Two-Commodity Multiroute Maximum Flow Problem

Transcription

1 Two-Commodity Multiroute Maximum Flow roblem Donglei Du R. Chandrasekaran August, 005 Abstract We consider a two-commodity multiroute maximum flow problem in an undirected network a generalization of the standard two-commodity maximum flow problem. An efficient combinatorial algorithm, which always guarantees a quarter-integer solution when the capacities are integers, is devised to solve a special case based on a novel extension of the augmentingpath technique. artial results are reported and the difficulties in applying the augmenting-path technique to the general case are explained. Keywords: augmenting-path, multiroute flow, two-commodity Corresponding Author: Faculty of Administration, University of New Brunswick,.O. Box 00, Fredericton, NB, Canada, E3B 5A3, ddu@unb.ca. Research supported in part by NSERC grant Department of Computer Sciences, University of Texas at Dallas, Richardson, TX 75083, U.S.A., chandra@utdallas.edu.

2 Introduction The roblem roviding high-level survivability is critical for computer, telecommunication, transportation and other networks. Traditional network flow models are vulnerable to arc failures because flows are routed along single paths. There has been recent interest in extending the traditional network flow model to a multiroute version that can survive arc failures up to a certain level [, 3,, 6, 7, 3, ]. Consider a network with specified source-destination pairs. In the traditional flow model, flow is sent along single paths between given source-destination pairs, and lost whenever arcs on the path fail. This routing method is vulnerable to even one arc failure. The multiroute flow model differs from the traditional one in that flow must to be routed along a multiroute channel that is, a channel that consists of m arc-disjoint paths rather than a single path from the source to the destination, where m is any given positive integer. Multiroute channels naturally provide protection against arc failures up to m. Existing research has focused on the one-commodity multiroute maximum flow problem [, 3,, 6, 7, 3, ]. The major concern of this research, however, is a two-commodity multiroute maximum flow problem, to be formally defined after some notations. For any finite set S, let S denote the cardinality of S, and let R S + denote the set of non-negative real column vectors with dimension S. Definition. Two-commodity multiroute maximum flow problem (m,m )-MF: Suppose G = (V,E) is an undirected network with vertex set V, edge set E, and edge capacity vector u R E +. There are two pairs of source-destination vertices {s,t } and {s,t } in V, and two positive integers m,m Z +. () An elementary m q -route (q =, ) from s q to t q consists of a set of m q arc-disjoint s q -t q paths in G. An edge (i,j) E is contained in an elementary m q -route if (i,j) is contained in some s q -t q path therein. () An m q -route flow (q =, ) from s q to t q is an assignment of non-negative weights to elementary m q -routes. The value of the an m q -route flow is the sum of the weights assigned.

3 (3) An (m,m )-route flow consists of an m - and an m -route flow, where the sum of weights assigned to the elementary m - and m -routes containing edge (i,j) is no more than its capacity u ij. The value of an (m,m )-route flow is the sum of the values of the m - and m -route flows. The problem is to find an (m,m )-route flow of maximum value. Let q (q =, ) denote the set of all elementary m q -routes from s q to t q. Let q ij (q =, ) denote the set of all elementary m q -routes from s q and t q containing edge (i,j) E. If weight x q p (q =, ) is assigned to each p q, then (m,m )-MF can be formulated as a linear program: max p x p + p x p () p ij x p + p ij x u ij, (i,j) E; () x p 0, p ; (3) x p 0, p. () The above linear program is solvable in polynomial time based on the Ellipsoid Algorithm [0] although not an efficient one. The idea is to provide a polynomial-time subroutine that solves the separation problem of the linear program that, given a solution, either proves the solution is feasible or finds a violated constraint. Herein, we solve the dual program of ()-(). The separation problem of the dual program is the minimum cost m q -route problem (q =, ), where the cost of each arc is the dual variable associated with that arc. We thus need verify only whether the minimum cost m q -route between s q and t q is at least. The problem is solvable in both weakly and strongly polynomial time; it can be viewed as a unit-capacity minimum cost flow problem, for which many efficient algorithms are available (Chapters 9- of []). In ()-() above, we define flows on the elementary m - and m -routes using the route-flow form. In traditional network flow models, flows can also be defined using the edge-flow form. An advantage of edge-flow over route-flow form is that its specification involves only a polynomial

