Wang, Cheng Marshal, "Comments on two direct methods in linear programming." (2004). Electronic

University of Windsor University of Windsor

Scholarship at UWindsor

Scholarship at UWindsor

Electronic Theses and Dissertations Theses, Dissertations, and Major Papers

2004

Comments on two direct methods in linear programming.

Comments on two direct methods in linear programming.

Cheng Marshal Wang

University of Windsor

Follow this and additional works at: https://scholar.uwindsor.ca/etd

Recommended Citation Recommended Citation

Wang, Cheng Marshal, "Comments on two direct methods in linear programming." (2004). Electronic Theses and Dissertations. 1479.

https://scholar.uwindsor.ca/etd/1479

C om m ents on Two D irect M ethods in Linear P rogram m ing

by

Cheng(M arshal) W ang

A Thesis

S u b m itte d to the F a cu lty o f G raduate Studies and Research th ro u g h th e D e p a rtm e n t of M athem atics and S ta tistics

in P a rtia l F u lfillm e n t o f the Requirements fo r the Degree o f M aster o f Science a t the

U n ive rsity o f W in d so r

W in d so r, O nta rio, Canada

1 * 1 Library and Archives Canada Published Heritage Branch 395 Wellington Street Ottawa ON K1A 0N4 Canada Bibliotheque et Archives Canada Direction du Patrimoine de I'edition 395, rue Wellington Ottawa ON K1A 0N4 Canada

Your file Votre reference ISBN: 0-612-96387-X Our file Notre reference ISBN: 0-612-96387-X

The author has granted a non­

exclusive license allowing the

Library and Archives Canada to

reproduce, loan, distribute or sell

copies of this thesis in microform,

paper or electronic formats.

The author retains ownership of the

copyright in this thesis. Neither the

thesis nor substantial extracts from it

may be printed or otherwise

reproduced without the author's

permission.

L'auteur a accorde une licence non

exclusive permettant a la

Bibliotheque et Archives Canada de

reproduire, preter, distribuer ou

vendre des copies de cette these sous

la forme de microfiche/film, de

reproduction sur papier ou sur format

electronique.

L'auteur conserve la propriete du

droit d'auteur qui protege cette these.

Ni la these ni des extraits substantiels

de celle-ci ne doivent etre imprimes

ou aturement reproduits sans son

autorisation.

In compliance with the Canadian

Privacy Act some supporting

forms may have been removed

from this thesis.

While these forms may be included

in the document page count,

their removal does not represent

any loss of content from the

thesis.

A b stra ct

In th is thesis we w ill analyse the tw o algorithm s fo r linear p ro g ra m m in g (L P ) presented by S tojkovic and S ta n im iro vic [15] in 2001. One o f the m ethods, which th e authors call the “ m in im a l angles m eth od” (M A m ethod) was designed to determ ine eithe r an o p tim a l extrem e p o in t or an extreme p o in t adjacent to an o p tim a l extrem e point.

U n fo rtu n a te ly, the theorem upon w hich the M A m ethod is based is n o t valid. T h is was shown by L i [10] in 2004 w ith tw o counterexamples. We w ill show th a t one of the counterexamples its e lf is n o t valid, and w ill provide an alternate, valid counterexample. We w ill also provide a careful stu d y of the M A m eth od to see i f there is a class o f LP, where it can be applied. T h is leads to a m eth od we call the “ active cone m ethod” w h ich can be used for

2

variable LPs.

A ck n ow led gem en t

s

The research th a t has gone in to th is thesis has been th o ro u g h ly enjoyable. T h a t enjoym ent is largely a result o f the in te ra c tio n th a t I have had w ith my supervisors and w ith the people I have know n th ro u g h o u t m y studies a t the U n i­ versity o f W in d so r. I feel very privileg ed to have worked w ith m y co-supervisors, D r. R ichard J. Caron and D r. T im T raynor, to each o f w hom I owe a great debt o f g ra titu d e fo r th e ir patience, in sp ira tio n , and friendship. D r. C aron and D r. T raynor have ta u g h t me a great deal about the field o f O p e ra tio n a l Research b y sharing w ith me the jo y o f discovery and investigation th a t lies w ith in the heart o f research. Also, m y profound g ra titu d e goes to D r. Yash P. Aneja, Dr. Sang-Chul Suh, D r. A b d o Y . A lfa k ih , and D r. G uoqing Zhang fo r acquainting me w ith m any other aspects o f O p e ra tio n a l Research.

T he D e p a rtm e n t o f M athem atics and S tatistics and the O p e ra tio n a l Research G roup have provided an excellent environm ent fo r m y research. Thanks to X iaow u Ke, John B a tta g lia , X in g Jiang, H u jia n Hua, Fengling Shi, H u im ing Song and O lusakin Em m anuel for th e ir suggestions.

C o n te n ts

A b stra ct

iii

A ck n ow led gm en ts

iv

List o f Figures

vii

1 In tro d u ctio n

1

1.1 O verview and O u tlin e o f T h e s is ...

1

1.2 N o ta tio n and D e fin it io n s ... 2

2 T h e M in im al A n gles M eth o d o f S tojkovic and S tanim irovic

5

2.1 I n tr o d u c tio n ... 5

2.2 Comm ents on the M A m e th o d ... 7

2.3 C o n je c tu r e s ... 10

2.4 C o n c lu s io n ... 14

3 T h e A c tiv e C one M eth o d

15

3.1 I n tr o d u c tio n ... 15

3.2 T he A c tiv e Cone M eth od in

R 2

... 25

3.3...C o n c lu s io n ... 32

4

S im p lification M eth o d s b ased on G am e T h eory

33

4.1 I n tr o d u c tio n ... 33

4.2 Ite ra te d Removal o f R e d u n d a n c y ... 36

4.3 Examples ... 41

4.4 Removal o f Redundancy based on M ixe d S tr a te g ie s ... 44

4.5 A lg o rith m and E x a m p le s ... 46

4.6...C o n c lu s io n ... 51

5 C onclusion

52

L ist o f F ig u r e s

F ig u re

1

: A graphical solution o f L P ...

6

1

I n tr o d u c tio n

1.1

O verview and O utline o f T h esis

T h is thesis is concerned w ith linear program m ing (L P ). A n L P problem is one o f choosing the value fo r a set o f variables th a t w ill m axim ize a linear fu n c tio n o f those variables, called the objective function, subject to a set of linear re strictio n s on those variables, called the constraints. A choice o f the variables w hich satisfies a ll the re strictio n s is said to be feasible, and the set o f a ll feasible choices is called the feasible region. W ith th e development o f applications o f LP, the size o f L P problem s th a t require solution, i.e., the num ber o f constraints and variables, has become very large, for example, hundreds of thousands o f constraints and variables. M a n y com mercial L P solvers preprocess, fo r example, remove redundancies, to increase the size o f problem th a t can be solved. T he paper by S tojkovic and S ta n im iro vic [15] w hich is the to p ic o f th is thesis, presented m ethods fo r the preprocessing o f LPs.

T he firs t m eth od in [15], named the “ m in im a l angles m eth od” (M A m ethod), was designed to d ire c tly determ ine eithe r an o p tim a l extrem e p o in t or an ex­ trem e p o in t adjacent to an o p tim a l extrem e p o in t in a class o f LPs. F rom the graph o f an L P problem in tw o variables th a t has no redundant constraints (a redundant co n stra in t is one th a t can be removed w ith o u t changing the set o f allowable or feasible choices fo r the variables) it seems th a t th e constraints whose gradients have m in im a l angles w ith the gradient o f the objective func­ tio n are p ro b a b ly the constraints active a t the o p tim a l solution. S to jko vic and S ta n im iro vic proposed a theorem based on th is idea w hich is the basis fo r th e ir M A m ethod.

w ith another counterexample, th a t even when the equations are consistent, the theorem is s till n o t valid.

