An Alternative Approach for Solving Bi-Level Programming Problems

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1 American Journa of Operations Research, 07, 7, ISSN Onine: ISSN rint: An Aternative Approach for Soving Bi-Leve rogramming robems Rashmi Bira, Viay K. Agarwa, Idrees A. Khan, Vishnu Narayan Mishra * Department of Mathematics, Sir adampat Singhania University, Bhatewar, India Department of Mathematics, Mody University of Science and Techonoogy Lakshmangarh, India Department of Mathematics, Integra University, Lucknow, India Appied Mathematics & Humanities Department, Sardar Vaabhbhai Nationa Institute of Technoogy, Surat, India How to cite this paper: Bira, R., Agarwa, V.K., Khan, I.A. and Mishra, V.N. (07) An Aternative Approach for Soving Bi- Leve rogramming robems. American Journa of Operations Research, 7, Received: October, 06 Accepted: May, 07 ubished: May 7, 07 Copyright 07 by authors and Scientific Research ubishing Inc. This work is icensed under the Creative Commons Attribution Internationa License (CC BY.0). Open Access DOI: 0.6/aor May 7, 07 Abstract An agorithm is proposed in this paper for soving two-dimensiona bi-eve inear programming probems without making a graph. Based on the cassification of constraints, agorithm removes a redundant constraints, which eiminate the possibiity of cycing and the soution of the probem is reached in a finite number of steps. Exampe to iustrate the method is aso incuded in the paper. Keywords Linear rogramming robem, Bi-Leve rogramming robem, Graph, Agorithm. Introduction Mutieve programming is deveoped to sove the decentraized panning probem in which decision makers are often arranged within a hierarchica administrative structure. The bi-eve programming probem is a hierarchica optimization probem in which a subset of the variabes are constrained to be soution of a given optimization probem parameterized by the remaining variabes. The inear bi-eve programming probem, which is a specific case of the Mutieve programming probem with a two eves structure is a set of nested inear optimization over a singe poyhedra region. Two decision makers are ocated at different hierarchica eves, each independenty controing ony one set of decision variabes, and with different and perhaps conficting obectives. The hierarchica optimization structure appears normay in penty of appication when ower eve moves are controed by upper eve decisions. Transpor- *Corresponding author, L. 67 Awadh uri Coony Benigan, hase-iii, Opposite-I.T.I., Faizabad, U.., India.

2 tation, management, panning and optima design are the few appication fieds of bi-eve programming probems. In mathematica terms, in bi-eve programming probems it is required to find a soution to the upper eve probem such that g( xy, ) 0, ( ) min, F xy xy, where y for each vaue of x, is the soution of the ower eve probem: such that hxy (, ) 0. min f ( xy y, ) The ower eve probem is aso referred as the foower s probem. In a simiar way, the upper eve probem is aso caed the eader s probem. The origina formuation for bi-eve programming probem appeared in 97, in a paper authored by J. Bracken and J. McGi [], athough it was W. Cander and R. Norton [] that first used the designation bi-eve or mutieve programming. However, it was not unti the eary eighties that these probems started receiving the attention they deserve [] [] [5] [6]. Motivated by the game theory of H. Stackeberg [7], severa authors studied bi-eve programming probems intensivey and contributed to its proiferation in the mathematica programming community. Since 980, a significant efforts have been devoted to understanding the fundamenta concepts associated with bi-eve programming. Various versions of the inear bi-eve programming probem are presented by [8] [9] [0] []. At the same time, various agorithms have been proposed for soving these probems. One cass of techniques inherent of extreme point agorithms and has been argey appied to the inear bi-eve programming probems because for this probem, if there is a soution, then there is at east one goba minimizer that is an extreme point []. Two other casses of agorithms are branch and bound agorithm and compementariy pivot agorithms [] []. A survey on the inear bi-eve programming probems has been written by O. Ben-Ayed [5]. The compexity of the probem has been addressed by a number of authors [6] [7] [8]. It has been proved that even the inear bi-eve programming probem where a the invoved functions are affine, is a strongy N-hard probem [9] [0]. In this paper, an attempt has been made to deveop a method in which constraints are anayzed, and used for soving two-dimensiona inear bi-eve programming probems. Constraints have been cassified broady in two categories; we have named them as concave constraints and convex constraints.. Fundamenta rincipes We define two types of constraint casses for the proposed method, which ay the foundation of this agorithm. Considering the norma to be towards the haf pane region not satisfied by constraints, we define the foowing: Concave Constraints: -constraints whose norma make anges with the x-axis 0