4 number of variables, thus most existing algorithms are based on the edge-flow form. The same is true for the multiroute version that we have proposed. Definition. Consider q {, }. () An m q -route edge-flow of value F q /m q is given by vector f q R E +, satisfying flow conservation constraints (5), capacity constraints (6), and extra capacity constraints (7): f q ij = j V F q, i = s; 0, i s, t; F q, i = t. (5) 0 f q ij u ij (i,j) E; (6) 0 f q ij F q m q, (i,j) E. (7) () An (m,m )-route edge-flow (f,f ) of value F /m + F /m is a pair of m -route edgeflow f of value F /m and m -route edge-flow f of value F /m. Here all flows are algebraic meaning positive value for f q ij implies the q commodity flow in the direction from i to j and negative flow in the reverse direction. Given any m q -route flow x q R q + (q =, ), an m q -route edge-flow f q R E + with the same value is obtained by taking f q ij = p q x q p. The other direction and proof, first established ij by Kishimoto [3], is much more involved: Theorem.3 (Kishimoto [3]) Given any m q -route edge-flow f q R E + of value F q /m q (q =, ), we can construct an m q -route path-flow with the same value. An alternative proof of the above result is given by Aggarwal and Orlin []. The equivalence of edge-flow and route-flow implies that (m,m )-MF can also be modeled 3

5 as a linear program in the edge-flow form: max F + F (8) m m f q ij = j V F q, i = s; 0, i s, t; q =, ; F q, i = t. f ij + f ij u ij (i,j) E; (0) f q ij F q m q (i,j) E and q =,. () The above linear program is of polynomial size and the coefficients of the constraint matrix are bounded by m and m, therefore it is solvable in strongly polynomial time using the technique of Tardos [0]. The algorithm is inefficient, however, because of its poor worst-case complexity. We have devised an efficient, strongly polynomial, combinatorial algorithm detailed below. Related Work Given the lack of research on multiroute flow problems of more than one commodity, we review results for one-commodity cases. The one-commodity multiroute maximum flow problem is introduced by Kishimoto and Takeuchi [5], and Kishimoto et al [7], who have provided a strongly polynomial combinatorial algorithm that solves a sequence of the ordinary maximum flow problems and implies a max-flow/min-cut relationship. Aggarwal and Orlin [], who have given an improved strongly polynomial combinatorial primal-dual algorithm for the same problem, simplified and extended Kishimoto and Takeuchi s theory. They also propose a weakly polynomial algorithm based on binary search, which results in a number of O(log( V U max )), where U max = max (i,j) E u ij. Both Kishimoto s and Aggarwal and Orlin s algorithms are based on reducing the multiroute maximum flow problem to a sequence of ordinary maximum flow problems by parameterization. This is not coincidental: Du and Chandrasekaran [6] show that both algorithms can be unified into a common framework based on Newton s method. They [6, 7] also devise an augmenting-path algorithm by exploiting the combinatorial properties of the problem, the foundation for the algorithm presented here. We note that (m,m )-MF is also related to the standard two-commodity maximum flow problem (,)-MF using our notation. Hu [] first proposes an augmenting-path algorithm, obtaining the following results: () Extremal roperty: there always exists an optimal solution (9)

6 such that one of the two commodities is maximized; () Half-integer roperty: if the capacities are integers, there always exists a half-integer optimal solution; and (3) Max-flow/min-cut roperty: maximum flow equals minimum cut. The main idea of Hu s algorithm is to fix the optimal value F of the first commodity so as to improve the value F of the second commodity without changing F. The edge flows may change, of course, when the first commodity is recirculated. Itai [] presents a strongly polynomial implementation of Hu s algorithm by using Dinic s layer network idea [5]. Sakarovitch [8] has proved that the problem is also solvable by applying transforming variables to the standard single-commodity maximum flow problem. New Contribution The general (m,m )-MF is difficult to attack for the purpose of efficient combinatorial algorithms. In this research, we focus on the special (, )-MF (m =,m = ). An immediate idea is to use the parameterization technique of the single-commodity multiroute maximum flow problem [, 6, 3]. This idea, however, fails here, because it leads to solving a twocommodity flow problem with both sum (0) and individual capacity constraints (), for which no efficient combinatorial algorithm exists. Instead we extend the augmenting-path algorithms of Hu [] and Du and Chandrasekaran [6, 7] to solve (, )-MF. Our algorithm always results in a quarter-integer solution when the capacity function is an integer, but the analysis of our algorithm is challenging because of the lack of an explicit max-flow/min-cut relationship. Finally, we give partial results on (, )-MF (m = m = ), a direct extension of (, )-MF, and explain why the augmenting-path technique fails for this problem. The rest of our discussion is organized as follows. In Section, we solve (, )-MF by presenting an augmenting-path algorithm. In Section 3, we deal with (, )-MF and point out the difficulties in extending the augmenting-path technique to the problem herein. We conclude in Section. 5