In the s p irit o f try in g to determ ine the set o f assum ptions under w hich the S tojkovic and S ta n im iro vic theorem is valid, we propose tw o conjectures. U n­ fo rtu n a te ly, we show th a t even these are invalid. In a more p o sitive vein, we do manage to develop an “ active cone m eth od” , w h ich can be th o u g h t o f as an extension o f the M A m ethod o f S tojkovic and S tanim iro vic, w hich is va lid in R 2. T h is is given in C hapter 3. O ur active cone m eth od exploits the fact th a t a t an o p tim a l extrem e p o in t, the gradient o f the objective fu n c tio n is in the cone spanned by the gradients of active constraints.

T he M A m eth od proposed by S tojkovic and S ta n im iro vic requires th a t the L P be free o f redun dant constraints. Thus, in th e ir paper th e y also present a m ethod, based upon results about two-person zero-sum games, to detect and re­ move redun dant constrains as well as redundant variables. A re d un dant variable is one th a t w ill always have a value o f zero in any feasible solution. In th e game theory paradigm , th is is equivalent to the e lim in a tio n o f s tric tly dom inated rows and columns. In th e ir paper, S tojkovic and S ta n im iro vic o n ly consider pairwise dominance, i.e., one row or colum n s tric tly d o m in a tin g another row or column, and th e y o n ly consider a single pass, th a t is, th e y do n o t use ite ra tio n to consider fu rth e r elim inations. We show th a t by using ite ra tio n and th a t by the consideration o f m ixed strategies, we can considerably im prove upon th e ir results. For example, we show an L P problem w ith 7 constraints and

8

variables th a t can be sim p lifie d to 2 constraints and 2 variables. In contrast, using the S tojkovic and S ta n im iro vic m ethod no redundancies can be found.

In C h a p te r 5, we conclude the thesis w ith a statem ent o f its co ntribu tions, as well as directions fo r fu tu re work.

1.2

N o ta tio n and D efin ition s

We use th e LP form :

m ax : z(x) — cTx

T ( !)

s.t.: a- x < b i , i =

1

, . . . , m,

where cT = ( c i , . . . , c „ ) , x T = { x x, . . . , x n) , a j = (aa , • • •, ain ), and “ m ax” refers to “ m axim ize” and “ s .t.” means “ subject to ” . We use “ T ” to denote m a trix and vector transposition.

We ca ll z {x ) the objective fu n c tio n and we refer to “ a j x < b " as the z-th constrain t in the set o f a ll constraints. T he vector aj is the gradient o f the z-th constrain t fu n ctio n , and c is the gradient o f the objective fu n ctio n .

I f th e co n stra in t set includes the non n e g a tiv ity constraints, as is often the case, we w ill w rite the L P as

m ax : z(x ) = cTx s.t.: a j x < b i , z = l , . . . , m , (

2

) Xj >

0

, j = l , . . . , n . D e f in it io n 1.1 7Z = { x : a j x < = 1 , . . . , m } is the fe a s ib le r e g io n fo r ( ! ) • D e f in it io n 1.2 We say th a t x is a fe a s ib le s o lu tio n to (1) if x € 1Z.

D e f in it io n 1.3 We say th a t x* is an o p t im a l s o lu tio n to (1) i f x* £ TZ and cTx* > cT x fo r x G 71.

In w h a t follows, we assume th a t x* is an extreme p o in t o f th e feasible region. (Recall th a t under the assum ption th a t there is a subset o f size n o f the set o f a ll constrain t gradients th a t is lin e a rly independent, the LP w ill have an extreme p o in t o p tim a l so lution.) We denote th e o p tim a l value o f th e ob je ctive fu n c tio n by

2

* = cr x*.

D e f in it io n 1 .4 [7] T h e fc-th co n stra in t is r e d u n d a n t w ith respect to the con­ s tra in t set defining TZ, i f 7Zk = { x : a j x < b i , i = l , . . . , m , z ^ k } — 7Z. A constraint th a t is non-redundant is said to be n e ce ssa ry.

a redundant constrain t. The follow ing lemma, w hich follows easily fro m the defin itio n , gives a characterization o f a necessary constrain t th a t w ill be used la ter in th is thesis.

Lem m a 1.1

The k -th constraint is necessary i f and only i f there exists a point x € Tlk with x f i l Z . That is, a j x < bi,V i k and a j x >

bk-D efin ition 1.5 [7]

L e t x £ 1Z. I f a j x = bx, th e n the z-th co n stra in t is said to be

an a ctiv e con strain t at

x.

D efin ition 1.6 [2]

L e t a be a nonzero vector in

R"

and le t b be a scalar. Then the set { x €

R n

: ar x —

6

} is called

a hyperplane.

For convenience, in L P problem (

1

) we call { x €

R n

: a j x —

6

, }

th e h yp erp lan e

i, since it corresponds to the constrain t i: a j x < bi.

D efin ition 1.7

[12] T he tw o d is tin c t extreme points Xi and x 2 o f 7Z are

adja­

cent

if every x on th e line segment jo in in g x x and x 2 has th e p ro p e rty th a t if x = A£i + (1 — X )x 2,

0

< A <

1

then x x and x 2 m ust themselves be on th e line segment jo in in g x x and x 2.

D efin ition 1.8 [7]

T he dual o f L P (1) is m

m in : {bTy : ^ = c, y > 0 } (3) t= i

where bT = (bu . . . , bm) and y T = (yx, . . . , ym).

T h eorem 1.1 [2] S tron g D u a lity T h eorem .

I f a linear programming prob­ lem has an optim al solution, so does its dual, and the respective optim al values are equal.

T heorem 1.2 [2] W eak D u a lity T h eorem .

I f x is feasible f o r the p r im a l

(1) and y is feasible f o r the dual (3), then cTx < bTy.

T heorem 1.3 [2] Farkas’ lem m a.

Let { a x, . . . , ak} be a set o f vectors in

R n

and let c be a vector in

R n.

Then, exactly one o f the fo llo w in g two alternatives holds:

2

T h e M in im a l A n g le s M e th o d o f S to jk o v ic a n d S ta n ­

im ir o v ic

2.1

I n t r o d u c t i o n

In th is chapter we present the M in im a l Angles (M A ) m eth od proposed by S tojkovic and S tanim iro vic. We discuss th e ir m ain theorem and present coun­ terexamples w hich invalidate the theorem. W e propose tw o conjectures, hoping to fin d a theorem , s im ila r in s p irit, th a t is correct, b u t we show th a t even these are invalid.

To m o tiva te the M A m ethod, we firs t consider the fo llow ing example.

E xam ple 2.1

We have the LP

- y

< 0

T he unique so lu tio n is x* — (29.23 , 17.69)T and constraints (1.1) and (1.2) are the o n ly active constraints a t x*. In Figure

1

, we see th a t the angles between

the p e rtin e n t p a rts o f th e ir m ain theorem using the n o ta tio n o f th is thesis and

invalid, we refer to it as the S. Theorem, where “ S.” refers to the in itia l o f the last names S to jko vic and S tanim iro vic.

T h e S. T h e o r e m

2.1

[15, page 419] Consider the m axim iza tion problem (2) with no redundant constraints and w ith feasible region 71. Define the set

m ax : z = x + y s.t.: 3x —y < 70 x + 4 y < 1 0 0 (1.1) ( 1.2) (1.3) V < 2 0 —X <

0

the gradients o f co n stra in t functions (

1

.

1

), (

1

.

2

) and c are th e smallest o f the angles between a ll the other constrain t gradients and c. T h is is the basis fo r the M A m eth od proposed by N .V . S tojkovic and P.S. S tanim iro vic. We reproduce

«>3

“V -*►

Figure 1: A graphical solution of LP.

