A Combinatorial Polynomial Algorithm for the Linear Arrow-Debreu Market

A Combnatoral Polynomal Algorthm for the Lnear Arrow-Debreu Market Ran Duan Max-Planck-Insttut für Informatk Saarbrücken, Germany duanran@mp-nf.mpg.de Kurt Mehlhorn Max-Planck-Insttut für Informatk Saarbrücken, Germany mehlhorn@mp-nf.mpg.de ABSTRACT We present the frst combnatoral polynomal tme algorthm for computng the equlbrum of the Arrow-Debreu market model wth lnear utltes. Our algorthm vews the allocaton of money as flows and teratvely mproves the balanced flow as n [Devanur et al. 2008] for Fsher s model. We develop new methods to carefully deal wth the flows and surpluses durng prce adjustments. In our algorthm, we need O(n 6 log(nu)) maxmum flow computatons, where n s the number of persons and U s the maxmum nteger utlty, and the length of the numbers s at most O(n log(nu)) to guarantee an exact soluton. Prevously, [Jan 2007] has gven a polynomal tme algorthm for ths problem, whch s based on solvng a convex program usng the ellpsod algorthm.. INTRODUCTION We provde the frst combnatoral polynomal algorthm for computng the model of economc markets formulated by Walras n 874 [2]. In ths model, every person has an ntal dstrbuton of some goods and a utlty functon of all goods. The market clears at a set of prces f every person sells ther ntal goods and then uses ther entre revenue to buy a bundle of goods wth maxmum utlty. We want to fnd the market equlbrum n whch every good s assgned a prce so that the market clears. In 954, two Nobel laureates, Arrow and Debreu [2], proved that the market equlbrum always exsts f the utlty functons are concave, whch s why we call ths model Arrow-Debreu market. But, ther proof s based on Kakutan s fxed pont theorem and s nonconstructve. Snce then, many algorthmc results studed the lnear verson of ths model, that s, all utlty functons are lnear. The frst polynomal algorthm for the lnear Arrow-Debreu model was found by Jan n 2007 [9]; t s based on solvng a convex program usng the ellpsod algorthm and smultaneous dophantne approxmaton. Before that, Devanur and Vazran [5] gave an approxmaton scheme for computng the Arrow-Debreu model wth runnng tme O( n4 log n ), mprovng [0]. Recently, Ghyasvand and Orln [8] mproved ɛ ɛ the runnng tme to O( n (m+n log n)), where m s the number of pars (, j) such that buyer has some utlty for pur- ɛ chasng good j. Many combnatoral algorthms consder a smpler model proposed by Fsher (see [3]). Esenberg and Gale [6] reduced the problem of computng the Fsher market equlbrum to a concave cost maxmzaton problem and thus gave the frst polynomal algorthm for the Fsher market by ellpsod algorthm. The frst combnatoral polynomal algorthm for an exact lnear Fsher market equlbrum s gven by Devanur et al [4]. They use the maxmum flow algorthm as a black box n ther algorthm. When the nput data s ntegral, ther algorthm needs O(n 5 log U + n 4 log e max) max-flow computatons, where n s the number of buyers, U the largest nteger utlty, and e max the largest ntal amount of money of a buyer. If we use the common O(n 3 ) max-flow algorthm (see []), ther runnng tme s O(n 8 log U +n 7 log e max). Recently, Orln [] mproved the runnng tme for computng the lnear Fsher model to O(n 4 log U +n 3 log e max) and also gave the frst strongly polynomal algorthm wth runnng tme O(n 4 log n). Our results. In ths paper, we extend the method n [4] to fnd a combnatoral algorthm for the lnear Arrow-Debreu market. W.l.o.g., we can assume that each of the n persons has only one unt of goods, whch s dfferent from the goods other people have. When the prce of that goods s p, the person wll have a budget of p amount of money equvalently. It can be shown that each person only buys ther favorte goods, that s, the goods wth the maxmum rato of utlty and prce. We construct a graph on the nodes of buyers and goods as well as equalty edges representng each buyer s favorte goods, and then fnd a maxmum flow n t. In the case of market clearng prces, all the buyers and goods wll be saturated. There wll otherwse be surplus, so we need to ncrease the prces of some goods n order to decrease the surplus. As n [4], we use the noton of balanced flow, whch s a maxmum flow that balances the surplus of buyers. It s helpful for fndng a cut n the resdual graph. When we ncrease the prces of some goods, the buyers n possesson of these goods wll obtan more budget. Unlke [], where the flows stay unchanged durng prce adjustment, we also