7 (, )-MF In this section, a new augmenting-path algorithm is developed to solve (, )-MF in an undirected network. Our strategy generalizes the augmenting-path algorithms for the standard twocommodity (, )-MF [] and for the single-commodity multiroute maximum flow problem [6, 7]. The balance of discussion focuses on edge-flows, so from now on we refer to any multiroute edge-flow as m -route flow, m -route flow, or (m,m )-route flow for any positive integers m and m, which should not in context pose confusion. In Section., we first prove that Hu s extremal property holds for (, )-MF, paving the way for our augmenting-path type algorithm. We describe and analyze our algorithm in Section.; we illustrate the algorithm by example in Section.3.. Extremal roperty Lemma. There exists an optimal (, )-route flow (f,f ) with values ( F, F /) for (, )- MF such that F = F, where F is the maximum value of the first flow when the second is ignored. roof: If not, suppose F < F, and that after ignoring the second commodity, there is an augmenting-path from s to t. Along this path, we can increase a certain amount of -route flow, but cannot locate a path with ( F, F /) because of the presence of a second commodity on some arcs of the path. Let e = (a,b) and e = (c,d) be the first and last such arcs respectively. Case. There is one s -t path passing through both e and e carrying second commodity. Because each unit flow of the second commodity is sent along -arc-disjoint-path, there exists an arc-disjoint s -t path which carries the same amount of -route flow; see Figure (a) for an illustration, where : s a e b c e d t is the dotted line; : s t is the dashed line. Reducing appropriate δ units of the -route flow on each of and allows sending δ units 6

8 s s s a e e b c d t s ' ' e e a b c d t ' t t (a) Case (b) Case Figure : Extremal property for (, )-MF of -route flow along the following path: s a e b c e d t. Case. There are two s -t paths and passing through e = (a,b) and e = (c,d), respectively. We have, in like manner, arc-disjoint paths and ; see Figure (b) for an illustration, where : s e a b t and : s t are the dotted lines; : s e c d t and : s t are the dashed lines. Notice that and cannot be coupled in every (, )-route flow; otherwise, we can adjust the flow along the following paths to increase the total value of F + F /, contradicting the previously held assumption that ( F, F /) is the optimal value. Therefore reducing an appropriate number of δ units of the -route flow on each of,, and permits sending δ units -route flow along the following paths. s a s e c d t ; s a e b t d t. This finishes the proof. Figure shows that this result is not symmetrical. It is therefore crucial to begin by maximizing the first commodity in Lemma.. As is evident, maximizing the second commodity prevents - route flow from being sent: we get 3/ although the optimal value is. 7

9 s s t t Figure : Asymmetry of the extremal property for (, )-MF. Main Algorithm and Analysis First, some preliminary definitions and notations: Definition. Consider any feasible (, )-route flow f = (f,f ) with values ( F, F /), where f is a -route flow with value F for the first commodity and f is a -route flow with value F / for the second commodity. () A forward (directed) graph G + (f) = (V,A + (f)) is defined as follows: arc (i,j) A + (f) if and only if δ + ij = u ij f ij f ij > 0. A backward (directed) graph G (f) = (V,A (f)) is defined analogously: arc (i,j) A (f) if and only if δ ij = u ij + f ij f ij > 0. () An arc (i,j) is tight if f ij = F /. (3) A -path from s to t consists of two s -t paths, not necessarily arc-disjoint. An arc on the -path is called a double arc if it is a shared arc; a single arc otherwise. The shortest -path from s to t contains the fewest arcs, where each double arc is counted twice. See Figure 3 for an illustration of a typical -path. () A bridge of the forward and backward graphs is an arc, the deletion of which leaves no directed s -t path. (5) For any directed graph G = (V,A), and an arc set H A, denote G\H as the graph obtained from G by deleting all the arcs in H. 8

10 s i t (b) j s i j t (a) s i t j Figure 3: Illustration of -path: (a) is a -path, containing two single paths (b) and (c), where (i,j) is a double arc. (c) The following simple fact for the forward and backward graphs is useful in the flow-updating process at Step of algorithm AUG later. roposition.3 Suppose we want to direct a flow value pattern of (0, δ) that is, send δ of second flow without changing the value of the first. Let a be a rational number within the interval [, ].. Each arc in the forward graph admits a flow value pattern of (aδ,aδ), allowing aδ first and second flows to be sent simultaneously from s to t and s to t respectively.. Each arc in the backward graph admits a flow value pattern ( aδ,aδ), allowing aδ first and second flows to be sent simultaneously from t to s and s to t respectively.. We now state the augmenting-path algorithm AUG. Algorithm AUG Input. An instance of the (, )-MF Output. An optimal (, )-route flow Step. Ignoring the second commodity, find a maximum -route flow f with value F. Use any available algorithm for the standard one-commodity maximum flow problem with capacity vector u R E + (e.g. []). 9

