Network Flow Problems Luis Goddyn, Math 408

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1 Network Flow Problems Luis Goddyn, Math 48 Let D = (V, A) be a directed graph, and let s, t V (D). For S V we write δ + (S) = {u A : u S, S} and δ (S) = {u A : u S, S} for the in-arcs and out-arcs of S respectiely. For V we write δ + () instead of δ + ({}). Ma s, t-flow Problem An s, t-flow is a rational-alued ector = ( a : a A) such that (δ + ()) (δ ()) = for all V (D) {s, t}. (As usual, (δ + ()) is a short notation for a δ + () a.) Let u = (u a : a A) be a capacity function. An s, t-flow is feasible if a u a for all a A. The alue of an s, t-flow, denoted al(), is the quantity (δ + (t)) (δ (t)) = (δ (s)) (δ + (s)), which is the net rate of flow from s to t for. The maimum s, t-flow problem is do determine, for a gien capacitated network (D, s, t, u), a feasible flow of maimum alue. This may be written as a linear program. (L) ma (δ + (t)) (δ (t)) (δ + ()) (δ ()) =, V (D) {s, t} a u a, a A(D) Additionally, one may require that be an integral solution. Let S V (D) be such that s S and t S. Then S is called an s, t-cut. It is intuitiely oious (and easy to proe) that the alue of any feasible s, t-flow in D is bounded aboe by the total capacity of all arcs in δ (S), which we write as cap(s) = u(δ (S)). That is, for any s, t-cut S and any feasible s, t-flow, 2 Ford-Fulkerson Algorithm al() cap(s). (). Input an instance (D, s, t, u) of the maimum s, t-flow problem. Initialize the flow ector to the zero flow =.. Form the auiliary directed graph H = (V, B) with V (H ) = V (D). For each arc u A(D) we add the arc u to B if u < u u, and add the arc u to B if < u. 2. Find an s, t-path P in H. If none eists, then go to step 4.. Augment the flow along P as much as is possible. That is, we replace each u by u + ɛ if u P u ɛ if u P otherwise u Where ɛ is the largest possible rational number such that the new flow is still feasible. 4. Let S V be those ertices which are reachable from s in the final auiliary directed graph H. Output the and the subset S and STOP. It was shown by Ford and Fulkerson (956) that for any finite network D with rational capacities, this algorithm must eentually stop. Howeer, it is possible that this algorithm can take an eponential number of steps before it stops. Een worse, if the capacities u a are allowed to be irrational, it is possible that the algorithm does not stop at all! Fortunately, J. Edmonds and R. Karp found (in 972) that by always choosing P in step 2 to be an s, t-path haing the fewest number of edges, the algorithm stops after a polynomial number of iterations.

2 Suppose that the algorithm stops with and S. Then eery arc a δ (S ) is saturated ( a = u a), and eery arc b δ + (S ) is empty ( b = ). Thus al( ) = cap(s ). In iew of inequality (), the fact this algorithm works proes the following Ma Flow Min Cut theorem. Theorem For any network (D, s, t, u) we hae ma{al() : is an s, t-flow in D} = min{cap(s) : S is an s, t-cut in D} Secondly, if u is integer-alued, then eery flow found in step is integer-alued. It follows that there eists a maimum s, t-flow which is integer alued. Theorem 2 For any ma flow problem (D, s, t, u), if u is integer-alued then there is an integer alued optimum flow. Thus the integer program associated with (L) is no harder to sole than (L) itself. Bipartite matching is closely related to network flows. For eample, let G be a bipartite graph with ertices B R, and suppose we wish to find a matching in G haing the largest possible cardinality. We can sole this problem either by modifying it into a Minimum Weight Bipartite Perfect Matching problem, and using a preiously presented algorithm, or by modifying it into a Ma Flow problem as follows. Let D = (V, A) be a directed graph obtained from G as follows. First we direct each edge in G from B to R. Then we add two new ertices s and t, and add all arcs from s to the ertices in B, and add all arcs from the ertices in R to t. All arcs a A(D) are gien capacity u a =. We use the Ford-Fulkerson algorithm to find a Ma Flow for (D, s, t, u). By Theorem 2, we can assume is integer alued, and thus is,-alued. Let M be those arcs from B to R which get alue in. It is easy to see that M corresponds to a matching in G and that the M = al(). This correspondence is reersible, so M must yield a maimum matching in G. Min Cost Flow Problem Here we hae a directed graph D = (V, A), a capacity function u = (u a : a A), a cost function c = (c a : a A), and a demand function b = (b : V (D)). A flow is feasible if at each erte V, (δ + ()) (δ ()) = b. Thus a erte is a source if b < and a sink if b >. Note that no feasible flow eists unless V b =. The cost of a flow is c = a A c a a. The Min Cost Flow problem for (D, u, c, b) is to find a feasible flow of minimum possible cost. This is clearly a linear program. (L2) min c (δ + ()) (δ ()) = b, V (D) a u a, a A(D) We may write (L) and (L2) more compactly as follows. Let M be the erte-arc incidence matri for D. That is, the rows of M are indeed by V (D), the columns are indeed by A(D), so M = (m a : V, a A). Here if is the head of a m a = if is the tail of a otherwise. Let M be the matri obtained from M by deleting rows s and t, where s, t are ertices of D. Let I be the identity matri. We may write the aboe linear programs as follows. (L) ma al() (L2) min c M = I u M = b I u 2

