Generalizations of Ski-Rental

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1 Gnraliations of Ski-Rntal Hisham Harik, Jo Chung Octobr 23, 25 1 Th Ski Rntal Problm Suppos you want to start a nw hobby: skiing. In th ski shop, thy giv you th options to ithr rnt skis or buy skis. If you know a priori how many days you will b skiing, thn you can asily comput whthr you should rnt or buy right from th first day. This is an xampl of offlin problms all information is givn at onc, vn futur information. In th ral world, you would not know bforhand how many days you will b skiing. So, you nd to dcid on whthr to rnt or buy skis basd on past information only. This is an xampl of onlin problms you ar only givn past information. You hav no knowldg of th futur, and you hav to dcid in th prsnt. Algorithms that solv onlin problms ar calld onlin algorithms. A rasonabl qustion to ask is: how dos an onlin algorithm compar to its offlin countrpart? On way to assss this is by solving th ratio of th cost of th onlin algorithm to th cost of th offlin algorithm smallr ratio mans bttr onlin algorithm. This ratio is calld th comptitiv ratio. Mathmatically, an algorithm is c-comptitiv if thr xists a constant α such that: 1.1 Offlin Algorithm E(C A (I)) c Cost OPT (I) + α Hr is an xampl of an offlin algorithm. Suppos that th ski shop chargs $2/day to rnt a pair of skis and $3 to buy a pair. Sinc w ar working in an offlin stting, th numbr of days you will ski is known bforhand. Hr is an algorithm that dtrmins th action to tak to minimi th total cost: if 2 d 3, thn always rnt ls, buy on th first day d stands for th numbr of days you will ski. Lt th instanc b th numbr of days you ski. Thn, givn this algorithm and an instanc I d, th optimal cost is Cost OPT (I d ) = min(2 d,3) So, say that you will ski for xactly 1 days, thn you should rnt, sinc Cost OPT (I 1 ) = min(2 1,3) = min(2,3) = 2. If you will ski for xactly 2 days, thn you should buy, sinc Cost OPT (I 2 ) = min(2 2,3) = min(4,3) = 3. 1

2 1.2 Dtrministic Onlin Algorithm Th offlin algorithm can b considrd as a lowr bound of th onlin algorithm sinc th offlin algorithm is givn all information whil th onlin algorithm is givn partial information. So, an onlin algorithm cannot do bttr than Cost OPT (I d ). Latr, w will s that thr ar bttr lowr bounds. A st of all dtrministic onlin algorithms that solv th ski-rntal problm can b dfind as {A x x } whr x is th numbr of days to rnt skis bfor buying. In othr words, A x rnts skis up to x days thn buy. Hr is th function that calculats th cost givn th numbr of days, d, of skiing: { 2 d if d < x Cost(A x,i d ) = 2 x + 3 othrwis Justapos this function with Cost OPT (I d ) = min(2 d,3). Notic that if d is lss than x, thn both th onlin algorithm and th optimal solution hav th cost of 2 d. This is th bst cas for our onlin algorithms whr th comptitiv ratio is 1. Howvr, it is mor maningful to considr th worst cas of th onlin algorithm. Mor spcifically, w want to dtrmin th algorithm that givs th bst worst cas. For that, w us C(A x,i d ) = 2 x + 3 whr d x. Lts look at an xampl of a dtrministic onlin algorithm. Suppos w choos A 5 for our algorithm. What is th worst cas that can happn? In othr words, what is th comptitiv ratio of this algorithm? It is intuitiv to choos d = 6 as th worst cas for A 5. A 5 tlls you to rnt for 5 days, buy on th 6th day, and thn you will nvr go back to ski again aftr th 6th day. Th offlin solution is Cost OPT (I 6 ) = min(6 2,3) = 12. A 5 s solution is Cost(A 5,I 6 ) = = 4. Th comptitiv ratio for A 5 is at last 4/12 or approximatly That mans that w paid about 3.33 tims th optimal amount. So, what is th smallst comptitiv ratio for any dtrministic onlin algorithm? If th skis cost p dollars pr day to rnt and q dollars to buy, thn it is optimal to rnt for q/p days, thn buy. Substituting p and q into th quations, w gt Cost(A q/p,i q/p+1 ) = q p p + q = 2 q. And Cost OPT (I q/p+1 ) = min(( q + 1) p,q) = q. p So, th comptitiv ratio is 2 q/q = 2. And, if w lt c(a x ) b th comptitiv ratio of algorithm A x. Thn A q/p givs th bst comptitiv ratio out of all dtrministic algorithm bcaus c(a x ) = (p x + q) min(p x,q) 2. 2

