Near Optal Onlne Algorths and Fast Approxaton Algorths for Resource Allocaton Probles Nkhl R Devanur Kaal Jan Balasubraanan Svan Chrstopher A Wlkens Abstract We present algorths for a class of resource allocaton probles both n the onlne settng wth stochastc nput and n the offlne settng Ths class of probles contans any nterestng specal cases such as the Adwords proble In the onlne settng we ntroduce a new dstrbutonal odel called the adversaral stochastc nput odel, whch s a generalzaton of the d odel wth unknown dstrbutons, where the dstrbutons can change over te In ths odel we gve a 1 O approxaton algorth for the resource allocaton proble, wth alost the weakest possble assupton: the rato of the axu aount of resource consued by any sngle request to the total capacty of the resource, and the rato of the proft contrbuted by any sngle request to the optal proft s at ost O 2 logn/ where n s the nuber of resources avalable There are nstances where ths rato s 2 / log n such that no randozed algorth can have a copettve rato of 1 o even n the d odel The upper bound on rato that we requre proves on the prevous upper-bound for the d case by a factor of n Our proof technque also gves a very sple proof that the greedy algorth has a copettve rato of 1 1/e for the Adwords proble n the d odel wth unknown dstrbutons, and ore generally n the adversaral stochastc nput odel, when there s no bound on the bd to budget rato All the prevous proofs assue that ether bds are very sall copared to budgets or soethng very slar to ths In the offlne settng we gve a fast algorth to solve very large LPs wth both packng and coverng constrants We gve algorths to approxately solve wthn a factor of 1 + the xed packng-coverng proble wth γ logn/δ 2 O oracle calls where the constrant atrx of ths LP has denson n, the success probablty of the algorth s 1 δ, and γ s a paraeter whch s very slar to the rato descrbed for the onlne settng We dscuss several applcatons, and how our algorths prove exstng results n soe of these applcatons Mcrosoft Research, 1 Mcrosoft Way, Redond Eal: nkdev@crosoftco Mcrosoft Research, 1 Mcrosoft Way, Redond Eal: kaalj@crosoftco Coputer Scences Dept, Unversty of Wsconsn-Madson Eal: balu2901@cswscedu Part of ths work was done whle the author was at Mcrosoft Research, Redond Coputer Scence Dvson, Unversty of Calforna at Berkeley Eal: cwlkens@csberkeleyedu Part of ths work was done whle the author was at Mcrosoft Research, Redond
1 Introducton The results n ths paper fall nto dstnct categores of copettve algorths for onlne probles and fast approxaton algorths for offlne probles We have two an results n the onlne fraework and one result n the offlne settng However they all share coon technques There has been an ncreasng nterest n onlne algorths otvated by applcatons to onlne advertsng The ost well known s the Adwords proble ntroduced by Mehta et al [MSVV05], where the algorth needs to assgn keywords arrvng onlne to bdders to axze proft, subject to budget constrants for the bdders The proble has been analyzed n the tradtonal fraework for onlne algorths: worst-case copettve analyss As wth any onlne probles, the worst-case copettve analyss s not entrely satsfactory and there has been a drve n the last few years to go beyond the worst-case analyss The predonant approach has been to assue that the nput satsfes soe stochastc property For nstance the rando perutaton odel ntroduced by Goel and Mehta [GM08] assues that the adversary pcks the set of keywords, but the order n whch the keywords arrve s chosen unforly at rando A closely related odel s the d odel: assue that the keywords are d saples fro a fxed dstrbuton, whch s unknown to the algorth Stronger assuptons such as d saples fro a known dstrbuton have also been consdered Frst Result A key paraeter on whch any of the algorths for Adwords depend s the bd to budget rato For nstance n Mehta et al [MSVV05] and Buchbnder, Jan and Naor [BJN07] the algorth acheves a worst case copettve rato that tends to 1 1/e as the bd to budget rato let s call t γ tends to 0 1 1/e s also the best copettve rato that any randozed algorth can acheve n the worst case Devanur and Hayes [DH09] showed that n the rando perutaton odel, the copettve rato tends to 1 as γ tends to 0 Ths result showed that copettve rato of algorths n stochastc odels could be uch better than that of algorths n the worst case The portant queston snce then has been to deterne the optal trade-off between γ and the copettve rato [DH09] showed how to get a 1- O copettve rato when γ s at ost O 3 n logn/ where n s the nuber of advertsers and s the nuber of keywords Subsequently Agrawal, Wang and Ye [AWY09] proved the bound on γ to O 2 n log/ The papers of Feldan et al [FHK + 10] and Agrawal, Wang and Ye [AWY09] have also shown that the technque