ncrease the flows to those goods by the same rato. So, some buyers wll have more surplus, and some wll have less. We can show that n some cases, the new flow s more balanced, that s, the l 2-norm of the surplus vector wll decrease by a factor of Ω(/n 3 ). In other cases, some prces wll ncrease by a factor of + O(/n 3 ). Snce we can bound the number of the latter type of prce adjustment teratons by a polynomal of n and log U, the total runnng tme s also polynomal. 2. MODEL AND DEFINIITONS We make the followng assumptons on the model as n Jan s paper [9]:. There are n persons n the system. Each person has only one good, whch s dfferent from the goods other people have. The good person has s denoted by good. 2. Each person has only one unt of good. So, f the prce of good s p, person wll obtan p unts of money when sellng ts good. 3. Each person has a lnear utlty functon P j ujzj, where z j s the amount of good j consumed by. 4. Each u j s an nteger between 0 and U. 5. For every, there s a j such that u j > 0. (Everybody lkes some goods.) 6. For every j, there s an such that u j > 0. (Every good s lked by somebody.) 7. For every proper subset P of persons, there exst P and j / P such that u j > 0. All these assumptons, wth the excepton of the last, are wthout loss of generalty. The last assumpton mples that all the equlbrum prces are nonzero [9], and t s only useful for the next secton. In Secton 4, we wll dscuss more about the last assumpton. Let p = (p, p 2,..., p n) denote the vector of prces of goods to n, so they are also the budgets of persons to n. In ths paper, we denote the set of all buyers to be B = {b, b 2,..., b n} and the set of all goods to be C = {c, c 2,..., c n}. So, f the prce of goods c s p, buyer b wll have p amount of money. For a subset B of persons or a subset C of goods, we also use p(b ) or p(c ) to denote the total prces of the goods the persons n B own or the goods n C. For a vector v = (v, v 2,..., v k ), let: v = v + v 2 +... + v k be the l -norm of v. v = p v 2 + v2 2 +... + v2 k be the l2-norm of v. 3. THE ALGORITHM Our algorthm resembles [4], whch fnds a balanced flow and ncreases the prces n the actve subgraph. But, n the Arrow-Debreu model, when we ncrease the prces of some good, the budget of buyer wll also ncrease. So, we need to fnd a careful way to prevent the total surplus from ncreasng. Construct a flow network G = ({s, t} B C, E G), where s s the source node and t s the snk node, then B = {b,..., b n} denotes the set of buyers and C = {c,..., c n} denotes the set of goods. E G conssts of: Edges from s to every node b n B wth capacty p. Edges from every node c n C to t wth capacty p. Edges from b to c j wth nfnte capacty f u j/p j = α. Call these edges equalty edges. So, our am s to fnd a prce vector p such that there s a flow n whch all edges from s and to t are saturated,.e., (s, C B t) and (s C B, t) are both mnmum cuts. When ths s satsfed, all goods are sold and all of the money earned by every person s spent. In a flow f, defne the surplus r(b ) of a buyer to be the resdual capacty of the edge (s, b ), and defne the surplus r(c j) of a good j to be the resdual capacty of the edge (c j, t). That s, r(b ) = p P j fj, and r(cj) = p j P fj, where fj s the amount of flow n the edge (b, c j). Defne the surplus vector of buyers to be r(b) = (r(b ), r(b 2),..., r(b n)). Also, defne the total surplus to be r(b) = P r(b), whch s also P j r(cj) snce the total capacty from s and to t are both equal to P p. For convenence, we denote the surplus vector of flow f by r (B). In the network correspondng to market clearng prces, the total surplus of a maxmum flow s zero. 3. Balanced flow As n [4], we defne the concept of balanced flow to be a maxmum flow that balances the surpluses of buyers. (However, unlke n ther paper, the surpluses of goods can be postve here, whch are not supposed to be balanced, so the balanced flow s not necessarly unque.) Defnton. In the network G of current p, a balanced flow s a maxmum flow that mnmzes r(b) over all choces of maxmum flows. For flows f and f and ther surplus vectors r(b) and r (B), respectvely, f r(b) < r (B), then we say f s more balanced than f. The next lemma shows why t s called balanced. Clearly, every person s happest wth goods j, whch maxmzes the rato u j/p j. Defne ts bang per buck to be α = max j{u j/p j}. The classcal Arrow-Debreu [2] theorem says that there s a non-zero market clearng prce vector. Lemma. [4] If a b 0, =, 2,..., k and δ δ, where δ, δ 0, =, 2,..., k, then: P k = (a, b, b 2,..., b k ) 2 a+δ, b δ, b 2 δ 2,..., b k δ k 2 δ 2.

