On Disjoint Path Pairs with Wavelength Continuity Constraint in WDM Networks

Reid Andersen Fan Chung Arunabha Sen Guoliang Xue Department of Computer Science and Engineering

University of California, San Diego. CA 90005 Email: {randerse. fan} @math.ucsd.edu

Department of Computer Science and Engineering Arizona State University, Tempe. A 2 85287

Email: {asen. xue} @asu.edu

Abstract-In a WDM optical network, each fiher link can carry a certain set of wavelengths A =

{,\,. X?, . . . ~ ,AIr.}. One scheme for tolerating a single link failure (or node failure) in the network is the path prolection scheme, which establishes an active path and a link-dajnint (or nude-disjnint) backup path, so that in the event of a link failure (node failure) on the active path, data can be quickly re-routed through the backup path. We consider a dynamic scenario, where requests to estahlish active-hackup paths between a specified source- destination nude pa i r arrive sequentially. I f a link-disjoint (nude-disjoint) active-hackup path pair can he found at the time of the request, the paths are established; otherwise, the request is hlucked. In this scenario, at the time a request arrives, not every fiber link wi l l have al l 1V wavelengths availahle for new cal l estahlishment, as some of the wavelengths may already have been allocated to earlier requests and communication through these paths may s t i l l be in progress. We assume that the network nudes do not have any wavelength converters. This paper studies the existence of a pair o f link-disjoint (node-disjoint) active-hackup paths satisfying the wavelength continuity constraint hetween a specified source-devtination node pair. First we prove that hoth the link-disjoint and ncde-disjoint versions of the prohlem are NP-Complete. Then we f ixus on the link-rlisjoint version and present an apprx imat i im algorithm and an exact algorithm fnr the problem. Finally, through our experiniental evaluations, we denllrnstrate that our approximation algorithni produces near-<tptimal solutions in almost al l of the instances of the prohlenl in a fraction of the time required hy the exact algorithm.

I . INTRODUCTION

Survivability of high bandwidth optical networks has emerged as an important area of research in recent years due to its tremendous importance as a national and international infrastructure for moving large volumes of data. Failure of any part of this infrastructure, either due to natural causes or nialicious attacks is bound to have a significantly barge adverse impact on the economy. In the last few years researchers have been examining survivability issues in WDM networks [I]. [21. 151, [71.

1141. [17]. [ IX] . [19], [20]. 1221. L231, 1271. 1311. 1321, 1331. Two techniques, protectiorl at the WDM layer and rrsrorutioii at the IP layer have einerged as the two nuin contenders for lault management in optical networks [21], [26]. Protection refers to pre-provisioned M u r e recovery (usually hardware based) whereas restoration refers to more dynamic recovery (usually software based) [IO]. [9:1. Between the two sc.hemes. protection is typ- ically lasler and normally offers single link lailure and sinyle node failure protection. The protection schemes can be divided into link protection. purl? protecfior? [231. [24], and purfiul pufk protrcfiorr schemes [I I], [301. Path prolection can be fwther subdivided into sfiured path jirotectiori and dedicated path profectiori [231. [241. In the shared path protection scheme. spare capacity is shared among various backup paths whereas in dedicated path protection, sparc capacity is rcservcd lor each individual source-to-destination path and no sharing of this capacity is allowed [12].

Fix. 1. Active niitl backup paths in a WDM network xus link-disjoint.

In optical networks. cuts in fibers are considered to be one of the most coininon failures, while failures of

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routers are also possible. In the dedicated path protection wavelength scheme. an alternate path is lnaintained in a Stand- We refer to the problem under study in this paper a.. by mode Sor every source-destination path used for the Disjoint putli Problem wider Wuveleiigth CoiitirruiQ data transmission. These paths are referred to the Construint (DPPWCC). Informally, it can be stated as secondap or backup path and the priiiiur? or active follows: Given a set A(e) A of wavelengths available path. respectively. Fig. I shows a WDM network (with on each link e, is it possible to establish an active and a 8 nodes and 10 links) and 2 existing connections. The hackup path hetween a specified source-destination node active and the hackup paths between the nodes a and d pair satisfying the wavelength continuity constraint? are (1.-6-d on wavelength X I and cr-f- t l on waveleng(h A?. AS discussed earlier, for link survivability (node respectively. The active and the backup paths between survivability, respectively), the active and the backup the Ilodes f and h We f-g-11 on WaVekngth and f- paths should be link-disjoint (node-disjoint, respec- ri-h on wAVdWglh XI. ~eSpectiVely. Clearly, in order tO tivcly). Thcrcforc there is a link vcrsion (LDPPWCC) tolerate any single link (node. respectively) Failure. the and a node version (NDPPWCC) of the DPPWCC backup path should not be sharing any fiber link (node. problem. me following approach may be considered for respectively) with its corresponding active path. Thus solving the 1)PPWCC problem. First. consider a network the backup path should he link-disjoint (node-disjoint. comprising of only those l i nk where wavelength XI is respectively) with the active path. available for call establishment. If this network h&s two