Assume that the set V contains I positive elements and denote those elements by vh , . . . , vk .

(c) I n the case I > n, let xq be the solution to the system o f equations

a Jpx = bip, p = l , . . . , n , (5)

where the indices i i , . . . , i n correspond to n m axim al and positive values selected fro m the set V , then either ( i) x 0 = x * , o r ( ii) x 0 and x* are adjacent extreme points.

(d) I n the case 0 < I < n we consider the system

°%x = bip, p = I , . . . , I , (

6

)

where the indices i \ , . . . , i i correspond to positive values in the set V . A basic solution x 0 f o r (2) can be obtained by setting (n — I) o f the variables in (6) to zero and then solving f o r the rem aining variables. Provided that the remaining (n — I) o f the variables are uniquely determined, Xq and x* are adjacent.

called “ w arm s ta rt” , and i t would p o te n tia lly decrease the num ber o f iteration s required to solve the problem.

2.2

C om m en ts on th e M A m eth od

T he firs t th in g to note about the S. Theorem 2.1 is th a t w h ile the prob­ lem (

2

) contains non n e g a tiv ity constraints, the authors do n o t include them in the ca lcu la tio n o f the m in im a l angles. T h is seems to suggest an assum ption by S tojkovic and S ta n im iro vic th a t non n e g a tiv ity constraints w ill be inactive at any o p tim a l solution. T h is is clearly incorrect, fo r example, consider the s itu a tio n where cT = ( —

1

, —

1

, . . . , —

1

) in w hich case a ll non n e g a tiv ity con­ stra in ts { x j >

0

) w ill be active at the o p tim a l so lu tio n x* =

0

, provid ed th a t it is feasible. T h is suggests an im m ediate m o d ific a tio n to include th e non nega­ t iv it y constraints in the d e te rm in a tio n o f V (4 ). Since Xj > 0 can be w ritte n as —e j x <

0

where Cj is the j -th u n it vector, we now re s tric t our a tte n tio n to the L P m odel (1).

We are n o t th e firs t to discuss problem s in the S. Theorem 2.1. In [10], W ei L i reported th e fo llow ing counterexamples.

E xam ple 2.2

[10]

C ou n terexam p le to S. T h eorem 2.1 (c)

E x a m p le 2 .3 [ 10] C o u n te r e x a m p le to S. T h e o r e m 2.1 (d) Z = — X i —X 2 + 100^3 03 c-t- 1—1 O O + x 3 < 100 (3.1) x 2+ 100x3 < 10050 (3.2) x 2 < 100 (3.3) - X i < 0 ~X2 < 0 ~ x 3 < 0

In these tw o counterexamples, some elements o f c are non-positive, and the o p tim a l so lu tio n fo r b o th is the extreme p o in t (0, 0,100). Clearly, non n e g a tiv ity constraints are active a t the o p tim a l extreme p o in t. In a p riv a te com m unication [16], the corresponding author, N .V . S tojkovic, in dica ted th a t in consideration o f L i’s counterexamples, the S. Theorem 2.1 should be changed to include the co n d itio n th a t c >

0

.

L e t’s lo o k more closely at example 2.2. T he co n stra in t x \ > 0 together w ith constrain t (2.1) im plies th a t < 100. C o m bining th is w ith co n stra in t(2 .5 ) im ­ plies th a t

5

x

2

+

^3

< 600, which then im plies th a t constrain t (2.4) is redundant. As th is co n stra in t plays a role in th e d e te rm in a tio n o f th e set V , and since the hypothesis o f S. Theorem

2.1

is th a t there are no redun dant constraints, example

2.2

is not a v a lid counterexample.

E xam ple

2 .4 Consider the LP X \ TO. 1^2 + O .IX

3

X i <

10

(4.1) x x +

0

.

2

x

2

<

12

(4.2) X x +0.3x2 < 14 (4.3) O.laq +

0

.

1

x

2

+ X 3 <

12

(4.4) - X x <

0

(4.5) ~X2 <

0

(4.6) - x 3 <

0

(4.7)

We firs t show th a t a ll constraints are necessary. Lem m a 1.1 means th a t, for each constrain t, we o n ly need fin d a p o in t th a t violates th a t co n stra in t alone. In the fo llow ing ta b le each row gives constrain t index i, a p o in t x w hich only violates th a t co n stra in t, and the left-hand-side value a j x showing th e v io la tio n as it is greater th a n th e right-hand-side £+ i X a j x 1 (

12

,

0

,

0

)

12

2

( 10, 13, 0 )

12.6

3 (

0

, 60,

0

) 18 4 ( 0, 0, 500 ) 500 5 ( - 50 0 , 0, 0 ) 500

6

( 0, -500, 0 ) 500 7 ( 0, 0, -500 ) 500 We now calculate th e set

V = { v u v2, v 3, v 4} = {1,1.0002,0.9866,0.20797}

2.3

C on jectu res

In th is section, we a tte m p t to fin d variatio ns o f S. Theorem 2.1 th a t are correct. We firs t conjecture th a t perhaps the constraints corresponding to the tw o largest positive values in V are active at x*.

C on jectu re 2.1

Consider S. Theorem

2.1

applied to L P

(2)

w ith c >

0.

Sup­ pose th e set V contains l ( >

0)

positive elements. W ith o u t loss of generality, suppose Vi > v2 > . . . > Vi >

0.

L e t x* be an o p tim a l solution. I f / >

2,

then the constraints th a t correspond to V\ and t>

2

, are active a t x*.

U n fo rtu n a te ly, we have a counterexam ple fo r th is conjecture. R eferring to Figure 2 we see th a t the two largest values in V correspond to constraints 1 and

2

and co n stra in t

2

is n o t active a t x*.

Figure 2: In R 2, a counterexample to Conjecture 1.

W h ile the next example shows th a t C onjecture

2.2

is in va lid , we w ill prove, in C hapter 4, Lem m a 3.5, th a t it does hold in R 2.

E xam ple 2.5

A counterexample to C onjecture

2.2:

the o p tim a l extrem e p o in t is not on the hyperplane w hich corresponds to the largest value in th e set V. Consider the L P Xi + X2 + O.IX3 Xi - 0 . 1 x 3 < 4.8 (5.1) 2x\ - x 2 < 5 (5.2) - x x +

2^2

< 5 (5.3) O .laq + X3 < 4.5 (5.4) -X x < 0 (5.5) ~X2 < 0 (5.6) - X3 < 0 (5.7)

We firs t show th a t a ll constraints are necessary.

i X a j x

1

( 5, 5, 0 ) 5

2

( 5, 0, 4 )

10

3 (

0

,

100

,

0

)

200

4 (

0

,

0

,

100

)

100

5 ( - 5 ,

0

,

0

) 5

6

( 0, -5, 0 ) 5 7 ( 0, 0, -4.8 ) 4.8 We now calculate the set

V = { v u v 2, v ^ v ^ } = {0.9851,0.4472,0.4472,0.1990}

We see th a t x* = ( 5 , 5 , 4)T , at w hich constraints 2, 3, and 4 are active, is n o t on hyperplane

1

as predicted.

C onjecture 2.3

I n the S. Theorem 2.1 (c), i f I

>

n, and i f the system of

equations (5) has a solution xq, then either (i) xq = x * , o r ( ii) xq and x* are adjacent extreme points.