Proof. (a + δ) 2 + 2aδ + δ 2 2 δ 2 + 2a(δ k (b δ ) 2 a 2 = k b δ = k δ ) δ 2. = Lemma 2. [4] In the network G for a prce vector p, gven a maxmum flow f, a balanced flow f can be computed by at most n max-flow computatons. Proof. In the resdual graph G f w.r.t. to f, let S B C be the set of nodes reachable from s, and let T = (B C) \ S be the remanng nodes. Then, there are no edges from S B to T C n the equalty graph, and there s no flow from T B to S C. The buyers n T B and the goods n S C have no surplus w.r.t. f, and ths holds true for every maxmum flow. Let G be the network spanned by s S t, and let f be the balanced maxmum flow n G. The f can be computed by n max-flow computatons. (Corollary 8.8 n [4] s applcable snce (s S, t) s a mncut n G.) Fnally, f together wth the restrcton of f to s T t s a balanced flow n G. The surpluses of all goods n f are the same as those n f snce we only balance the surplus of buyers. 3.2 Prce adjustment As n [4, ], we need to ncrease the prces of some goods to get more equalty edges. For a subset of buyers B, defne ts neghborhood Γ(B ) n the current network to be: k = Γ(B ) = {c j C b B, s.t.(b, c j) E G}. Clearly, there s no edge n G from B to C \ Γ(B ). In a balanced flow f, gven a surplus bound S > 0, let B(S) denote the subset of buyers wth surplus at least S, that s, B(S) = {b B r(b ) S}. Lemma 3. In a balanced flow f, gven a surplus bound S, there s no edge that carres flow from B \ B(S) to Γ(B(S)). Proof. Suppose there s such an edge (b, c j) that carres flow such that b / B(S) and c j Γ(B(S)). Then, n the resdual graph, there are drected edges (b k, c j) and (c j, b ) wth nonzero capactes n whch b k B(S). However, r(b k ) S > r(b ), so we can augment along ths path and get a more balanced flow, contradctng that f s already a balanced flow. From Lemma 3, we can ncrease the prces n Γ(B(S)) by the same factor x wthout nconsstency. There s no edge from B(S) to C\Γ(B(S)), and the edges from B\B(S) to Γ(B(S)) are not carryng flow, and hence, there wll be no harm f b 2 they dsappear from the equalty graph. If there are edges (b, c j) and (b, c k ) where b B(S), c j, c k Γ(B(S)), then u j/p j = u k /p k. Snce the prces n Γ(B(S)) are multpled by a common factor x, u j/p j and u k /p k reman equal after a prce adjustment. However, the goods n C \ Γ(B(S)) wll become more attractve, so there may be edges from B(S) to C \ Γ(B(S)) enterng the network, and the ncrease of prces needs to stop when ths happens. Defne such a factor to be (S), that s, (S) = mn{ uj p j pk u k b B(S), (b, c j) E G, c k / Γ(B(S))}. So, we need O(n 2 ) multplcatons/dvsons to compute (S). When we ncrease the prces of the goods n Γ(B(S)) by a common factor x (S), the equalty edges n B(S) Γ(B(S)) wll reman n the network. We wll also need the followng theorem to prevent the total surplus from ncreasng. Theorem. Gven a balanced flow f n the current network G and a surplus bound S, we can multply the prces of goods n Γ(B(S)) wth a parameter x >. When x mn {p /(p r(b )) b B(S), c / Γ(B(S))} and x (S), we obtan a flow f n the new network G of adjusted prces wth the same value of total surplus by: j f j x fj f c = j Γ(B(S)); f c j / Γ(B(S)). f j Then, the surplus of each good remans unchanged, and the surpluses of the buyers become: 8 x r(b ) f b B(S), c Γ(B(S)); >< r ( x)p (b ) = + x r(b ) f b B(S), c / Γ(B(S)); (x )p + r(b ) f b / B(S), c Γ(B(S)); >: r(b ) f b / B(S), c / Γ(B(S)). We call these knds of buyers type to type 4 buyers, respectvely. Proof. Snce the flows on all edges assocated wth goods n Γ(B(S)) are multpled by