Linear ProgrammingReading: CLRS, Ch. 29 or reference

CSE 6331 AlgorithmsSteve 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 b

x x x

f x x x a x a x a x

L

L

L L

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 b

f x x x b

L

L

L

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 n

x x x

c x c x c x

L

L

1

subject to linear constraints

for 1 . n

ij j ij

a x b i m

3

1 2

1 1 2 2

1

find real rnu

1 , 1

1

1

, ,

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 n

b i m

c j n

x x x

c x c x c x

a x b i m

x

L

L

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 b

a a a x b

a a a x b

c x

Ax b x 0

K

K

M M O M M M

K

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 a

a a a

a a

m

n

m n

a

b y

x c

A

K K

K K

K

K

M M O M

K

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 n

m m mn

y b y b y b

a a a

a a ay y y c c c

a a a

y b

y A c y 0

L

K

KK K

M M O M

K

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 x

x x x

L

L

ion: a solution that optimizes the objective

function.

Optimal objective value *:

A linear program may be

or ( is nonempty or empty)

feasible infeasible

(feasible)

bounded u or

f

g

g ( * is finite onbound r infied nite)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

Example

x x

x x

x x

x x

x x

nonnegativity const ra ints

9

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

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 Programs

L L

zL

L

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 L

L

x

e

solution to with objective value , and

vice versa.

L z x

11

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.

12

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

K K

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 b

a x b

a x b a x b

13

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

x

x x x x x

c x c x c x

a 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

K K

K K

14

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

Example

x x x x

x xx x

x xx x

x xx

x

15

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 to

7 7

7

2 4

h

0

:

3 3 3

x x x

x x xx x

x x x x x

x x x x

x x

x

2

1 2 2

1 2 2

7

2 2 4

, , 0

x

x x x

x x x

16

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

7

7

Renaming ,

2 2

as , yie

4

, , 0

lds:x

x x x

x x x

x x x

x x x

x x x

x x x

17

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 L

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

L

18

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 x

y a a a b

y a a a

K

K

K 2

1 2

1 2

m m m mn m

n

b

y a a a b

c c c

M M M O M M

K

K

19

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 x

x x

x x

1 2 3

1 2 3

1 2 1 2 3

4 1

2 2 1

1

, 0 , , 0

y y y

y y y

x x y y y

20

1 2

1

2

3

1 2 4

4 2 12

1 1 1

1 1

Example

x x

y

y

y

21

If is a feasible solution to the standard maximum

LP ( , , ) and a feasible solution to its dual, then

Theorem.

Proof.

.

,

,

Weak Duality Theorem

T T

T T

T T T T

x

A b c y c x y b

Ax b x 0 y Ax y b

y 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 and

dual L

Corollary

Ps, respe

.

Cor

ctiv

ollary.

ely, and

T T

T T

c x y b

x y

c x y b, then

and are optimal solutions to their respective problems.x y

22

A feasible solution to the primal linear program

( , , ) is optimal iff there is a feasible solution to its dual

linear program such

Theore

that . In this ca

m.

se, is

Strong Duality Theorem

T T

x

A b c y

c x y b y also

optimal.

23

Primal

feasible bounded feasible unbounded infeasible

f.b. yes no noDual

f.u. no no yes

i no yes yes

24

1 2 1 2

1 2

11 1 12 2 1,

wh

Fi to

and

, , , , , , ,

, , ,

nd

maximize

subject

o

ere

t

LP in equality form

T

T Tn m m

Tn m

n

x x x b b b

c c c

a x a x a

x

c x

Ax b x 0

x b

c

K K

K

K 1

21 1 22 2 2,

1 1 1 2 ,

m n m

n m n m n

m m m n m n m m

x b

a x a x a x b

a x a x a x b

K

M M M

K

25

Assume rank( ) .

Let be a matrix formed by choosing columns out of

the columns of .

If B is non-singular and if all the variables not associated

with the columns of are se

Basic Solutions

A m

B m

m n A

n

B

t to zero, then the solution to

the resulting system of equations is called a .

The variables associated with the columns

basic solution

basic variables;

in are

called the other variables are im x B

non-basic

variables basic

feasib

. If a basic solution is feasible, it is a

. The number of basic solutions is

( )!at most .

! !

le solution

m nm

m nC

m n

26

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 points

C x x C

x x C

x C

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

ces

ex set.

as i f

)

c

x x C x x

x x x

of an LP (in the equality form)

are of the feasible regions.

If the LP is feasible bounded, t

easible solutions

extreme points

optimum solut

hen at least one

occurs at an extreme poion int.

27

Simplex Method

A system of linear inequalities defines 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.

28Source: Linear programming, Thomas S. Ferguson, UCLA

v̂

1 2

Find

maximize

to

subject t and

and ( , , , ) to

Standard maximum form:

Equivalent sl

o

Find

ac

m

k for :

Converting standard form to slack form

T

Tms s s

x

c x

Ax b x 0

x s K

(or minimize )

and ,

may be written a

maximize

subjec

t to

s .or

T T

c x c x

Ax s b x 0 s

Ax s b s b Ax s Ax b

0

29

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 x

s a a a b

s a a a

z

b

s a a

s Ax b

c x

K

K

K

K

M M M O M M

2

1 0m mn m

m n

a b

z c c c K

K

30

1 2

1 2

and , :

Basic variables: , , ,

Non-basic variab

minimiz

les: , , ,

The solution i

e

subje

s a, b

ct

as

to

Basic and non-basic variablesT

m

n

B s s s

N x x x

c x

s Ax b x 0 s

x 0 s b

0

K

K

ic solution.

basic feasible s If , the solution is a .