3 4 The Dual LP We write the dual of (L2) using dual ariables y = (y : V ) and z = (z : A). (D2) ma V b y b A u z y + y w z c, for all A z The student should practice deriing (D2) from (L2). Economically, we interpret y to be the ector of prices of the commodity being shipped. That is, to buy a unit at erte one would hae to pay y dollars. We calculate, for each arc, a new quantity c := c + y y w called the reduced cost of the arc. The reduced cost c is thus the net cost incurred by buying a unit at, shipping it along (at a cost of c dollars), then selling it again at erte w. Using c = (c : A), we can simplify an optimal solution (y, z) to (D2) as follows. Since u, the objectie function of (D2) is maimized when each z is as small as possible subject to z and z c y + y w = c. Thus there is an optimal solution (y, z) with z = ma{, c }. z * u c c z ersus c * ersus c in an optimum solution. For such (y, z), the complementary slackness conditions for (L2,D2) are:. If >, then y + y w z = c (which is equialent to z = c and c ). 2. If z u > (that is, if c < ), then = u. We rewrite these and include the feasibility constraint u.. If c >, then = ( is an unprofitable arc) 2. If c <, then = u ( is a profitable arc). If c =, then u ( is an equality arc). These hae a reasonable economic interpretation, if c > (so shipping anything along is unprofitable), then should be empty. If c <, then it is profitable to ship more along u, thus u should be capacitated (otherwise you could sell more units at a lower price at w). If c =, then the quantity flowing along does not affect one s profit. If all three conditions hold at all, then the commodities are being transported at maimum efficiency. Here is an eample of an optimal flow and dual solution y. Note that we omit z since it can be calculated from y. To test optimality, we simply calculate c for each arc and check that, 2, and hold for the pair (, c). * u,-4 < < < < 2 < < 6, < < 2,8 c c

4 5 A Primal Dual Algorithm for Min Cost Flow With input (D, u, c, b) this algorithm stops with a min cost feasible flow, and a price ector y such that reduced costs c satisfy, 2 and aboe, proided c is conseratie, and a feasible flow eists. Inariants: At any point, we hae a flow = ( : A) and prices y = (y : V ) such that complementary slackness conditions, 2, and hold for all arcs A, where c = c + y y w are the reduced costs determined by y. Plan: We will repeatedly modify and y, maintaining,2 and, until we obtain primal feasibility. That is, we stop when, for each V, 4. (δ + ()) (δ ()) = b. At this point, is a min cost flow, by complementary slackness. Algorithm Summary:. Input points (D, u, c, b). Find initial and y satisfying, 2 and.. While 4 is not satisfied, do the following. (a) (-change) If there is an -augmenting path P of equality arcs, then augment along path P, until either an arc becomes full or empty, or either the -source or the -sink of P becomes feasible. (b) (y-change) Otherwise, find the set R of nodes reachable from the -sources. Increase y for all V R until some arc bumps in condition or Output (, y) and stop. Algorithm Details: There are many details and undefined terms to describe. An arc is an equality arc if the reduced cost c =. Step : Set =. If no arc-cost c is negatie, then set y = (why does this work?). If c has negatie entries, but is still conseratie, then we can find such a y ia a shortest path algorithm (details omitted). Step : First we make an auiliary graph H,y = (V, A ). For each arc A we put A * if < and c = u w A if > and c =. Thus A has or 2 arcs for each equality arc in A: a forward arc, if uncapacitated a backward arc, if nonempty. We present on the right the preious eample (D, u, c, b,, y) with one -alue and one y-alued changed (in boldface). Also drawn is the corresponding auiliary graph H,y. Second we find and indicate on H,y the -sources and - sinks. Each erte V is either an -source, an -sink or is -feasible depending on whether (δ + ()) (δ ()) is greater than, less than or equal to b. That is, is an -source if it not enough fluid is currently leaing, and is an -sink if it not enough fluid is currently entering. We write the alue of b (δ + ())+(δ ()) on each erte of H,y, and indicate -sources and -sinks with boes and circles. Third we look for an -augmenting path P. An - augmenting path is a path in H,y from any -source to any -sink. To do this, we find the set R of nodes which are reachable from any of the -sources ia a directed path in H,y. If R contains an -sink then we find a path P and go to Step., else we go to Step.2 with R. It this eample, there is no -augmenting path. 4 c < < 2 6, -source -sink c -feasible < < < < Equality arcs are bold -, < <, 4 (D,u,c,b,,y,c) H,y 2,4

5 Step. (-change): Here we augment, in the usual way, the flow along an -augmenting path P in D. We augment the flow along P until either -change: Some forward arc in P becomes full, Some backward arc in P becomes empty, The -source in P becomes -feasible, or The -sink in P becomes -feasible. +d +d -d Step.2 (y-change): Here we increase the prices y for eery erte V R. This change will reduce c for all arcs δ (R), increase c for all arcs δ + (R), and leae all other reduced costs unchanged. We increase these prices until condition or 2 is about to be iolated. That is, either Some underfull arc δ (R) (ie. < u ) has its reduced cost reduced to zero. Some nonempty arc δ + (R)(ie. < ) has its reduced cost increased to zero. y -change: Augment flow One of these must happen (why?). This results in R < V-R new equality arcs crossing in or out of R. We illustrate the Steps. and.2 schematically aboe, and complete our eample below. R H,y V-R c < < 2 6, c < <, -4 < < < u < <, 2,8 (D,u,c,b,,y,c) y y+d Decreased from to -4 No longer equality arc. Increased from 4 to 8 Increased to (bump) New equality arc Net Iteration: < < H,y - -augmenting path found < < 2 6, < <, -4 < < 2, 8, Flow augmented (D,u,c,b,,y,c) This flow is feasible (no -sources or -sinks), and is thus optimal. 5