x, the surplus of each good n Γ(B(S)) remans zero. Only the surplus of type 2 buyers decreases because the flows from a type 2 buyer b are multpled by x, but ts budget p s not changed. The flow after adjustment s x(p r(b )). We need ths to be at most p, so x p /(p r(b )) for all type 2 buyers b, and n f, the new surplus r (b ) = ( x)p + xr(b ). Snce both money and flows are multpled by x for a type buyer, hs surplus s also multpled by x. For a type 3 buyer b, hs flows are not changed, but hs money s multpled by x, so the new surplus s xp (p r(b )). After each prce adjustment, n the new network, we wll fnd a maxmum flow by augmentaton on the adjusted flow f and then fnd a balanced flow by Lemma 2. Ths wll guarantee that when the surplus of a good becomes zero, t wll not change to non-zero anymore. Thus, the prces of the goods wth non-zero surplus wll not be adjusted. Property. The prces of goods wth non-zero surpluses reman unchanged n the algorthm.

3.3 Whole procedure The whole algorthm s shown n Fgure, where K s a constant we wll set later. In ths secton, one teraton denotes the executon of one entre teraton nsde the loop. We wll dscuss the roundng and termnaton condtons n Secton 3.4. In the frst teraton, we constructed a balanced flow f n the network where all prces are equal to. In the equalty graph, we have at least one edge ncdent to every buyer. The total surplus wll be bounded by n, actually n as at least one good wll be sold completely. From Theorem, n the executon of the algorthm, the total surplus wll never ncrease. To ensure that the algorthm wll termnate n a polynomal number of steps, we wll requre the followng lemmas. From Property, the prces of goods wth surplus stay one durng the whole algorthm, so there s stll a good wth prce one n the end. And, we need to bound the largest prce: Lemma 4. The prces of goods are at most (nu) n. Proof. It s enough to show that durng the entre algorthm, for any non-empty and proper subset Ĉ of goods, there are goods c Ĉ, cj / Ĉ such that p/pj nu. So, when we sort all the prces n decreasng order, the rato of two adjacent prces s at most nu. Snce there s always a good wth prce, the largest prce s (nu) n. If Ĉ contans goods wth surpluses, then ther prce s. The clam follows. Let ˆB = Γ(Ĉ) be the set of buyers adjacent to goods n Ĉ n the equalty graph. If there exst b, c j s.t. b ˆB, c j / Ĉ and u j > 0, let c k Ĉ be one of the goods adjacent to b n the equalty graph, and then u j/p j u k /p k. So, p k /p j u k /u j U. If there do not exst such b, c j, then there s no flow between ˆB and C \ Ĉ, and there s b k / ˆB, but c k Ĉ. Otherwse the persons whose goods are n Ĉ wll not lke any goods not n Ĉ, contradctng assumpton (7). Let B = {j b j ˆB, c j Ĉ} and B = {j b j ˆB, c j Ĉ}. Then, there s a j B wth p j p(b )/n, and hence, p k p(b ) = p(ĉ) p({j bj ˆB, c j Ĉ}) p( ˆB) p({j b j ˆB, c j Ĉ}) = p(b ) np j. The nequalty of the second lne holds snce goods n Ĉ have surplus 0 and all of the flows from ˆB go to Ĉ. By Lemma 9, we can round to the exact soluton when the algorthm termnates. To analyze the correctness and runnng tme, we need the followng lemma: Lemma 5. After every prce adjustment by x, the l 2-norm of the surplus vector r(b) wll ether be multpled by a factor of + O(/n 3 ) when x = + Kn 3, or be dvded by a factor of + Ω(/n 3 ). Note that by Lemma 8, the roundng procedure can only ncrease r(b) by a factor of +O(/n 4 ) snce S ɛ/(e n). (We wll leave the dscusson of Lemma 8 later, so we can gnore t n the analyss.) Theorem 2. In total, we need to compute O(n 6 log(nu)) maxmum