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Finding Disjoint Routes in Telecommunications Networks with Two Technologies Author(s): Anne de Jongh, Michel Gendreau and Martine Labbe Source: Operations Research, Vol. 47, No. 1 (Jan. - Feb., 1999), pp. 81-92 Published by: INFORMS Stable URL: http://www.jstor.org/stable/222895 . Accessed: 09/05/2014 12:10 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . INFORMS is collaborating with JSTOR to digitize, preserve and extend access to Operations Research. http://www.jstor.org This content downloaded from 194.29.185.252 on Fri, 9 May 2014 12:10:31 PM All use subject to JSTOR Terms and Conditions

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Page 1: Finding Disjoint Routes in Telecommunications Networks with Two Technologies

Finding Disjoint Routes in Telecommunications Networks with Two TechnologiesAuthor(s): Anne de Jongh, Michel Gendreau and Martine LabbeSource: Operations Research, Vol. 47, No. 1 (Jan. - Feb., 1999), pp. 81-92Published by: INFORMSStable URL: http://www.jstor.org/stable/222895 .

Accessed: 09/05/2014 12:10

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

INFORMS is collaborating with JSTOR to digitize, preserve and extend access to Operations Research.

http://www.jstor.org

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Page 2: Finding Disjoint Routes in Telecommunications Networks with Two Technologies

FINDING DISJOINT ROUTES IN TELECOMMUNICATIONS NETWORKS WITH TWO TECHNOLOGIES

ANNE DE JONGH Universite Libre de Bruxelles, Bruxelles, Belgium

MICHEL GENDREAU Universite de Montreal, Montreal, Quebec, Canada

MARTINE LABBE Universite Libre de Bruxelles, Bruxelles, Belgium

(Received July 1995; revisions received June 1996, October 1997; accepted October 1997)

We consider networks in which a cost is associated with each arc or edge and a transition cost is associated with each node. This last cost is related to the presence of two technologies on the network and is incurred only when a flow enters and leaves the corresponding node on arcs of different types. The problem we consider consists in finding two node disjoint paths with minimum total cost. We show that it is strongly NP-complete. Then we propose two heuristics, study their worst case behavior, provide a lower bounding procedure based on Lagrangean relaxation, and finally embed those elements in a branch and bound procedure.

Due to the rapid development of new equipments and communication protocols, telecommunications net-

works are constantly evolving. It is thus common to en- counter networks where two technologies coexist for the same type of traffic. For instance, in Europe, the transmis- sion network for voice communications uses both the Syn- chronous Digital Hierarchy (SDH) and the Plesiochronous Digital Hierarchy (PDH) technologies. In such networks, traffic flowing from an origin to a destination will often need to be routed over links of both technologies. This traffic will in most cases incur not only link transmission costs (which are generally different for the two technolo- gies) but also transition costs, whenever it passes from a link of one technology to a link of the alternate technol- ogy. One of the most important concerns of telecommuni- cation operators is to ensure the reliability of their network. This is often achieved by providing more than a single path for routing the traffic between any given origin- destination pair. Indeed, if two arc disjoint paths are pro- vided and the traffic is split evenly between them, no more than 50% of the demand will be lost, should a single link fail. Moreover, if these paths are node disjoint, then the traffic will be similarly protected from single node failures.

The purpose of this paper is to analyze the so-called bifurcated routing problem of finding a minimum cost pair of disjoint paths between an origin and a destination in a network. Four variants of the problem can be considered: the two paths must be arc or node disjoint and transition costs are incurred or not. It should be noted that capacity considerations are not taken into account since in most fiberoptic networks, link capacities are large enough so

that they can be neglected. Considering that most telecom- munication networks are made up of two-way links, one may want to tackle the problem separately for directed and for undirected networks. In fact, this is unnecessary, and we may restrict our attention to directed networks by re- placing each undirected link with a pair of arcs in opposite directions; only one of these two arcs will ever belong to an optimal pair of arc or node disjoint paths, if arc costs are positive.

Bifurcated routing problems have attracted little atten- tion. In the absence of transition costs, finding a pair of node or arc disjoint paths from a given origin to a given destination reduces easily to a minimum cost flow problem in an appropriate network. Obviously, when such pairs of paths are needed from a given origin to many destinations, the above method can be repeated for each destination separately. However, Suurballe and Tarjan (1984) propose a faster procedure, which determines all those path pairs in one single Dijkstra-like calculation. Li et al. (1990, 1992) study difficult bifurcated routing problems. In the first paper, the cost of the longer path of the pair is mini- mized. In the second one, each path corresponds to the routing of a different commodity so that each arc is en- dowed with a cost depending on the path to which it be- longs. Finally, Perl and Shiloach (1978) study the complexity of finding two disjoint paths between two dif- ferent sources and two different sinks.

The paper is organized as follows: in ?1, we present the four variants mentioned above and ??2 to 5 are devoted to the single (strongly) NP-complete problem of finding two

Subject classifications: Communications: telecommunications networks. Network/graphs, flow algorithms: disjoint paths. Programming, integer: algorithm for bifurcated routing with transition costs.

Area of review: TELECOMMUNICATIONS.

Operations Research 0030-364X/99/4701-0081 $05.00 Vol. 47, No. 1, January-February 1999 81 0 3 1999 INFORMS

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82 / DE JONGH, GENDREAU, AND LABBE

node disjoint paths with transition costs. We propose heu- ristics and study their worst case behavior in ?2, ?3 is devoted to a Lagrangean relaxation, an enumerative algo- rithm is described in ?4 and computational experiments are presented in ?5. Conclusions and suggestions for fur- ther research are stated in ?6.

