Some reverse location problems

European Journal of Operational Research 124 (2000) 77±88

www.elsevier.com/locate/dsw

Theory and Methodology

Some reverse location problems Jianzhong Zhang a

a,*

, Zhenhong Liu b, Zhongfan Ma

b

Department of Mathematics, City University of Hong Kong, Tat Chee Avenue, Hong Kong, People's Republic of China b Institute of Systems Science, Academia Sinica, Beijing, People's Republic of China Received 1 April 1998; accepted 1 January 1999

Abstract The general location problem in a network is to ®nd locations of facilities in a given network such that the total cost is minimum. It is very often to have its reverse case, in which the facilities have already been located on the network, and the problem is how to improve the network within a given budget so that the improved network is as ecient as possible. In this paper we present a strongly polynomial algorithm for a reverse location problem in tree networks. The method can be extended to solve reverse two- or more-location problems. Ó 2000 Elsevier Science B.V. All rights reserved. Keywords: Facility location; Tree network; Minmax criterion; Minimum cut

1. Introduction Location theory in network deals with the problem of ®nding locations of facilities in a network, such that the total cost, including set-up cost and transportation cost, is a minimum. Such problems exist extensively and have wide applications, such as locations of ®re stations, hospitals, schools, shopping centers and warehouses. It is also possible to meet a reverse case, in which the facilities have already been located in a network and cannot be moved to a new place, and the problem is how to improve the network within a

* Corresponding author. Tel.: +852 2788 8662; fax: +852 2788 8463. E-mail address: [email protected] (J. Zhang).

given budget, so that the improved network works as ecient as possible. Usually we need to use some budget to improve a network. Since the costs of shortening distinct edges by one unit may be di€erent, it is required to determine which edges to be shortened and by how much. Such decisions depend also on the measurement of eciency. Usually MinSum or MinMax are adopted as objectives for measuring the eciency of facility locations. MinSum is to minimize the total weighted distance between the facility and users; MinMax is to minimize the longest weighted distance between the facility and users. Berman et al. [2,3] discuss respectively the location problem with MinSum and MinMax objective functions. We use the MinMax measure in this paper. That is, we consider how to shorten the `lengths' of some edges to improve the network

0377-2217/00/$ - see front matter Ó 2000 Elsevier Science B.V. All rights reserved. PII: S 0 3 7 7 - 2 2 1 7 ( 9 9 ) 0 0 1 2 2 - 8

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J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

system so that the longest distance from the facility location to other nodes is minimum. Here `lengths' can be, for example, the travelling times. So, shortening `lengths' can be realized by various means such as to widen roads, to enhance road quality, or to upgrade travelling tools, etc. We now state the problem. Let N ˆ …V ; A; a; l; w; B† be a network, where V and A are respectively the node set and edge set of N ; a; l; w are functions from A to R‡ , and B is a positive scalar representing the available budget. For each edge …i; j† 2 A, lij is its current length, aij is its minimum permissible length, and wij is the cost of reducing the length by one unit. Let s 2 V be the location of the facility in the network. This paper considers the following problem: how to reduce the lengths of edges, subject to the budget limitation and the lower bound request, so that the longest distance from s to other nodes in the modi®ed network is minimum? We denote this problem by Reverse Location Problem (RLP). Note that recently there are many papers discussing inverse optimization problems, see [4±11,13±21], in which people want to adjust the parameters (lengths, cost coecients, capacities, etc.) of an optimization model as little as possible so that a given feasible solution can become an optimal solution. The problem that we are concerned with in this paper is closely related to that type of inverse problems, but is somehow di€erent because here we do not demand node s to be the center (optimal solution) of the location problem after reducing some lengths. To distinguish our problem from those inverse problems, we use the term reverse problem in this paper. The following result was given in [3]. Theorem 1. The decision problem of RLP in a bipartite network is NP-complete. From the theorem it can be seen that the reverse location problem hardly has a polynomial algorithm in general network, unless P ˆ NP . Berman et al. [3] discusses the reverse location problem in tree networks and formulates it as a linear program, but does not give any concrete algorithm. We consider in this paper the problem in tree networks and give a polynomial order combinatorial algorithm with complexity O…n2 log n† in