3 1.3 Randomid Onlin Algorithm W showd that thr is a dtrministic algorithm for th ski-rntal problm that givs a comptitiv ratio of 2. Howvr, it is possibl to gt a comptitiv ratio bttr than 2 by using a randomid algorithm. Lts s a simpl xampl of a randomid algorithm that givs a comptitiv ratio lss than 2. Suppos you us A 1 with probability.5 and A 15 with probability of.5, thn th xpctd cost is: E(Cost(A,I d )) =.5 Cost(A 1,I d ) +.5 Cost(A 15,I d ) Expanding this quation givs us 2 d if d 1 E(Cost(A,I d )) =.5(5) +.5(2 d) if 1 < d 15.5(5) +.5(6) if d > 15 To solv for th comptitiv ratio c(a), plug in all possibl intgr solution for d 15 into th quation and us d = 16 for all d > 15 sinc that maximis c(a). In this xampl, th xpctd cost is maximid whn d = 16. So, w gt th following: c(a) = max( E(Cost(A,I d)) Cost OPT (I d ) ) c(a) = E(Cost(A,I 16)) Cost OPT (I 16 ) c(a) = 55/3 < 2 Hr, w choos a dtrministic algorithm randomly basd on a probability distribution. Th probability distribution is Prob(A 1 ) =.5, Prob(A 15 ) =.5, and vrything ls =. Sinc th instanc I must b fixd bfor th algorithm can run, th advrsary must choos th worst instanc for th randomid algorithm without any knowldg of th raliation of th random variabls usd by th algorithm. Th advrsary wants to find th largst comptitiv ratio c such that E(Cost(A,I A )) c Cost OPT (I A ) for all randomid algorithm A. Thr ar two prspctivs to th ski-rntal problm. On is from th advrsary s point of viw. Th advrsary wants to choos an instanc I that maximis th comptitiv ratio. Anothr prspctiv is from th skir s point of viw. Th skir wants to slct an algorithm that minimis th comptitiv ratio. Ths two playrs hav xact opposit intrsts. Lts say th skir found th bst algorithm that minimis th comptitiv ratio. W will call this algorithm A bst. Th advrsary will choos th bst instanc I with th following formula: c = supe p ( Cost(A bst,i) I I Cost OPT (I) ) If th skir dos not know th bst algorithm that solvs for th minimal comptitiv ratio c, thn th skir would us th following: c = inf sup E p ( Cost(A ),I Cost OPT (I) ) I I 3