11 A S ( f ) (, ) A D ( f ) s t (, ) (a) Forward -path (, ) A S ( f ) A D ( f ) (, ) A S ( f ) (, ) s t (b) Backward -path (, ) A S ( f ) Figure : Illustration of augmenting scheme for (, )-MF Step. Fix the edge flows of the first commodity, and find a maximal -route flow ˆf with value ˆF /. Use any algorithm for the one-commodity -route maximum flow problem with capacity vector û = {u ij f ij } ij E R E + (e.g. [7]). Let f = ˆf and F = ˆF. Step 3. f = (f,f ) is a (, )-route flow with values ( F, F /). Step 3.. If there is no directed s -t path in either G + (f) or G (f), stop with an optimal (, )-route flow f; otherwise denote A T +(f) and A T (f) to be sets of tight bridges of G + (f) and G (f), respectively. If A T +(f) = A T (f) = (that is, no new tight bridges exist), go to Step 3.; otherwise let G + (f) := G + (f)\a T (f) and G (f) := G + (f)\a T (f) and proceed to Step 3.. Step 3.. If an elementary -route in G + (f) exists, find the shortest one. If, on the other hand, at least one bridge in G + (f) exists, find the shortest -path (not necessarily an arc-disjoint one). This process generates a forward -path in the forward graph. Repeat the same for the backward graph to generate a backward -path see Figure for an illustration. Let A +(f) and A (f) be the set of arcs on the forward and backward -paths; merge A +(f) and A (f) to form the augmenting scheme. Step. In the forward -path, A +(f) = A S +(f) A D +(f), where A S +(f) and A D +(f) are respectively the single and double arcs of A +(f) in the forward -path; analogously A (f) = A S (f) A D (f) in the backward -path. Further denote A S T + (f), A D T + (f), A S T (f), and A D T (f) as the tight arcs in A S +(f), A D +(f), A S (f), and A D (f) respectively see Figure for an illustration. Flows on any arc belonging to A +(f) and A (f) are updated using the 0

12 following formulas: (i,j) A S +(f) : fij := fij + δ, fij := fij + δ; () (i,j) A D +(f) : fij := fij + δ, fij := fij + δ; (3) (i,j) A S (f) : fij := fij δ, fij := fij + δ; () (i,j) A D (f) : fij := fij δ, fij := fij + δ; (5) F := F, F := F + δ. (6) In the above δ = min{δ,δ,δ 3,δ,δ 5,δ 6 } (7) where Go to step. δ = min ( (i,j) A S T + (f) ) A D T ( (f) A S T (f) ) A D T + (f) ( F δ = min (i,j) A S T + (f) A S T (f) f ij ( ) δ 3 = min uij f (i,j) A S + (f) ij fij ; ( ) δ = min uij f (i,j) A D + (f) ij fij ; ( ) δ 5 = min uij + f (i,j) A S (f) ij fij ; ( ) δ 6 = min uij + f (i,j) A D (f) ij fij. ) ; ( ) F f ij ; In the algorithm above, Step is self-explanatory. The optimal value of the -route flow can be fixed in the rest of the algorithm because of the extremal property (Lemma.). Step generally yields a maximal (locally optimal) solution for the second commodity unless the -route flow is altered. Step 3 is designed to locate an augmenting scheme by allowing adjustment of the -route flow. This step, if successful, will result in flow update in preparation for Step. Once the flow is updated, further -route augmenting may not require alteration of -route flow. Return to Step is all that s needed; the process is repeated until an optimal solution is found.

13 Elaboration of Steps 3. and 3. may be necessary. The purpose of Step 3. is locating a pair of forward and backward -paths in the forward and backward graphs; once identified, the -paths can be merged to form an augmenting scheme Figure. However, when the pair of -paths is merged, the same arc may occur in both, and we need to guarantee that a tight arc can occur in both -paths only when it is not a single arc in either of the -paths. Otherwise, there exists a tight arc (i,j) in both -paths and it is a double arc in at least one of the pair of -paths. Let the current flows be (f,f ) with values ( F, F /). To increase the flow value of the second commodity by δ, the flow on arc (i,j) is updated by either fij + 3δ, assuming (i,j) is single in one of the -paths or fij + δ, if in both. Because (i,j) is a tight arc, we have f ij = F /, suggesting that extra capacity constraints (7) cannot be guaranteed for the new flow. To find such a pair of -paths, Step 3. adopts an iterative process. First note that bridges in the forward or backward graph must be on every -path from s to t. With any iteration, we therefore update the forward (backward) graph by deleting all tight bridges of the backward (forward) graph. This updating may result in creation of new tight bridges in the graphs, leading to the next iteration. Iteration is terminated only when () there is no s -t path in either graphs in which case the algorithm stops with no augmenting scheme; or () no new tight bridges are created in which case the algorithm proceeds with Step 3. to find the augmenting scheme. Note that the number of iterations is bounded by E, the limit for the number of tight bridges in G + (f) and G (f). Step 3. s method resembles that of Du and Chandrasekaran [6, 7] for the one-commodity multiroute maximum flow problem. As previously remarked [6, 7], simply choosing the shortest -path in the forward or backward graph will not guarantee strongly polynomial implementation, unlike it does with the standard maximum flow problem [5, 8]. Du and Chandrasekaran [6, 7] suggest choosing the shortest elementary -route whenever possible. The idea also works for the two-commodity case (Theorem.7), as will be shown in the proof.