1. FOUR BIFURCATED ROUTING PROBLEMS

Let G = (V, A) be a directed graph with node set V of cardinality 1< = n. The arc set A is partitioned into two disjoint subsets A1 and A2 containing the arcs of the first and the second technology respectively. To each arc (i, j) of A is associated a nonnegative cost cij, which represents the corresponding link transmission cost (this cost depends generally on the technology of the link). To each node i E

V are associated two nonnegative transition costs f+ and if. The first cost f7+ is incurred when a selected path enters node i on an arc (j, i) E A1 and leaves i on an arc (i, k) E A2. Similarly, fi is incurred when arcs (j, i) and (i, k) belong to a selected path with (j, i) E A2 and (i, k) E

A1. These costs J+ and fi correspond thus to the opera- tion of some equipment required to transfer the traffic between links of different technologies.

Let node 1 represent the origin and node n the destina- tion of some traffic demand. The bifurcated routing problem is defined as finding two disjoint paths from node 1 to node n and with minimum total cost. Four bifurcated rout- ing problems can be distinguished according to the fact that paths must be arc or node disjoint and that transition costs are taken into account at intermediate nodes of the paths or not. Three of these problems are easily solvable and rather well known. We present them briefly here for the sake of completeness.

First, if the two paths must be arc disjoint and if all transition costs are equal to zero, the problem is obviously equivalent to finding a minimum cost flow of value 2 from node 1 to node n in G where each arc is endowed with a unit capacity. Suurballe and Tarjan (1984) propose an al-

gorithm for that problem that is especially useful if pairs of paths must be found from node 1 to many different destinations.

In the second case, the two paths must be arc disjoint and nonnegative transition costs are taken into consider- ation. Hence, the cost of a pair of paths is given by the sum of the costs of their arcs plus the transition costs at the nodes of those paths where the incident arcs belong to different setsA1 andA2. Again, this problem reduces to a minimum cost flow of value 2 from node 1 to node n but in a modified network defined as follows. Each node i of G is split into two nodes il and i2 as shown in Figure 1. All arcs (i, j) E A1 are replaced by arcs (i1, jI) and all arcs (i, j) E A2 are replaced by arcs (i2, 12). The costs of all these arcs keep the same value as the original ones. Finally, for all i E V, we create two new arcs (i1, i2) and (i2, il) with respective costs ]+ and f. All arcs have unit capacity.

In the third case, a pair of node disjoint paths is sought for and transition costs are zero. The solution of this prob- lem is a minimum cost flow of value 2 from node 1 to node n in a network with unit capacity arcs where each node i is duplicated as shown in Figure 2. All arcs that enter (leave) node i are replaced by arcs that enter node i1 (leave node i2). A single arc (i1, i2) with zero cost is added between every pair of nodes.

The fourth and last problem, in which the two paths must be node disjoint and transition costs are not all zero, merits special attention not only because it is relevant in the context of emerging technologies but also because it is more challenging from a resolution point of view, as shown in the following theorem.

Theorem 1. The problem (2-NDPP) of finding two node disjoint paths of minimum cost in a network containing arcs of two technologies and transition costs is strongly NP-Hard.

Proof. The problem (2 - NNDPP) can be formulated as the following decision problem:

+

f.

f1

initial network transformed network

Arcs belonging to A1 - - - * Arcs belonging to A2

- -> transition costs

Figure 1. Two arc disjoint paths with transition costs.

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DE JONGH, GENDREAU, AND LABBE / 83

(DP) Instance: G = (V, A) a directed graph A =A1 UA2 such thatA1 nfA2 = 4 a nonnegative cost cij for each vi, vj E A, two nonnegative transition costs fi+ and fi on each node vi E V, a source node s and a sink node t, a nonnegative integer K,

Question: Do there exist two node disjoint paths P1(s, t) and P2(s, t) with total cost not larger than K?

Clearly, (DP) belongs to the class NP. To verify that it is strongly NP-complete, we show that the 3-Satisfiability Problem (3 - SAT) reduces to (DP). Let 3-SAT be de- fined as:

(3-SAT) Instance: n variables x1, . .. , x, m clauses C1,..., Cm, each clause Cj being a set of three literals xi or i, a truth assignment T: {xi} -- {true, false}, a clause Cj is said to be satisfied if it contains one literal xi such that T(xi) = true or one literal xi such that T(Xi)= false.

Question: Does there exist one truth assign- ment that simultaneously satisfies the m clauses?

Given an instance of (3 - SAT), let us build the follow- ing directed network. Let pi be the number of occurrences of the variable xi in the m clauses. For each variable xi, we build a subgraph called a lobe as shown in Figure 3. All the arcs of these lobes are of technology 1 and of cost zero, and the lobes are connected with each other in series (see Figure 4). For each clause Cj, we add two nodes yj and zj. The arcs (s,y1), (zj,yj11),j = 1, ... , m - 1 and (Zm, t) are all of technology 2. To connect the clauses to the variables, we add arcs of technology 2 as follows:

* Add (yj, u') and (ut, zj) if the kth occurrence of the variable xi is xi and belongs to clause Cj.

* Add (yj, iU) and (iU, zj) if the kth occurrence of the variable xi is xi and belongs to clause Cj.

For example, the network corresponding to the instance

(Xl VX2VX3)A(O2VX3Vx4)A(C1VX2VX4)A(X2Vx3VX4)

is depicted in Figure 5. To each arc we associate a cost equal to zero, we set all transition costs equal to 1 and K = 0. This network, containing arcs of technologies 1 and 2, is constructed in polynomial time.

It suffices now to show that there exists a truth assign- ment T that satisfies the m clauses simultaneously if and only if one can find two node disjoint paths from the source to the sink of this network, of total cost zero. By construction, the first arcs of both solution paths have to be of different technology (otherwise they have a node or an arc in common). Thus, the total cost zero will be ob- tained if one of the paths uses only arcs of technology 1 and the other one uses only arcs of technology 2.

Suppose first that a truth assignment T exists. Then the path that passes through the upper part of the lobe i, i = 1, . .. n if T(xi) = false and through its lower part if T(Xi) = true is chosen as the first path. Note that it uses only arcs of technology 1. For the second path, we take

Figure 3. Lobe associated to the variable xi.

initial network transformed network

- arcs belonging to A - - - * arcs belonging to A2

- -~ arc of cost 0

Figure 2. Two node disjoint paths without transition costs.