of [DH09] can be extended to other onlne probles The frst an result n ths paper s the followng 3-fold proveent of prevous results: Theores 2-4 1 We gve an algorth whch proves the bound on γ to O 2 logn/ Ths s alost optal; we show a lower bound of 2 logn 2 The bound apples to a ore general odel of stochastc nput, called the adversaral stochastc nput odel Ths s a generalzaton of the d odel wth unknown dstrbuton, but s ncoparable to the rando perutaton odel 3 It apples to a ore general class of onlne probles that we call the resource allocaton fraework A foral defnton of the fraework s presented n Secton 22 and a dscusson of any nterestng specal cases s presented n Secton 7 Regardng the bound on γ, the reoval of the factor of n s sgnfcant Consder for nstance the Adwords proble and suppose that the bds are all n [0,1] The earler bound ples that the budgets need to be of the order of n/ 2 n order to get a 1 copettve algorth, where n s the nuber of advertsers Wth realstc values for these paraeters, t sees unlkely that ths condton would be et Whle wth the proved bounds presented n ths paper, we only need the budget to be of the order of log n/ 2 and ths condton s et for reasonable values of the paraeters Furtherore, n the resource allocaton fraework, the current hghest upper bound on γ s fro Agrawal, Wang and Here k s the nuber of avalable optons see Secton 22 and n typcal applcatons lke network routng, k could be exponental n n, and thus, the factor saved by our algorth becoes quadratc n n We note here that so far, all the algorths for the d odel wth unknown dstrbuton were actually desgned for the rando perutaton odel It sees that any algorth that works for one should also work for the other However Ye [AWY09] and equals O 2 n logk/ 1
we can only show that our algorth works n the d odel, so the natural queston s f our algorth works for the rando perutaton odel It would be very surprsng f t ddn t One drawback of the stochastc odels consdered so far s that they are te nvarant, that s the nput dstrbuton does not change over te The adversaral stochastc nput odel allows the nput dstrbuton to change over te The odel s as follows: n every step the adversary pcks a dstrbuton, possbly adaptvely dependng on what the algorth has done so far, and the actual keyword n that step s drawn fro ths dstrbuton The copettve rato s defned wth respect to the optu fractonal soluton for an offlne nstance of the proble, called the dstrbuton nstance, whch s defned by the dstrbuton In Secton 22, where we defne the dstrbuton nstance, we also prove that the optal fractonal soluton for the dstrbuton nstance s at least as good as the coonly used benchark of expected value of optal fractonal soluton, where the expectaton s wth respect to the dstrbuton A detaled descrpton of ths odel, how the adversary s constraned to pck ts dstrbutons and how t dffers fro the worst-case odel s presented n Secton 22 Second Result Another portant open proble s to prove the copettve rato for the Adwords proble when there s no bound on γ The best copettve rato known for ths proble s 1/2 n the worst case Nothng better was known, even n the stochastc odels For the specal case of onlne bpartte atchng, n the case of d nput wth a known dstrbuton, recent seres of results acheve a rato of better than 1-1/e, for nstance by Feldan et al [FMMM09] and Bahan and Kapralov [BK10] The best rato so far s 702 by Manshad, Gharan and Saber [MGS11] The sae onlne bpartte atchng has been recently studed n the rando perutaton odel by Mahdan and Yan [MY11] and by Karande, Mehta and Trpath [KMT11] The best rato so far s 0696 by Mahdan and Yan [MY11] The second result n ths paper s that for the Adwords proble n the adversaral stochastc nput odel, wth no assupton on γ, the greedy algorth gets a copettve rato of 1 1/e aganst the optal fractonal soluton to the dstrbuton nstance Theore 5 The greedy algorth s partcularly nterestng snce t s a natural algorth that s used wdely for ts splcty Because of ts wde use, prevously the perforance of the greedy algorth has been analyzed by Goel and Mehta [GM08] who showed that n the rando perutaton and the d odels, t has a copettve rato of 1 1/e wth an assupton whch s essentally that γ tends to 0 Thrd Result Charles et al [CCD + 10] consdered the followng offlne proble: gven a lopsded bpartte graph G = L, R, E, that s a bpartte graph where = L R = n, does there exst an assgnent M : L R wth j, Mj E for all j L, and such that for every vertex R, M 1 B for soe gven values B Even though ths s a classc proble n cobnatoral optzaton wth well known polynoal te algorths, the nstances of nterest are too large to use tradtonal approaches to solve ths proble The value of n partcular s very large The approach used by [CCD + 10] was to essentally desgn an onlne algorth