14 We show how finding the shortest -path in Step 3. is a matter of discovering the shortest elementary -route in a modified graph. All bridges in the forward graph must, obviously, be on every -path from s to t. arallel arcs may be used to duplicate bridges for the purpose of constructing a new forward graph, then Suurballe and Tarjan s algorithm ([9]) applied to find the shortest elementary -route in O( E + V log V ). The shortest -path in the backward graph may be found the same way. The shortest -paths in the original forward and backward graphs may be recovered by re-merging the parallel arcs. rovided that the initial flow is feasible, we begin by establishing that the updated flow at any iteration in algorithm AUG is feasible as well. Lemma. Let (f ij, f ij) be a feasible (, )-route flow with values ( F, F /). Then the (, )- flow ( f ij, f ij ), defined in ()-(5), is a new feasible (, )-flow with values ( F, F / + δ). roof: Evidently the flow conservation constraints (5) hold. Capacity (6) and extra capacity constraints (7) follow by examining the following sixteen mutually exclusive and exhaustive cases. Case. (i,j) A +(f), (i,j), (j,i) / A (f); or (i,j), (j,i) / A +(f) and (i,j) A (f):.. (i,j) A S +(f);.. (i,j) A D +(f);.3. (i,j) A S (f);.. (i,j) A D (f). Case. (i,j) A +(f) A (f):.. (i,j) A S +(f) A S (f);.. (i,j) A S +(f) A D (f);.3. (i,j)a D +(f) A S (f);.. (i,j) A D +(f) A D (f) Case 3. (i,j) A +(f) and (j,i) A (f): 3.. (i,j) A S +(f), and (j,i) A S (f); 3.. (i,j) A S +(f), and (j,i) A D (f); 3.3. (i,j) A S +(f), and (j,i) A S (f); 3.. (i,j) A S +(f), and (j,i) A D (f) Case. (j,i) A +(f) and (i,j) A (f):.. (j,i) A S +(f), and (i,j) A S (f);.. (j,i) A S +(f), and (i,j) A D (f);.3. (j,i) A D +(f), and (i,j) A S (f);.. (j,i) A D +(f), and (i,j) A D (f) 3

15 Only Case.3 is shown in its entirety; the others are similar. We demonstrate the capacity constraints (6) in the following. f ij δ + f ij + 3δ u ij ; f ij δ f ij 3δ u ij ; f ij + δ + f ij + 3δ u ij ; f ij + δ f ij 3δ u ij. From the definition of δ, we have δ δ AD + (f) u ij f ij f ij; therefore the first one follows since the original flow satisfies the capacity constraints. The third one follows similarly, since we have δ δ AS (f) u ij + f ij f ij. The other two are obvious. To demonstrate the extra capacity constraint (6): f ij F + δ, the following statements must first be shown: f ij 3δ F + δ; f ij + 3δ F + δ. The first one is obvious. The definition of δ implies δ δ F f ij. Therefore the second one follows in that the original flow satisfies its extra capacity constraints. Next, the correctness of algorithm AUG must be proven. Theorem.5 If the algorithm AUG terminates, it results in an optimal solution. roof: Suppose algorithm Aug terminates for some (, )-route flow f = (f,f ) with values ( F, F /). As remarked earlier, for a given flow f, Step 3. of Algorithm Aug iteratively updates the forward and backward graphs, say, K times before Aug terminates. Denote the updated forward graphs as G +(f),,g K +(f), and let A T + (f),,a T K + (f) be respectively the tight bridges therein; analogously, G (f),,g K (f) and A T (f),,a T K (f) are defined for the updated backward graphs. By the updating rule in Step 3., for any k =,,K, we have G k+ + (f) = G k +(f)\a T k (f); G k+ (f) = G k (f)\a T k + (f).