flows, and the length of numbers s bounded by O(n log(nu)). Thus, f we use the common O(n 3 ) max-flow algorthm (see []), the total runnng tme s O(n 0 log 2 (nu)). Proof. By Lemma 4, every prce can be multpled by x = + for O(log Kn 3 +/Kn 3(nU) n ) = O(n 4 log(nu)) tmes, so the total number of teratons of the frst type s O(n 5 log(nu)). The total factor multpled to r(b) by the frst type teratons s ( + O(/n 3 )) O(n5 log(nu)). At the begnnng, r(b) n. When the algorthm termnates, r(b) < ɛ =, so the number of second 4n 4 U type teratons s bounded by 3n log +Ω(/n 3 ) ( ɛ n( + O(/n 3 )) O(n5 log(nu)) ) = O(n 5 log(nu)). Thus, the total number of teratons performed s bounded by O(n 5 log(nu)). Snce we need to compute n max-flows for the balanced flow n every teraton, we need O(n 6 log(nu)) maxmum flow computatons n total. By Lemma 4 and Lemma 8, the prces are ratonal numbers (nu) n and wth denomnator U n = 4n 9 U 4n. Thus, the length of the numbers to be handled s bounded by O(n log(nu)). Note that max-flow computatons only need addtons and subtractons. We perform multplcatons and dvsons when we scale prces and when we set up the max-flow computaton n the computaton of balanced flow. The numbers of multplcatons/dvsons s by a factor n less than the numbers of addtons/subtractons, and hence, t suffces to charge O(n log(nu)) per arthmetc operaton. Next we wll prove Lemma 5. When we sort all the buyers by ther surpluses b, b 2,..., b n, b s at least r(b) /n (where r(b) s the total surplus). So, for the frst n whch r(b ) > + /n, we can see r(b j ) r(b + ) r(b j+ + /n for j <, ) so r(b ) r(b )( + /n) n > r(b) /(e n). When such an does not exst, each r(b ) s larger than r(b) /(e n), and all goods n Γ(B) must have zero surplus because the flow s otherwse not maxmum. Thus, there are goods that have no buyers, and hence, ether new equalty edges emerge, or x reaches + (condton (3a) below). Kn 3 From the algorthm, n every teraton, x satsfes the followng condtons: () x + Kn 3.

Intally set p = for all goods ; Repeat Construct the network G for the current p, and compute the balanced flow f n t; Sort all buyers by ther surpluses n decreasng order: b, b 2,..., b n; Fnd the frst n whch r(b ) r(b + > + /n, and = n when there s no such ; ) Let the surplus bound S = r(b ) and obtan B(S), Γ(B(S)), (S); (B(S) = {b, b 2,..., b }) Multply the prces n Γ(B(S)) by a gradually ncreasng factor x > untl: (Let f be the flow correspondng to x whch s constructed accordng to Theorem.) New equalty edges emerge (x reaches (S)); OR the surplus of a buyer B(S) and a buyer / B(S) equals n f ; OR x reaches + K n 3 Round the prces n Γ(B(S)) accordng to Lemma 8 wth = 4n 9 U 3n ; Untl r(b) < ɛ, where ɛ = ; 4n 4 U 3n Fnally, round the prces accordng to Lemma 9 to get an exact soluton. Fgure : The whole algorthm (2) In f, r (b) r (b ) for all b B(S), b / B(S). Here, r (b) s the surplus of b w.r.t. f, the flow correspondng to x by Theorem. (3) If x < +, the followng possbltes arse: Kn 3 (a) There s a new equalty edge (b, c j) wth b B(S), c j / Γ(B(S)). By Lemma 6 below, we can obtan a flow f n whch ether r (b ) = r (b ) p j, or there s a b k / B(S) wth r (b ) = r (b k ) (same as (b)). (b) When x satsfes the second requrement n the algorthm, t satsfes: there exsts b B(S) and b / B(S) such that r (b) = r (b ) n f. Lemma 6. If there s a new equalty edge (b, c j) wth b B(S), c j / Γ(B(S)), we can obtan a flow f from f n whch ether r (b ) = r (b ) p j, or there s a b k / B(S) wth r (b ) = r (b k ). Proof. Let B B \ B(S) be the set of buyers wth flows to c j n f, and let w be the