In this paper we study the hlbwing problem: Con- link-disjoint (node-disjoint, respectively) paths between sider a WDM network where each fiber can carry Mi the specified source-destination node pair, then we have wavelengths , X w ) . A lightpath WI. a solution for the DPPWCC problem >. If this network 1151 is established between a SoUrce-deStination node does not have two link-disjoint (node-disjoint, respec- pair when a request for such a path arrives and ap- tively) paths between the specified source-destination propriate network resources are available. For reasons node pair. then this process can be repeated for other of survivability. k)lloWing the dedicated path protection wavelengths X2 and X3 etc. If one of them finds the strategy, we try to establish a primary (active) as well as: a disjoint paths, we have a solution for the DPPWCC secondary (hackup) path. If sufficient network resources problem. It may be noted that if any one of the attempts are availahle at the time the request arrives. the active- succeeds. both the active and the backup paths will be backup path pair is established. otherwise the call request establishaj using the SUWE wavelength. However, it may is blocked. We assume that the nodes do not have wave- be noted that it is no[ ilecessuF that the active and length converters and a a consequence. the aclive path the backup paths use the Same wavelength. It may be must maintain the Same wavelength throughout the entire possible to establish the active path using wavelength path. The Same is true for the backup path, although Xz and the hackup path using wavelength XI, as in the it way he using a different wavelengh. In the WDM case of connection between nodes f and h in Fig. 1. research community, this is known as the wuveleiigtli n i s situation is certainly more complex than the one contiriuir); construint. We point out that computing a pair where both the active path and the backup path use of active-backup paths wilh shared proleclion is mOre the Same wavelength. As will be seen in this paper. complicated than the case with dedicated protection. under this sifuation both the I.DPPWCC problem and Therefore the NP-completeness of the prohleill with the NDPPWCC problem are intractable. dedicated protection &es a strong indication of the of this paper is.to specifically tackle hardness of the problem with shared protection. Ihe DPPWCC problems. We show that the disjoint

Consider a situation where a request to establish path coInputation probleIll with wavelength continuity an active-backup path pair arrives at time T. At time constraint is NP-complete, a coIllrllonly held belief in T, several active-backup path pairs may already be in file WDM researckl c(>mInunity. ~ " s forInal cxistcncc. Accordingly. not cvcry fihcr link Will have proof. validates the study of heuristics and integer linear all Mi wavelengths available for the establishment of prograInIning (ILP) formulations. we also design an the new paths. It is possible that link I may have only enhancd version coin1i>onIy used active path wavelengths {XI . A,} available. link 2 may have only first heuristic and present simulation results showing

that our new heuristic finds optimal solutions in 99.8%

available. and so on.

= ( X l ? X a ,

me

'TIE use uf wavclrnqth cwvritrrs is cuiisidcrcd expensive in cu~rrnt WUM networks. Also. as s1mu.n in the appendix. the disjoint pntlls prohlems c m be solved in polynomial lime in WDM networks with wavelength converters. while the disjoint paths problems are

'The existence of a pair of disjoint paths on IRe sume wnwlenglll can he solved in polynomial time usins Suurhalle's algonlhm [281,

NP-complete in WDM networks withoul wavelength convcrlers. [29].

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of the cases tested. while using only a fraction of the time requircd by the integcr linear programnung hased algorithm. ' h e rest of the paper is organized as follows. Section I1 provides some hackground in related areas. Section 111 states the problems in a formal setting. Sec- tion IV presents the complexity result for the prohlems. Sections V and VI provide an approximate and an exact solution for the link version of the LDPPWCC problem. Section VI1 compares the results between the exact and the approximate solutions and Section VI11 concludes the paper.

11. RELATED WORK Although several researchers in the last few years have

published a significant nuuiher of papers on survivability issues in WDM optical networks [I]. [21. [51. [71, 1141. 1161. [ IX] . [191. [22], [23], [27]. to the hest of our knowledge, the topic OC this paper, the complexity of the disjoint paths problem with wavelength continuity constraint. remains open. Although many researchers in the area helieve that the problem is NP-complete [6]. [3]. there is no formal proof available in the literature. A major contribution of this paper is to provide for the first time a formal proof that the prohlein indeed is NP- Complete [XI .

Recently. Hu [ 121 puhlishcd NP-Comnplctcness results related to diverse routing in optical mesh networks [3]. Although at a first glance it inay appear that the problems discusscd in 1121 arc the samc as the problems disc.ussed in this paper. they are in het significantly different. In this paper we consider the case where each link can carry only a certain suhsct of thc wavclcngths, and ask whether a disjoint pair of active-hackup lightpaths satisfying the wavelength continuity constraint can he established bctwccn a specified pair of nodes. In thc prohlein studied in [12]. the logical topology of the network is given as part of the input. This implies that lightpaths havc already hccn established hctwccn the appropriate pairs of nodes. The objective of the problem studied in [I21 is not to try establish a new active- backup lightpath pair. but to use the already esruhlished lighfpatlrs to find link-disjoint active-backup paths he- tween the specified .source-destination node pair. sub,ject to the constraint that failure of a single physical link would not disrupt hoth the active and the hackup paths. Even in the fiber disjoiirt purl! pr06lein where only link failures are considered, the objective is not to estahlish a new pair o f disjoint lightpaths. hut to find a pair of disjoint paths using the existing lightpaths. Moreover, the wavelength continuity constraint. which plays a key role in the prohlerii under consideration in this paper. has no role in the prohlein considered in [121.

In order to compute a pair of link-disjoint active- backup lightpaths. the current literarure uses what is known as the slrorresr tictive porh j r x t (APF) heuris- tic 161. [131. which first computes a shortest lightpath as a candidate for the active path. then finds the shortest lightpath that is link-disjoint from the candidate path. Computational studies show that the AI'F heuristic is quite effective in practice [61. hut there is no perforiiiance guarantee Tor the APF heuristic. 'Thie ILP formulation or the prohlem can he used to find optimal solutions. hut solving an ILP may take exponential time in the worst case. Thus ILPs 'are usually used only to solve problems known to he NP-Complete or problems with unknown complexity.