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84 / DE JONGH, GENDREAU, AND LABBE

')S ''I(S

Figure 4. Clauses connected to each other.

(s, Yi) and (Zm, t) as first and last arcs. For each clause C1, there exists one literal xi (or xi) such that T(xi) = true (or T(xi) = false). This implies that one can find a subpath containing the arcs (yj, u') and (uk, zj) or (yj1 ik) and (Uk, zj) and that is node disjoint from the first path. There exists at least m such subpaths, and linking them one after the other gives a second path, made up only arcs of tech- nology 2.

Conversely, assume there exist two node disjoint paths from s to t, one on technology 1 and the other on technol- ogy 2. One of them, say P1(s, t), must necessarily pass through the lobes. If it passes through the upper part of lobe i, set T(xi) = false, else set T(xi) = true. Since the second path, P2(s, t), is node disjoint from the first one, it must pass sequentially through arcs (s, Y1), (zj, Yj+ l),] =

1,..., m and (Zm, t) for j. Moreover, for each j, there exists one k such that P2(s, t) contains the arcs (yj, u') and (Uk, zj) or (yj1 i,) and (yj1 ',). As these two paths are node disjoint, the arcs (yj, u') and (uk, zj) belong to the second path if T(xi) = true, while in the case where T(xi) = false, the arcs (yj, -') and (i4, zj) belong to P2(s, t). In both cases, it follows from the construction of the lobes, that the clause Cj is satisfied as it contains at least one variable xi (or xi) such that T(xi) = true (or T(xi) = false). D

Note that the construction we use in the proof of Theo- rem 1 is based on the lobe notion introduced in Even et al. (1976). Furthermore, it is also valid for the undirected case (by simply replacing every directed arc by an undirected link in the constructed network). Before closing this sec- tion, let us present a detailed formulation of the problem

Figure 5. Network for the given instance.

of finding two minimum cost node disjoint paths in a net- work with transition costs. This formulation is based on a transformation of the graph shown in Figure 1.

Let { 1, ..., n } be the set of the nodes of the network, where 1 represents the source and n the sink. Let ci1 be the routing costs on the arcs, fi be the transition costs at node i for the change from technology 1 to 2, and f- be the transition costs at node i for the change from technology 2 to technology 1.

The variables are defined as follows:

[1 if arc(i, j) E A 1 U A2 is used in one X=ij of the two paths,

0 otherwise, 11 if at node i, a path passes from technology 1

zj = to technology 2, 0 otherwise,

I1 if at node i, a path passes from technology 2 Z= to technology 1,

O otherwise.

The variables z do not need to be defined for the source and the sink, since there is no change from one technology to another at these two nodes.

With these notations, the problem (2 - NDPP) can be formulated as:

n-1 n-1

min CijXij + Z Jtz[ + ZE fizi x,z (i,j)41A UA2 i=2 i=2

subject to:

Xzij z 1 i = 2 n, ...,n-1, j:(i,j)EAI UA2

E xji + z i - x x1j - Zi+ = O, i = 2, ... ., n - 1, j:C(j,i)sA I j: (ij)e A I

E xji + zi+ - E xij - zi- = O, i = 2, ... ., n - 1, j: (j,i)(=A 2 j:(i,j)(=A,_

E x ij - E xjl 2, j: (1,j) eA I UA 2 j: (j,l A 1) - UA 2

E X nj E Xjn - -2, j: (n,j)A F- UA 2 j: (j,n)A F- UA 2

x jj, z it, Z i- E {O0, 1}1, i E V\{ 1, n}1, (i, j) A1 IU A2 (1)

Constraints (1) force the two paths to be node disjoint between the source and the sink. Equations (2) and (3) are flow conservation constraints for the technologies 1 and 2 respectively. Note that we have to account explicitly for those paths which change from technology 1 to technology 2 (variable zj) and conversely (variable z,). Constraints (4) and (5) specify that there must be two paths, i.e. the flow must be of value 2. Figure 1 provides an illustration of the flow relationships in this model.

2. POLYNOMIAL TIME HEURISTICS

In this section, we develop three heuristics to find two node disjoint paths between the source and the sink of a network with transition costs. The worst-case performance of the third heuristic is established, and we show that it is tight.

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DE JONGH, GENDREAU, AND LABBE / 85

-~ 6

Figure 6. Usefulness of the connectivity test.

Heuristic 1

Step 1:

Transform each node of the network, except the source and the sink, as shown in Figure 2.

Step 2:

Compute two minimum cost node disjoint paths in this transformed network.

Step 3:

Add the transition costs to the solution when necessary, i.e. each time the flow passes from one technology to an- other in the solution of Step 2.

The addition of the transition costs at the end of the procedure implies that this heuristic can give very bad re- sults (see ?5). The second heuristic, described next, has proved to perform better as it presents the advantage of taking the transition costs into account from the beginning.

Heuristic 2

Step 1:

Transform each node of the network, except the source and the sink, as proposed in Figure 1.

Step 2:

Determine a first minimum cost path from the source to the sink. (Ties are broken in favor of the path with the smallest number of arcs and transitions. If ties still subsist, a single technology path is chosen.)

Step 3:

Delete all the intermediate nodes of this first path from the graph, as well as all the arcs that are incident to these nodes.

Step 4:

If the remaining graph is connected (i.e. if there is at least one path from the source to the sink), determine as second path a minimum cost path. (Ties are broken in favor of the path with the smallest number of arcs and transitions. If ties still subsist, a single technology path is chosen.)

Note that the connectivity test of Step 4 is necessary and that this heuristic may fail to find a solution. As an exam- ple, consider the network depicted in Figure 6, where the source and the sink are respectively the nodes 1 and 6. Suppose that the first minimum cost path, obtained in the

second step of the heuristic, is 1-2-4-6. Applying Step 3 will delete nodes 2 and 4, and all the arcs of the network, except (1, 3) and (5, 6). This makes it impossible to find a second path between 1 and 6. To circumvent this difficulty, we introduce a third heuristic in which Heuristic 1 is exe- cuted whenever the connectivity test fails.