largest surplus of a buyer n B \ B(S). Run the followng procedure (f denotes the current flow n the algorthm): unchanged for type 4 buyers. Note that the surplus of a type or 2 buyer cannot be smaller than the surplus of any type 3 or 4 buyer. From Theorem and condton (2), we nfer that the total surplus wll not ncrease, type 2 and 3 buyers wll get more balanced, and r (b) = x r(b) for type buyers b, so r (B) x r(b) = ( + O(/n 3 )) r(b). In (3a), there s a new equalty edge (b, c j). After the procedure descrbed n Lemma 6, f there s no b k / B(S) such that r (b ) = r (b k ), then r (b ) = r (b ) p j (p j ). For all b k / B(S), r (b ) > r (b k ), and r (b k ) = r (b k ) + δ k, where δ k 0 and P b k / B(S) δ k p j. Because r(b) n, r(b) 2 n 2. By Lemma, r (B) 2 r (B) 2 p 2 j x 2 r(b) 2 x 2 r(b) 2 n 2 r(b) 2 = ( Θ(/n 2 )) r(b) 2. Augment along (b, c j) gradually untl: r (b ) = w or r (c j) = 0; If r (b ) = w then Ext; For all b k B n any order Augment along (b, c j, b k ) gradually untl: r (b ) = max{r (b k ), w} or f (b k, c j) = 0; Set w = max{r (b k ), w}; If r (b ) = w then Ext. Durng the procedure, the surplus of b decreases but cannot become less than the surplus of a buyer n B \ B(S), so condton (2) holds. In the end, f r (b ) = w, then there s a b k B \ B(S) s.t. r (b ) = r (b k ); otherwse, c j has no surplus, and the flow to t all comes from b, so r (b ) = r (b ) p j. From Theorem, the surpluses n f wll ncrease for type and 3 buyers, wll decrease for type 2 buyers, and wll stay So, we have r (B) = ( Ω(/n 2 )) r(b). In (3a), after the procedure descrbed n (3a), f there exsts b k / B(S) such that r (b ) = r (b k ), then we are n a smlar stuaton as (3b), possbly wth an even smaller total surplus. So, we can prove ths case by the proof of (3b). In (3b), let u, u 2,..., u k and v, v 2,..., v k be the lst of orgnal surpluses of type 2 and 3 buyers, respectvely. Defne u = mn{u }, v = max{v j}, so u u for all, and v j v for all j, and u > ( + /n)v. After the prce and flow adjustments n Theorem, the lst of surpluses wll be u δ, u 2 δ 2,..., u k δ k and v + δ, v 2 + δ 2,..., v k + δ k (here δ, δ j 0 for all, j), and there exst I, J such that u I δ I = v J +δ J, where u I δ I s the smallest among u δ, and v J + δ J s the largest among v j + δ j by condton (2). Snce the surpluses of type edges also ncrease, we have

P δ P j δ j, δ I P δ, and δ J P j δ j. Compute: (u δ ) 2 + (v j + δ j) 2 ( u 2 + vj 2 ) j j = 2 u (u v) (u v) u δ + 2 j δ + v j v jδ j + δ j δ (u I δ I) δ (u v) max{δ I, δ J} (u v)2 2 < 2(n + ) 2 u2. δ 2 + j δ 2 j δ (u δ ) + j δ + (v J + δ J) j δ j(v j + δ j) Let w, w 2,...w k be the lst of surpluses of type buyers; all of them are e u. After prce adjustment, the surpluses wll be x w, x w 2,...x w k from Theorem. Compute: (xw ) 2 ( + Kn 3 )2 w 2 w 2 2 + ( Kn + 3 K 2 n ) 6 ne2 u 2 = w 2 2 + ( Kn + 2 K 2 n 5 )e2 u 2. Let K = 32e 2, then the change to the sum of squares of surpluses for type 2 and 3 buyers s less than u 2 = 8n 2 4 e 2 u 2. The total change to r(b) 2 s: Kn 2 ( 2 Kn 2 + K 2 n 5 )e2 u 2. Snce u e r(b) for all buyers b, nu2 e 2 r(b) 2. Snce the change s negatve, we have: r (B) 2 r(b) 2 + ( 2 Kn 2 + K 2 n 5 ) n r(b) 2 = r(b) 2 2 Kn 3 r(b) 2 + K 2 n 6 r(b) 2 = r(b) 2 ( Kn 3 )2. Thus, Lemma 5 s proved. 