n the d odel: choose vertces fro L unforly at rando and assgn the to vertces n R n an onlne fashon The onlne algorth s guaranteed to be close to optal, as long as suffcently any saples are drawn Therefore t can be used to solve the orgnal proble approxately: the onlne algorth gets an alost satsfyng assgnent f and only f the orgnal graph has a satsfyng assgnent wth hgh probablty The thrd result n ths paper s a generalzaton of ths result to get fast approxaton algorths for a wde class of probles n the resource allocaton fraework Theore 6 Probles n the resource allocaton fraework where the nstances are too large to use tradtonal algorths occur farly often, especally n the context of onlne advertsng Foral stateents and a ore detaled dscusson are presented n Secton 23 The underlyng dea used for all these results can be suarzed at a hgh level as thus: consder a hypothetcal algorth called Pure-rando that knows the dstrbuton fro whch the nput s drawn and uses an optal soluton wrt ths dstrbuton Now suppose that we can analyze the perforance of Pure-rando by consderng a potental functon and showng that t decreases by a certan aount n each step Now we can desgn an algorth that does not know the dstrbuton as follows: consder the sae potental functon, and n every step choose the opton that nzes the potental functon Snce the algorth nzes the potental n each step, the decrease n the potental for ths algorth s better than that for Pure-rando and hence we obtan the sae guarantee as that for Pure-rando For nstance, for the case where γ s sall, the perforance of Pure-rando s analyzed usng Chernoff bounds The Chernoff bounds are proven by showng bounds on the expectaton of the oent generatng functon of a rando varable Thus the potental functon s the su of the oent generatng functons for all the rando varables that we apply the Chernoff bounds to The proof shows that n each step ths potental functon decreases by soe ultplcatve 2
factor The algorth s then desgned to acheve the sae decrease n the potental functon A partcularly pleasng aspect about ths technque s that we obtan very sple proofs For nstance, the proof of Theore 5 s extreely sple: the potental functon n ths case s sply the total aount of unused budgets and we show that ths aount n expectaton decreases by a factor of 1 1/ n each step where there are steps n all On the surface, ths technque and the resultng algorths 1 bear a close reseblance to the algorths of Young [You95, You01] for derandozng randozed roundng and the fast approxaton algorths for solvng coverng/packng LPs of Plotkn, Shoys and Tardos [PST91], Garg and Köneann [GK98],Flescher [Fle00] In fact Arora, Hazan and Kale [AHK05] showed that all these algorths are related to the ultplcatve weghts update ethod for solvng the experts proble and especally hghlghted the slarty between the potental functon used n the analyss of the ultplcatve update ethod and the oent generatng functon used n the proof of Chernoff bounds and Young s algorths Hence t s no surprse that our algorth s also a ultplcatve update algorth It sees that our algorth s closer n sprt to Young s algorths than others A basc dfference of our algorth fro ths prevous set of results s that n all these works, every sngle teraton of the algorth nvolves changng the entre soluton vector x, whle our algorth changes only a sngle coordnate of the vector x per teraton In other words, our algorth uses the specal structure of the polytope P j n gvng a ore effcent soluton It s possble that our algorth can also be nterpreted as an algorth for the experts proble In fact Mehta et al [MSVV05] asked f there s a 1 o1 copettve algorth for Adwords n the d odel wth sall bd to budget rato, and n partcular f the algorths for experts could be used They also conjectured that such an algorth would teratvely adjust a budget dscount factor based on the rate at whch the budget s spent Our algorths for resource allocaton proble when specalzed for Adwords look exactly lke that and wth the connectons to the experts fraework, we answer the questons n [MSVV05] n the postve Organzaton The rest of the paper s organzed as follows In Secton 2, we defne the resource allocaton fraework, the adversaral stochastc odel and state our results forally as theores We also dscuss one specal case of the resource allocaton fraework the adwords proble and forally state our results In Secton 3, we consder a splfed n-ax verson of the resource allocaton fraework and present the proofs for ths verson The other results buld upon ths sple verson In Secton 4 we gve a fast approxaton algorth for the xed coverngpackng proble Theore 6 The 1 O copettve onlne algorth for the resource allocaton fraework wth stochastc nput Theore 2 s n Secton 5 The 1 1/e copettve algorth Theore 5 for the Adwords proble s n Secton 6 