E xam ple 2.6

A counterexample to the S. Theorem

2.1

(c) w ith th e existence assumption: th e firs t three hyperplanes w ith m in im a l angles have a common p o in t w hich is neither an o p tim a l extreme p o in t or an extrem e p o in t adjacent to an o p tim a l extreme p o in t. Consider the L P

X \ + X2 + 0.1x3 X i T _{%2 ~} 0.4x3 _{< 12} (6.1) l.l& C i + 0.8x2 — 2x3 < 3.86 (6.2) 0.8x! + 1.18X2 - 2x3 < 3.86 (6.3) 2 x i — _{X 2} <

10

(6.4) - x x + 2X2 < 10 (6.5) O.lxx + 0.1x2 T

£3

< 28 (6.6) - X i <

0

(6.7) - x 2 <

0

(6.8) - X 3 < 0 (6.9) We firs t show th a t a ll constraints are necessary.

i X a j x 1 ( 10, 10, 8 ) 16.8 2 ( 5, 0, 0.07 ) 5.76 3 ( 0, 5, 0.07 ) 5.76 4 ( 22, 0, 25 ) 44 5 ( 0, 22, 25 ) 44 6 ( 10, 10, 500 ) 502 7 ( -

10

,

0 ,0

) 10 8 (

0

, -10, 0 ) 10 9 ( 0, 0, -1.93 ) 1.93 We now calculate the set

We see th a t x* — (10,10, 26)T , w hich is the com m on p o in t o f hyperplanes 4, 5, and

6

. T he common p o in t o f hyperplanes 1,

2

, and 3 is xq — ( 7 , 7 , 5)T . The common p o in t o f hyperplanes 1, 4, and 5 is (10,10, 20)T . x 0 and x* are n o t on any hyperplanes defined by the given constraints.

From th is example, we can see w h y the p ro o f [15, page 420-421] o f the S. Theorem

2.1

is invalid. The authors assume th a t constraints active a t x*, w ill be the same constraints active a t a solution to the m odified L P m a x {c Tx : a j x < 1111, i = 1 , . . . , m } . We show th a t th is is false.

In exam ple 2.6, the m odified L P has the constraints

Xx + %2 ~ 0 . 4 x3 < 1 . 4 6 9 7 (6 '.1 ) 1 . 1 8 x i + 0.8x2 — 2x3 < 2 . 4 5 6 1 (6 '.2 ) 0.8x i + 1 . 1 8x2 — 2x3 < 2 . 4 5 6 1 (6'.3) 2xx — x 2 < 2 . 2 3 6 1 (6'.4) — X \ + 2x2 < 2 . 2 3 6 1 (6'.5) O .laq + 0.1x2 + _x3- < 1.0100 (6 '.6 )

The o p tim a l solution set for th e m odifie d L P are th e elements o f th e set { x E R

3

: x i + x 2 = 1.8016, x

3

— 0.8298 }, a t w hich constraints 1 and

6

are active. The unique o p tim a l solution fo r th e o rig in a l L P has active constrain ts 4, 5, and 6.

E xam ple 2.7

Consider the

LP

m ax :

10

x i + _{x 2} s.t.: Xi —

0

.

8£2

<

2

(7.1) - x 2 <

1

(7.2) —

0

.

8

x

1

+ _{x 2} <

2

(7.3) - X i <

0

(7.4) ~ x 2 <

0

(7.5) We firs t show th a t a ll constraints are necessary.

i X a j x

1

( 15, 14 ) 3.8

2

(

2

,

0

)

2

3 (

0

,

100

)

100

4 ( -2.5, 0 ) 2.5 5 (

0

, -

1

)

1

We now calculate the set

V = { v u v2, v 3} = {7 .1 8 4 0 ,6 .3 6 4 0 ,-5 .4 6 6 1 }

We see th a t x * = (1 0 ,10)T , w hich is the com mon p o in t o f hyperplanes 1 and 3, and v 3 is negative.

2.4

C onclusion

In th is chapter, we were concerned w ith the M A m ethod in [15]. T h e S. Theo­ rem 2.1 upon w hich th e M A m eth od is based is n o t valid. T h is was shown by L i

3

T h e A c t iv e C o n e M e th o d

3.1 I n t r o d u c t io n

The M A m ethod is m o tiva te d by the graphical so lution to L P problems. T h is m ethod analyzes the relatio nship between the active constraints and the angles between the gradients o f the constrain t functions and th e gradient o f the objective fu n c tio n in order to fin d the o p tim a l solution. In C h apter

2

, we have shown th a t th is m ethod is n o t valid. In th is chapter we present an m ethod, w hich we call the A ctive Cone M eth od, to solve LPs w ith 2 variables. T h is m ethod was m o tiva te d by our stu d y o f th e M A m ethod. We begin w ith some p re lim in a ry results and definitions.

D e f in it io n 3.1 The cone A s p a n n e d

b y

{ a X). . . , a t } is given by t

A = { x = '■ ^ 0, V i}. i = l

We w rite A = C { a i , . . . , a^} or A = C { a x : i —

1

, . . . , £}.

D e f in it io n 3.2 [

8

] A d ire ctio n d o f a set S is an e x tr e m e d ir e c t io n if it cannot be w ritte n as a positive com bin ation o f tw o d is tin c t directions o f S. T h a t is, there does not exist directions dx, d

2

o f S, d x

7

^ d2, and a Xl a

2

>

0

such th a t d = a xd x -f- a

2

d2.

D e f in it io n 3.3 T h e cone A is fr a m e d

b y

{ a i , . . . , a/.}, i f each a* is an extrem e d ire ctio n o f A and if A = C { a i , . . . , a*,}

D e f in it io n 3 .4 Consider the L P problem (1). A m in im a l c o n e is a cone containin g c, fram ed by some gradients o f co n stra in t functions, such th a t no other cone containin g c fram ed by gradients o f constrain t fu n ctio n s is p ro p e rly contained in th is cone.

D e f in it io n 3 .6 Consider the L P problem (1). I f c € . .. , a ik}, we say th a t w = Y^kj = i is the b- v a lu e o f t h e c o n e . Setting Xi fo r i e { i i , . . . , i fc} Vi

= s

( V ^

0

fo r i E {

1

, . . , , m } \ { i u

we see th a t y T = ( y i , . . . , ym) is a feasible so lu tio n to the d u a l problem (3) and th a t the b-value o f the cone C { a i 1, . . . , a ^ } is the corresponding objective value o f (3). T he follow ing statem ents are restatem ents o f the d u a lity theorems given in Section 1.2.

(1) T he b-value o f an active cone at an o p tim a l extreme p o in t is th e o p tim a l value o f (

1

);

(2) The lv a lu e o f an active cone is less th a n or equal to th a t o f any cone w hich is fram ed by gradients o f co n stra in t fu nctions and contains c.

I f we can o b ta in a ll m in im a l cones dire ctly, then by calcu la tin g th e b-values o f these cones, we can id e n tify the active cone fro m a ll m in im a l cones. Therefore, we can solve the L P problem . F irs t, we investigate th is idea in R 2. To do so, we propose the follow ing lemmas and theorems.

L e m m a 3.1 I f j is a necessary constraint and i f l Z ^ t y , then there exists a p o in t x ' £ 71 such that a j x' — b j.

P r o o f: Since and since constrain t j is necessary, there exists x \ € TZ C IZ j, such th a t a j x x < bj and there exists x 2

6

H jJ R ,, such th a t a j x 2 > bj. Then, 3A, 0 < A < 1, x ' — Aaq + (1 — X )x 2, such th a t a j x ' = bj and x ' € 7Zj. Therefore, x ' £ 1Z and a j x 1 = bj. Q E D

L e m m a 3 .2 Let u \, u 2 and u$ be non-zero vectors in TZn . Let u

3

be a ele­ ment o f C { u i , u 2}. Then C { u i , u 2} = C { u i , u 3} U C { u3, u 2} , and C { u i , u 3} D

£ { ^

3

, ^

2

} = C { u 3}.

P r o o f: We can w rite

Let x G C { u i , u 2}. Then

x = PiUi + p 2u 2, Pi

> 0, A > 0,

(9)

We want to show either

x = A itii + \ 2u 3, Ai >

0

, A

2

>

0

,

or

x - f i i u 3 + i i 2u 2, i i \ >

0

, 112 >

0

.