Heuristic 3

Step 1:

Execute Steps 1 to 3 of Heuristic 2.

Step 2:

If the remaining graph is not connected then go to Step 3, else determine as second path a minimum cost path. Stop.

Step 3:

Execute Heuristic 1.

We now introduce some notations that will be useful to establish the worst-case behaviour of Heuristic 3.

We denote by: H', 1 = 1, 2, 3, the solutions given by Heuristic 1, HI and H2, 1 = 1, 2, 3, the two node disjoint paths given

by Heuristic 1, CH', and CH2, 1 = 1, 2, 3, the costs of HI and H2, CH' = CH, + CH2, 1 = 1, 2, 3, the total cost of the

solution given by Heuristic 1, 01 and ?2, the two optimal paths, CO, and C02, the costs of 01 and ?2,

CO = CO1 + C02, the cost of the optimum solution, cl , the maximum arc cost of technology 1, c2, the maximum arc cost of technology 2, Tmax the maximum transition cost, cmin, the minimum arc cost. Before giving the worst-case bound of Heuristic 3, let us

prove the following result:

Lemma 1. If at least one of the optimal paths is node disjoint from path H21, then CH2 = CO (thus CH3 = CO).

Proof. Assume that the path ?2 is node disjoint from Hl. By construction of the first heuristic path, we know that

CH1 CO1. (2)

Now, since path ?2 does not have any node in common with H21, the remaining graph of Step 4 is connected and contains ?2- Again, since H2 is a minimum cost path in that graph, we obtain that

CH22 S CO2. (3)

Combining (7) and (8) implies that CH2 = CO since 01 and 02 constitute an optimal solution. D

From Lemma 1, we know that if CH2 * CO, then both optimal paths have at least one node other than s or t in common with Hb . Since those two paths 01 and ?2 must be node disjoint, this implies that H12 has at least four

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86 / DE JONGH, GENDREAU, AND LABBE

nodes. Otherwise CH2 = CO thus CH3 = CO, and there is no need to compute a worst-case bound for Heuristic 3.

Theorem 2. Let n : 4 and cl :- c2.

1. If the remaining graph of Step 2 of Heuristic 3 is con- nected, then the solution of Heuristic 3 is H2 and (a) if n : 6 then CH3/CO is bounded by

1 (n -3)cmax 2if maxmax +2Tmax, 2 + 6cmin4if C+Lnax>C max2ma)

1 3

2 lC+aLn 2 36cmax max) otherwise 2 T6c min

(b) if n = 5 then CH3/CO is bounded by

i1 Cmax + l~MaX{ max, Cmax + Tmaxl . 1 2 +T }'3c 2 + 6cmin i jmaxjc maxv Cmax max 3 min

I 1, otherwise;

(c) if n = 4 then CH3/CO is bounded by

+ Cimax if cmax 3c min, 2 C min

1, otherwise.

2. If the remaining graph of Step 2 of Heuristic 3 is not connected, then the solution of Heuristic 3 is H' and CH3/CO is bounded by

1 +(n - 2)Tmax ncmin

Proof. 1. Note that CH3 : CO. Since CH31 = CH2 is equal to the cost of a shortest path, CO : 2CH3 and

CH3 + CH3 (4) CO 2 2CH1

Now, as H31 contains at least three arcs (otherwise CH3 =

CO),

CH3 :3cmin (5)

It remains to find an upper bound on CH3 and verify that the resulting bound on CH3/CH-1 is larger than or equal to 1 in each case.

(a) If n : 6, H2 contains at most n - 3 arcs. Assume that H2 contains p arcs of technology 1 and q arcs of technology 2, with p + q - n - 3, and that r transitions occur on H32. Then,

3 + rTm CH3 PClmax + qcmax + max

-P lmax +qcmax+ 2 min{p, qnTmax

pcmax +qcmax+ 2 mincp, n2-p-3}Tmax

Since c1 l c2a, the largest value of the right hand side of the last inequality is obtained forp =

n -3 if c' > c2 + 2Tm and forp = F(n -

3)/21 otherwise. Replacingp by those expressions in (11) and combining it with (9) and (10) yields the desired result.

(b) If n = 5, H3 contains at most two arcs. Thus,

CH23<c max + max{c max, C max + Tmax } (7)

If the right hand side of the inequality is strictly smaller than 3cmin, then CH2 < CH31, a contradic- tion. Else, replacing this expression in (9) and com- bining with (10) yields the desired result.

(c) If n = 4 and CO < CH3, H2 has only one arc. Thus,

CH < C max (8)

If cl < 3cmin, then CH32 < CH31, a contradiction. Else, replacing this expression in (9) and combining with (10) yields the desired result.

2. Assume that the solution of Heuristic 3, H3, contains p arcs. Let CH3 designate the cost of the solution H3 without the transition costs, and let CO_ designate the cost of the optimal solution without the transition costs. As

CH3 BPCmin, (9)

we have

CO- _ > PCmin, (10)

thus

CO :PCmin. (11)

Moreover,

CH3 < CO + (p - 2) Tmax, (12)

thus

CH3 (P - 2) Tmax (13) CO

+ PC min

(3

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DE JONGH, GENDREAU, AND LABBE / 87

technology 1

- - - - - - -technology 2

Figure 7. Network for Theorem 3, case 1 (a).

This last expression is maximum forp = n, which com- pletes the proof. D

The next theorem shows that these worst-case bounds are tight:

Theorem 3. Let n : 4 and cl :- c2.