111. PROBLEM FORMULATION

A WDM network is modeled by an undirected graph G = (V> E: A). where 1/ is the set of vmices . de- noting iiodes in the network; E is the set of edges. denoting links (or optical fibers) in the network: A = (A,> A2,. . . AJTJ} is the set of waveler!gtl!s and A ( E ) C A is the set of wavelengths availahle at link e '. We will use vertices and nodes interchangeably. as well as edges and links. We will use 7 1 and nr to denote the number of nodes and links, respectively, and use W IO

denote the total number of wavelengths a link may carry. Throughout this paper, we will assunie that the network C: is connected. Therefore ni. 2 n - 1. I n our model, each undirectcd link ( U , U ) in the network represents a hidirectiondl link connecting '11 and U. Whenever a link is used hy a connection. it is occupied in hoth directions.

Dejrrifioii I : A lightpath P ( s , t l A) between nodes s E V and t E V on wavelength A € A is an s-t path T ( . S > 1 ) in G which uses wavelength X on every link of path A(S ; t ) , a(s, t ) is callcd the basepath of P ( s , t? A).

Since each link has many wavelengths. several light- paths m y pass through the Same link. The hilure of a particular link e inay affect many existing connections- all the connections whose lightpaths use link e.

Link-Disjoint Path Pair with Wavelength Continuity Constraint (LDPPWCC)

Iiisrunce: A graph G = (V. E ; IV. A), where PIT rep- resent3 the total number of wavelengths a link can carry, A(e) C i l = {XI ~. . . ~ XW} represents the set of

'Nore t i n t A(e) i s a function of tune as the set of ovailahle wavelengths at n pmticulnr link vnrics with time. In this paper. we we interested in the esrnhlishment of a new connection at the time instance T when the request comes in. Therefore we issume that ,\(e) is the set of available wavelengths on link e ;It tllnc T .

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available wavelengths on link c E E: a source node s and a destination node t . Question: Is il possible to eslablish lwo link-disjoint paths from s to t. such thal both the paths satisfy the wavelength continuity conslraint?

Node-Disjoint Path Pair with Wavelength Continuity Constraint (NDPPWCC)

resents the total number of wavelengths a link can carry, A(e) C A = { X I ? . . . ~ X n . } represents the set of available wavelengths o n link c E E ; a source node s and a destination node t. Qiresriorr: Is it possible to cstablish two node-disjoint paths from s to t. such that hoth the paths satisfy the wavelength continuity conslraint'?

Since our objective is to establish only two paths. and each path is required to use same wavelength throughout. at most two wavelengths will be needed for path estab- lishment. When both the paths can be cstahlishcd using the same wavelength. the problem reduces to finding two disjoint paths in a graph. polynonlial time solutions for which are already well known 12x1. 1291. Accordingly, in this paper we concentrate on the scenario where both the paths cannot he established using the same wavelength, Instead of considcring all Mf wavelengths, wc will only consider IWO different wuvelerig/lrs ut U time. We will use XR and X n to denote the two different wavelengths under consideration. This leads lo the following two problems.

fns/Unce; A graph G = (V. E? PV: A). where bV rep-

Link-Disjoint Path Problem with Wavelength Conti- nuity Constraint and 2 Wavelengths (LDPPWCCZW) hisrurrcr: A graph G = (V. E: A). associated with each edge e E E, is A(e) C { X R , X ~ } , indicating the set of frcc wavclcngths (channels) associatcd with link e; a source node s and a destination node t. Question: Is it possible to establish two link-disjoint paths from s to t , such that one of the paths uses the wavelength XR and the other uses AB'?

Node-Disjoint Path Prohlem with Wavelength Conti- nuity Constraint and 2 Wavelengths (NDPPWCCZW) histance; A graph G = (V. E: A). associated with each edge e E E, is A(e) C {Xn,X5} , indicating the set of free wavelengths (channels) associated with link e; a source node s and a destination node t. Qirestion; Is it possihle to eslahlish two node-disjoint paths from s to b. such lhat one of the paths uses the wavelength XR and the other uses AB'?

The following lemma shows that the simplification from I V wavelengths to 2 wavelengths does not lose any I nenerality.

Leinniu 1; LDPPWCC (:an be solved in polyrrottriul time .if arid only if LnPPWCC2 W cut1 be solved it1 poly- riorriinl /itire. NDPPWCC can he solved in poly~iot~~iul t i r w if urd only if NDPPWCL'ZW can he ,solved in polyrioniial tittie.

PROOF. LDPPWCCZW is a special casc of LDPP- WCC. Therefore LDPPWCC can he solved in poly- nomial time only if LDPPWCC2W can he solved in polynomial time.