3.4 Roundng and termnaton condton In ths secton, we wll show how to round the prces to ratonal numbers wth denomnators of length O(n log(nu)). Also, we need the roundng process to obtan an exact market equlbrum when the surplus s very small. Here, we defne the equalty graph F on B C of undrected equalty edges between buyers and goods, and we consder every connected component n ths undrected equalty graph. δ j Lemma 7. In a connected component Ψ contanng k goods n the equalty graph, f we know that p j s a ratonal number wth denomnator N, where c j Ψ C, then all the prces of goods n Ψ C are ratonal numbers wth denomnator N U k. Proof. Fnd a tree that connects all the goods n Ψ. The tree wll contan k + k edges f t contans k buyers. Then, we can get k lnear ndependent equatons p j/u j = p j /u j when both (b, c j) and (b, c j ) are tree edges. Together wth the equaton p j = I/N for some nteger I, we can see that all the prces of goods n Ψ have denomnator N U k. Gven an nteger, we call a connected component n the equalty graph consstent f t has a good whose prce s a ratonal number wth denomnator. Then, by Lemma 7, the prces of goods n a consstent connected component are ratonal numbers wth denomnator U n. Lemma 8. In a balanced flow f n the network, gven > nu n and a surplus bound S n 5 /, f all the connected components n (B\B(S), C \Γ(B(S))) are consstent, we can adjust the prces n Γ(B(S)) so that all connected components n the equalty graph are consstent. In the adjusted flow f by Theorem, r (B) = r(b) (+O( )), S where r (B) s the surplus vector n f. Proof. The procedure s shown below: Set B = B(S); Repeat Multply the prces n Γ(B ) by x > untl: A prce n Γ(B ) has denomnator ; OR a new equalty edge emerges; Update the flow f by Theorem ; Remove new consstent components from B Γ(B ); Untl B =. Snce all the prces change by at most /, the changes to the total surplus of type 2 buyers s at most n/ < S, so the surplus of a type 2 buyer s stll postve. In addton, the surplus added to each type or 3 buyer s at most /, so r (B(S)) 2 r(b(s)) 2 + 2 r(b(s)) + n 2 r(b(s)) 2 + 2 S r(b(s)) 2 + n S 2 2 r(b(s)) 2 = r(b(s) ( + O( S )). Durng the algorthm, we can see that all the connected components n (B \ B, C \ Γ(B )) are consstent snce we move the new consstent components to t. When we fnd new equalty edges connectng B and C \ Γ(B ), some nodes n B Γ(B ) wll connect to (B \B, C \Γ(B )), so these nodes can be removed. When a prce n Γ(B ) has denomnator

, the component contanng t wll become consstent, so the loop wll run for at most n tmes. In each loop, we need to compute O(n 2 ) multplcatons/dvsons, so the runnng tme for ths roundng procedure s less than the computaton of a balanced flow. Lemma 9. When the total surplus s < = ɛ n 4n 4 U a flow f, we can obtan an exact soluton from the 3n current equalty graph. Proof. Add the edge (b, c ) for each person to the equalty graph F to obtan F. For a connected component of F, the sum of the prces on both sdes are the same. For every component Φ of F wth no surplus node, ncrease ts prces by a common factor untl a new equalty edge emerges; ths wll unte two components. Repeat ths untl all components n F have a surplus node. We may assume w.l.o.g. that F becomes connected by ths process. Otherwse, the followng argument can be appled ndependently to each component of F. The total surplus s stll less than ɛ. The followng roundng procedure wll be performed on these revsed prces. Denote the set of connected components n the equalty graph (after the adjustments of the prevous paragraph) by Λ = {Ψ k }. For each component Ψ k n F, fnd a spannng tree T k n t, then wrte the followng equatons: p j/u j = p j /u j, (b, c j), (b, c j ) T k. Snce we have one such equaton f c j and c j are connected by one b, we can have Ψ k C lnear ndependent equatons. The total number of lnear ndependent equatons for all components n F