Several specal cases of the resource allocaton fraework are consdered n Secton 7 Secton 8 concludes wth soe open probles and drectons for future research 2 Prelnares & Man Results 21 Resource allocaton fraework We consder the followng fraework of optzaton probles There are n resources, wth resource havng a capacty of c There are requests; each request j can be satsfed by a vector x j that s constraned to be n a polytope P j We refer to the vector x j as an opton to satsfy a request, and the polytope P j as the set of optons The vector x j consues a,j x j aount of resource, and gves a proft of w j x j Note that a,j, w j and x j are all vectors The objectve s to axze the total proft subject to the capacty constrants on the resources The followng LP descrbes the proble: axze j w j x j st, j a,j x j c j, x j P j 1 For the case of sall γ It s not clear f ths dscusson apples to the case of large γ, that s to Theore 5 3
We assue that we have the followng oracle avalable to us: gven a request j and a vector v, the oracle returns { a,jx the vector x j that axzes vx j aong all vectors n P j Let γ = ax j c },j { w jx j W be the noton }j correspondng to the bd to budget rato for Adwords Here W s the optal offlne objectve to the dstrbuton nstance, defned n Secton 22 The canoncal case s where each P j s a unt splex n R K, e P j = {x j R K : k x j,k = 1} Ths captures the case where there are K dscrete optons, each wth a gven proft and consupton Ths case captures ost of the applcatons we are nterested n, whch are descrbed n Secton 7 All the proofs wll be presented for ths specal case, for ease of exposton The co-ordnates of the vectors a,j and w j wll be denoted by a, j, k and w j,k respectvely, e, the k th opton consues a, j, k aount of resource and gves a proft of w j,k For an exaple of an applcaton that needs ore general polytopes see Secton 75 We consder two versons of the above proble The frst s an onlne verson wth stochastc nput: requests are drawn fro an unknown dstrbuton The second s when the nuber of requests s uch larger than the nuber of resources, and our goal s to desgn a fast PTAS for the proble 22 Onlne Algorths wth Stochastc Input We now consder an onlne verson of the resource allocaton fraework Here requests arrve onlne We consder the d odel, where each request s drawn ndependently fro a gven dstrbuton The dstrbuton s unknown to the algorth The algorth knows, the total nuber of requests The copettve ratos we gve for resource allocaton probles wth bounded γ are wth respect to an upper bound on the expected value of fractonal optal soluton, naely, the fractonal optal soluton of the dstrbuton nstance, defned below Consder the followng dstrbuton nstance of the proble It s an offlne nstance defned for any gven dstrbuton over requests and the total nuber of requests The capactes of the resources n ths nstance are the sae as n the orgnal nstance Every request n the support of the dstrbuton s also a request n ths nstance Suppose request j occurs wth probablty p j The resource consupton of j n the dstrbuton nstance s gven by p j a,j for all and the proft s p j w j The ntuton s that f the requests were drawn fro ths dstrbuton then the expected nuber of tes request j s seen s p j and ths s represented n the dstrbuton nstance by scalng the consupton and the proft vectors by p j To suarze, the dstrbuton nstance s as follows axze p j w j x j st j n the support, j p j a,j x j c j, x j P j We now prove that the fractonal optal soluton to the dstrbuton nstance s an upper bound on the expectaton of OPT, where OPT s the offlne fractonal optu of the actual sequence of requests Lea 1 OPT[Dstrbuton nstance] E[OPT] Proof: The average of optal solutons for all possble sequences of requests should gve a feasble soluton to the dstrbuton nstance wth a proft equal to E[OPT] Thus the optal proft for the dstrbuton nstance could only be larger The copettve rato of an algorth n the d odel s defned as the rato of the expected proft of the algorth to the fractonal optal proft for the dstrbuton nstance The an result s that as γ tends to zero, the copettve rato tends to 1 In fact, we gve the alost optal trade-off Theore 2 For any > 0, we gve an algorth such that f γ = O algorth s 1 O Theore 3 There exst nstances wth γ = 2 logn/ then the copettve rato of the 2 logn such that no algorth can get a copettve rato of 1 o2 2 The proof of ths theore s obtaned by a odfcaton of a slar theore for rando perutatons presented n [AWY09] 4