Now assume that LDPPWCC2W can be solved in polynomial time. To solve LDPPWCC. we can use Suurballe's algorithm lo see if the two link-disjoint paths can be established using the same wavelength A. for some X E A. This can he done in polynomial time since there are at most M i wavelengths and Suurballe's algorithm runs in polynomial time. If such a pair of paths can he found in this way. we have a positive answer to LDPPWCC. If such a pair of paths cannot he found in this way, then there docs not exist a pair of link-disjoint paths both using the same wavelength. Next. we can solve W ( W - 1)/2 instances of LDPPWCC2W. using all combinations of XR and AB with XR; A 5 E A. This process will also take polynomial time, if LDPPWCC2W can he solved in polynomial time. There exists a pair of link-disjoint paths if and only if we can find such a pair for one of the W ( L V - 1)/2 LDPPWCC2W instances. Therefore if LDPPWCC can he solved in polynomial timc if LDPPWCC2W can be solved in polynomial time. We can prove for the node version of the problems

LDPPWCC2W (NDPPWCC2W. respectively) can he viewed as a path finding prohlem where each link is colored using either red. green or blue. where the colors red. green and blue indicate that wavelength XR, Xg. and both. respectively, are available for path cstablishnient on that link. We want to find two link-disjoint (node- joint. respectively) paths from node s to node t such that onc path (callcd the red path) uses only red and green links and the other (called the bhre path) uses only blue and green links. Formally, the problem can be stated as follows:

~

similarly. 0

Link Dis.joint Path Problem in Graphs with Red, Green, Blue Links (LDPPRGB)

his/arice; A graph G = (V. E ) , where each link e E E is colored red. blue or green; a source node s and a destination node t .

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Qursriori: 1s i t possible to establish two link-disjoint paths from s to t . such that one of the paths uses only the red and green links and the other uses the blue and green links'?

Node Disjoint Path Problem in Graphs with Red, Green, Blue Links (NDPPRGB) hrsr~rncr; A graph G = (V. E) . where each link e E E is colored red. blue or green; a source node s and a destination node t . Qur.s/ion; Is i t possible to establish two node-disjoint paths from .$ to t. such that one of the paths uses only the red and green links and the othcr uses the bluc and green links'?

Note that LDPPRGB is equivalent to LDPPWCC2W and NDPPRGB is equivalent 16 NDPPWCC2W. In the following section we will prove that both LDPPRGB and NDPPRGB are NP-complete. Therefore the prob- lems LDPPWCC2W and NDPPWCC2W are also NP- Complete. This in turn implies that the problem LDP- PWCC and NDPPWCC are NP-Complete.

IV. COMPLEXITY ANALYSIS

We will consider a graph whose edges 'are colored by one of the three colors: red. green and blue. Recall that a red path is a path whose links are either red or green, and a hlitr path is a path whose links are either blue or green.

Tlrrorrrri 2: The prohleni LDPPRGB is NP-

PROOF. Clearly LDPPRGB is in W. as one can verify efficiently whether two paths are link-disjoint. In the rest. we will prove that the problem is also NP-hard.

We give a polynomial time reduction from 3SAl to LDPPRGB. Since 3SAT is NP-hard. this will prove the theorem. The reduction maps a 3-CNF formula c,b (an instance of the 3SA1 problem) to an LDPPKGB instance

is satisfiable if and only if G contains a red s-t path p R and a blue s-t path p B such that p R and pB are link-disjoint in G.

We describe the mapping in two steps. while illus- trating the process with the sample 3 S A T instance @ =

Complete:

t ) so that

(zlV~L.2V:i.-()A(z1Vn:2V:c:$ zlV"*V:I:.I)A(:~1 v"2vz:$). Step 1. Let L = {zl 1;. . . :cl> % I } be the set of

literals made from the variables {ZI ~ 2 2 ; . . . ; . % I } in $4. Let C = {Cl!. . . ~ C k } be the set of clauses in $4. Note that we use 1 to denote the number of vxkbles and use k to denote the number of clauses.

We define a graph GI in the following way: Create vertices s and t. For each clause-literal pair (Ci, x i ) .

create two vertices t i i ; , and u i j l and a green edge con- necting them. For each clause-literal pair (ci [WO vertices rii,;o and 'uzio and a green edge connecting them. For each variable %;. create two vertices 'ti,j and 'u j

and red edges ( u j ; 711 jl). (.ti; '11, I j o ) . ( ,u j , IIJ;;,). (I:; ~ r:kjo). as well as red edges ( v ~ , ~ : t i i + ~ j , ) and (V~;O;.II~.+.,,~) for

~ k - 1. Finally we add red edges ( . s 'uI) . ( t i ( ; t ) and red edges (.i:j, q + l ) for j = 1: 2; . . . ~ 1 - 1. The graph G1 corresponding to the sainple fixmula is shown in Fig. 2.

Fig. 2. GI corrrspandiny to the sample 3SAT instance

From the construction of G1 and Fig. 2 we can see the following facts: n e r e are 2' different red s-t paths in GI. Let / l R be any red s-l path. For any variable x ; . /JR

either goes h o u g h all of the vertices corresponding to literal xj and none of thc vertices corresponding to litcral Z j , or gnes through all of the vertices corresponding to literal :E; and none of the vertices corresponding to literal " j .

Step 2. Next we add k + 1 vertices and Gk + 2 blue edges to GI in the following way to obtain graph G2: Add vertices ul>. . . ,uik and blue edges (9: .WO) and

Let Ci be the %th clause and ai be one of the three literals in Ci. If R; is :c j , we add two blue edges ( ~ i i ~ i - l ~ ~ ~ ~ ~ , l ) and (,iii<+i;1); if ai is .%;. we add two blue edges (,IIJ-,: o i j o ) and (~Iu~.~ ~ 7 ~ ~ 0 ) . Therefore (ik blue edges are added in this way. The graph GY corresponding to the sample formula is shown in Fig. 3.

( l l f k : t ) .