Section 2. We then propose further a variation of the reverse location problem and consider how to extend the method to the case of multi-facility in Section 3. 2. Reverse location problem in tree network Let N ˆ …V ; A; a; l; w; B† be a tree network, s 2 V be the location of the facility, and V ; A; a; l; w and B be de®ned as above. We can use the budget B to shorten some edges. The problem is to determine which edges to be shortened and by how much, so that in the resulting network the longest distance from s to other nodes is minimum subject to the budget limit. In order to formulate this problem as a linear programming we regard N as a directed tree network rooted s, i.e. assign each edge a direction, such that N becomes a directed tree with s as its root. To avoid using too many symbols, we still use N ˆ …V ; A; a; l; w; B† to denote the corresponding directed tree network. Then the reverse location problem can be formulated as the following linear program …P † which is di€erent from that in [3]: …P †

min

v

s:t: uj ÿ ui ‡ xij ˆ lij

for all …i; j† 2 A;

v ÿ ui P 0 for all i 2 V ; xij 6 lij ÿ aij for all …i; j† 2 A; X wij xij 6 B;

…1† …2† …3† …4†

…i;j†2A

xij P 0

for all …i; j† 2 A;

…5†

us ˆ 0; where xij represents the length to be shortened on arc …i; j† and ui is the length of the shortest path from s to node i in the resulting network. Note that since the network is a directed tree, fui g can be uniquely determined by fxij g and us ˆ 0 from Eq. (1), and v equals the maximum value of fui g. So, each feasible solution of …P † depends only on a set of xij satisfying Eqs. (3)±(5). In order to simplify the model …P †, we may assume that all aij ˆ 0, because if aij > 0 for some arc …i; j† 2 A, then the arc …i; j† can be replaced by

J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

two arcs …i; i0 † and …i0 ; j†, and de®ne lii0 ˆ lij ÿ aij , li0 j ˆ aij , wii0 ˆ wij and wi0 j ˆ 1. Therefore …P † becomes the following form: …P1 † s:t:

min

uj ÿ ui ‡ xij ˆ lij for all …i; j† 2 A; v ÿ ui P 0 for all i 2 V ; xij 6 lij for all …i; j† 2 A; X wij xij 6 B; …i;j†2A

xij P 0 for all …i; j† 2 A; us ˆ 0; and the corresponding network is denoted by N ˆ …V ; A; l; w; B†. Furthermore, to simplify the description and the proof of the algorithm for solving reverse problem …P1 †, we reformulate the network N ˆ …V ; A; l; w; B† to a standard form in the following way. Let Q be the set of all end nodes of N , and for i 2 Q, let l…i† be the length of the unique path from s to i in N . De®ne d ˆ maxfl…i† j i 2 Qg:

A1 ˆ f…i; t † j i 2 Qg;

lij ˆ lij lit ˆ d ÿ l…i†



min

…9†

for …i; j† 2 A; for …i; t † 2 A1 :

for …i; j† 2 A; for …i; t † 2 A1 for …i; t † 2 A1 





and d ÿ l…i† ˆ 0; and d ÿ l…i† > 0:

ut 

s:t: uj ÿ ui ‡ xij ˆ lij ; for all …i; j† 2 A ;

…6†

xij 6 lij ;

…7†

for all …i; j† 2 A ;

Lemma 2. The feasible solutions fui ; v; xij g of …P1 † satisfying the condition v ˆ maxfui j i 2 V g and the feasible solutions of …P 0 † have one to one correspondence with same objective value. Proof. For any feasible solution fu0i ; v0 ; x0ij g of …P1 †, with v0 ˆ maxfu0j j j 2 V g, we de®ne  ui ˆ

u0i v0

if i 6ˆ t ; if i ˆ t ;

if …i; j† 2 A; x0ij d ÿ l…i† ÿ …v0 ÿ u0i † if …i; j† 2 A1 :

u0i ˆ ui ; v 0 ˆ ut  ; x0ij ˆ xij ;





Note that in Eq. (8), we de®ne 1
0 ˆ 0 (this 1 shall be replaced by a big positive number M when we implement our algorithms). It is easy to see that N  has the following properties: (a) For any i 2 V  , the lengths of all paths from i to t are equal; (b) For any i 2 V  and i 6ˆ t , there is exactly one path from s to i. These properties will be used frequently in the following proofs.

It is easy to verify that fui ; xij g is a feasible solution of …P 0 † with the same objective value. Conversely, for any feasible solution fui ; xij g of …P 0 †, we can de®ne a feasible solution fu0i ; v0 ; x0ij g of …P1 † with the same objective value as follows:

Then N ˆ …V ; A ; l ; w ; B† is called the standard form of N . Consider the following problem …P 0 † over N  : …P 0 †

…8†

for all …i; j† 2 A ;

xij P 0;

xij ˆ

A ˆ A1 [ A;

8  < wij ˆ wij w  ˆ 1 : it wit ˆ 0

wij xij 6 B;



Introduce a new node t and de®ne



…i;j†2A

us ˆ 0:

v

V  ˆ V [ ft g;

X

79

for i 2 V ; for …i; j† 2 A:

This completes the proof of Lemma 2.