1. If the remaining graph of Step 2 of Heuristic 3 is con- nected, then (a) if n : 6 there exists a network for which CH3/CO

equals

1 n-3)cmnax fc >C +2 T 2 + 6c min if c lmax > max + 2Tmaxv

C + max + J(cmax + (n -2)Tmax) 1 + 2 2~~6 )Ta) otherwise; 2 6c min

(b) if n = 5 there exists a network for which CH3/CO equals

1 C max + maX{Cmax, Cmax + Tmax} . 1 2 Tma 3 2 6cmin if max{cmax Cmax + Tmax} m

(c) if n = 4 there exists a network for which CH3/CO equals

1 Cmnax*1 -+

C If Cmax 3cmin 2 c min

2. If the remaining graph of Step 2 of Heuristic 3 is not connected, then there exists a network for which CH3/CO equals

+ (n - 2)Tmax ncmin

Proof. 1. If n : 6 then two cases must be considered:

(a) If cmax> c2nax + 2Tmax, consider the graph depicted in Figure 7 where the value of all transition costs is Tmax = cmin. All arcs of technology 1 or 2 in the upper part of the graph (resp. in the lower part) are of cost cmin (resp. cMax). In this case, H31 is the upper path [1 - 2 - 3 - n] of cost 3cmin and H2 is the path [1 -4 - 5 -.. - (n - 1) - n] of cost (n - 3)c' .. The optimal paths are [1 - 2 - n] and [1 -

- - 1 o 2 3

technology 1

- technology 2

Figure 8. Network for Theorem 3, case 1 (a).

3 - n] of total cost CO = 6cmin, and the lower bound is reached. If cl S c2 + 2Tmn, consider the graph depicted in Figure 8, where again all transition costs are Tm. = cmin. All arcs on the upper part of the graph are of cost Cmin, while tech- nology k arcs on the lower path are of cost Ckax. An argument similar to the one used in the previous case yields the desired result.

(b) If n = 5 and max{cl ., c 2 + Tmax} : 3cmin, then consider the network depicted in Figure 9, where Tmax = cmin. All arcs, except (s, 3) and (3, t) are of

cost Cmin. The arc cost of (s, 3) equals c'. If c c

2 + Tm., then the arc (3, t) is of cost clax and of technology 1. Otherwise, its cost is c2ax and its technology is 2. H3 is the path [s - 1 - 2 - t] of cost 3cmim' and H3 is the path [s - 3 - t] of cost C1ax + max{clax, c2 + Tmaxj. The optimal paths Cmax max~

max

1 2

technology 1

- technology 2

Figure 9. Network for Theorem 3, case 1 (b).

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88 / DE JONGH, GENDREAU, AND LABBE

technolcgy 1

? - - - - - - - technology 2

Figure 10. Network for Theorem 3, case 1 (c).

01 and ?2 are [s - 1 - t] and [s - 2 - t] of total cost 6cmin, and the bound is reached.

(c) If n = 4, consider the network depicted in Figure 10 where again all transition costs are Tmax = Cmin.

All arcs are of cost cmin, except (s, t) whose cost is cmax1 The first heuristic path, Hl3, is [s - 1 - 2 - t] of cost 3cmin and the second one, H3, is [s - t] of cost Cmax. The optimal paths 01 and ?2 are [s -

1 - t] and [s - 2 - t] of total cost 6cmin and the bound is reached.

2. Consider the graph depicted in Figure 11, where all transition costs are the same (Tmax) and all routing costs are equal to cmin. In this case, the first path given by heuristic 2, H 2, is [1 - 2 - (n - 1) - n] and the remaining graph is not connected. The following step of Heuristic 3 consists thus in applying Heuristic 1. Let H3 be the path [1 - 2 - * - Ln/2I - n] in an alternating sequence of technologies 1 and 2, beginning with the arc (1, 2) of technology 2. Let H 3 be the path [1 - (Ln/2I + 1) - ... (n - 1) - n] in an alternating se- quence of technologies. The total heuristic cost, CH3 is thus ncmin + (n - 2)Tmax. The optimal paths 01 and ?2

are [1 - (Ln/2I + 1) - . .. (n - 1) - n] and [1 - 2

- * * - Ln/2I - n], each of them on arcs of the same technology. The total cost CO is ncmin and the bound is reached. F]

3. LOWER BOUNDS

To obtain a lower bound on the optimal value of (2 - NDPP), we consider a Lagrangean relaxation in which constraints (1) are dualized. Let A1 = An = 0 and Ai D 0, i = 2, .. ., n - 1. After transformation, we obtain:

D(A) = min L (x, z, A)

s.t.(2) - (6),

where

L(x, z, A) = E (cj + Ai)x (i,j)eA UA2

nt-I n-I

+ E (fiz i +fi zi - E Ai. i=2 ~~~~~i=2

This lower bound D(A) is easily obtained by finding a pair of arc disjoint paths of minimum cost.

To strengthen the lower bound, we introduce the follow- ing capacity constraints on the nodes:

E xi1 + z[+ 1, i = 2, .. ., n - 1, (14) j:(i,j)EA I

E xi + zi- 1, i= 2, ......... . . I,n - 1 (15) j: (inj) F-4

These constraints are redundant with the constraints of the problem (2 - NDPP). Indeed, in (2 - NDPP), constraint (1) implies that at most one arc can leave node i. If that arc belongs to Al (resp. A2), then z+ = 0 (resp. z7 = 0) and 0](jj) 2 = 0 (resp. Ej:(i,j)FA x1j = 0) so that (19) and (20) hold. However, when added to constraints (2)- (6), they impose that if the paths have a node i in common, then one path, say P1, must enter and leave i on arcs of the

____ 2 ____ 3 3Li 1=s -K

-20 -1 n-i n=t

- technology 1

- - - - - - - -- technology 2

Figure 11. Network for Theorem 3, case 2.

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same technology, and the other path, p2, must enter and leave i on arcs of the other technology. Furthermore, the stronger lower bound D'(A) = min{L(x, y, A):(2)-(6), (19), (20)} can be obtained by solving a minimum cost flow of value 2 from node 1 to node n in a modified network defined as follows. Each node i of G is divided into four nodes il to i4 as shown in Figure 12. All arcs (i, j) E Al (resp. A2) with costs (cij + Ai) are replaced by arcs (i2, Il) (resp. (i4, 13)). The cost of (i4, il) equals Ji, that of (i2, i3)fi and those of (i1, i2) and (i3, i4) equal 0. All arcs of the transformed network have unit capacity.