F irs t, since u 3

7

^ 0, then a i , a 2 are n o t b o th 0. I f a i — 0, th a t is, u 3 — a 2u 2, then s u b s titu tin g

u 3 u 2 = — a 2 in (9), we get x = P1U1 + — u 3, P i > 0 , — >

0

. OL2 Oi2 So, x e C { u i ,u 3}.

Sim ilarly, if a 2 = 0, then x G C { u 3, u 2}. Now, if a.i

7

^ 0 and a 2 ^ 0, then fro m (

8

),

u 3 — a 2u 2 Ui = Oil u 3 - OiUi u 2 = --- . ct2 S u b s titu tin g these in (9), we have

n u 3 ol2u2 Pi ( oiiP2 — a.2Pi

x = P i --- b P2U2 = — U3 + ( --- ) u 2. (

10

)

Ol Oi Cti

Q I Q U ‘Z ~ a l Ul ,® 2P l — OiiP2. p2 , , x = P i U i + P 2---= ( --- )Ui + — u 3, (11)

Oi2 Oi2 Oi2

I f

then (

10

) shows, x £ C { u 3,U2\. I f n ot, then

a 2pl ~ a l@2 >

0

.

so th a t x € C { u i , u 3} by (11).

Now, we w ill show C { u i ,u 3} f l C { u

3

, u 2} = C { u 3}.

I f in (

8

) a i — 0, a

2

> 0, then u 3=

02

^

2

- T h a t means u

3

is in the same d ire ctio n as u2\ hence, C { u 3, u 2} = C { u 3}. Therefore C { u i , u 3} f l C { u 3,U2} = C { u 3}. S im ila rly we can prove, when a i > 0 , a 2 = 0, th a t C { u i , u 3} = C { u 3}.

I f Oil >

0

, Oi2 >

0

, suppose x is in b o th C { u i , u 3} and C { u 3, u 2} , then in (

10

)

& l p 2 - (3\O i2

>

0

,

and in (

11

) a 2@l ~~ @2a l hi

0

) so th a t b o th are equal to 0. In (10) A X = — u 3 011 and in (

11

) A x

= —

u 3, 012 which means C { u i , w3} f l C { u 3, u 2}

=

C { u 3}. Q ED

Lem m a 3.3

I n the L P problem (1), i f there are no redundant constraints, then all the gradients o f constraint fu n ctio n s have different directions.

Proof:

Suppose to the co n tra ry th a t there exist i , j such th a t, a,- a _*3a i n i i N i r i f b' > A

(say ).

A l l

11

A then fo r a ll x such th a t a j x < bj, a% a T bj b: — nX = ,, P , X < t,— it < r,— ; A l l I l l ' l l l A I I I A

L em m a 3.4

Let u i, u

2

and u 3 be non-zero vectors in 1Z2. Assume that u \, u 2 and u 3 are pairwise independent and let u 3 be an element o f C { u i , u 2} . Then

R 2 =

C { u i , u 2} U C { u i , - u 3} U C { - u 3, u 2}.

P r o o f: I f x (ft C { w i , —u 3}, then for

x = a xu i + a 2( - u 3), (12)

we have 3 cases: (1) ot\ < 0 , a

2

> 0 ; (2) cti > 0 , a

2

< 0 ; and (3) a x < 0 , a 2 < 0 .

Since u 3 G C { u i , u 2} and u 3 has different d ire ctio n from U\ and u 2,

u 3 = A i« i + A2u 2, where Ai > 0 , A

2

> 0 . (13)

In case (1), re w rite (13)

1

a

2

u i = w-w

3

- T~u 2- (14) Ai Ai

S u b s titu tin g th is equation in (12), we have

x = - a i ^ u

2

+ ( a

2

- y - ) ( - w 3).

A i Ai

T h a t means x G C { —u 3, u 2}.

In case (2), x G C { u i , u 3} C C { u i , u 2} by (12) and Lem m a 3.2. We can re w rite (13) as

( \

1

U2 = T - ( - W i ) + — u 3.

A2 A

2

T h a t means u 2 G C { —U i , u 3}. In case (3), by Lem m a 3.2, x G C { —u \ , u 3} = C { —u i , « 2} U C { u 2, u 3}. R e w rite (14),

1

/ \

^2

- U i = - r - ( - « 3) + t - ^

2

-

A i A i

T h a t means —u i G C { u 2, —u 3}. B y Lem m a 3.2, C { u 2, — u 3} = C { u 2, —u 1} U C { —Ui, —u 3}. Therefore, either x G C { —U i , u 2} C C { u 2, — u 3} or x G C { u 2, u 3} C C { u x , u 2}. Hence, we have proved th a t if x £ C {v ,i, — u 3} then x € C { —u 3, u 2} or x G C { u i , u 2}. Q E D

T h eorem 3.1

Let 1Z

= {

x : a j x < bi, i

G

1 } be a nonempty region in

R n.

Suppose that there are no redundant constraints in { a j x

<

bi

:

i

G

/

}.

Let x

G

1Z, and suppose that a j x < bk f o r some index k

G

I . Also, let

P r o o f: Assume I ( x ) = { A , • • •, i p}- B y hypothesis, x is such th a t:

a j x < bk,

a j.x = bi,, j = l , . . . , p . B y Lem m a 3.1, we can choose feasible x ' so th a t

a j x ' = bk, a j.x ' < b^, j = 1 Then,

a j (x

-

x ') < bk - bk = 0

,

al ( x ~ x 1) > b{j - b i

.

=

0

, j =

1

, . . . ,p.

Let A = [ fljj . . . ctip ] . The system A T y > 0 , a [ y < 0 has a s o lu tio n y = ( x —x !). Therefore, by Farkas’ lemma, there is no so lu tio n to

v

a k = Y l where >

0

, j =

1

, . . . ,p. 3=1

T h a t means ak £ C { a ^ : j = 1 , . . . ,p } . Q E D

We now prove, th a t when n —

2

C onjecture

2

is valid.

L em m a 3.5

I n the L P problem (1), let n —

2,

c

7

^

0

and ai ^

0,

f o r i =

1

, . . . , m. Assume that there are no redundant constraints and that the L P problem has an o ptim al solution, say x * , that is an extreme point. Then

T T

a 1. c a, c

a l x* = b k, where - - - = max

||afc|| | j

11

Proof:

T he p ro o f is by co n tradiction. To s im p lify the proof, we assume c , a i , i =

p o in t, x* is the common p o in t of exactly tw o hyperplanes corresponding to two active constraints [3], say the 2-nd and the 3-rd. In th is active cone C { a 2, a 3] a t x*, c = A

2

a

2

+ A

3

a3, where A

2

> 0 , A

3

> 0.

Suppose a x in C { —c, a2}, th a t means a i = P i { —c ) + P 2a2, where Pi >

0

, p2 > 0. B y Lem m a 3.3, a i and a

2

have different directions, therefore Pi p 0. Since v i > 0 , P2

7

^ 0. We have

a i = /?i ( —c) + P2CL2, where Pi >

0

, P2 >

0

. (15)

M u ltip ly in g (15) by c, we get

a j c = - P i + p 2a jc . (16)

M u ltip ly in g (15) by a2, then m u ltip ly in g by a j c. we get

a j a 2 ■ a j c = — P ic J a2 ■ a j c + p 2a jc . (17)

S u b tra ctin g (16) fro m (17), we get

a j a 2 ■a j c — a j c =

—

Hc ^ i r

• S im ila rly, we can get

0 a j a 2 — a j c ■ a^c l - | | CTa2|p -Since v x > 0 and Pi > 0, then

a j a 2 ■ a j c > a j c > 0. (18)

I f a j c > 0, since a j a 2 < 1, then a j c > a j a 2 ■ a j c > a jc . T h is co n tra d icts th a t a j c > a j c . I f a j c < 0, since P2 > 0, th e n a j a 2 > a j c ■ a j c . M u ltip ly a j c on

b o th side, we get a j a 2 - a j c < a j c ■ | | a jc

||2

< a j c . T h is co n tra d icts (18). S im ila rly, axin C { —c ,a 3} does n o t hold.