In the following for simplicity we assume that the tree network N ˆ …V ; A; l; w; B† has already been expressed as the standard form, i.e. N  ˆ N . In this paper instead of using a general LP solver to solve the above problem …P 0 †, we present a combinatorial algorithm which is strongly polynomial.

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ui can be calculated from fxij ; us g. If the algorithm stops at step 4, the optimum solution can be de®ned as follows: xij ˆ lij ÿ lk‡1 ij ; us ˆ 0, and other ui can again be obtained from xij and us by Eq. (6) easily.

Algorithm 1. Step 0: Put k 0, N k N , Bk B, lkij lij ; k wij : wij Step 1: Regard wkij as the capacity of arc …i; j†. Let Rk be the minimum s ÿ t cut of N k , which can be found by the algorithm for maximum ¯ow problems in a s ÿ t planar network (Chapter 8 of [1]). The capacity of Rk is denoted by wk …Rk † ˆ P k …i;j†2Rk wij . Step 2: If wk …Rk † ˆ 1, stop, an optimum solution has been obtained. Otherwise, calculate d1 ˆ Bk =wk …Rk †;

Remark 2. wk‡1 …Rk‡1 † P wk …Rk † for k ˆ 1; 2; . . . ; as wk‡1 P wk : Example. To illustrate Algorithm 1, we give an example. A given tree network N is shown in Fig. 1, where the node s is the given location of the facility, and the pair of numbers beside each edge …i; j† represents …lij ; wij †. For instance, edge …2; 5† has length 3 and the cost to reduce per unit of length is 2. Note that in fact there were an edge between nodes 1 and 12 and an edge between 6 and 11, with the lengths 4 and 2, respectively, and the lower bounds for their lengths are both 1. We have used the method explained at the beginning of this section to split the edge …1; 12† into two edges …1; 4† and …4; 12†, and the edge …6; 11† into …6; 10† and …10; 11†. Assume the total budget is B ˆ 17. We orientate each edge to make the graph a directed tree network rooted at s, and then change the network into the standard form N 0 , see Fig. 2, so that Algorithm 1 can be used to obtain an optimal solution. The minimum s ÿ t cut of N 0 can be found by a maximum ¯ow algorithm, which is R0 ˆ f…3; t †; …12; t †; …7; t †; …5; 8†; …6; 9†; …6; 10†g, with a capacity w0 …R0 † ˆ 4. We use the formulas in Step 2 of Algorithm 1 to obtain d0 ˆ d02 ˆ 1, that

d2 ˆ minflkij j …i; j† 2 Rk g

and decide the decrement d ˆ minfd1 ; d2 g. Step 3: Let  k lij ÿ d if …i; j† 2 Rk ; ˆ lk‡1 ij otherwise: lkij Put Bk‡1 Bk ÿ dwk …Rk †. Step 4: If d ˆ d1 , stop, the budget is used up and an optimum solution has been obtained. Otherwise we de®ne  1 if lk‡1 ˆ 0; k‡1 ij wij ˆ wij otherwise: Let N k‡1 ˆ …V ; A; ; lk‡1 ; wk‡1 ; Bk‡1 † k k ‡ 1, return to Step 1.

and

put

Remark 1. If the algorithm stops at step 2, the optimum solution of …P 0 † can be determined uniquely by setting xij ˆ lij ÿ lkij ; us ˆ 0 and other

Fig. 1.

J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

81

Proof. We choose from all optimum solutions of …P 0 † the one fx0ij ; u0i g with the maximum value P 0 …i;j†2R xij . We shall prove that such a solution must satisfy the requirement. Let Pi be the unique path from s to node i, N …i† be the subnetwork consisting of all paths from i to t in N and A…i† be the arc set of N …i†. Obviously N …s† ˆ N and N …t † ˆ ft g. It is easy to see from constraint (6) that if fxij ; ui g is a feasible solution of …P 0 †, then for any q 2 V , along any two paths Pq1 and Pq2 from q to t we have that X X …lij ÿ xij † ˆ …lij ÿ xij † ˆ ut ÿ uq ;