3 in the range [0, π]. Convex Constraints: -constraints whose norma make anges with the x-axis in the range [π,π]. Concave and Convex constraints defined here are other than non-negativity constraints.various types of constraints on the basis of the above definition are given in the tabe beow: constrained type a i x + b i y c i S. No. sign of a i b i c i Cass of constraint concave concave concave concave concave concave convex convex convex convex convex 0 convex The form of bi-eve inear programming probem considered here is of the foowing type: ( ) max or min f xy, where y soves () x x ( ) max or min f xy, () y y ax i + by i ci () xy, 0 () It can be observed easiy that inducibe region, for the finite soution is one among foowing two cases: ) a part of the ine of concave constraints; ) a part of the ine of convex constraints or part of the x-axis. Reason behind this observation is the fact that in (), the contro is ony on the y variabe, therefore for a given x, if () is to be maximized in the positive direction of the y-axis, then the extreme point wi be a point on a ine of concave constraint as shown in Figure, and if () is to be minimized in the positive direction of the y-axis, then the extreme point wi be a point on the ine of convex constraint or on the x-axis as shown in Figure. Whie deaing with this method of soving probems we come across two types of redundant concave constraint and one type of redundant convex constraint. A concave constraint which is redundant when no convex constraints are considered is one type of redundant concave constraints, is a ine of such type of

4 Figure. Location of extreme point in case of maximization or minimization probems. redundant constraint as shown in Figure, hereafter represented as RCC. A concave constraints which is redundant when convex constraints are aso considered with concave constraints is another type of redundant concave constraints, is a ine of such type of redundant concave constraint as shown in Figure, hereafter represented as RCC. One type of redundant convex constraints used for the proposed method is redundant when no concave constraint are considered, hereafter represented as RCX. RCC are removed in two steps, at the first step those RCC are removed which can be identified ust by inspection, it can be easiy seen that a concave constraint having ine i y= mx i + c i is RCC with respect to concave constraint having ine y= mx + c if mi > m and ci > c. After remova of such RCC et the concave constraints sustained are those whose ines are represented by y= mx+ c, y= mx+ c,,, y= mx + c, p y = mp x+ cp, where m > m > > m > > mp and c < c < < c < < cp. From Figure we can observe that, and are three ines of constraint such that their sopes m, m and m respectivey and their intercepts with y- axis c, c and c respectivey foow reation m > m > m and c < c < c and none of the three are RCC but under the same condition constraint with the ine is RCC with respect to constraint having ines and. Such RCC can be removed by finding out the x-coordinates for point of intersection of the ine of concave constraints. x coordinate x for point of intersection of and c c is given by x =, simiary the x coordinate x of point of m m intersection of and is given by c c x =, for constraint having the m m ine not to be RCC with respect to constraint having ines and we

5 Figure. Redundant constraint. must have x < x. In case constraint having the ine is redundant with respect to constraint having ines and, repace constraint having the ine by constraint having the ine and constraint having the ine by c c c c constraint having the ine and find out x = and x =, m m m m thus one by one a the redundant concave constraints can be removed. Here it is assumed that x and corresponding y coordinate y are non-negative otherwise constraint having the ine become RCC and we repace constraint having the ine by constraint having the ine and so on. Whie finding x i to check redundancy for concave constraints we aso find corresponding y coordinates y i. In this process after is obtained if we come across yi 0 for a positive x i, the concave constraint having ( xi, y i ) as termina point is considered to be the ast non-redundant concave constraint. It is to be noted that a ine segment parae to y-axis cannot be a part of inducibe region, therefore whie removing RCC, if a concave constraint parae to the y-axis having ine p is encountered then point of intersection of the ine of concave constraint ust before p and p is considered to be the termina point of the ast nonredundant concave constraint making the reaction set. RCX can be removed in the simiar way as RCC, for this et i y= mx i + c i and y= mx + c be two ines of convex constraint, such that m i < m and c i < c then constraint having ine i is RCX with respect to convex constraint having ine. After remova of such RCX et the convex constraint eft are those having ines y= mx + c, y= mx + c,, y= mx + c,, p y = m px+ c p, where m < m < < m < < m p and c > c > > c > > c p. Such RCX can be removed by finding out the x-coordinates of the point of intersection of ines of convex constraint, x-coordinate x for point of intersection of and is given by c c x =, simiary x coordinate m m