Therefore, by Lem m a 3.4, a x is in C { a 2, a 3}. However, th is contradicts Theorem 3.1. Q E D

P r o o f: F irs t, if there exists j £ { 1 , . .. , m } such th a t dj is in th e same d irectio n as c; th a t is,

a j

lajll llc l

then C { d j } is the m in im a l cone containing c. I f constrain t a3 is n o t active at an o p tim a l extrem e p o in t x*, a j x * < bj. B y Lem m a 3.1, we can choose a feasible p o in t o f the L P problem x ' so th a t a j x ' = b j. Therefore,

T *

IMI

T *

IMI

l

IMI

T / T /

c x = Ti— n aj x < T T T \bi = Ti— V\ai x = c x ■

I M I

I N I

I N I

T h is contradicts th a t x* is an o p tim a l extrem e p o in t. Hence, C { d j } is an active cone a t x*.

Suppose there is no d j w hich is in the same d ire c tio n as c. Assume the m in im a l cone containin g c is fram ed by tw o gradients o f co n stra in t functions, denoted a x and a2. B y Lem m a 3.3, any a* £ A = { a x, . . . , a f c } \ { a x, a2} has a different d ire c tio n from ax and a2. We w ill show th a t no o th e r gradient ax € A belongs to C { a

1

, o 2}. Suppose a* £ C { a i , a 2}. Then, by Lem m a 3.2, since c is in the cone C { a i , a 2}, it is in either C { a x, a i } or C { a i , a 2}- W ith o u t loss of generality, suppose c is in C { a x, a i} . I f C { a i , a j } = C { a x, a 2}, then

a

2

= a xa x + a

2

aj, where ckx > 0 , a 2 > 0. (19)

Since a* £ C '{a 1, a2}, we have

a,i = (3xa i + /?

2

a2, where f3x > 0 , (32 > 0. (20)

From (3X x (19) — a x x (20), we get

(Pi + cxiP2)a2 — (ax + a2Pi)di.

This co ntradicts th a t ax and a2 have different directions. Therefore C ' { ax, a j } is p ro p e rly contained in C { a x, a2}. T h is co ntradicts the d e fin itio n o f a m in im a l cone. Hence, ax ^ ( 7 { ax, a 2}.

We w ill show th a t constraints 1 and 2 are b o th active a t an o p tim a l extreme p o in t x*.

tw o hyperplanes corresponding to tw o active constraints [3]. Assume constraint 3 is another active constrain t at x*. C { a

2

, a 3} is the active cone at x*. Now, a i i

0

-

2

! o

,3

and c a ll have different directions. We have proved a

3

is n o t in the m in im a l cone C { a i ,

0

,

2

}. Also by Theorem 3.1, since co n stra in t

1

is n o t active a t x*, a\ is n o t in C { a

2

,a

3

}.

In the m in im a l cone C { o i , a

2

}, c = AiGq + A

2

CI

2

, where Ai > 0 , A

2

> 0. In the active cone C { a

2

, a3} , c = /q a

3

+ P

2

O

2

, where p i > 0 , p

2

> 0. Combine these tw o equations, we get

A i d i + X2CL2 — H 1 ^ 3 + h 2 & 2 , A i , A2, d 1, h2 > 0 . I f f i2 — A

2

> 0, then Mi . ^ 2 — A

2

Ol — T~fl3 ^---- \ a

2

; Ai Ai T h is contradicts th a t a i is n o t in (

7

{ a

2

, a3}. I f A

2

— P

2

> 0, then Ai A

2

— p

2

a

3

— a i 3--- (Z

2

, h i h i T h is contradicts th a t a

3

is not in C { a i , a

2

}.

I f A

2

= h

2

, then A

1

O

1

= hiQ-

3

- T h is contradicts the fact th a t a\ and a

3

are not in the same d irectio n. Therefore, we proved th a t constrain t 1 is n o t active and constraint

2

is active at x* does not hold.

S im ilarly, th a t co n stra in t 1 is active and co n stra in t 2 is n o t active a t x* does no t hold.

Now, suppose a t x*, constraints

1

and 2 are b o th n o t active. For convenience, suppose th a t constraints 3 and 4 are active at x*. So far, we have tw o cones containing c: th e m in im a l cone C {

01

,

02

} and the active cone C { a

3

,a

4

}. We have proved th a t a

3

and a

4

are b o th n o t in the m in im a l cone C { a i , a2}. B y Theorem 3.1, since constraints 1 and 2 are b o th n o t active a t x*, Gp and a.2 are

b o th not in the active cone C { a

2

, a3}.

Since C { a i , contains c and no aj is in the same d ire ctio n as c, c = A j d i + A

2

Q

2

) Ai >

0

, A

2

>

0

.

Since C { a 3, a4} contains c and no a j is in the same d ire ctio n as c,

C om bining th ese tw o equ ations, we get

A1O4 + X2Q.2 — M 1&3 + /h2ft4, A i, A2, fix, H2 > 0. (21)

I n R 2, since a3 and a

4

are linear independent, we have

a\ = a ia,3 + a

2

a4, for some an, a

2

(22)

and

a

2

= /?ia

3

+ /?

2

a4, fo r some /?1; p 2. (23)

S u b s titu tin g these tw o equations in (21), we have

( A ia i + A2/?i — H i ) as + (A ia

2

+ A

2

/3

2

~ ^2 )aA — 0. (24)

B y linear independence, the coefficients o f a3 and a

4

are 0. Since Aiaq + X2p i = Hi > 0, a t least one o f oq and Pi is positive. S im ila rly, a t least one o f a

2

and P2 is positive. Since ax and a

2

are b o th not in C { a

3

,a 4}, one o f and a 2 is

negative and one o f Pi and p 2 is negative. There are o n ly 2 cases: (1) Gq < 0 , 0.2 > 0, Pi > 0 , P2 < 0,

(2) Qfi > 0 , a 2 < 0, p i < 0 , p 2 > 0.

Consider th e case (1), in (24), Aiaq + X2p i > 0 and A

4

a

2

+ A2p 2 > 0. Then

yielding

~ P 2a

1

= - a i P 2a3 - a

2

/3

2

a4, and a 2a2 = a 2P ia 3 + a 2p 2a4.

A d d these tw o equations together,

(ct2Pi — cniP2)a 3 = — p2&i T Q(

2a 2

here — p 2 > 0.

Since a

3

is n o t in (

7

(

01

,

02

}, a 2p i - aq

/?2

< 0. T h is contradicts (25). A

2

> ^ ( —1Q;i) > ^

Therefore we have

a 2Pl > Q1P2 (25)

S im ila rly, case (2) does not hold. Therefore, we proved th a t constraints 1 and

2

are b o th n o t active at x* does not hold.

In summary, constrain t 1 and 2 are b o th active at x*. Thus, a m in im a l cone is an active cone. Q E D

3.2

T h e A c tiv e C one M eth o d in R 2

Based on Theorem 3.2, we propose the active cone m eth od in

R 2.

I f we fin d th e m in im a l cone containing c, then we o b ta in the active constraints at an o p tim a l extreme p o in t.

Consider L P (1) in

R 2

and assume th a t there are no redun dant constraints and th a t there is an o p tim a l solution.