16 Assume, on the contrary, f is not an optimal solution; then a flow value pattern of (0, δ) is still allowed that is, send δ of second flow without changing the value of the first. This assumption, however, leads to a contraction, as shown below. The following claim is crucial. Claim.6 If a flow value pattern of (0, δ) is allowed, then at least δ units of -route flow must be increased along any tight bridge in A T k + (f), and decreased along any tight bridge in A T k (f) for any k =,,K. Since there is nothing to prove when a tight bridge set is empty, we therefore assume, without affecting the later argument, that A T k + (f) and A T k (f) for any k =,,k. We prove the claim by induction on k. For the basis step k =, because of symmetry, we show only the above claim for tight bridges in A T + (f) at least δ units of -route flow must be increased along any arc in A T + (f). Note that s and t are -arc connected in G +(f) since A T + (f). Consider any bridge (i,j ) A T + (f). Let S be the set of vertices in G +(f) that can be reached from s without passing (i,j ) in G +(f). Then in the original undirected network G, (S, S ) is an s -t cut, the edges of which consists of three mutually exclusive parts: (i,j ); A + = {(i,j) (S, S ) : δ + ij = 0}, that is, edges absent from G +(f) in the construction; A = {(i,j) ( S,S ) : δ + ji > 0,δ+ ij = 0}, that is, edges in G +(f), but going from S to S. 5

17 Let (δi j,δi j ), (δ,δ ), and (δ,δ ) be the total of the - and -route flows across A + A + A A (i,j ), A +, and A respectively. We then have the following constraints for these quantities: δ i j δ; (8) δ A + δ A δi j + δ A + δi j + δ A + + δ A + + δ A δ A δ A = 0; (9) = 0; (0) = 0; () = δ. () Note that (8) follows because (i,j ) is a tight bridge; that (9) and (0) follows from the definition of sets A + and A ; that () and () follows from the non-optimal assumption. Therefore δ i j = δ A δ A + = δ A + δ A + = δ δ i j δ, where the three equalities follow from (), (9) and (0), and () respectively, and the inequality follows from (3). Suppose the claim were correct for up to k. We will now show that it is also correct for k +. Because of symmetry, we show only the above claim for tight bridges in the A T k+ (f) at least δ units of -route flow must be decreased along any tight bridge in A T k+ (f). Note that s and t are -arc connected in G k+ (f) since A T k+ (f). Consider any bridge (i k+,j k+ ) A T k+ (f). Let S k+ be the set of vertices in G k+ (f) that can be reached from s without passing (i k+,j k+ ) in G k+ (f). Then in the original undirected network G, (S k+, S k+ ) is an s -t cut, the edges of which consists of four mutually exclusive parts: (i k+,j k+ ); A + k+ = {(i,j) (S k+, S k+ ) : δ ij = 0}, that is, edges absent from G (f) in the construction, going from S k+ to S k+ ; A k+ = {(i,j) ( S k+,s k+ ) : δ ji > 0,δ ij S k+ to S k+ ; = 0}, that is, edges in Gk+ (f), but going from 6

18 k l= AT l + (f), that is, tight bridges deleted from G (f) in the previous k iterations. For the last part, note that not all arcs of k l= AT l + (f) contribute to the construction to the cut (S k+, S k+ ); for any (i,j) k l= AT l + (f), we have either i,j S k+, or i,j S k+, or (i,j) (S k+, S k+ ), or (j,i) S k+, S k+ ). Obviously first two cases have no contribution; we prove that the last case cannot happen, which implies only the third case contributes. Suppose on the contrary, we have (j,i) (S k+, S k+ ) for some (i,j) k l= AT l + (f). Then δ + ij > 0 and fij = F / > 0. This implies that u ij + fji fji > 0, and hence (j,i) G k+ (f). So we have i S k+ according to the construction of set S k+, a contradiction to the assumption i / S k+. { ( k (Sk+ Therefore the last part only contains A ( k) = (i,j) l= AT l + (f)), S } k+ ). Let (δi k+ j k+, δi k+ j k+ ), (δa ( k),δa ( k) ), (δ,δ ), and (δ,δ be the total of the A + k+ A + k+ A k+ Ak+) - and -route flows sent along (i k+,j k+ ), A ( k), A + k+ and A k+ respectively. These quantities satisfy the following constraints: δ A + k+ δ A k+ δ i k+ j k+ δ; (3) δ A + k+ = 0; () δ A k+ = 0; (5) δi k+ j k+ + δa ( k) + δ δ = 0; A + k+ A k+ (6) δi k+ j k+ + δa ( k) + δ δ = δ; A + k+ A k+ (7) δa ( k) δ A ( k) ; (8) δa ( k) T δ A ( k), (9) where (3) and (9) follow because of the tightness of arc (i k+,j k+ ) and arcs of A ( K) ; () and (5) follows from the definition of A + k+ and A k+ ; (6) and (7) follow from the non-optimal 7

