select the row (or column) with the variable having the lowest subscript, or if there are no variables, with the varia

xx y ble having the lowest

subscript.

51

References

• T.S. Ferguson, “Linear Programming: A Concise Introduction,” http://www.math.ucla.edu/~tom/LP.pdf .

• Lecture notes: http://www.math.cuhk.edu.hk/~wei/LP11.html

52

http://www.math.ucla.edu/~tom/LP.pdfhttp://www.math.cuhk.edu.hk/~wei/LP11.html

Linear Programming�Reading: CLRS, Ch. 29 or references�Slide Number 2Slide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Slide Number 12Slide Number 13Slide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Slide Number 26Slide Number 27Slide Number 28Simplex MethodSlide Number 30Slide Number 31Slide Number 32Slide Number 33Slide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Slide Number 43Slide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48Slide Number 49Slide Number 50Slide Number 51References