s n Λ. Snce there s no flow between components, for each component Ψ k n F, the money dfference between buyers and goods n Ψ k s only the surplus dfference. So, we can wrte b B Ψ k p c C Ψ k p = ɛ k, Ψ k. Here, ɛ k (postve or negatve) comes from the surpluses of goods and buyers, so P ɛ k 2ɛ. If b and c belong to dstnct connected components Ψ j and Ψ k, the coeffcent of p s + n the equaton of Ψ j, n the equaton for Ψ k, and 0 n all other equatons. If b and c belong to the same connected component, the coeffcent of p s zero n all equatons. Assume now that there s a proper subset of the equatons that s lnear dependent. Then, f b or c belongs to one of the components n the subset, both of them do. However, the subset of components s a proper subgraph of F, and hence, there s at least one such that only one of b or c belongs to the subset of components. Thus, we have Λ ndependent equatons. Snce there s a good c wth non-zero surplus, we have p =. Thus, the current prce vector p s the soluton of these lnear equatons Ap = n whch A s nvertble. Consder the followng n lnear equatons wth ɛ k removed: p j/u j = p j /u j, (b, cj), (b, c j ) T k p p = 0, Ψ k b B Ψ k c C Ψ k p =, r(c ) > 0. They can be denoted by Ap =, so there s also a unque soluton. The soluton wll be ratonal numbers wth a common denomnator D nu n by Cramer s rule. Snce < 2ɛ, the dfference p p of solutons of each prce s 2n 3 U 2n at most 2ɛ nu n = by Cramer s rule. The dfference between any two numbers of denomnators D, D nu n s a ratonal number of denomnator D D < n 2 U 2n, whch s larger than 2 p p. Snce p s a ratonal number wth denomnator D nu n, we can get p by roundng p to the nearest ratonal number of denomnator nu n. Ths can be done by contnued fracton expanson, whch needs O(n log 2 D) = O(n 3 log 2 (nu)) tme by Theorem 3.3 n [7]. We can also compute D = det(a) drectly and round every prce to the nearest ratonal wth denomnator D or solve the lnear equatons Ap =. By Theorem 5.2 n [7], computng the determnant of a matrx of dmenson n wth entres U takes Õ(n4 log U) tme, and solvng Ap = also takes Õ(n4 log U) tme. Now, all the prces p are of the form q /D, where q, D are ntegers and D nu n a common denomnator. So, p q = ɛ, n whch D 2n 3 U 2n D ɛ D =. Construct the flow network G for the new prces q = (q n, q 2,..., q n). 2n 3 U 2n 2n 2 U Consder any b B and c j, c k C and assume u j/p j u k /p k. Then, u jq k u j(p k D + ɛ ) u k p jd + u jɛ u k (q j + ɛ ) + u jɛ u k q j + (u k + u j)ɛ < u k q j +, and hence, u jq k u k q j snce u jq k and u k q j are ntegral. We conclude that the edges n G are all n G. Denote the sze of the cuts (s, B C t) and (s B C, t) n G by Z, whch s an nteger. Then, the sze of ths cut n G s (Z nɛ )/D. If there s another cut n G of sze Z, t s also a cut n G, and ts sze n G s ɛ 2nɛ + 2n = Z/D D D wll have total surplus > 3ɛ Z D D, so the maxmum flow n G > ɛ. Thus, (s, B C t) and (s B C, t) are both mn-cuts n G, so the prces reach a market equlbrum. 4. GENERAL CASE Here, we consder the case whch does not satsfy the assumpton (7) n Secton 2. We draw the lnkng graph of persons n whch there s a drected edge from to j ff u j > 0. If the graph s strongly connected, then the case satsfes assumpton (7). Otherwse, we can shrnk each connected component nto one vertex, then the graph wll be a DAG, and we can fnd a topologcal order of strongly connected components: P, P 2,...P k, n whch there are only edges from a lower order to a hgher order. We use the algorthm n Secton 3 to compute the equlbrum for all

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