If

olu

in addition , the solution is .

tion

optimum

b 0

c 0

31

32

1

1 11 1 1 1

1

1

1

s.t. and , :

1 1

where 0 initially

If

Max

0, we can pivot ar

imi e

u

z

o

Pivot OperationT

n

j n

i in i

m m mj mn m

j n

j

ii j

i

jx

s

z c x v

x x

s a a a b

a a b

s a a a b

z c c c v v

a

a

s Ax b x 0 s 0

K K

K K

M M M M M

K K

M M M M M

K K

K K

nd , switching the roles

of and

ij

i j

a

s x

33

1 1 , 1 1 , 1 1

, 1 , 111 1 1

Dividing both sides of the equation

by , we have an equivalent equation:

ij

ij

ij ij ij ij ij i

i i j j i j j in n i

i j i ji in ij j

jn

i j

ij

a x a x a x a x b

a aa a bx x x

a

a

a a a a ax

sx

a

s x

L L

L L

11 1

.

Substituting this into every other equation yields:

kj i kj ink k kn n

ij ij

kj ik i

i

kj

j

jij

a ax

a

a a as a x a x

a a

a bb b

a

L L

34

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 x

a a a

s

a

a

a a bx

bs a a b

a a

a a a bs a a b

a a

z c c c v

a

a

a

a a

a

a

K K

K K

M M M M M

K K

M M M M M

K K

K K

35

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 p

c 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 2

3 1 1 2 3 2 1 2

0 4 3 0 4 3

2

1 1

Find

minimize

subject

to

an

Standard minimum

form:

Equiva

to

Find

d

and ( , , , )

lent slack fo m:

t

r

Converting minimum form to slack form

T

T T

Tns s s

y

y b

y A c y 0

y s K

minimize

subje

o

and ct o t ,

T

T T T

y b

s y A c y 0 s 0

36

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

T e h

n

n

n m

n

m m m mn m

m

T T T T

n

s s s

y a a a b

y a a a b B s s s

N y y y

y a a a b

c c c

s y A c y b

K

K

K KM M M O M M K

K

K

, basic solution.

basic feasible soluti

solution is a

If , the solution is a .

If and , the solution i

o

s a

n

optimum solut onn i

y 0 s c

c 0

b 0 c 037

38

1

So we display both the primal and the dual in the

same tableau.

Pivoting for the dual follows the same rule:

r p r p

c q c p q rc p

p

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 x

y a a a b

y a a a b

y a a a b

z c c c

K

K

K

M M M O M M

K

K

39

1 2

1 11 12 1 1

2 21 22 2 2

Pivot madly until we suddently find that all entries in the last

row and last column (exclusive of the corner) are nonnegative.

ˆˆ ˆ ˆ

ˆˆ ˆ ˆ

ˆ

The Pivot Madly Method

n

n

n

m

r r r

t a a a b

t a a a b

t

K

K

K

M M M O M M

1

1

1

1 21

1

0 for , ,

ˆ for , ,

ˆ ˆ for , ,ˆˆ ˆ

0 fo

Solution for primal:

Solution for dual:

ˆOp

r , , 0ˆ ˆ

timal vaˆ ˆ

lue

j j n

i i i m

j j n j

m m mn mi i m

m n

x x r r

x b x t t

y c r r ca a a b

y y t tc c c

vv

K

K

KK

KK

40

1 2 3

11

2 32

4

1 43

2 3

01

4,

Max solution for primal:

Min solution for

34

1

dual:

Optimal value 1

0 03

2, 45 2 4 16

6

Example

x y yx

yx x

x

yy y

xy y

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

r

t 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

r

t 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 1

0 4 4 4 0 2 4

0 3 1 7 7 2 3 1

5 1 4 2

3

3 2

2

3

3

1

2

1

2

0 5 4 2

5 2 4 2 1 0 5 4 0

i

r r r r r r r r

t t

t t

t t

t 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

v

v b

r p r p

c 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 b

k 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 1

1 2 4 4

2 1 3 1 2

1 4 2 0

5 2 4 2 0

1 2

4

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 r

t

t

t

t

1 2 3

1

2

3

4

0 1 2 4

1 3 3 1

2 2 3 1

5 1 4 0

r r r

t

t

t

t

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

b

b b

r p r p

c q c c

p

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 such that 0. Among those with

0, choose

row

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 3 1 6

0 3 2 4 4

2 0 3 1 2

5 1 4 2 3

1

i ja j

r r r r

t

t

t

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 x

x x

x x x

y y y y

y y

y y

3

1 2 3

1

2 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 2

1 0 3 2 1 0 2

2 1 1 2 1 1 1

1 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:

3

1

x x x x y x

y y

y y

y x

x y y

y xx

x

[

2 3

1 2 3

0, 1 3, 2 3

l

0,

Minimum solution for dua

Optimal value for bot

1 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

x

x 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