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. , . , , . , . . , , , , , , . ,

.... .... ........... -..: .............................. I# ., .................................... it>" I

_._._._,_ Rad e d r s - Green edpp ............ Blue&+

Pis. 3. G3 corrcsponding to the sample 3SAT instance.

Froin the construction of Gz as shown in Fig. 3 we can see the following facts: There are I" different blue ,s-t paths in G2. Let p R be any red s-t path in G2 and p B bc any blue s-t path in Gz which is link-disjoint with pR. For any variable :E?, if p B goes through any of the vertices corresponding to literal :c,. then p R niust go through all of the vertices corresponding to literal 2,. Siniilarly, if p R goes through any of the vertices corresponding to literal E, . then p R must go through all of the vertices corresponding to literal zj.

It is clear that graph GZ can be constructed from formula 4 in polynomial time, since G2 contains 4k1 + 21 + k + 3 nodes. 2k,l+:31 t 1 red edges, 2k l green edges and 6k + 2 hlue edges.

We now prove the desired result that (CZl s, t ) con- tains a pair of link-disjoint red and hlue s-t paths if and only if 4 is sdtisfiahle. Let us first assume that (C2; s, t ) contains link-disjoint paths p R and p', where pK goes through only red and green edges and that that p B goes through only blue and green edges. We will define a truth assigninent o f the variables f :

{ : C I , ~ Z ? ... >zl} { T R U E : F A L S E } sothat4istrue under this assignment.

Let zj be any literal in 4. if p R goes through any of the verticcs corresponding to literal :cr. then p R niust go throuyh all of the vertices corresponding to literal :Cj. In this case, we assign f(zj) = TRUE.

If p R goes through any of the vertices corresponding to litcral Zj, then p R must go through all of the vertices

corresponding to literal xj. In this case. we assign

If p B does not go through any of the vertices corre- sponding to litcral z, or litcral Zj, then p R goes through either all of vertices corresponding to literal xj or all the vertices corresponding to literal 4. In the former case. we set f(z;) = FALSE. In the latter case. we set f(z;) = TRUE. In short, p R goes through the vertices wi and the edges ( i i , i j l , v i j l ) (with f(:cj) = TRCIE) and edges ( i c v 0 ~ uVo) (with f (q) = FALSE) in the truth assignment. Since the blue path p R goes through

with the above truth assignment.

f(.q) = FALSE.

the vertices U J ~ , ~ I I J ~ , . . . ;uiC in that order. 4 is satisfied

. . . . . . . . . . . . . . . . . . . . . 2f I . . . . ......... , ....

- , - , - ! Rodpnth - Blue p l h

Fig. 4. Link-disjoint R B paths to truth nssigluncnt satisfying d.

Fig. 4 illustrates the truth assignment corresponding to the blue path p R : S - W O - ~ I I I O - I : I L O - ~ ~ U I - ~ I , ~ I O - ~ I ! ~ ~ O -

U ~ I L ~ ~ ~ -,ti321 - w s - ~ i i ~ ~ 1 - ~ t ~ ? ~ l - ~ u ~ 4 -t and the red path p H : S - u l -'if,111-'lJ1,, - 1 1 ~ ~ ~ - 1 : 2 1 1 - 7 1 . 3 1 ~ - 2 1 ~ ~ ~ - 7 1 ~ ~ ~ 1 ' u ~ 1 1 - u 1 - 7 i . 2 -

U , ~ ~ - 1 ! , ~ ~ - ' 1 i ~ ~ ~ ~ - 1 1 ~ ~ ~ - 7 i ~ ~ ~ ~ - v ~ ~ ~ - 1 l ~ ~ ~ - ' 1 1 ~ ~ ~ - v ~ - ' i l . g - ' 1 ~ ~ 13 I - ~ ! ~ ~ ~ - ~ t ~ ~ ~ ~ - i ~ ~ ~ ~ - ~ ~ ~ , ~ ~ : ~ ~ - w ~ ~ ~ - i i . ~ y ~ - ~ r ~ ~ , - i : ~ ~ - t . Since p R goes through the vertices '11,110, ~ 1 1 0 . 71210. V ~ I O which correspond to the literal :El. we set f(m) = FALSE in the truth assignment. Since p' goes through the vertices ' U : , J I , ~ 3 2 1 , i iqa l . 'tidZl which correspond to the literal z2, we set f(:cz) = I'RCIE in the truth assignment. For this example. tiR does not go through any of the vertices corresponding to the literals z3 and 63. We can assign f(s:,) as don't care in the truth assignment. Since p R goes through the vertices corresponding to the literal :c3. we set f(:ca) = FALSE in the truth assignment.

One can easily verify that d evaluates to II'HCiE with this truth assignment.

f be a truth assignnient thal satisfies @. We will show that there exist a red s-t path p R and a blue ,s-t path p B which are link-disjoint in G2.