means we should shorten the length of each arc in the cut by 1, and the available budget is correspondingly reduced from B0 ˆ 17 to B1 ˆ 13. After shortening the lengths of the arcs in cut R0 , network N 0 becomes N 1 which is shown in Fig. 3. The minimum s ÿ t cut of N 1 is R1 ˆ f…3; t †; …12; t †; …2; 5†; …2; 6†g, with a capacity w1 …R1 † ˆ 5. As d1 ˆ d12 ˆ 1, the length of each arc in R1 should be shortened by 1, and the remaining budget becomes B2 ˆ 8. After these operations, network N 1 becomes 2 N , see Fig. 4. The minimum s ÿ t cut of N 2 is R2 ˆ f…3; t †; …1; 4†; …2; 5†; …2; 6†g, whose capacity is w2 …R2 † ˆ 8. Now, d2 ˆ d21 ˆ d22 , and thus the length of each arc in R2 can again be shortened by 1. The algorithm then stops as the budget has been used up: B3 ˆ 0. The optimum solution can be obtained by calculating all lij ÿ l3ij , which is x14 ˆ x69 ˆ x6;10 ˆ 1, x25 ˆ x26 ˆ 2, and other xij ˆ 0. The resulting network is shown in Fig. 5, and the optimum value ut , i.e., the longest distance from s to other nodes of the network N , is now 7 against the previous value 10. We now turn to the justi®cation of the algorithm. In the following lemma we suppress the subscript or superscript k as this lemma shall be repeatedly used for di€erent k.

…i;j†2Pq1

…i;j†2Pq2

and thus X X xij ˆ xij : …i;j†2Pq1

…10†

…i;j†2Pq2

Let h ˆ minf x0ij j …i; j† 2 Rg ˆ x0pq . If h P d, then the lemma is true. Otherwise, x0pq < d 6 lpq : We consider the subnetwork N …q†. If N …q† contains a q ÿ t cut R such that x0ij > 0 for all …i; j† 2 R , then we can de®ne 8 0  < xij ÿ e if …i; j† 2 R ;  0 xij ˆ xpq ‡ e if …i; j† ˆ …p; q†; : 0 otherwise; xij

Lemma 3. Let R be a minimum s ÿ t cut of N and d be de®ned as in Algorithm 1. Then there exists an optimum solution fx0ij ; u0i g of …P 0 †, such that x0ij P d for all …i; j† 2 R if B > 0 and w…R† < 1.

where e is a small positive number which ensures that x0ij ÿ e P 0 for all …i; j† 2 R , and x0pq ‡ e 6 lpq .

Fig. 2.

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Fig. 3.

Fig. 4.

Fig. 5.

J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

So, x satis®es constraints (7) and (9). We know that X wij ; …11† wpq 6 …i;j†2R 

for otherwise R could not be a minimum s ÿ t cut as we can replace the arc …p; q† by R and reduce the capacity of R. Eq. (11) ensures that ! X X  0 wij xij ˆ wij xij …i;j†2A

…i;j†2A

‡ e wpq ÿ

X

! wij

6 B;

…12†

us

i.e., x meets Eq. (8). Set ˆ 0. Then for any node r 6ˆ t , we determine ur uniquely by using Eq. (6) with xij being replaced by xij . For t , we know that if a path from s to t contains …p; q†, it must have one arc in the q ÿ t cut R . On the other hand, if an s ÿ t path does not contain …p; q†, as the original network has no cycle, the path cannot contain any arc in R . These two facts ensure that along any path from s to t , the sum of xij equals the sum of x0ij . So, ut can be determined by using Eq. (6) along any path from s to t and get the same result: ut ˆ ut . So far we veri®ed that fxij ; ui g is also an optimum solution of …P 0 †. However, as R \ R ˆ ;, X X xij ˆ x0ij ‡ e; …i;j†2R

and us ˆ 0. For small e, obviously x P 0. Consider any …i0 ; j0 † 2 R0 . By the de®nition of x0pq , x0i0 ;j0 P x0pq :

…13†

1 This arc is in a path from m to t , say Pmt  . The de®nition of …r; m† and previous proof imply that 2 there is a path from m to t , say Pmt  , along which 0 xpq is the only nonzero component of x0 . By Eq. (10), X x0ij ˆ x0pq : …14† 1 …i;j†2Pmt 

…i;j†2R 

83

Eqs. (13) and (14) mean x0i0 ;j0 ˆ x0pq 1 and x0ij ˆ 0 for other …i; j† 2 Pmt  . Therefore,

x0i0 ;j0 < d 6 li0 ;j0 ; which ensures that for small e, xi0 ;j0 ˆ x0i0 ;j0 ‡ e 6 li0 ;j0 ;