Step 1:

C alculate C-2 a l2 ai2 Wo = — , U>i = ci a n an T o o if c

2

> 0 —oo if c

2

<

0

f T o o if ai2 > 0 ti I f Ci = 0, w 0 = I f an = 0 ,W i = { —oo if ai2 <

0

Since we assume c ^ 0 and a,t

7

^ 0, when Ci = 0 and at\ — 0, c

2

and a

i2

are not equal to

0

.

S te p

2:

For each i, let U be the num ber o f quadrant th a t c or at falls in. Specifically, set

1

if Gji >

0

, ai2 >

0

2

if a ii <

0

, a

i2

>

0

3 if Gji < 0, ai2 < 0

4 i f an > 0, ai2 < 0 and s im ila rly set t Qfo r c.

are a ll o p tim a l points.

I f there is no aj in the same d ire ctio n as c. L e t ap be the firs t at afte r c in counterclockwise order and aq be the firs t in clockwise order. Then, the constraints p and q are active at an o p tim a l extreme p o in t.

Y

0

2nd quadrant

OO A

_{: 3rd quadrant}

*

-~fl-counterclockwise order

_OO OO

1st quadrant

clockwise order.

4th quadrant

<-... — OO 0 X

counterclockwise order

clockwise order

Figure 3: Counterclockwise order and clockwise order.

Referring to F ig u re 3, we provide the fo llow ing definitions:

D efin ition 3.7

In

R 2,

fo r any tw o vectors a\ and a2 w h ich are in the same quadrant, if W\ < w 2, we say th a t a i is

before

a

2

in counterclockwise order and a

2

is

after

a-y in counterclockwise order.

D efin ition 3.8

In

R 2,

fo r any three vectors a0, a\ and a2, we say th a t a\ is

closer

to ao th a n a

2

in counterclockwise order, when we m eet one o f the follow ing cases:

(1) I f the thre e vectors are in the same qua drant and a\ is a fte r a0 and before a2.

do-(3) I f a

2

is in a different quadrant th a n a0, ax and do are in th e same quad­ rants, and ax is afte r

do-(4) I f a,2 is in a different quadrant th a n ao, a i and a

2

are in th e same quad­

rants, and a x is before a2.

(5) I f three vectors are a ll in different quadrants and t 2 > t i > t 0.

(

6

) I f three vectors are a ll in different quadrants, t 0 > t 2, and t x > to or t 2 > t i .

In Step 3, to fin d ap w hich is the firs t after c in counterclockwise order is to fin d ap w h ich is closer to c th a n a ll the other a*(? £ {

1

, . . . , m } \ { p } ) in counterclockwise order. In the process o f fin d in g ap, we set the firs t gradient which is not in the same d ire ctio n as c to be ap. T hen we com pare ap w ith each other ai w hich is n o t in the same d ire ctio n as c. I f new a* is closer to c th a n ap in counterclockwise order, then we use the new a, to replace ap, otherwise we keep the cu rre n t ap. A fte r com paring a ll the gradients o f co n stra in t functions, we can say th a t ap is the firs t afte r c in counterclockwise order. T he process o f fin d in g aq w h ich is the firs t after c in a clockwise order is sim ilar.

Here we w ill ju s tify th a t in Step 3 if there is an dj in the same dire ctio n as the c then any p o in t on the line segment described is an o p tim a l solution. In the p ro o f o f Theorem 3.2, we have obtained the conclusion th a t a ll o p tim a l solutions are on th e hyperplane corresponding to the j - t h co n stra in t if aj is in the same d ire c tio n as the c. We o nly need to prove th a t x x w h ich is the p o in t of the intersection o f constraints j and p is a feasible p o in t. S im ila rly, x 2 w hich is the p o in t o f th e intersection o f constraints j and q is a feasible p o in t. Therefore, on the co n stra in t aj any po in ts between x x and x 2 are a ll o p tim a l points.

Suppose x x is n o t feasible and violates the co n stra in t s. In Step 3, aj £ C { a p, a q} , where av ap and aq are pairw ise independent, and ax £ C { a p, a q}, if i ^ p , q , j- In p a rtic u la r, as C { a p,a q}. B y Lem m a 3.4, as £ C { a p, —a j } or ds £ d t}d qi —d j } .

For the p o in t x x, we have a j x i = bj, a j x x = bp, and a j x i > bs. B y Lem m a 3.1, we can choose feasible x ' on the constrain t p so th a t a j x ' < bj, a j x ' = bp, and a j x ' < bs. Then, a j (x x — x ') > bj — bj = 0, —a j { x x — x') = —bp + bp = 0, and —a j [ x x —x ') < bs — bs = 0. Now, le t A = [aj, —ap], y = x i —x', th e n th e system A J y > 0, — a j y < 0 has a solution. B y Farkas’ Lemma, A x = — as, x > 0 has no solution. T h a t means A i( —ap) + A2a j = — as, Ai > 0, A

2

> 0 has no solution. T h is contradicts (26).

S im ila rly, if as E C { a q, —a j } , we can deduce a c o n tra d ictio n as well. There­ fore, X\ is feasible. Q ED

The fo llow ing program im plem ents the active cone m ethod in R 2. T h e com­ p u ta tio n a l cost o f th is a lg o rith m is 0 ( m ) .

(The numbers refer to the remarks follow ing the a lg o rith m ). (01) C alculate w 0 and set t 0

(02) Set p = 0; q = 0; j= 0 For i = 1 to m

(03) C alculate u>i and set U

(04) I f j = 0 and ti = to and Wi = w o j = i Else I f p =

0

(05) p = i] q = i Else Counterclockwise(p, i) Clockwise(g, i) E nd if E nd if E n d for

Subroutine: C ou n terclock w ise(p ,

i )

(06) Case t 0 — t p I f wp > Wq

T h is subrou tine is sim ila r to the subroutine o f Counterclockw ise(p, i) so it is n o t included.

R em arks:

(01) W e calculate wo and to for

c-(

02

) Set p, q and j to be 0. p and q w ill store the index o f the co n stra in t which is the firs t after c in counterclockwise order and a clockwise order respectively. j w ill store the index o f the constrain t, if it exists, w hich is in the same d ire ctio n as c.

(03) Each c o n s tra in t’s Wi and U are calculated one by one in the “ fo r . . . endfor” loop.

(04) I f the curre nt constraint i is in th e same dire ctio n as c, then the index num ber i is stored in j .

(05) I f p — 0, we set p — i. L a te r on, afte r the new i- t h c o n s tra in t’s Wi and U are calculated, in the subroutine, if new a* is closer to c th a n cu rre n t ap in counterclockwise order, then we w ill set p = i.

(06) For c, curre nt p and new i, we w ill consider 3 cases: to — tp, to < tp and to tp.

(07) W h e n to = t p and wp > w 0, o n ly i f aj is in the same qua d ra n t and Wi is between wq and wp, a* is closer to c th a n curre nt ap in counterclockwise order.

(08) W hen t 0 = t p and wp < w 0, a c tu a lly ap is before c in counterclockwise order. Except the s itu a tio n th a t a; is in the same quadrant and Wi is between w 0 and wp, a, is closer to c th a n curre nt ap in counterclockwise order.

(09) In the case th a t to < t p, a, w ill replace ap in 3 cases th a t a* is in the same quadrant w ith c and afte r c or a* is in the same quadrant w ith ap and before ap, or ^ is in another qua drant between c and ap.

(10) I f ai is in th e same qua drant as c and w t > w 0, is a fte r c in counter­ clockwise order. Therefore ai is closer to c th a n current ap in counterclockwise order.

(12) I f t p > ti > to, ai is in the quadrant between c’s quadrant and ap s quad­ rant. Therefore a, is closer to c th a n curre nt ap in counterclockwise order.

(13) In th e case th a t to > t p, if a, is in the same quadrant w ith c and after c, or ^ is in the same quadrant w ith ap and before ap, then ai is closer to c th a n current ap in counterclockwise order.