Note that any red s-t. path will contain the edges (.S.II.I), ( v l > t , ) , ( U J - ~ J I , , ) for j = 1:2 path from 'ti, LO for j = 1>2: ... : 1 . II f (z1) = FALSE, we define the segment of p R from 'uj-c~j

to be s-ii 1-11 11 1-'i:1,1-i~211-.1:21 I - . . . -iic11-u,;~1 -11, . Other- wise. we define the segment of p R from I I ; - ' u ~ to he s- '1i.1-'11110''u11"-11.~10-'1I~1"- - u , - ~ O - t i c l ~ - ~ u ~ . We define the blue s-t path p B to contain the blue edges ( s w O ) , (,iuC: t ) and one green edge and two blue edges for each clause in the sample 3SAT instance $. Let Ci be the ith clause. Since C: is TRUE under truth assignment f . at least one of the three literals of Ci is TRUE. Let :cj ( Z j , respectively) he one suc.h literal. The blue path p B contains the blue edges (wi-t)t i ,*j~). ( .u i j~ ;wi) and the green edge ( , L I Z J ~ ~ , U ~ ~ ~ ) (the blue edges (wi-,+ij"), ( .ui jo , , I & ) and the green edge ( i ~ i j ~ ? v i j ~ ) , respectively).

Consider the sample the sample 3SAT instance f#J again. The truth assignment j ( r 1 ) = TRUE. f(.%2) = TRUE, f(m) = FALSE satisfies 4. Refer to G2 in Fig. 3. the corresponding blue path is p B : .s-11~0-11~11~-'1!1;~0-u:1 -U291-V221 -u12-113 11-4r;j1 l - ' u l ~ - ' l I , ~ ~ 1 - vTz1-u!d-t. and the corresponding red path is pR:

'If L 2 0 - 1 : ~ ~ ~ - l 1 ~ ~ ~ - 7 : ~ ~ ~ - ' i 1 5 2 o " u : ~ ~ ~ - ' l 1 , * ~ - l I ~ ~ ~ - 1 ! ~ - ' 1 1 : ~ - ' 1 , 1y,-

TO show the converse. assu11ie d is sdlistiabk and let

s-11, l-ll, 1 10-7: 1 1"-'!12 10-'l,'2 10-'lL( 1O-'L'3 1o-'U4 10-'U110- L! 1-112-

1113 1 -'l1231-1!2Sl -'I1331 -'L'3:$ 1 -'lf,4:31 -Vi131 -113-t.

Since p B docs not go through any vcrticcs com- sponding to literal whose value is F A L S E and the 11,i-?i.i segment of p R goes through only those vertices corrcsponding to a literal whose value is not TRUE, p B and p R are link-disjoint. This proves that LDPPRGB is NP-hard. Since LDPPRGB is both in NP and NP-hard, it is NP-Complctc. 0

77ieoruiL 3: The problem NDPPRGB is NP- Complete. PROOF. Again we prove this using reduction from 3SA'll Given a 2-CNF formula d. we constmct the Same graph G2 as in the proof of Theorern 2. It is clear from the graph construction that a red s-t path p' and a blue s-t path p A are link-disjoint if and only if they are node- disjoint. It then follows the argument of the proof of Theorem 2 that NDPPRGB is also NP-hard. It is easy to verify whether Iwo paths are node-disjoint, Therefore

Given that the problem is NP-Complete. it is appropri- ate to study heuristic algorithms and ILP solutions. In the rest of this paper. we will concentratc on LDPPWCC and

NDPPRGB is NP-Complete. 0

study both heuristic algorithms and ILP based optimal solutions.

v. APPROXIMATE SOI.UTION USING HEURISTIC TECHNIQUE

We consider the situation that some active-backup paltis are in existence while a new conneciion request with source node s and destination node t arrives. Recall that each link can carry up to II' channels (W different wavelengths). A channel is free if it is not used by an active or a backup path. A channel is active if it is used by an ac.tivc path. A channel is reserved

of the network status.

is described in the following.

Algorithm 1 APF (G, s, t) step-l Find an .s-t lightpath AP on free channels using

the minimum number of links. step2 Let G' be a copy of G. Remove every active

or reserved channel. Remove every free channel using a link on AP.

step3 Find an s-t lightpath H P in G' using the mini- nium number of links.

Step-4 IF both A P and B P can he found, output the two paths. Otherwise, the request is blocked.

a backup path. we aSSullle that we haVC full knowledge

The coininonly used active path tirst (APm heuristic

The advantage of heuristic APF is that it is simple and runs fast. It the activc path .4P is chosen correctly. the heuristic will find the link-disjoint path pair. However, there are cases where the heuristic APF inay fail to find a pair of link-disjoint paths. even when such a pair exists. Fig. 5 shows an example for which APF Fails.

_._._._._ - ., ,,... ...,,

Link with only A ) availahle Link with both A1 and A. available Link with only A? availahle

Pig. 5. An example for which APP foils

In the example shown in Fig. 5 . APF will find the path s-z-u:-t (on XI) as the candidate for the active path AP. However. after removing the links along this candidate active path, s and t is no longer connected in

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the network. merefore the heuristic APF cannot find a pair of link-disjoint s-t path. Clearly. we can use .S - ' I - IJ -

Irr-t (on X1) as the active path and use s - q p z - t (on XI) as the backup path. In the following, we will present an enhanced active path first heuristic (called AE'FE) which can avoid pitfalls like this. In the description of APFE. M represents a very big riurriber which is greater than any integer and ,i . M > j . M for integers i > j .

Algorithm 2 APFE (G, S, t) Step-1 Find an .s-t lightpath 4 P on free channels using . .

the minimum number of link?. Set cost = m. step2 Let C' he a copy of C. For every active or

reserved channel, assign a cost of 00. For every free channel on a link on A P , assign a cost of M . For every other free channel, assign a cost of 1.