Just like Eq. (11), as R is a minimum s ÿ t cut, we know that X wij 6 wrm ; …i;j†2R0

from which we obtain X

…i;j†2R

which is against the de®nition of fx0ij g. This contradiction implies that there is a path P  from q to t satisfying x0ij ˆ 0 for all …i; j† 2 P  , from which and Eq. (10) we have x0ij ˆ 0 for all …i; j† 2 A…q†. Now we consider the unique path Psp from s to p in N . There are two cases: Case 1: There is an arc …i; j† in Psp such that x0ij > 0: Let …r; m† be the last arc of Psp with x0rm > 0. Clearly the subnetwork N …m† contains N …q†. De®ne R0 ˆ R \ A…m†, which is a m ÿ t cut of N …m†. Put 8 0 0 < xij ‡ e if …i; j† 2 R ;  0 xij ˆ xrm ÿ e if …i; j† ˆ …r; m†; : 0 otherwise; xij

8…i0 ; j0 † 2 R0 :

…i;j†2A

wij xij

X

!

wij x0ij …i;j†2A

ˆ

0 ÿ e@wrm ÿ

X

1 wij A 6 B:

…i;j†2R0

So, x meets constraints (7)±(9). By the same argument as before, we know that all ui , including ut , can be uniquely determined to meet constraint (6) and ut ˆ u0t . So, fxij ; ui g is an optimum solution of …P 0 †. But as …r; m† 62 R and …p; q† 2 R0 , X

xij >

…ij†2R

X

x0ij ;

…ij†2R

which contradicts the de®nition of x0 . Case 2: x0ij ˆ 0 for all …i; j† 2 Psp .

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J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

In this case, again by Eq. (10) and the de®nition of x0pq , we know that x0 must be such a vector:  0 xpq for …i; j† 2 R; x0ij ˆ 0 otherwise: Since x0pq < d, we de®ne  d if …i; j† 2 R; xij ˆ 0 otherwise;

k‡1 Proof. By the de®nitions of fxk‡1 g and ij ; ui   fxij ; ui g, it is easy to verify that they are feasible solutions of …Pk‡1 † and …Pk †, respectively. As the optimum values of …Pk † and …Pk‡1 † are, respectively, ukt and ut , by the assumptions of the lemma, we have that

ˆ ukt ut ˆ ut 6 uk‡1 t

and

ukt 6 ut :

and us ˆ 0. Then by a similar argument we know that all ui can be obtained from us and fxij g uniquely, and fxij ; ui g is a feasible solution of …P 0 †. But its optimal value

ˆ ut , which imply Therefore ut ˆ ukt and uk‡1 t k‡1 ; u g is an optimum solution of …Pk‡1 † that fxk‡1 ij i and fxij ; ui g is an optimum solution of …Pk †. 

ut ˆ u0t ÿ …d ÿ x0pq † < u0t ;

We are now ready to prove the ®nite convergence of the algorithm for ®nding a solution of the reverse location problem.

which contradicts the optimality of x0 . This completes the proof of Lemma 3.  As we know, after the kth iteration of the algorithm, we shall obtain from network N k ˆ …V ; A; lk ; wk ; Bk † the next one N k‡1 ˆ …V ; A; lk‡1 ; wk‡1 ; Bk‡1 † if the computation does not stop. In fact the problem …P 0 † depends only on l; w, and B. When we replace l; w and B in problem …P 0 † by lk ; wk and Bk , respectively, and the resulting LP problem is denoted by …Pk † which is the reverse location problem over network N k , and after completing the kth iteration, we obtain a feasible solution f~xkij ; u~ki g of …Pk † in which  k d if …i j† 2 Rk ; k ~xij ˆ 0 otherwise; and u~ki can be obtained from f~xkij g and u~ks ˆ 0. Lemma 4. Let fxkij ; uki g be an optimum solution of …Pk † satisfying that xkij P dk for all …i; j† 2 Rk : Put ( xkij ÿ dk if …i; j† 2 Rk ; ˆ xk‡1 ij otherwise; xkij k‡1 uk‡1 ˆ uki for i 2 V . Then fxk‡1 g is an optii ij ; ui mum solution of …Pk‡1 †. Conversely, if fxij ; ui g is an optimum solution of …Pk‡1 †, then fxij ; ui g is an optimum solution of …Pk †, where  xij ‡ dk if …i; j† 2 Rk ;  xij ˆ xij otherwise;

and ui ˆ ui for i 2 V .