19 assumption; (8) follows from the induction hypothesis. So δi k+ j k+ = δ δ δ A k+ A + A k+ ( k) = δ δ δ A k+ A + A k+ ( k) = δ + δ i k+ j k+ + δ A ( k) δ A ( k) δ + δ + δ A ( k) δ A ( k) = δ. The three equalities follow from (6), () and (5), and (7) respectively; the inequality follows from (3), (8), and (9). This finishes the claim. Armed with the above claim, we are ready to establish the contradiction. At termination, there is no s -t path in either G K +(f) or G K (f). Because of symmetry, we consider only backward graph G K (f). Let S K be the set of vertices in G K (f) that can be reached from s in G K (f). Then in the original undirected network G, (S K, S K ) is an s -t cut, the edges of which consists of three mutually exclusive parts: A + K = {(i,j) (S K, S K ) : δ ij = 0}, that is, edges absent from G (f) in the construction, going from S K to S K ; A K = {(i,j) ( S K,S K ) : δ ji > 0,δ ij = 0}, that is, edges in GK (f), but going from S K to S K ; K l= AT l + (f), that is, tight bridges deleted from G (f) in the previous K iterations. Similar as before, the cut only consists of edges in A (<K) = (i,j) { ( K (SK l= AT l + (f)), S } K ). The situation is similar to that of the induction step of Claim.6, but there is no arc (i k+,j k+ ). We define analogous notations: (δa (<K),δA (<K) ), (δ,δ ), (δ,δ A + K A + K A K AK). By Claim.6, we know the total -route flow along the arcs of A (<K) increases units of flow by a certain amount: δ A (<K). To maintain -route flow, we must decrease units of -route flow along arcs of A + K and A K. However, by the definitions of A+ K and A K, we must also decrease units of -route flow along them. Because all arcs in A (<K) are tight, we can increase 8

20 at most δ A (<K) units of the -route flow along the A (<K) -arcs, therefore the total -route flow across the cut (S K, S K ) is no more than 0, a contradiction to the non-optimal assumption. Last is analysis of the time bound of algorithm AUG. Theorem.7 Algorithm AUG delivers an optimal solution in T time, where T is the time to solve a one-commodity -route maximum flow problem. The algorithm in [6, 7], for example, has T = O( V A + V A log V ). roof: Consider a flow f generated at the beginning of Step 3 at an arbitrary state. Let G + (f) be the forward graph generated by Step 3.. Let α + (f) denote the number of arcs of the shortest -paths in G + (f) (double arcs being counted twice) and β + (f) be the number of arcs in the union of the shortest -paths in G + (f). Let f be the updated flow generated from f at Step, so that analogous quantities of α + ( f) and β + ( f) in G + ( f) (the forward graph generated at the end of Step 3. for flow f) may be defined. Du and Chandrasekaran [6, 7] have shown the following to be true: Lemma.8 [6, 7] (i) α + ( f) α + (f); (ii) If α + ( f) = α + (f), then β + ( f) < β + (f). The above result is equally applicable to the backward graph, therefore both α + ( f) and α ( f) are non-decreasing and at least one of the two decreases strictly with each update of the flow. The worst-case scenario takes the time of two runs of the one-commodity -route maximum flow problem, the solution s overall time bound..3 An Illustrative Example of Algorithm AUG Figure 5(a) illustrates the process of algorithm AUG. The number on each arc indicates its capacity. Figure 5(b) shows what results at the end of Step of algorithm AUG, where the first and second numbers are, respectively, the flows of the first and second commodities, and the third number 9

21 6 3 s s t t (a) 3 s s 6 5 t t (c) 6 (-,) 5 (-,) (0,) (0,) s (,) (-,-) (,) t (0,) (0,) 3 (-,) (-,) (-,) s (e) 6 s (0,0,) (6,0,6) (,0,) (,0,) t (0,0,) (,0,) 3 (0,0,6) (,0,) (,0,) 5 (0,0,) (,0,) (b) 3 (0,0,) (,0,) (,0,) s (0,0,) s s 6 s (,,3) (-0.5,-0.5,) (,-,) (d) (-0.5,0.5,) (,0,) (0,-,) (5.5,0.5,6) (,0,) (0,,) (,,6) (.5,-0.5,) (.5,0.5,) t (.5,0.5,) (f) (,0,) (,0,) (.5,0.5,) 6 t t s 6 t s 3 s s 3 s t t t t (g) (h) Figure 5: Illustration of algorithm AUG on an example 0