Step3 Find a minimum cost s-t lightpath B P in G'. step-4 IF ilP and B P are link-disjoint THEN

A P and BP are the active and backup paths

the connection request: STOP

STOP failure

Set cosl to the cost of B P and let AP represent BP. goto Skp.3.

for

ELSEF the cost B F is at least cost THEN

ELSE

ENDIF

We note that the heuristic APFE will find a pair of link-disjoint paths whenever the heuristic APF can find such a pair. However. APFE may find a link-disjoint path pair even when APF fails to do so. A P E tries to reduce the number of shared links in the pair of active-backup paths iteratively. When that number cannot be reduced during one iteration, the heuristic stops and blocks the connection request. We note that the heuristic A P E is not guaranteed to find a pair of link-disjoint paths when one exists.

For the example network shown in Fig. 5 . APFE will first find the path s-:c-,w-t (on XI) a the candidate for the active path A P . It then find the path . s - ~ ~ r - i ~ - u i - t (on X1) as a candidate for the backup path BP. Note that A P and BP shares a common link ( , I & t ) . The heuristic then discards A P and finds s-2-y-z-t (on XI) as the active path AP. This time. AP and B P are link-disjoint so APE finds the pair of link-disjoint paths successfully. However. there are examples for which APFE fails. For performance studies. we present an integer linear programming (KP) formulation in the next section and present simulation results comparing APF, A P E and ILP on sample network topologies.

VI. EX:\CT SOLUTION USING ILP

To evaluate the performance of the the heuristic al- goriiluns. we lormulate he lollowing ILP Cor LDP- PWCCZW with wavelengths XR and X g and source- destination nodc pair s and 1.

For each undirected link ( . t i 1 .IJ) E E , define R(rr, 1 ) ) = 1 if XR is available on ( U ? I ) ) and R ( ~ L , . 7)) = 0 otherwise. Similarly, define B(u; 71) = 1 if X g is available on ( 1 1 ; ,(I)

and U(?,; , I ) ) = 0 otherwise. Define a [unction 6,t(.) on V such that dZt(s) = 1. &(t) = - I , 6,,(v) : 0 for every othcr nodc 71 E V . For cach undircctcd link (.u, I ! ) E E. we associate four 011 variables ~ ( , I I . : ,ti), 6(,1r, , U ) , r(a, ( 1 ) .

and b(.u, .If,). Note that r(.u, IJ) and r(.ci, 71,) are different although ( . u : u ) and ( v : u ) denote the same undirected edge. The Same can be said about ~ ( P L : v ) and 6 ( 0 , PI).

We will use the r(, tf .: ,f i) variables to define a red path from s to t and use the 6(71>71) VaridblCS to define a blue path from s to 1. The integer linear programming formulation is given in the following:

(P): mill 1 ,r(71, v) +r.(v:ri) + h(t1, 0 ) + b('fJ; I f . )

I t r .c l€E

r ( u ; 1 1 ) + P(U> 11) 5 R(,u, a) , V(.ci,, V ) E E; b ( , t r ; U ) + b ( ' U : '11) 5 B(u; IJ), v(lr> 7J) E E?

T ( u . L ! ) + , ~ ( . I L ; ' I ~ ) + 6 ( ~ . ~ : ) + ~ ( , I ! > ' I I ) 4 l>V('it,> U ) E E , , r ( ~ 1 1 . ~ ~ ~ 1 ) ~ , r ( . u , 7 ~ . ) , 6 ( . 1 1 ~ I L ) , ~ ( , I J > I ~ , ) 2 0> V ( ~ [ ; , I J ) E E.

r(ti,!,V)> h(tL;,fJ) E {Ol I}.

'lhc following is an explanation ofthc ILP formulation (P). Recall that R(rr,,t!) is I if and only if wavelength XR is available on link ( & j ) and B('ir> 11) is 1 if and only if wavelength ,\E is available on link ( i l j ) . Therefore a feasible solution to (P) corresponds to a red s-t lightpath on wavelength XR. defined by the nonzero variables r(,u,tJ), and a blue s-t lightpath on wavelength Xg. defined by the nonzero variables 6(7r; e) .

The constraint r (u ) w)+r(u, .ti,) 5 R(,tr: , U ) ensures that the red path can only use the links on which wavelength XR is available. The constraint h(rr , 71) + b(t!; 11,) 5 B(u, U!) ensures that the hlue path can only use the links on which wavelength X g is available. The constraint r ( n , 71) + I ( I ! > 1 1 ) + h(,ir, 11) + ~ ( I J , , t i ) 5 1 ensures that the red path and the blue path are link-disjoint.

Note that we have to solve up to \,li . (W - I) instances of (P) to solve problem 1,DPPWCC. since we need to loop over all possible wavelength conlbinations.

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The cases where A,, = A H can be solved using the polynomial time algorithm of Suurhalle [2XJ [291.

We choose to use this ILP formulation because (1) it is simpler to understand and to implement since only two wavelengths are involved; (2) [he sinaller the ILP. the easier to solve. as ILPs inay require exponential time to solve.

V11. KESLILTS AND DISCUSSION

To evaluate the effcctiveness of the heuristics and ILP based algorithms, we have compared them using the 20- node Arpanet topology (see Fig. 6) and the 33-node Italian National Network topology (see Fig. 7). All three algorithms are impleilienled on a SUN Ulua iliachine using prograinniing the language C and the optimization package CPLEX 6.5.

2

Fig. 6 . 20-Node ARPANET Topology.

