linear programming - nptel.ac.innptel.ac.in/courses/105108127/pdf/module_3/m3l6slides.pdf · 1 d...

D Nagesh Kumar, IISc Optimization Methods: M3L61

Linear Programming

Other Algorithms


Introduction & Objectives

Few other methods, for solving LP problems, use an entirely different algorithmic philosophy.– Khatchian’s ellipsoid method– Karmarkar’s projective scaling method

ObjectivesTo present a comparative discussion between new methods and Simplex methodTo discuss in detail about Karmarkar’s projective scaling method


Comparative discussion between new methods and Simplex method

Simplex AlgorithmKarmarkar’s Algorithm

Optimal solution point

Feasible Region

Khatchian’s ellipsoid methodand Karmarkar’s projective scaling method seek the optimum solution to an LP problem by moving through the interior of the feasible region.



1. Both Khatchian’s ellipsoid method and Karmarkar’s projective scaling method have been shown to be polynomial time algorithms. Time required for an LP problem of size n is at most anb , where a and b are two positive numbers.

2. Simplex algorithm is an exponential time algorithm in solving LP problems.Time required for an LP problem of size n is at most c2n, where c is a positive number



3. For a large enough n (with positive a, b and c), c2n > anb. The polynomial time algorithms are computationally superior to exponential algorithms for large LP problems.

4. However, the rigorous computational effort of Karmarkar’s projective scaling method, is not economical for ‘not-so-large’ problems.


Karmarkar’s projective scaling method

Also known as Karmarkar’s interior point LP algorithm

Starts with a trial solution and shoots it towards the optimum solution

LP problems should be expressed in a particular form



LP problems should be expressed in the following form:

:with1

:subject to Minimize T

0X1X

0AXXC

≥==

=Z

where

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nx

x

x

M

2

1

X

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

nc

cc

M2

1

C [ ] ( )n×= 1111 L1⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

mnmm

n

n

ccc

cccccc

L

MOMM

L

L

21

22221

11211

A 2≥nand



It is also assumed that is a feasible solution and

Two other variables are defined as: and .

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

n

n

n

1

1

1

0M

X 0min =Z

( )11−

=nn

r ( )n

n3

1−=α

Karmarkar’s projective scaling method follows iterative steps to find the optimal solution



In general, kth iteration involves following computations

a) Compute ( )[ ] T1TT CPPPPIC −−=p

where

⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

ADP

k

kDCC T=

( )( )

( )⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

nk

k

k

k

X

X

X

D

000

000

0020

0001

O

If , ,any feasible solution becomes an optimal solution. Further iteration is not required. Otherwise, go to next step.

0C =p



b)p

pnew r

C

CXY α−= 0

c)newk

newkk Y1D

YDX =+1

However, it can be shown that for k =0, k newnew

k new

=D Y Y

1D YThus, 1 new=X Y

1T

+= kZ XCd)

Repeat the steps (a) through (d) by changing k as k+1e)


Karmarkar’s projective scaling method: Example

Consider the LP problem:

0,,1

02 :subject to2 Minimize

321

321

321

32

≥=++=+−

−=

xxxxxxxxxxxZ

3=nThus,⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=

1

2

0

C [ ]121 −=A

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=

31

31

31

0M

X

( ) ( ) 61

1331

11

=−

=−

=nn

r ( ) ( )92

3313

31

=×−

=−

=n

nαand also,



Iteration 0 (k=0):

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

3/100

03/10

003/1

0D

[ ] [ ]3/13/20

3/100

03/10

003/1

1200T −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−== DCC

[ ] [ ]3/13/23/1

3/100

03/10

003/1

1210 −=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−=AD



Iteration 0 (k=0)…contd.:

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

111

3/13/23/10

1

ADP

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−×⎥⎥⎦

⎤

⎢⎢⎣

⎡ −=

30

03/2

13/1

13/2

13/1

111

3/13/23/1TPP

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡=

−

3/10

05.11TPP



( )⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=−

5.005.0

010

5.005.01TT PPPP


( )[ ]⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

=−=−

6/1

0

6/1T1TT CPPPPIC p

62)6/1(0)6/1( 22 =++=pC




⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

××

−

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

=−=

3974.0

3333.0

2692.0

6/1

0

6/1

62

61

92

31

31

31

0Mp

pnew r

C

CXY α

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==

3974.0

3333.0

2692.0

1 newYX [ ] 2692.0

3974.0

3333.0

2692.0

1201T =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×−== XCZ



Iteration 1 (k=1):

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

3974.000

03333.00

002692.0

1D

[ ] [ ]T1

0.2692 0 0

0 2 1 0 0.3333 0 0 0.6667 0.3974

0 0 0.3974

⎡ ⎤⎢ ⎥⎢ ⎥= = − × = −⎢ ⎥⎢ ⎥⎣ ⎦

C C D

[ ] [ ]3974.06666.02692.03974.00003333.00002692.0

1211 −=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×−=AD




1 0.2692 0.6667 0.3974

1 1 1

−⎛ ⎞ ⎡ ⎤= =⎜ ⎟ ⎢ ⎥⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

ADP

1

T

0.2692 10.2692 0.6667 0.3974 0.675 0

0.6667 11 1 1 0 3

0.3974 1

⎡ ⎤− ⎢ ⎥⎡ ⎤ ⎡ ⎤

⎢ ⎥= × − =⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

PP

( ) 1T T

0.441 0.067 0.492

0.067 0.992 0.059

0.492 0.059 0.567

−

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥⎢ ⎥−⎣ ⎦

P PP P




( ) 1T T T

0.151

0.018

0.132

p

−

⎡ ⎤⎢ ⎥

⎡ ⎤ ⎢ ⎥= − = −⎢ ⎥⎣ ⎦ ⎢ ⎥⎢ ⎥−⎣ ⎦

C I P PP P C

( )22 2(0.151) 0.018 ( 0.132) 0.2014p = + − + − =C

0

1 30.151 0.26532 1

1 3 9 6 0.018 0.34140.2014

0.132 0.39281 3

pnew

p

rα

⎡ ⎤⎢ ⎥ ⎡ ⎤ ⎡ ⎤×⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − = − × − =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎢ ⎥⎣ ⎦

CY X

C M




1

0.2692 0 0 0.2653 0.0714

0 0.3333 0 0.3414 0.1138

0 0 0.3974 0.3928 0.1561

new

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= × =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

D Y

[ ]1

0.0714

1 1 1 0.1138 0.3413

0.1561

new

⎡ ⎤⎢ ⎥⎢ ⎥= × =⎢ ⎥⎢ ⎥⎣ ⎦

1D Y




12

1

0.0714 0.20921 0.1138 0.3334

0.34130.1561 0.4574

new

new

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= = × =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

D YX1D Y

[ ]T2

0.2092

0 2 1 0.3334 0.2094

0.4574

Z

⎡ ⎤⎢ ⎥⎢ ⎥= = − × =⎢ ⎥⎢ ⎥⎣ ⎦

C X

Two successive iterations are shown. Similar iterations can be followed to get the final solution upto some predefined tolerance level


Thank You

linear programming - nptel.ac.innptel.ac.in/courses/105108127/pdf/module_3/m3l6slides.pdf · 1 d...

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