4 Givn an arbitrary algorithm A and an arbitrary instanc I d, calculating th comptitiv ratio is simpl. Lts first scal th cost of th ski st to 1 unit and th numbr of days whr th cost of rnting th skis quals th cost of buying th skis to 1 unit. Thn, Cost(A,I d ) Cost OPT (I d ) = +1 d if d if d and d > 1 d if > d > 1 1 (othrwis) Th first scnario is whn you rnt for days and thn buy th skis, but it s chapr to rnt thm vrytim. Th scond scnario is whn you rnt for days and thn buy thm, but it s chapr to just buy thm in th bginning. Th third scnario is whn you rnt thm vrytim but it s chapr to buy thm from th bginning. Th fourth scnario is whn your action is optimal. Notic that it is not conomical for th skir to buy skis aftr 1 unit, sinc th comptitiv ratio will only incras aftr 1 unit. Also, it is not conomical for th advrsary to pick th numbr of days you ski to b lss than 1 unit. Th comptitiv will always dcras whn th numbr of days th advrsary chooss go pass 1 unit. So, scnario 2 and 3 (whr d 1) ar not considrd in th probability distribution. Th advrsary, using scnario 1 and 2 abov, can maximi th lowr bound of th th comptitiv ratio with th following systm of diffrntial quations: q(u)du u q(u)du = 1 q(u) q(u)du c Th advrsary wants to maximi c. Solving th quations givs: c = { u q(u) = 1 u 1 if u < 1 if u > 1 On th othr hand, th skir, using scnario 1 and 2, can minimi th uppr bound of th comptitiv ratio with th following systm of diffrntial quations: u u p()d + p()d c u p()d = 1 p() Th skir wants to minimi c. Solving th quations givs: c = { ( ) p() = 1 if 1 if > 1 Not that th maximum lowr bound of th advrsary s comptitiv ratio is qual to th minimum uppr bound of th skir s comptitiv ratio. W just found th tight bound of 1. 4

5 2 Background matrial for TCP acknowldgmnt Anothr intrsting onlin problm is TCP acknowldgmnt problm. Th TCP acknowldgmnt is known to b a gnraliation of th ski-rntal problm, though it was always thought to b hardr. W will start by discussing th rsult from th Sidn papr that put a lowr bound on th TCP acknowldgmnt problm. W thn mov to xplain th algorithm prsntd in th Karlin- Knyon-Randall papr that has a comptitiv ratio of /( 1), and dmonstrat that th TCP acknowldgmnt problm is no mor complx than ski-rntal. According to th Von Numann / Yao princinpl, th bst possibl randomid algorithm has as a lowr bound th valu incurrd by all dtrministic alogithms ovr a distribution of problm instancs. Sidn argus that a gnral framwork for finding th right distribution is ndd and h gos on to dvis on. Without going into th dtails, Sidn dviss a gnral solution form, thn plugging paramtrs into th solution would giv you th right distribution. H provs a nw rsult, namly th TCP acknowldgmnt problm, using his mthod. Sidn provs a lowr bound for th TCP acknowldgmnt problm of /( 1). What is intrsting about th solution is that Sidn dscribs th problm as not-so-simpl. Shortly aftr that Karlin-Knyon-Randall papr prsnts th first optimal randomid onlin algorithm for th TCP acknowldgmnt problm, that is an algorithm that rachs th lowr bound limit provs in th Sidn papr. Furthrmor it dmonstrats that th problm is not as hard as first thought, spcifically not mor so than th ski-rntal problm. In th rmaindr of this papr w will go on to xplain th TCP acknowldgmnt problm, dscrib th algorithm dvisd by th authors, and thn prsnt an analysis of th algorithm that provs th comptitiv ratio of /( 1). 3 TCP acknowldgmnt (plas us figur on for clarifications) In lam man s trms, th TCP acknowldgmnt problm is that of trying to minimi th numbr of packts snd across a ntwork in ordr to hlp with rliving congstion. In cas of no lost or othrwis damagd packts, th minimum rquird packts snt ar xactly twic th numbr of packts ndd for transfrring th data, sinc vry packt snt has to b acknowldgd by th rcivr. Trying to rduc th total numbr of packts snt is through trying to minimi th numbr of acknowldgmnts snt through acknowldging svral data packts rcivd by a singl packt. Howvr, du to th congstion control mchanism of TCP, if th sndr dos not rciv an acknowldgmnt for his packt within a spcifid priod of tim, th sndr will rsnd his packt, and th ffort to rduc th numbr of total packts would hav bn foild. Dooly, Goldman and Scott combind both rstrictions into a singl quation that capturs both th numbr of acknowldgmnt packts snt and th latncy in snding th acknowldgmnt packts. Th latncy actually capturs th risk of having th data packt snt again by th sndr. Thus th cost incurrd ovr a squnc of n arrival tims a 1, a 2,..., a n, with k acknowldgmnt tims t 1, t 2,..., t k, is statd as th following: whr latncy(j) = t j 1<a i t j (t j a i ). Cost = acknowldgmnt cost + latncy cost = k + latncy(j) 1 j k 5