22 represents the capacity. The value of the the two commodities is (6, 0). The - and -route flows remain the same after Step. Current forward and backward graphs are shown in Figures 5(c) and 5(d). The dashed line Figure 5(c) and the dotted line in Figure 5(d) are, respectively, the corresponding forward and backward -paths constructed after Step 3. Figure 5(e) represents the augmenting scheme achieved by merging these two -paths, Figure 5(f) the new flows with value (6, /), obtained after Step. Figures 5(g) and 5(h) show the new forward and backward graphs. At this stage, no backward -path in the backward auxiliary graph exists; the optimal solution is shown in Figure 5(f). 3 (, )-MF We report in this section partial results of our study of (, )-MF, an immediate extension of (, )-MF. As may be noted for even this special case, the previously accepted technique for solving (, )-MF is inadequate. In Sections 3. and 3., examples show that extremal and maxflow/min-cut properties are not generally true for (, )-MF and explain why previous techniques should no longer be used to solve the general problem, (, )-MF. In Section 3.3, we demonstrate that the notion of extremal property remains valid if there is just one source or destination for (, )-MF. 3. No Extremal roperty The extremal property crucial for Hu s and our (, )-MF and (, )-MF algorithms no longer holds for (, )-MF. As is shown in Figure 6, no -route flow can be sent from s to t once the optimal -route flow from s to t of value 3/ is presented. However, the optimal flow has a value of, therefore two units of -route flow may be sent from s to t as follows: 0.5 units of flow along double path [s s t, s t ] and 0.5 units of flow along double path [s t t, s t ]. Two units of the second commodity from s to t may also be sent: 0.5 units of flow along double path [s s t, s t ] and 0.5 units of flow along double path [s t t, s t ].

23 s s t t (b) Figure 6: Violation of Extremal property for (, )-MF 3. No Max-Flow/Min-Cut Here we show that the max-flow/min-cut equality does not hold for (, )-MF, given single source s and two destinations t, t. We also define two distinct cuts. Definition 3. Consider any undirected, one-source network with two separate destinations. () A type- cut is a set of edges whose deletion disconnects s from t and t. The cut capacity of a type- cut may be described as the lesser of two quantities: (a) half the sum of the capacities of edges in the cut, and (b) the sum of the capacities of arcs, excluding the largest capacity in the cut. () A type- cut consists of a set of edges whose deletion leaves no double path from s to t or t. The cut capacity of a type- cut is half the sum of the capacities of arcs in the cut. The cut capacities are, clearly, the upper bounds of the maximum flow. Figure 7 shows that the max-flow/min-cut equality does not hold. The maximum flow is 8; the minimum cut capacity is min{, 0} = Extremal roperty for Single Source (, )-MF For (, )-MF with a single source or single destination (i.e., s = s or t = t in L 5 ), the extremal property may be proven.

24 t 0 s 0 t Figure 7: Violation of Max-flow/min-cut for single source (, )-MF Lemma 3. There exists an optimal solution with values ( F /, F /) for (, )-MF with single sources such that F = F where (assuming the second flow is ignored) F / is the maximum value of the first commodity. roof: If not, suppose F < F. Disregard the second commodity and locate the -path from s to t, which will permit δ of the -route flow to be increased along this path. If a path in ( F /, F /) cannot be found, it must be because of the presence of the second commodity on arcs of the path. Without loss of generality, assume this -path contains arc disjoint s-t paths U and L. Let e = (i k,j k ) and e = (c,d) be the last arcs with positive second commodity on U and L respectively. Case. If an s-t path passes through both e and e, let be the coupled path (Figure 8(a)), making the proper augmenting path easily verifiable: : s a e b U t, : s t c e d L t. Case. Assume two different s-t paths and passing through e and e respectively. Let 3 and be the corresponding coupled paths (Figure 8(b)). The following illustrates an augmenting scheme: : s a e b U t, : s 3 t b U t ; 3

25 U s t t ' L (a) Case a c e e b d s 3 U L t (b) Case a c e e b d t Figure 8: Extremal property for single source (, )-MF and This finishes the proof. 3: s c e d L t, : s t d L t. Concluding Remarks Certain open problems remain. First, there is no result at present for multi-commodity cases. Knowledge about the two-commodity problem is still relatively limited; here, we solve a special case. One two-commodity problem for prompt investigation is whether or not the augmenting-path technique can be extended to solve either (, )-MF or (,m)-mf. We believe the first step to be a matter of solving the special (, )-MF with a single source or destination, for which we have demonstrated the extremal property holds. References [] C. C. Aggarwal, and J. B. Orlin, On Multiroute Maximum Flows in Networks, Networks, 39() (00) 3-5. [] R. K. Ahuja, T. L. Maganti, and J. B. Orlin, Network Flows: Theory, Algorithms, and Applications, rentice Hall, Englewood Cliffs, NJ, 993.

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