Finally, the best lower bound LB is obtained by solving the Lagrangean dual

LB= maxD'(A) A_-o

with a subgradient method (e.g., see Beasley (1993) or Fisher (1981)). Computational experiments (see ?5) show that this lower bound is very tight.

4. BRANCH AND BOUND

We now present a Branch and Bound algorithm to find an optimal solution to (2 - NDPP). Upper bounds are given by the best value achieved by either heuristics H1, H2 and H3 described in ?2. Lower bounds are obtained by solving the improved Lagrangean dual LB described in ?3.

It may happen that the Lagrangean relaxation proce- dure terminates after a certain number of iterations with- out providing the optimal solution. Since the dualized constraints are inequalities, the best Lagrangean solution is either an infeasible one (two arc disjoint paths with some node k in common) or a feasible one (two node

disjoint paths) which is not optimal (cf. Beasley (1993)). In the latter case, there exists a node k for which the comple- mentary slackness condition

Ak( Xkj - 1 =0 (16) j:(k,j)GAi UA2

is violated. As the solution is feasible, we know that Ej (i,j)EA4UA2 xi; = 1 for all nodes i on the two node disjoint paths. It follows that node k does not belong to these paths and that Ak > 0. In either case, a valid branching scheme consists in creating two subproblems with additional con- straints on the flows leaving node k: in the first (resp. second) subproblem, only technology 1 (resp. 2) flows are allowed. Note that this may be implemented by deleting from the network the arcs for which flows are fixed to 0. When several candidate nodes k exist at a given step of the algorithm, the one closest to the source is selected for branching. The overall strategy for the branch and bound procedure is a depth-first search with the left child of every node being the one in which technology 1 arcs are deleted. It should be noted that after a certain number of branch and bound steps, and in particular when the network is sparse, arc deletions can make it impossible to find two arc disjoint routes between s and t. For this reason, before each branch and bound step, we add a test of 2-connectivity. If the remaining graph is not 2-connected, the current branch and bound node can be fathomed.

For a better understanding of the branch and bound scheme, consider the examples depicted in Figure 13 and Figure 14. The Lagrangean relaxation found some feasible solution which does not include node k. The positive value Ak was created during the Lagrangean procedure: at a cer- tain iteration, the node k was common to the two paths. The cost augmentation of the arcs leaving k with a value of Ak was sufficient to find a better solution. In order to pre- vent this situation, the current branch and bound node is divided into two children as shown in Figure 15.

Qg) k < positive dual values ,~ s , on the arcs leaving k

Figure 14. Unfeasible Lagrangean solution.

f technology 1

--- ~ ~ technology 2

- - - - transition cost

cost O

Figure 12. Transformation of the problem for the Lagran- gean subproblem.

k < 4 positive dual values

on the arcs leaving k

Figure 13. Feasible Lagrangean solution.

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90 / DE JONGH, GENDREAU, AND LABBE

current branch and bound node

first child: delete second child: delete all arcs of technology all arcs of technology 1 leaving k 2 leaving k

Figure 15. Division of the current branch and bound node.

Consider now Figure 14. The solution found in the La- grangean procedure has a crossroad node k. Dividing the current branch and bound node as shown in Figure 15 will forbid node k to be a crossroad in the next iterations.

5. COMPUTATIONAL EXPERIMENTS

The branch and bound procedure was programmed in Borland Pascal 7 and tested on 720 randomly generated networks, on a PS/2 57 486DX66 microcomputer.

These networks were created as follows. First, we place randomly n nodes into a square of side n, and choose among them a source node and a sink node. A two- connected network is created by taking the union of two disjoint spanning trees on the n nodes. The first spanning tree T1 is minimum in the complete graph Kn with edge lengths given by the Euclidean distances between pairs of nodes. The second spanning tree T2 is minimum in Kn\Tl. Some additional arcs are created if the graph is requested to be denser. It is a common fact in practice that, once an edge {i, j} is installed on the network, it can be traversed by flow in both directions, from i to j and from j to i. For this reason, we replace each edge {i, j} by two arcs (i, j) and (j, i) and work in the so-obtained directed network. Symmetric and asymmetric networks were considered. For the symmetric networks, the cost of an arc is 100 times its length, multiplied by a factor 0.7 for arcs of technology 2. In the asymmetric case, each arc has a cost which is ran- domly chosen in the interval [Fcij/21, F3cij/21]. The relative density of technology 2 arcs is around 40%. Three different densities are considered: 10%, 20% and 30% of the com- plete network. Finally, transition costs are either all the same or different for each node in each direction. How- ever, when all transition costs are the same, we tested the procedure with three different transition cost values for each network. These depend on a reference value I equal to the average edge cost in the two connected networks, i.e.,

- {i,J}ET UT2 Cij

2n + 2

The three transition cost values are given by al, where a =

1/4, 5/4, and 25/4. When all transition costs are different,

Table I Notation for the Computational Experiments

n: number of nodes d: density of the network a: the factor multiplying 1 to obtain the transition

costs, (a does not appear when the transition costs are not all the same)

H1: number of problems (over 10) where the first heuristic gave the optimal solution

H2: number of problems (over 10) where the second heuristic gave the optimal solution

H3: number of problems (over 10) where the third heuristic gave the optimal solution

Lag: number of problems (over 10) where the Lagrangean relaxation gave the optimal solution

H3 Gap: gap between the best heuristic value (CH3) and the optimal value (CO): Heur Gap = (CH3 - CO)/CO

a v: mean heuristic gap on 10 cases max: maximum heuristic gap on 10 cases Lag Gap: gap between the Lagrangean value (L) and the

optimal value (CO): Lag Gap = (CO -L) CO

a v: mean Lagrangean gap on 10 cases max: maximum Lagrangean gap on 10 cases Time, a v: mean time for the 10 problems (in seconds) Time, max: largest time for the 10 problems (in seconds) Bn largest number of nodes generated in the

branch and bound tree.

their values are randomly chosen in the interval [7/41, F251/41].