6 Anothr variant of this problm is through having wights for th packts, but for all practical purposs that formulation is quivalnt to th currnt on, sinc th arrival of a packt with wight two is quivalnt to th arrival of two sparat packts with unit wight. On could asily imagin a dynamic programming solution, in cas th squnc of packt arrivals is known to th algorithm, which would calculat th optimal cost for this problm. It would b through calculating th minimal total cost incurrd for acknowldging a packt dirctly aftr arrival by building th solution ovr th costs calculatd for th prvious packts. This proof is not cntral to proving th A algorithm but srvs as a practic run for proving th comptitivnss of th A algorithm. Howvr, th onlin problm is not as asy to solv. Th onlin algorithm dos not know th squnc of packts arrivals. Thus has th dilmma facd by th algorithm is whthr to wait for a nw packt to arriv bfor acknowldging or will that wait prov to b too long and latncy tim would b wastd (mayb vn that was th last packt snt). As w pointd out arlir, Sidn obtaind a lowr bound of /( 1) for th comptitiv ratio of randomid onlin algorithms, and in th nxt sction w will prsnt th algorithm that achivs that ratio from th Karlin-Knyon-Randall papr. 4 A and A algorithms (plas us figur two for clarifications) Prviously proposd algorithms probabilistically vary th amount of tim thy tolrat bfor snding an acknowldgmnt. Howvr, ths algorithms wr provn not to achiv th optimal comptitiv ratio of /( 1). Th algorithm A that will b prsntd in this sction will us a family of algorithms calld A, whr 1. First w will dfin A : Considring th last acknowldgmnt happnd at tim t i (with t starting at tim ), A would acknowldg at a tim t i+1 that rspcts th following quation: P(t i,t i+1)(t i+1 t i+1) =. Whr P(t,t ) is th st of packts arriving btwn tim t and tim t, and t i+1 is btwn t i and t i+1 (t i t i+1 t i+1). Thus, considring th last acknowldgmnt happnd at tim t i (with t starting at tim ), th algorithm A would chos to acknowldg at th first tim t i+1 availabl that would nsur at last units of latncy hav passd sinc th last acknowldgmnt. Notic on thing about this algorithm, th algorithm considrs only th packts arriving sinc th last acknoldgmnt till th hypothtical tim t i+1. Randomid algorithm A chooss a valu for btwn ro and on, according to th probability dnsity function p() = 1, and thn runs A on it. Th main dificulty is to analy th algorithm, so as a practic run th authors dcidd to analy th A 1 algorithm. 5 Practic run on A 1 (plas us figur thr for clarifications) Th authors go about proving th 2-comptitivnss of th A 1 algorithm in two stps. Th first stp is through proving that thr is at last an optimal algorithm that snds an acknowldgmnt btwn any pair of succssiv A 1 acknowldgmnt. Th scond stp rlis on th first stp for proving that th A 1 algorithm has a 2-comptitiv ratio. Th proof for th first stp is through contradiction. Considr thr is no optimal solution with acknowldgmnt btwn vry pair of A 1 algorithm acknowldgmnt. Tak any squnc for th optimal solution that dos not acknowldg 6