Also, our algorth works when the polytope P j s obtaned as an LP relaxaton of the actual proble 3 To be precse, suppose that the set of optons that could be used to satsfy a gven request corresponds to soe set of vectors, say I j Let the polytope P j I j be an α approxate relaxaton of I j f for the proft vector w j and for all x j P j, there s an oracle that returns a y j I j such that w j y j αw j x j and for all, a,j y j a,j x j Gven such an oracle, our algorth acheves a copettve rato of α O Theore 4 Gven a resource allocaton proble wth an α approxate relaxaton, and for any > 0, we gve an algorth such that f γ = O then the copettve rato of the algorth s α O 2 logn/ Proof:Sketch Consder the proble n the resource allocaton fraework defned by the relaxaton, that s request j can actually be satsfed by the polytope P j The optal soluton to the relaxaton s an upper bound on the optal soluton to the orgnal proble Now run Algorth 1 on the relaxaton, and when the algorth pcks a vector x j to serve request j, use the roundng oracle to round t to a soluton y j and use ths soluton Snce the sequence of x j s pcked by the algorth are wthn 1 O of the optu for the relaxaton, and for all j w j y j αw j x j, ths algorth s α1 O copettve In fact, our results hold for the followng ore general odel, the adversaral stochastc nput odel In each step, the adversary adaptvely chooses a dstrbuton fro whch the request n that step s drawn The adversary s constraned to pck the dstrbutons n one of the followng two ways In the frst case, we assue that a target objectve value OPT T s gven to the algorth 4, and that the adversary s constraned to pck dstrbutons such that the fractonal optu soluton of each of the correspondng dstrbuton nstances s at least OPT T or at ost OPT T for nzaton probles The copettve rato s defned wth respect to OPT T In the second case, we are not gven a target, but the adversary s constraned to pck dstrbutons so that the fractonal optu of each of the correspondng dstrbuton nstances s the sae, whch s the benchark wth respect to whch the copettve rato s defned Note that whle the d odel can be reduced to the rando perutaton odel, these generalzatons are ncoparable to the rando perutaton odel as they allow the nput to vary over te Also the constrant that each of the dstrbuton nstances has a large optu value dstngushes ths fro the worst-case odel Ths constrant n general ples that the dstrbuton ust contan suffcently rch varety of requests n order for the correspondng dstrbuton nstance to have a hgh optu To truly sulate the worst-case odel, n every step the adversary would chose a deternstc dstrbuton, that s a dstrbuton supported on a sngle request Then the dstrbuton nstance wll sply have copes of ths sngle request and hence wll not have a hgh optu For nstance consder onlne bpartte b-atchng where each resource s a node on one sde of a bpartte graph wth the capacty c denotng the nuber of nodes t can be atched to and the requests are nodes on the other sde of the graph and can be atched to at ost one node A deternstc dstrbuton n ths case corresponds to a sngle onlne node and f that node s repeated tes then the optu for that nstance s just the weghted by c degree of that node If the adversary only pcks such deternstc dstrbutons then he s constraned to pck nodes of very hgh degree thus akng t easy for the algorth to atch the We refer the reader to Secton 7 for a dscusson on several probles that are specal cases of the resource allocaton fraework and have been prevously consdered Here, we dscuss one specal case the adwords proble 221 The Adwords proble In the d Adwords proble, there are n bdders, and each bdder has a daly budget of B dollars Keywords arrve onlne wth keyword j havng an unknown probablty p j of arrvng n any gven step For every keyword j, each bdder subts a bd, b j, whch s the proft obtaned by the algorth on allocatng keyword j to bdder The objectve s to axze the proft, subject to the constrant that no bdder s charged ore than hs budget Here, the resources are the daly budgets of the bdders, the requests are the keywords, and the optons are once agan the bdders The aount of resource consued and the proft are both b j For ths proble, wth no bounds on γ, we show that the greedy algorth has a copettve rato of 1 1/e For our results for the adwords proble wth bounded γ, see Secton 71 3 There ay be trval ways of defnng P j such that ts vertces correspond to the actual optons The otvaton for allowng non-trval relaxatons s coputatonal: recall that we need to be able to optze lnear functons over P j { } { 4 a,j x In ths case, we requre γ = ax j c wj x j,j OPT, e, OPT T takes the place of W T }j n the defnton of γ 5