Ten different networks were generated for each combi- nation of size (number of nodes), density and transition cost value. Computational results are reported in Table II for symmetric networks and Table III for asymmetric net- works. Each line summarizes results for 10 similar prob- lems. The meaning of the column headings is explained in Table I.

The results reported in Tables II and III clearly indicate that the difficulty in solving the problem increases with a, i.e., with the relative importance of the transition costs, and that it decreases as the network density increases. Note that we also did some experiments on networks with higher density (40% to 70%): for almost all such networks the Lagrangean relaxation provided the optimal solution and no branching was needed. Furthermore, the average computational time was almost always below one second, so that we did not feel the need to include those experi- ments in Table II and Table III. Moreover, considering symmetric or asymmetric costs for the arcs and randomly generated transition costs does not have any significant impact.

On the other hand, Tables II and III show that heuristic H2 performs better than H1 since it gave the optimal solu- tion in 86.25% of the cases (86.94% for the symmetric networks and 85.55% for the asymmetric ones), while H1 gave it in only 4.30% of the cases (4.72% for the symmet- ric networks and 3.88% for the asymmetric ones). The

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DE JONGH, GENDREAU, AND LABBE / 91

complete heuristic, H3, gave the optimal solution in 87.61% of the cases (88.06% for the symmetric networks and 87.15% for the asymmetric ones). The Lagrangean relaxation appears very efficient: it gave the optimal solu- tion in 85% of the cases (84.17% for the symmetric net- works and 85.83% for the asymmetric ones), and when it failed, the maximum number of branch and bound nodes to explore was 7 (root included). The worst resolution time for all the tests was 1003.87 seconds, for an asymmetric network whose number of nodes was 80, whose density was 10% and whose transition cost was of 51/4. It is also inter- esting to note that neither the Lagrangean relaxation nor the heuristics gave the optimal solution in only 30 of the generated problems.

Additional tests were carried out on the backbone net- work provided by BELGACOM, the Belgian telephone company. Composed of 49 nodes and 122 undirected

edges (i.e., 244 arcs) this network has a density of about 10%. We tested the procedures for each of the 260 origin- destination pairs defined on the network. The mean time to solve one problem was of 12.6 seconds, with a maximum time of 60.48 seconds. The first heuristic, H1, found the optimal solution in only 6.3% of the cases, and the second one, H2, in 87.1% of the cases. This result is comparable to the percentage obtained for random networks, and con- firms the good performance of heuristic H2. The Lagran- gean relaxation however gave the optimal solution in only 56.1% of the cases. The maximum gap for the relaxation was 0.286 and the mean gap 0.02537. The maximum gap for the heuristics was 0.52055 and the mean gap 0.01222. Finally, we observed once again the efficiency of the branch and bound scheme: in spite of some larger gaps for the relaxation, the maximum number of branch and bound nodes to explore over all cases was 7.

Table II Tests on Symmetric Randomly Generated Networks (cij = cji)

H3 Gap Lag Gap Time

n d a H' H2 Hj3 Lag av max av max av max Bn

40 0.1 0.25 3 9 10 8 0.0000 0.0000 0.0270 0.1474 5.70 27.79 3 40 0.1 1.25 3 9 9 10 0.0005 0.0053 0.0000 0.0000 0.24 0.28 1 40 0.1 6.25 0 8 8 9 0.0426 0.2410 0.0120 0.1201 3.21 29.77 3 40 0.1 0 7 7 7 0.0993 0.5103 0.0263 0.1402 9.76 55.26 5 40 0.2 0.25 2 8 10 10 0.0000 0.0000 0.0000 0.0000 0.27 0.55 1 40 0.2 1.25 0 10 10 7 0.0000 0.0000 0.0045 0.0246 9.40 44.11 5 40 0.2 6.25 0 6 6 9 0.0420 0.2846 0.0057 0.0568 7.22 67.07 3 40 0.2 1 10 10 9 0.0000 0.0000 0.0016 0.0159 2.14 19.17 3 40 0.3 0.25 1 7 7 9 0.0005 0.0042 0.0012 0.0124 3.38 20.54 3 40 0.3 1.25 0 10 10 10 0.0000 0.0000 0.0000 0.0000 0.23 0.27 1 40 0.3 6.25 2 9 9 10 0.0036 0.0364 0.0000 0.0000 0.24 0.28 1 40 0.3 0 10 10 10 0.0000 0.0000 0.0000 0.0000 0.24 0.28 1

60 0.1 0.25 2 9 9 9 0.0025 0.0246 0.0069 0.0689 6.31 58.28 3 60 0.1 1.25 0 9 9 10 0.0000 0.0000 0.0000 0.0000 0.52 0.55 1 60 0.1 6.25 0 6 6 6 0.0122 0.0490 0.0215 0.1062 35.74 123.19 5 60 0.1 0 8 8 10 0.0105 0.0748 0.0000 0.0000 0.50 0.55 1 60 0.2 0.25 1 8 9 10 0.0024 0.0237 0.0000 0.0000 0.74 2.70 1 60 0.2 1.25 0 10 10 8 0.0000 0.0000 0.0030 0.0242 10.08 55.09 3 60 0.2 6.25 1 8 8 8 0.0204 0.1750 0.0043 0.0333 17.50 110.45 5 60 0.2 0 10 10 8 0.0000 0.0000 0.0075 0.0689 9.43 53.66 3 60 0.3 0.25 0 8 8 10 0.0015 0.0098 0.0000 0.0000 0.55 0.66 1 60 0.3 1.25 1 10 10 7 0.0000 0.0000 0.0026 0.0137 10.37 50.31 5 60 0.3 6.25 0 9 9 9 0.0013 0.0127 0.0010 0.0100 5.15 46.58 3 60 0.3 0 10 10 9 0.0000 0.0000 0.0010 0.0100 4.74 42.79 3