6 x of point of intersection of and is given by c c x = m m, for con- straint having the ine not to be RCX with respect to constraint having the ine and constraint having ine we must have x < x. In case constraint having the ine is redundant with respect to constraint having the ine and constraint having the ine, repace constraint having the ine by constraint having the ine and constraint having the ine by constraint having the ine and find out c c c c x = and x m m =, thus one by m m one a the redundant convex constraints can be removed. Non-redundant concave constraints obtained after remova of RCC as dis- cussed above are such that constraint having ine k is nearer to y-axis than constraint having ine if m k < m. Aso during this process we have obtained, coordinates of corners made by a non-redundant concave constraint ines, after remova of RCC, except the starting point of first non-redundant concave con- straint ine and the termina point of ast non-redundant concave constraint ine in genera. To obtain the starting point of first non-redundant concave constraint ine, find its intersection first with the y-axis if the coordinate so obtained is ( 0, y 0) then this is the required point, otherwise we find its intersection with x-axis. To obtain the termina point of ast non-redundant concave constraint ine we find its intersection with the x-axis, if the coordinate so obtained is ( x 0,0) then this is the required point, otherwise termina point is unbounded.. Agorithm Method and agorithm in case inducibe region is a part of concave constraints ine is given beow. A simiar method and agorithm can be given in case inducibe region is a part of convex constraints ine. Step : Remove RCC from a concave constraints and find y i for a x i obtained during the process of remova. Find RCX from a convex constraints. Step : Find starting point of first non-redundant concave constraint ine, and the termina point of ast non-redundant concave constraint ine which may be part of inducibe region. Step : Check if end points of first non-redundant concave constraint ine satisfy a non-redundant convex constraint ines or not, if they do so go for of and so on, to check the same, otherwise there may be one of the foowing three cases: ) There is at east one non-redundant convex constraint not satisfied by both in this case constraint having ine become RCC otherwise become part of boundary of the feasibe region, and we move to constraint having ines,,. ) There is at east one non-redundant convex constraint not satisfied by but satisfied by. Let be one such ine of convex constraint, then find the

7 intersection point of and and shift to this point of intersection, with this new if is again ine of such constraint do the same, and so on, ti a such constraints are exhausted, then move to constraint having ine,,. ) There is at east one non-redundant convex constraint not satisfied by but satisfied by. Let constraint having ine then find the intersection point of p and and shift p be one such convex constraint, to this point of intersection, with this new if constraint having ine q is again such constraint do the same, and so on, ti a such constraints are exhausted, in this case become the ast point to be considered for soution. Step : The points,, satisfying a non-redundant convex constraints, obtained from step are used to find optima soution by putting its vaue in the obective function (). If there are some feasibe points than in either of the foowing two cases we may have unbounded soution ) No concave constraint exist ) No constraint of the type, 5 and 7 as given in the tabe is present in the probem. If there is at east one of the convex constraints not satisfied by any of the point,, then there is a case of no feasibe soution.. Exampe where y soves max z = x+ y x max z = x+ y y 5x+ 5y 5 (5) y.5 (6) x+ y (7) x y (8) 8x y (9) x (0) xy, 0 Soution: As per the cassification (5), (6), (7) and (0) are concave constraints and (8) and (9) are convex constraints, inducibe region is a part of concave constraints. Step : art of ration reaction set are y = x+, y = 0x+ 9 and y = x+ 8. None of the concave constraint is RCC and none of the convex constraint is RCX. ( x, y ) = (,9 ) ( x, y ) = ( 8,9 ). Step : = ( 0,) = (, ). Step :,, satisfy both (8) and (9). Step : z for a the,, are z ( 0,) = 6, z ( ),9 = 7, 8,9 5 8 z, = 7. z ( ) = and ( ) 5

8 z, = 7. Therefore soution is ( ) max z = x+ y x where y soves max z = x+ y y 5x+ 5y 5 () y.5 () x+ y () x y () 8x y (5) x (6) xy, 0 Soution: As per the cassification (5), (6), (7) and (0) are concave constraints and (8) and (9) are convex constraints, inducibe region is a part of concave constraints. Step : art of ration reaction set are y = x+, y = 0x+ 9 and y = x+ 8. None of the concave constraint is RCC and none of the convex constraint is RCX. ( x, y ) = (,9 ) and (, ) ( 8,9 ) Step : = ( ) = ( ). x y =. 0,, Step :,, satisfy both (8) and (9). Step : z for a the,, are ( 0,) 6, 8,9 5 8 z, = 7. z ( ) = and ( ) Therefore soution is z (, ) = Concusion z,9 = 7, z = ( ) The proposed method is based on the anaysis of constraints. Unike the tra- ditionay used method of finding optimum such as interior point method or simpex method in which search is made by moving aong the boundary of the feasibe region, an attempt made in this paper conveys that by propery expoiting the properties of constraints there is a possibiity of deveoping a method which soves the probem in finite number of steps efficienty. Acknowedgements We thank the Editor and the referee for their comments. This support is greaty appreciated. References [] Bracken, J. and McGi, J. (97) Mathematica rograms with Optimization robems in the Constraints. Operation Research,, [] Cander, W. and Norton, R. (977) Mutieve rogramming and Deveopment o- 6

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