For each network topology. we construct Y WDM networks, obtaincd by having 5 . 10 and 20 wavclcngths per. link and network loads of 25%. 50% and 75%. By a network with ,5096 load, we mean that 50% of the channels in the network are marked as cither active or reserved. 'Clearly, the higher the network load, the high the blocking probability. For each of the 18 WDM networks, wc considcr all possible sourcc-dcstination node pairs and try to find a pair of link-disjoint source- destination lightpaths using the optiinal algorithm and the two heuristics. For example. for the 9 WDM net- works generated from the Arpanet topology. we consider all I!J0 = 20*(20-1)/2 possible source-destination piiirs as the connection requests.

For each connection request. there are four possible outcomes: . NNN: meaning none of the three algorithms found

a pair of link-disjoint paths; . NNY: meaning both heuristics failed to find a pair of link-disjoint paths but the I12P exact algorithm found one;

v Fig I . 33-Node Italian National Network.

NYY: meaning the first heuristic Filed to find a pdir of link-disjoint paths hut the second heuristic and the ILP exact algorithm both found one: YYY: meaning all three algorithms found a pair of link-disjoint paths.

Simulation results for networks corresponding to the 20-node Arpanet are shown in Table I. Simulation re- sulrs for nelworks corresponding to the 33-node Italian National Network are shown in Table 11.

As an cxamplc, thc cntrics in thc first row of Table I have the following meaning (from left to riyht): Maxi- m u m number of wavelengths per link is i: The network load is 25% The average time (in milliseconds) used by APF is 0.632: The average time (in nulliseconds) used by APFE is 1.158; The average time (in milliseconds) used by ILP is 17.579; In 176 cases. all thrcc algorithms were successful: In 1 I cases, "FE and ILP were both successful while AF'F was not; In 2 cases. ILP was successful whilc AFTE was not; In 1 case, none of thc three algorithms can find a path pair, hecause there was no pair of link-disjoint paths; Total number of connection requests was 190.

From Table I, we can see that A P E found a pair of disjoint paths in 26 of thc 28 cases for which APF failed. From Table 11. we can see that APFE found a pair of disjoint paths in 7:1 of the 81 cases for which APF failed. This shows that APFE is noticeably better than APF, while having a similar running time.

One can see that A P E uses only a fraction of the time used by ILP. Among [he 6Kj2 test cases shown in

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TABLE I SIMULATION RESULTS ON ARPA-NET

Wavelmglhs "er l i n k

'FABLE I1 SIMUL.4TION RESULTS ON ITALIAN NATIONAL NETWORK

Network Load(%) APF I APFE I ILPrexnct) I YYY I NYY 1 NNY I NNN "f C.ases

Average Time ( i n milliseconds) I Statistics for APE APW snd ILP Total number

Lhe two table$, APFE hiled to find a pair of p a h when one existed only 10 times. In other words. APFE found optinial solutions in 99.8% of the cases. while using no more lhan 6 milliseconds in any particular case.

VIII. CONCLUSION

In this paper, we presented a rigorous proof that finding a pair of link-disjoint or node-disjoint paths with thc wavclcngth continuity constraint in a WDM network is NI-Complete. To the best of our knowledge, this is the first rigorous proof for this problem. We then presented an enhanced active path first heuristic and an ILP hased algorithm. Computational results show that our enhanced active path first heuristic outperforms the commonly used active path first heuristic and finds optimal solutions in almost all cases, while spending only a fraction of the time uscd by Lhe ILP based algorithm. Since finding a path with dedicated protection is NP-Complete. wc bz- lieve that the inore complicated problem of finding a path with shared protection is also NP-Coniplete. Currently we are investigating the computational complexity of shard backup path provisioning and effective heuristics for that problem.

ACKNOWLEDGMENT

We would like to acknowledge the assistance of Bao Shen for the 1I.P implementations and Bin Han and kakesh Banka for the heuristic implementations. The research of Fan Chung was supported in part by NSF Grant$ DMS-0100472 and ITR-0205061, The research

ANI-0312635 and ARO grant DAADIY-00-1-0377. Of GUOliang XUe Was SUpporled in pan by NSF ITR grant

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APPENDIX

In this section, we will show that the problem of con- puting a pair of link-disjoint (or node-disjoint) active- backup paths without the wavelength continuity con- straint in a WDM network with wavelength convert-

Let G = ( V > E : A ) he the WDM network with n nodes, nr links, and a inaxiuiuui of I I ' wavelengths per link. Define thc u~avelerrgrh rrriiorz graph 6 of G by deleting every link in G for which there is no aVdilahk wavelengths. The wavelength union graph of the network in Fig. 8 is shown in fig. 9. Basically. a link of G is a link if the link has at least one available .wavelength.

CTS [25] at the network nodCS iS corllpUVdtiolUdlly easy.

ivI Fis. 8. A WDM Network.

Fig. Y. Wavelength union graph of the network in Pig. 8

Clearly G can he constructed from G in 0(11'h) time. We can then use Suurhalle's algorithm [28]. [29] to compute a shortest pair of link-disjoint (or nodc-disjoint) palhs connecting s and t (or to confirm the nonexistence of such a pair) in 6. If such a pair does not exist,

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lhen clearly there is no link-disjoint (node-disjoint) path pairs connecting s and t in G. Let ?rl and ~g bc the shortest pair of link-disjoint (or node-disjoint) s-t paths. For each link on T < or A.L. we can assign an available wavelength. Therefore 7 ~ 1 and h2 can be used as a pair of active-hackup paths in a WDM network with wavelength converters at the nodes.

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