Theore 5 The greedy algorth acheves a copettve rato of 1 1/e for the Adwords proble n the adversaral stochastc nput odel wth no assuptons on the bd to budget rato We note here that the copettve rato of 1 1/e s tght for the greedy algorth [GM08] It s however not known to be tght for an arbtrary algorth 23 Fast algorths for very large LPs Charles et al [CCD + 10] consder the followng proble: gven a bpartte graph G = L, R, E where = L R = n, does there exst an assgnent M : L R wth j, Mj E for all j L, and such that for every vertex R, M 1 B for soe gven values B They gave an algorth that runs n te lnear 5 n the nuber of edges of an nduced subgraph obtaned by takng a rando saple fro R of sze O log n n {B } 2, for a gap-verson of the proble wth gap When n {B } s reasonably large, such an algorth s very useful n a varety of applcatons nvolvng ad assgnent for onlne advertsng We consder a generalzaton of the above proble that corresponds to the resource allocaton fraework In fact, we consder the followng xed coverng-packng proble Suppose that there are n 1 packng constrants, one for each {1n 1 } of the for j=1 a,jx j c and n 2 coverng constrants, one for each {1n 2 } of the for j=1 b,jx j d Each x j s constraned to be n P j Does there exsts a feasble soluton to ths syste of constrants? The gap-verson of ths proble s as follows Dstngush between the two cases, wth a hgh probablty, say 1 δ: YES: There s a feasble soluton NO: There s no feasble soluton even f all of the c s are ultpled by 1 + and all of the d s s ultpled by 1 We note that solvng offlne an optzaton proble n the resource allocaton fraework can be reduced to the above proble through a bnary search on the objectve functon value Suppose as n [CCD + 10] that s uch larger than n Assue that solvng the followng costs unt te: gven j and v, fnd x j P j that axzes vx j Let γ = ax{ [n 1 ], j [] : a,jxj c } { [n 2 ], j [] : b,jxj d } Theore 6 For any > 0, there s an algorth that solves the gap verson of the xed coverng-packng proble wth a runnng te of O γ logn/δ 2 Applcatons to onlne advertsng: The atchng proble ntroduced by [CCD + 10] was otvated by the proble of coputng the avalable nventory for dsplay ad allocaton see the orgnal paper for detals In fact, the atchng proble was a splfed verson of the real proble, whch fts nto the resource allocaton fraework Moreover, such algorths are used n ultple ways For nstance, although the technque of Devanur and Hayes [DH09] was orgnally desgned to solve the purely onlne proble, t can be used n the PAC odel where the algorth can ake use of a predcton of the future arrval of requests see for nstance Vee, Vasslvtsk and Shanugasundara [VVS10] The key technque s to forulate an LP relaxaton of the proble and learn the optal dual varables usng the predcton, and these duals can then be used for the allocaton onlne Even f the predcton s not entrely accurate, we note that such an approach has certan advantages Ths otvates the proble of fndng the optal duals We observe that our algorth can also be used to copute near optal duals whch can then be used to do the allocaton onlne Many probles for nstance the Dsplay ad allocaton proble can beneft fro such an algorth A slar approach was consdered by Abras, Mendelevtch and Toln [AMT07] for the followng proble otvated by sponsored search auctons: for each query j, one can show an advertser n each of the K slots Each advertser bds a certan aount on each query j, and has a daly budget However, the cost to an advertser depends on the entre ordered set of advertsers shown called a slate, based on the rules of the aucton Gven the set of queres that arrve n a day whch n practce s an estate of the queres expected rather than the actual queres, the goal s to schedule a slate of advertsers for each query such that the total cost to each advertser s wthn the budget and axze a gven objectve such as the total revenue, or the socal welfare Ths proble s odeled as an LP and a colun-generaton approach s suggested to solve t Also, any coproses are ade, n ters of ltng the nuber of queres, etc due to the dffcultes n solvng an LP of very large sze We observe that ths LP fts n the resource allocaton fraework and thus can be solved quckly usng our algorth 5 In fact, the algorth akes a sngle pass through ths graph 6
Chernoff bounds We present here the for of Chernoff bounds that we use throughout the rest of ths paper Let X = X, where X [0, B] are d rando varables Let E[X] = µ Then, for all > 0, 2 µ Pr[X < µ1 ] < exp 2B Consequently, for all δ > 0, wth probablty at least 1 δ, Slarly, for all [0, 2e 1], X µ 2µB ln1/δ Pr[X > µ1 + ] < exp 2 µ 4B Consequently, for all δ > exp 2e 12 µ 4B, wth probablty at least 1 δ, X µ 4µB ln1/δ 3 Mn-Max verson In ths secton, we solve a slghtly splfed verson of the general onlne resource allocaton proble, whch we call the n-ax verson In ths proble, requests arrve onlne, and each of the ust be served The objectve s to nze the axu fracton of any resource consued There s no proft The followng LP descrbes t forally nze λ st, j,k j, k a, j, kx j,k λc x j,k = 1, j, k, x j,k 0 For ease of llustraton, we assue that the requests arrve d unknown dstrbuton n the followng proof At the end of ths secton, we show that the