7 btwn a pair of A 1 acknowldgmnt, without loss of gnrality lt us call thm t i and t i+1. Add an acknowldgmnt to th proposd optimal squnc btwn t i and t i+1, and spcifically at t i+1. Calculating th cost for th modifid optimal squnc w discovr that w hav addd a cost of on, by adding an additional acknowldgmnt, and w hav rmovd a cost of at last 1, sinc = 1. Thus th nw squnc is at last as good as th optimal solution, if not bttr. Thus w hav provn that thr is an optimal solution that has an acknowldgmnt btwn vry pair of A 1 acknowldgmnt. Using this rsult w will go on to prov that th A 1 algorithm has a 2-comptitivnss ratio. Th cost of th A 1 algorithm can b boundd in th following mannr: it includs th cost of th optimal solution plus th latncy that is incurrd by th A 1 algorithm and not th optimal solution. Thus w can writ th following quation: Cost(A 1 ) nopt + latncy(opt) + latncy(a 1 OPT) whr latncy(a 1 OPT) is th latncy incurrd by A 1 and not th optimal solution. Th numbr of acknowldgmnts don by th optimal solution sinc ach optimal acknowldgmnt can rmov at most, which in this cas is qual to on, binds that xtra latncy. Thus, Cost(A 1 ) nopt + latncy(opt) + nopt Cost(OPT) + nopt 2 Cost(OPT) This proof is not cntral to proving th A algorithm but srvs as a practic run for proving th comptitivnss of th A algorithm. 6 Analysis of A Algorithm Th objctiv of th A algorithm is to show that th comptitiv ratio of th TCP acknowldgmnt problm is qual to its lowr bound, 1. W larnd from bfor that any dtrministic algorithm must hav a comptitiv ratio of at last 2. So, A must b a randomid algorithm. Lts analy th cost of th A algorithm with rspct to th optimal algorithm OPT. Assum that OPT is givn. Aftr plotting OPT and A on th sam graph, w s that th ara mad by A btwn th lin of packt arrival and A is prcisly th ara cratd by OPT (Cost OPT ), minus th ara in OPT but not in A, plus th ara in A but not in OPT. Th ara in A not in OPT is at most pr acknowldgmnt in OPT by th dfinition of A. This givs an ara of at most n OPT (I). Lt E b th ara in OPT but not in A. And, lt th numbr of acknowldgmnt packts snt by A b n. Thn th cost of A givn any instanc I is: Cost A (I) n (I) + Cost OPT n OPT (I) + n OPT (I) E (I) Now, lts calculat th lowr bound for E. It is difficult to calculat a lowr bound of E gomtrically sinc th shap of th graph may vary basd on I and. Instad, w attack this problm by calculating th ara gaind by E whn w incras by an infinitsimal amount. Thn, w will scal up to 1 using intgration. Lt L(n,) b th minimum ara in OPT but not in A givn th numbr of acknowldgmnts n. W find that if u > v, thn Th proof of this is givn in [1]. L(n u,) (v )(n v n u ) + L(n v,) 7

8 Moving L(n v,) to th lft, and writ it in intgration form whr u = v + dv, w gt for any < t 1, Solving this givs E is prcisly L(n OPT,), so t dl(n v,) L(n t,) t (v )dn v n v dv (1 )n t E n v dv (1 )n t With E, w can now substitut this to th cost quation and find thn find th bst comptitiv ratio. Lt p() b th probability dnsity function for A. Thn, Cost A Cost OPT n OPT + p()(n + n OPT n w dw + (1 )n OPT )d w = Cost OPT + p()n d n w p()ddw If w lt p() = 1, thn = Cost OPT + (p() P())n d CostA 1 + (p() P())n d Cost 1 OPT n d = 1 So, th comptitiv ratio of th A algorithm is 1 Rfrncs whn p() = 1. [1] Karlin, Knyon, Randall, Dynamic TCP Acknowldgmnt and Othr Storis about /(-1), Algorithmica, 36(3), (Nw York: Springr-Vrlag, 23): [2] Sidn, Stvn S., A Gussing Gam and Randomid Onlin Algorithms, Procdings of th 32 nd Annual ACM Symposium on Thory of Computing, (Portland: STOC, 2):

cycle that does not cross any edges (including its own), then it has at least

cycle that does not cross any edges (including its own), then it has at least W prov th following thorm: Thorm If a K n is drawn in th plan in such a way that it has a hamiltonian cycl that dos not cross any dgs (including its own, thn it has at last n ( 4 48 π + O(n crossings Th