80 0.1 0.25 0 8 8 8 0.0019 0.0143 0.0039 0.0383 19.25 87.88 5 80 0.1 1.25 0 7 7 5 0.0143 0.0555 0.0106 0.0493 80.28 285.34 5 80 0.1 6.25 0 8 8 7 0.0046 0.0308 0.0094 0.0574 34.98 213.66 7 80 0.1 0 8 8 8 0.0028 0.0154 0.0006 0.0047 28.20 177.63 5 80 0.2 0.25 0 9 9 10 0.0006 0.0063 0.0000 0.0000 0.98 0.99 1 80 0.2 1.25 0 9 9 8 0.0190 0.1898 0.0011 0.0066 17.82 80.19 5 80 0.2 6.25 0 9 9 8 0.0012 0.0116 0.0000 0.0000 1.41 3.35 3 80 0.2 0 9 9 8 0.0311 0.3115 0.0016 0.0115 26.64 173.95 7 80 0.3 0.25 0 10 10 6 0.0000 0.0000 0.0012 0.0044 24.33 79.64 7 80 0.3 1.25 0 10 10 7 0.0000 0.0000 0.0038 0.0237 25.48 87.89 3 80 0.3 6.25 0 9 9 8 0.0008 0.0085 0.0053 0.0370 20.42 99.64 5 80 0.3 0 9 9 8 0.0008 0.0085 0.0053 0.0370 20.29 98.98 5

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ACKNOWLEDGMENTS

This work was supported in part by the Institut pour l'encouragement 'a la Recherche dans l'Industrie et l'Agriculture, by the National Fund for Scientific Research of Belgium, by the Natural Sciences and Engineering Re- search Council of Canada and by a Communaute franqaise de Belgique-Quebec grant for scientific cooperation. The authors thank two anonymous referees for their useful comments.

Table III Tests on Asymmetric Randomly Generated Networks

H3 Gap Lag Gap Time

n d a H1 H2 Hj3 Lag av max av max av max Bn

40 0.1 0.25 1 9 10 8 0.0000 0.0000 0.0034 0.0190 4.71 23.18 5 40 0.1 1.25 3 9 10 10 0.0000 0.0000 0.0000 0.0000 0.26 0.38 1 40 0.1 6.25 2 9 10 9 0.0000 0.0000 0.0063 0.0632 2.79 25.54 3 40 0.1 0 7 7 7 0.1046 0.4450 0.0206 0.1185 12.16 78.82 5 40 0.2 0.25 4 8 10 9 0.0000 0.0000 0.0011 0.0105 2.03 17.96 3 40 0.2 1.25 0 10 10 10 0.0000 0.0000 0.0000 0.0000 0.27 0.33 1 40 0.2 6.25 0 10 10 9 0.0000 0.0000 0.0033 0.0329 2.63 23.90 3 40 0.2 0 10 10 9 0.0000 0.0000 0.0059 0.0587 2.17 19.66 3 40 0.3 0.25 0 10 10 10 0.0000 0.0000 0.0000 0.0000 0.27 0.28 1 40 0.3 1.25 0 8 8 9 0.0041 0.0279 0.0001 0.0014 2.65 24.00 3 40 0.3 6.25 0 10 10 10 0.0000 0.0000 0.0000 0.0000 0.27 0.28 1 40 0.3 0 10 10 8 0.0000 0.0000 0.0013 0.0125 2.98 19.83 5

60 0.1 0.25 0 9 9 9 0.0018 0.0182 0.0043 0.0433 6.35 58.71 3 60 0.1 1.25 0 9 9 10 0.0000 0.0000 0.0000 0.0000 0.57 0.61 1 60 0.1 6.25 0 8 8 7 0.0063 0.0603 0.0188 0.0920 30.88 129.19 5 60 0.1 0 8 8 9 0.0126 0.0831 0.0018 0.0176 20.89 204.65 11 60 0.2 0.25 1 9 9 10 0.0003 0.0032 0.0000 0.0000 0.56 0.61 1 60 0.2 1.25 0 10 10 8 0.0000 0.0000 0.0028 0.0152 10.46 52.45 3 60 0.2 6.25 0 8 8 8 0.0156 0.1521 0.0057 0.0479 20.42 110.84 5 60 0.2 1 9 9 8 0.0011 0.0115 0.0100 0.0922 10.71 61.57 3 60 0.3 0.25 0 7 7 10 0.0176 0.1357 0.0000 0.0000 0.56 0.66 1 60 0.3 1.25 0 10 10 9 0.0000 0.0000 0.0015 0.0147 14.07 135.72 7 60 0.3 6.25 0 8 8 8 0.0016 0.0111 0.0057 0.0483 12.94 82.78 5 60 0.3 0 10 10 8 0.0000 0.0000 0.0020 0.0101 9.17 45.54 3

80 0.1 0.25 1 8 9 9 0.0043 0.0434 0.0032 0.0322 11.80 109.24 3 80 0.1 1.25 1 7 8 9 0.0188 0.0901 0.0015 0.0152 29.47 285.84 5 80 0.1 6.25 0 9 9 8 0.0104 0.1043 0.0055 0.0400 110.27 1003.87 23 80 0.1 0 6 6 8 0.0093 0.0404 0.0078 0.0686 44.46 310.93 7 80 0.2 0.25 0 8 8 9 0.0034 0.0212 0.0024 0.0235 9.40 84.86 3 80 0.2 1.25 0 8 8 7 0.0310 0.3102 0.0061 0.0417 41.68 239.20 7 80 0.2 6.25 0 7 7 8 0.0061 0.0513 0.0012 0.0082 18.93 93.98 3 80 0.2 0 6 6 8 0.0385 0.3109 0.0037 0.0259 18.23 82.89 3 80 0.3 0.25 0 9 9 9 0.0000 0.0001 0.0010 0.0099 8.02 71.07 3 80 0.3 1.25 0 8 8 7 0.0030 0.0289 0.0036 0.0243 21.91 95.52 3 80 0.3 6.25 0 9 9 7 0.0003 0.0029 0.0079 0.0424 29.11 99.58 5 80 0.3 0 8 8 8 0.0017 0.0166 0.0034 0.0295 25.51 177.19 5

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