proof holds for the adversaral stochastc nput odel also The algorth proceeds n steps Let λ denote the fractonal optal objectve value of the dstrbuton nstance of ths proble Let X t be the rando varable ndcatng the aount of resource consued durng step t, that s, X t = a, j, k f n step t, request j was chosen and was served usng opton k Let S T = T t=1 Xt be the total aount of resource consued n the frst T steps Let γ = ax,j,k { a,j,k c }, whch ples that for all, j and k, a, j, k γc Let φ t = 1 + St /γc For the sake of convenence, we let S 0 = 0 and φ 0 = 1 for all The algorth s as follows ALG Mn-ax In step t + 1, on recevng request j, use opton arg n k { } a,j,kφ t c Lea 7 For any 0, 1], The algorth ALG Mn-ax descrbed above approxates λ wthn a factor of 1+, wth a probablty at least 1 δ, where δ = n exp 2 λ 4γ We wll prove Lea 7 through a seres of leas, naely Leas 8, 9 and 10 Before we begn the proof, we gve soe ntuton Consder a hypothetcal algorth, call t Pure-rando, that knows the dstrbuton Let x j denote the optal fractonal soluton to the dstrbuton nstance Pure-rando s a non-adaptve algorth whch uses x j to satsfy request j, e, t serves request j usng opton k wth probablty x jk Suppose we wanted to prove a bound on the perforance of Pure-rando, that s show that wth hgh probablty, Pure-rando s wthn 1 + O of the optu, say λ Ths can be done usng Chernoff bounds: for each resource separately bound the probablty that the total consupton s ore than λ c 1 + O usng Chernoff bounds and take a unon bound Note that the Chernoff 7
bounds are shown by provng an upper bound on the expectaton of the oent generatng functon of the rando varables If we could show the sae bound for our algorth, then we would be done Let φ t = φt We wsh to upper bound the expectaton of φ t+1 Consder the state of the algorth after the frst t steps Let X denote the aount of resource consued n step t + 1 had we served the request at step t + 1 usng the Pure-rando algorth nstead of our algorth Then we show that the expectaton of φ t+1 s upper bounded by t+1 X φt 1 + γc, and the rest of the proof s along the lnes of the Chernoff bound proof Lea 8 For all t, Proof: φ t+1 = φ t+1 φ t+1 φ t φ t 1 + = 1 + Xt+1 γc φ t 1 + X t+1 X γc t+1 γc φ t 1 + t+1 X γc The frst nequalty s because the convexty of the functon 1 + x can be used to upper bound t by 1 + x for all x [0, 1], and X t ax j,k a, j, { k γc The second nequalty follows fro the defnton of our algorth as t } a,j,kφ chooses the opton that nzes t c Lea 9 For all T, E[φ T λ ] n exp T γ, where λ s the optal soluton to the LP Proof: Fro Lea 8, t follows that E [ φ t+1 φ t for all ] [ ] E φ t t+1 X 1 + γc φ t 1 + λ γ = φ t 1 + λ γ λ φ t exp γ and hence the lea follows snce φ 0 = n The second nequalty follows fro the fact that E[ ] λ c for all Ths s because requests are drawn d, and hence the optal value of the dstrbuton nstance s the sae for all te steps and s equal to λ X t+1 Lea 10 Proof: [ { } S T Pr ax > T ] c λ 1 + n exp [ { } S T Pr ax > T ] c λ 1 + Pr 2 T λ 4γ ] [φ T > 1 + T λ 1+ γ E[φ T ]/1 + T λ 1+ γ λ T n exp /1 + T λ 1+ γ γ 2 T λ n exp 4γ 8
The nequalty n the frst lne s because ax φ T φ T The rest s slar to proofs of Chernoff bounds The second lne follows fro Markov s nequalty, the thrd lne fro Lea 9 and the fourth lne s a well known algebrac nequalty whenever 0, 2e 1], and n partcular when 0, 1] Substtutng T = n Lea 10, we get Lea 7 Adversaral stochastc nput odel In the above leas, we assued that the requests are drawn d, e, we used the fact that E[ X t] λ c / for all t But n an adversaral stochastc odel, snce the dstrbuton fro whch a request s drawn changes each step, the optal objectve of the dstrbuton nstance also changes every step, e, t could be λ t at step t So, n the proof Lea 9, where we proved that E [ φ t+1 φ t for all ] λ φ t exp, γ we would nstead have E [ φ t+1 φ t for all ] φ t λ t exp γ But gven a target λ, we know the adversary s constraned to pck dstrbutons whose dstrbuton nstance has an optu objectve at ost λ recall that ths s a nzaton proble Therefore, we can upper bound φ t exp by φ t exp The rest of the steps n the proof rean the sae Thus, the adversary s not constraned to pck requests fro the sae dstrbuton at every te step All we requre s that, whatever dstrbuton t uses for drawng ts request, the correspondng dstrbuton nstance has an optu objectve value at ost λ, whch s the target value we a for In the followng sectons, we llustrate all our proofs n the d odel wth unknown dstrbuton and t s easy to convert the to proofs for the adversaral stochastc nput odel λ t γ 4 Mxed Coverng-Packng and Onlne Resource Allocaton 41 Mxed Coverng-Packng In ths secton, we consder the xed packng-coverng proble stated n Secton 23 and prove Theore 6 We restate the LP for the xed coverng-packng