Theorem 5. When Algorithm 1 terminates, we obtain an optimum solution of …P †. Proof. Algorithm 1 terminates in one of the following two cases: Case 1: The algorithm ends at Step 2 of an iteration, say k, that is, wk …Rk † ˆ 1. Then, there is a path Pst from s to t in network N k , such that wkij ˆ 1 for each arc …i; j† 2 Pst . That means, along this path, the length lkij of every arc has reached the lower bound so that Pst can not be further shortened any more. Equivalently speaking, there is no feasible solution fxij ; ui g of …P 0 † with ut < ukt , the length of Pst . Therefore the feasible solution given by the algorithm is optimum. Case 2: The algorithm ends at Step 4 of iteration k, that is, Bk ˆ dk wk …Rk †. In this case, by Lemma 3 there exists an optimum solution fxkij ; uki g of problem …Pk † with xkij P dk for all …i; j† 2 Rk . Since Bk ˆ wk …Rk †dk 6

X …i;j†2Rk

wkij xkij 6

X …i;j†2A

wkij xkij 6 Bk ;

it must be the case that  k d if …i; j† 2 Rk ; k xij ˆ 0 otherwise; which is exactly the feasible solution obtained by running the kth iteration of the algorithm.

J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

Furthermore, since fxkij ; uki g is an optimum solution of …Pk †, by Lemma 4 we can recursively obtain the optimum solutions of problems …Pkÿ1 †; …Pkÿ2 †; . . . by the formula given in Lemma 4, until we get the optimum solution fxij ; ui g of …P 0 † for network N ˆ N 0 , which is xij ˆ

k X lˆ0

xlij ; ui ˆ uki :

Proof. We ®rst show that the algorithm must stop within at most O…n† iterations. In fact, at each iteration either the length of an arc reaches the lower bound, or the budget is used up, in the former case an arc capacity becomes in®nite, in the latter case the algorithm terminates. Since the number of arcs is O…n†, the total number of iterations is at most O…n†. Second, we count the number of operations in each iteration. At each iteration we need to ®nd a minimum s ÿ t cut, which can be done in O…n log n† time for s ÿ t planar networks (see Theorem 8.8 of [1]). Hence the total complexity order of Algorithm 1 is O…n2 log n†.  3. Variations of the reverse location problem As we have seen, in problem …P1 † we minimize the longest distance under given budget B. Sometimes budget is not very rigid but the length of the longest path from the place of location s must not exceed an upper bound, say q. In this case, we have to consider the following problem …Q1 †: X min wij xij …Q1 † …i;j†2A

s:t: uj ÿ ui ‡ xij ˆ lij ;

us ˆ 0:

X

wij xij

s:t:

Theorem 6. Algorithm 1 solves problem …P † with a complexity order O…n2 log n†, where n is the number of nodes of the network N .

xij 6 lij ; xij P 0;

min

…i;j†2A

0

ui 6 q;

Note that this problem is a variation of the problem …P1 † which we discussed in last section. Changing the network N to the standard form, problem …Q1 † becomes the following (we still use N to represent the resulting network N  ) …Q0 †



85

i2V; …i; j† 2 A; …i; j† 2 A;

…i; j† 2 A;

uj ÿ ui ‡ xij ˆ lij ;

…i; j† 2 A;

ut 6 q; xij 6 lij ;

…i; j† 2 A;

xij P 0;

…i; j† 2 A;

us ˆ 0: It is not dicult to see that problem …Q0 † can be solved by an algorithm, say Algorithm 10 , which is very similar to Algorithm 1 except that dk1 should be re-de®ned as dk ÿ q, where dk denotes the length of any path from s to t in N k , and when this number becomes non-positive, i.e. when dk 6 q, the algorithm terminates. Therefore problem …Q0 † also has a strongly polynomial algorithm to solve. The above two problems …P1 † and …Q1 † can also be extended to the case of two or more locations in a tree network. In the following we only consider the problem with two locations. Also, we reasonably assume wij > 0 for each …i; j† in the original network. Let N ˆ …V ; A; l; w† be a tree network, and s and t be two locations in V . The reverse two-location problem corresponding to problem …Q1 † is how to shorten the lengths of the edges in N by using as less budget as possible so that in the resulting network for each node j the distance to one of s and t is not longer than a given upper bound q. We may call this problem as reverse two-location problem under the minimum budget criterion, which can be formulated as follows. Let Psj and Ptj be the paths between nodes s and j and between nodes t and j in tree network N , respectively, and xij be the length to be shortened on edge …i; j†, then the mathematical model of this reverse two-location problem is:

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…Q †

min

X

X

wij xij

…i;j†2A

s:t: min

8 < X :…i;j†2P

…i;j†2Ps;jl‡1

…lij ÿ xij †jr 2 fs; tg

r;l

l2V;

0 6 xij 6 lij ;

9 = ;

6 q;

…15†

…i; j† 2 A:

We now consider its solution. Let Tspq and Ttpq denote the subtrees containing s and t, respectively, of the network N n …p; q† for an edge …p; q† 2 Pst . Of course the two subtrees have no common node or edge. It is noted that the relation between the reverse two-location problem over N and two reverse single-location problems over Tspq and Ttpq is given by Theorem 7. Theorem 7. Let fxij g be an optimum solution of …Q †, then there exists an edge …p; q† 2 Pst with xpq ˆ 0 such that fxij j …i; j† 2 Tspq g is an optimum solution of …Q1 † over the subtree network Tspq , and fxij j …i; j† 2 Ttpq g is an optimum solution of …Q1 † over the subtree network Ttpq . Proof. We ®rst make the following observations: 1. Let …p; q† 2 Pst , then xpq ˆ 0 if X …i;j†2Psp

and X …i;j†2Ptq

…lij ÿ xij † 6

…lij ÿ xij † 6

X …i;j†2Ptp

X …i;j†2Psq

…lij ÿ xij †

…lij ÿ xij †:

This conclusion is easy to understand because if xpq > 0, then we can check that when we reduce xpq to 0, it will still meet the condition (15), but with a lower cost, which means fxij g is not an optimum solution, a contradiction. 2. There is an edge …p; q† 2 Pst such that xpq ˆ 0. In fact, let the path Pst ˆ …j1 ˆ s; j2 ; . . . ; jr ˆ t† from s to t in the order of j1 ; j2 ; . . . ; jr , then there must be jl such that X X …lij ÿ xij † 6 …lij ÿ xij † …i;j†2Ps;jl

and

…i;j†2Pt;jl

…lij ÿ xij † P

X …i;j†2Pt;jl‡1

…lij ÿ xij †:

Hence by (1), the edge …p; q† ˆ …jl ; jl‡1 † shall have xpq ˆ 0. Now, let 9 8 X = < X …lij ÿ xij † 6 l 2 V …lij ÿ xij † Vs ˆ ; : …i;j†2P …i;j†2P s;l

t;l

and Vt ˆ V n Vs . By conclusion (2), there is an edge …p; q† 2 Pst such that xpq ˆ 0. When we delete this edge, we obtain two disjoint subtrees Tspq and Ttpq . It is not dicult to see that all nodes of Vs are in the subtree Tspq , and Vt in Ttpq , and furthermore, fxij j …i; j† 2 Tspq g and fxij j …i; j† 2 Ttpq g are the optimum solutions of …Q1 †, respectively, over the tree networks Tspq and Ttpq :  It is obvious that if for an edge …p; q† 2 Pst , fxij g and fyij g are feasible solutions of …Q1 † over the tree networks Tspq and Ttpq , respectively, then 8 < xij if …i; j† 2 Tspq ; 0 xij ˆ yij if …i; j† 2 Ttpq ; : 0 if …i; j† ˆ …p; q†; is a feasible solution of …Q † for network N . Combining this fact with Theorem 7, we know that to solve problem …Q † over N it is sucient to ®nd an edge …p; q† 2 Pst such that the two optimum solutions of …Q1 † over the tree networks Tspq and Ttpq have the minimum sum of their objective values. Therefore, we can solve the reverse two-location problem …Q † over N by the following method: for each edge …p; q† 2 Pst , solve a pair of problems …Q1 † over the tree networks Tspq and Ttpq , respectively, and let the minimum values be fs …p; q† and ft …p; q†, respectively, and their sum be f …p; q† ˆ fs …p; q† ‡ ft …p; q†. Then we ®nd minff …p; q† j …p; q† 2 Pst g, say it is f …p ; q †. Now, the corresponding solutions of …Q1 † over the tree networks     p q Tsp q  and T , i.e. fxij j…i; j† 2 Tsp q g and fxij j…i; j† t  2 Ttp q g, together with xp q ˆ 0, form an optimum solution of …Q †. As a matter of fact, for each …p; q† 2 Pst , instead of solving two …Q1 † problems over the two tree networks, we only need to solve one single location reverse problem …Q1 † by using the following tech-

J. Zhang et al. / European Journal of Operational Research 124 (2000) 77±88

nique. Let the unique path between s and t in the tree network N be Pst . For each edge …p; q† in Pst , de®ne a new tree network Npq ˆ …V ; A; l; w†; where V ˆ V [ fs g;

A ˆ A 1 [ A2 ;

A1 ˆ A n f…p; q†g;  wij ˆ  lij ˆ

wij 1

lij 0

from each node j to the closer one between s and t in the resulting network is minimum. Similar to problem …Q †, this problem can be formulated as follows: …P  † s:t:

A2 ˆ f…s ; s†; …s ; t†g;

if …i; j† 2 A1 ; if …i; j† 2 A2 ;

min X

if …i; j† 2 A1 ; if …i; j† 2 A2 :

87

min v 8 < X :…i;j†2P

…lij ÿ xij † j r 2 fs; tg

r;l

wij xij 6 B;

0 6 xij 6 lij ;

9 = ;

6 v;

l2V;

…i; j† 2 A:

…i;j†2A

In tree network Npq , s is regarded as the single location. We then change Npq into the standard  . Now for a given q, we can use the folform Npq lowing algorithm to solve the reverse two-location problem …Q †. Algorithm 2 (for problem …Q †). Step 1: For each edge …p; q† 2 Pst , solve problem  by using Algorithm 10 . …Q0 † over the network Npq Let the minimum value be f …p; q†. Step 2: Find the edge …p ; q † 2 Pst which attains the minimum value, that is f …p ; q † ˆ minf f …p; q† j …p; q† 2 Pst g: 

Step 3: From the optimal solution x of the problem …Q0 † over Np q , obtain the optimum solution of …Q †, i.e. fxij j …i; j† 2 A1 g together with xp q ˆ 0 form an optimal solution of the problem …Q †. Remark 3. If for an edge …p; q† 2 Pst , in the optimum solution of …Q0 †, xij ˆ 0 for all …i; j† 2 Pst , then we do not need to solve problem …Q0 † for other …i; j† on Pst , as we already obtain the optimum solution of …Q †: this …p; q† is just …p ; q † and we can immediately go to Step 3. From this method we see that the reverse twolocation problem …Q † can be solved by a strongly polynomial algorithm of order O…n3 log n†, because Pst has at most n ÿ 1 edges, and Algorithm 10 has a complexity order O…n2 log n†, where n is the number of nodes of the network N . The reverse two-location problem corresponding …P1 † is that under given budget limit B, we want to shorten some edges so that the longest distance

Similar to Theorem 7, we know that problem …P  † has an optimum solution fxij ; v g such that there is an edge …p; q† 2 Pst satisfying xpq ˆ 0. Based on this fact, we can develop an algorithm for this type of reverse two-location problems in a similar way. Algorithm 3 (for problem …P  ††. Step 1: For each …p; q† 2 Pst , use Algorithm 1 to  and solve the problem …P 0 † over the network Npq  obtain the optimal value ut …pq†. Step 2: Find the minimum value minfut …pq† j …p; q† 2 Pst g ˆ ut …p0 q0 †; where …p0 ; q0 † is the edge on Pst to attain the minimum value. Step 3: From the optimal solution x of the problem …P 0 † over Np0 q0 , obtain the optimum solution of …P  †, i.e. fxij j …i; j† 2 A1 g and xp0 q0 ˆ 0 form an optimal solution of the problem …P  †. Algorithm 3 can solve the reverse two-location problem …P  † in O…n3 log n† time. Remark 4. The complexity order of Algorithms 2 and 3 can be reduced to O…n2 log2 n†; if the Fibonacci search method is used to select the edges on Pst for testing. 4. Conclusions In this paper, we discussed the problem of how to use a limited budget to improve a given location in a network most e€ectively so that the longest distance from the given location node to other

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nodes can be minimized, and the problem of how to use minimum funding to ensure that the longest distance from a given node to others will not exceed a given upper bound. Generally, people may regard the problems of adjusting the parameters of a model most e€ectively to make a given feasible solution possess some desirable properties as reverse (or inverse) optimization problems. In this sense, the two problems discussed in this paper can be called reverse location problems. We showed that these reverse location problems are in general NP-hard problems, and therefore it is unlikely to ®nd a polynomial order algorithm for the general case. In this paper we focus on tree network only. We can ®rst formulate the problem as a LP problem, and more than that, we are able to develop a strongly polynomial method which uses the minimum cut or maximum ¯ow algorithm as its main subroutine. This method can be easily extended to the multi-location cases as we have shown that a reverse multi-location problem can be transferred into several singlelocation reverse problems to solve. Some di€erent types of reverse location problems can also be addressed which we shall discuss in our forthcoming papers. Note that although in this paper we focus on the issue of shortening the distances from a given node to others, in fact we can also consider how to use a limited budget to increase the capacity of a network optimally. A work of this type is given in Ref. [12].

Acknowledgements The author gratefully acknowledges the partial support of Hong Kong University Grants Committee under its CERG grant. The authors are grateful to the referees for their valuable suggestions.

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