(14) In th e case th a t t 0 > t p, if U > t 0 or U < t p then is in the quadrant between c’s quadrant and ap s quadrant. Therefore at is closer to c th a n current ap in counterclockwise order. E x a m p le 3.1 m ax : O.laq + x 2 s.t. : (a\) X\ +

_^2

< 30 M X\ -f-

0.612

<

20

M —

2

aq —

^2

< - 2 9 (a4)

0

.

2

a;! + X2 < 28 (°5) 3aq +

0

.

6

X

2

< 4 0 (oe) Xi - 0.5x

2

< 5 (a?) - X x - X2 <

- 2 0

We use the active cone m eth od to solve th is problem : Wi and t and the objective fu n c tio n are calculated as following.

wo = c2/ c i =

10

to -

1

Wi = O

12

/O il =

1

t i =

1

w2 =

022/021

=

0.6

t* =

1

W3 =

032/^31

= 0.5 to — 2> Wi =

042/041

= 5 t A =

1

W5 =

052

/

051

=

0.2

£5

₌

1

w e = a

62

/ a

61

= - 0 .5 t

6

= 4 w7 =

072/071

= 3

£7

— 3

constraints are active at an o p tim a l extreme p o in t. T he com mon p o in t o f the hy­ perplanes corresponding to these tw o active constraints is x = (0.5556 27.8889), w hich is the o p tim a l solution in th is example.

The active set m ethod developed by M .J. Best and K . R itte r [1] in 1985 perform s the equivalent of sim plex pivots and investigates w hether the active cone exists at each extreme p o in t reached. T he active cone m eth od in

R 2

in trodu ced above need not perform sim plex pivots, since we have proved th a t a m in im a l cone is an active cone at an o p tim a l extreme p o in t, and we can o btain the m in im a l cone d irectly. In

R n,

(n > 3), by calcula ting the cone’s b-value, we can id e n tify the active cone fro m a ll m in im a l cones. O f course, the ca lcula tion o f a ll m in im a l cones w ill be costly as well.

3.3

C on clu sion

In th is chapter, we gave d efin ition s o f an active cone and a m in im a l cone for L P problems. T h e n we proved th a t in th e L P problem (1) w ith o u t redundant constraints in

R 2,

a m in im a l cone is an active cone at an o p tim a l extrem e point. Based on th is idea, we created an active cone m ethod to solve L P problem s in

4

S im p lific a tio n M e th o d s b a se d o n G a m e T h e o r y

4.1

In tro d u ctio n

In th is chapter we discuss the m ethods created by S tojkovic and S ta n im iro vic and based on Game Theory, fo r reducing the LPs. Before we discuss this m ethod, we introduce some n o ta tio n and d efin ition s th a t are standard in Game Theory.

The n o rm a l fo rm o f a fin ite zero-sum two-person game reduces to a m a trix D , w ith as m any rows as player I has strategies and as m any colum ns as player I I has strategies. T he payoff fo r player I, assuming th a t player I chooses his z-th strategy and player I I chooses his j - t h strategy, is the element dtJ in th e z-th row and th e j - t h colum n of the m a trix . T he payoff fo r player I I is —d^.

D efin itio n 4.1

[

6

] In any game,

payoffs

are numbers w hich represent th e mo­ tiva tio n s o f players. Payoffs m ay represent p ro fit, qu a n tity, ’’ u tility ,” or other continuous measures (card ina l payoffs), or m ay sim p ly ra n k th e d e s ira b ility of outcomes (o rd in a l payoffs).

D efin ition

4 .2 [13, page 13] T h e element dij, w hich is b o th th e largest in its colum n and the smallest in its row, if it exists, is called a

sad d le p oin t.

(A strategy p a ir at a saddle p o in t w ill be in Nash E q u ilib riu m [

6

].)

D efin itio n 4.3 [13,

page

25]

In a m a trix

D,

we say th a t th e z-th row dom inates the fc-th row (or the k - th row is dom in ated by the z-th row) if

dij > dkj for a ll j .

W ith the fo llo w in g a d d itio n a l cond ition :

d^ > dkj fo r at least one j ,

E x a m p le 4 .1 [13, page 14] T he follow ing m a trix game has a saddle p o in t in th e

2

-n d row and the

2

-nd column.

' 5 1 3 ' 3 2 4 —3 0 1

We note th a t the 3-rd strategy for player I is s tric tly d o m in ated by the 2- nd strategy, since each element in the 3-rd row is less th a n each element in the 2-nd row. As a ra tio n a l player, player I w ill never choose th e 3 -rd strategy. Therefore, th is strategy may be disregarded to s im p lify the o rig in a l game. Also, the 3-rd strategy for player I I is s tric tly dom inated b y the 2-nd strategy, since each element in the 3-rd colum n is great th a n each element in th e 2-nd column.

L em m a 4.1

[13, page 15] Let v \ = m a x (m in c L ); v'n = m in ( m a xg L ) ; then

i j j i

v'i < v 'n

D efin itio n 4 .4 [13,

page

16]

A

m ixed stra teg y

fo r a player is a p ro b a b ility d is trib u tio n on the set o f his pure strategies. In case the player has o nly a fin ite num ber, m, o f pure strategies, a m ixed strategy reduces to an m -vector,

m

P = (Pi, • ■ • ,Pm) , satisfying p{ > 0, £ Pi = 1. i

—1

L e t P denote the set o f a ll m ixed strategies fo r player I, and le t Q represent the set o f player I I ’s m ixed strategies.

Suppose th a t player I and I I are playin g a m a trix game D . I f player I chooses the m ixed strategy p = ( p i , . . . ,Pm)T , and player I I chooses q

=

(<?i,

...,

q

n)T,

then the expected payoff w ill be

D (p , q) = p TDq.

Now, name v j and V / j as the values o f the game to player I and player II, respectively. Here:

Vi = m a x m in p 7D .j v n = m in m a x D i.g , v i < Vn,

p & P j q £ Q i

T h e o r e m 4.1 [13, page 18] ( T h e M in im a x T h e o re m ) I n a zero-sum, m a trix game, the values o f the game to the two players are equal. T ha t is

v i = vu

-l i we dea-l w ith m ixed strategies, we fin d th a t p-layer I ’s g a in -f-lo o r is precise-ly equal to player I I ’s loss-ceiling. The common value, v * , o f these tw o numbers is called th e value o f the game. A strategy p*, w hich is the o p tim a l so lu tio n to

m m

m a xim ize {u : sf2 / pi dij > v, j = 1 , . . . , n, P i > 0, i = 1 , . . . , m , y~]pj — 1}

2=1 i

is optim al for player I in the sense th a t there is no strategy w hich w ill give h im a higher expectation th a n v against every strategy o f player II. I f q* is the o p tim a l so lu tio n to

71 71

m in im iz e {u : ^ dijqj < v , i = 1 , . . . , m, q3 > 0, j = 1 , . . . , n, y g 7- — 1}

j= i j

(27) then q* is o p tim a l fo r player I I in the same sense.

Let yj = q j / v , then th e LP problem (27) can be transform ed as the follow ing L P problem :

m a x : Z ] = i V j

s.t.: E " = i dijVj < 1 i = 1 , . . . ,m . (28) y j > 0 j = l , . . . , n

Consider the fo llow ing L P problem : n

m ax : z (x ) = E CjXj — c x, Cj > 0, j = 1 , . . . , n,

s.t.: E a ijX j = a j x < b i , bi > 0, i = 1 , . . . , m, (29) j' =i

X j > 0 , j = 1 , . . . , n,

L e t d^ — aij/biC j and y3 = CjXj, then the L P problem (28) and the L P problem (29) are equal. T h a t means, solving the L P problem (29) is equivalent to determ ining th e o p tim a l strategy o f th e m a trix game given by D .