proble here λ γ, j,k a, j, kx j,k c, j,k j, k b, j, kx j,k d x j,k 1, j, k, x j,k 0 The goal s to check f there s a feasble soluton to ths LP We solve a gap verson of ths proble Dstngush between the two cases wth a hgh probablty, say 1 δ: YES: There s a feasble soluton NO: There s no feasble soluton even f all of the c s are ultpled by 1 + and all of the d s are ultpled by 1 For convenence of descrpton, we refer to the quanttes ndexed by j as requests, those ndexed by as resources { a,jx and those ndexed by k as optons The paraeter γ for ths proble s defned by γ = ax j c },j { bjx j d },j As before, the algorth proceeds n steps In each step, the algorth saples a request unforly at rando fro γ lnn/δ the total of possble requests We wll prove that f the nuber of saples T Θ, then the algorth 2 solves the gap verson wth probablty at least 1 δ Snce the te taken for servng any gven request s one by γ lnn/δ takng the te consued by a sngle oracle call to be one, ths proves that the total run-te s O Ths 2 proves Theore 6 9
Let X t, Xt, S t t be as defned n Secton 3 Let Y be the rando varable ndcatng the aount of deand satsfed durng step t, that s, Y t = b, j, k f n step t, request j was chosen and was served usng opton k Let t Ỹ denote the aount of deand satsfed durng step t by the optal algorth for the dstrbuton nstance of ths proble Let V T = T t=1 Y t Let φt = η c 1 + 2 S t γc 1 + 2γ T t, where η c = 1 + 2 1+ 2 T γ Let ψ t = η d1 2 V t γd 1 As before, we let S 0 = 0 and V 0 2γ T t, where η d = 1 2 1 2 T γ Let φ t = φt, let ψt = ψt and Φt = φ t + ψ t = 0 The algorth s as follows ALG Packng-Coverng arg n k Gven request j n step t + 1, use the opton 1 1 + 2γ φ t a, j, k 1 c 1 2γ ψ t b, j, k d S At the end of T steps, the algorth checks f ax T c < T 1 + 2 and f n V T d > T 1 2 If true, the algorth answers YES Else t says NO We now proceed to prove that, whenever the real answer s YES, the algorth says YES wth a hgh probablty Leas 11 and 12 prove ths case Lea 11 For a YES nstance E [ Φ T ] Φ 0 Proof: Slar to the proof of Lea 8, we have 1 + Φ t+1 φ t t+1 X 2γc 1 + 2γ + ψ t 1 Ỹ t+1 2γd 1 2γ E [ Φ t+1 φ t, ψ t for all ] φ t 1 + 2γ + 1 + 2γ ψ t 1 2γ = Φ t 1 2γ where the nequalty follows fro the fact that, when the real answer s YES, E[ X t] c Snce the above sequence of nequaltes holds for every t, the lea follows t and E[Ỹ ] d for all Lea 12 For a YES nstance [ S T Pr ax T c 1 + ] 2 Proof: As n proof of Lea 10 [ S T Pr ax T c 1 + ] 2 [ + Pr n [ φ T Pr η c V T T d 1 ] 2 Φ 0 1 ] E[φ T ] η c where the nequalty n the frst lne follows fro φ T φ T for all, and the next lne follows fro Markov s nequalty Slarly, we have [ V T Pr n T d 1 ] [ ψ T 2 Pr 1 ] E[ψ T ] η d η d Thus the su of these probabltes s at ost E[φ T ] + E[ψ T ] = E[Φ T ], whch s at ost Φ 0 fro Lea 11 η c 1 + 2γ T + η d 1 n Observe that Φ 0, the falure probablty equals n exp + exp 2 T 16γ 2 T 8γ 2γ T, whch s upper bounded by γ logn/δ If T = O, we have the falure probablty to be at ost δ Thus 2 10
Lea 12 proves that the algorth ALG Packng-Coverng says YES wth a probablty at least 1 δ when the real answer s YES We now proceed to prove that when the real answer s NO, our algorth says NO wth a probablty at least 1 δ, e, Lea 13 For a NO nstance, f T Θ [ S T Pr ax c γ logn/δ, then 2 < T 1 + 2 & n Proof: Let S denote the set of requests sapled Consder the followng LP V T > T d 1 ] 2 < δ nze λ 1, λ a, j, kx j,k T c j S,k, λ + j S,k b, j, kx j,k d j S, x j,k 1 k j, k, x j,k 0 λ 0 T If the above LP has an optal objectve value at least, then our algorth would have declared NO We γ lnn/δ 2 2 T now show that by pckng T = Θ probablty at least 1 δ Ths akes our algorth answer NO wth a probablty at least 1 δ Consder the dual of LP 1:, the above LP wll have ts optal objectve value at least T 2, wth a axze β j + T ρ α 2 j S j S, k, β j a, j, k b, j, k α ρ c d α + ρ 1, α, ρ 0 j S, β j 0 The optal value of LP 1 s equal to the optal value to LP 2, whch n turn s lower bounded by the value of LP 2 at any feasble soluton One such feasble soluton s α, β, ρ, whch s the optal soluton to the full verson of LP 2, naely the one wth S = [], T = Thus, the optal value of LP 1 s lower bounded by value of LP 2 at α, β, ρ, whch s = j S β j + T ρ α 3 For proceedng further n lower boundng 3, we apply Chernoff bounds to j S β j In order to get useful Chernoff bounds, we frst prove that βj resdes n a sall nterval Consder the full verson of LP 2, e, S = [] and T = In ths verson, snce accordng to the second constrant the optal soluton ust satsfy α + ρ 1, t follows that 0 βj γ accordng to the frst constrant Further, let τ denote the optal value of the full verson of LP 2 Recall that snce we are n the NO case τ Now, because of the constrant α + ρ 1, t follows that 11
ρ α 1 and thus t follows that j β j τ + 1 2 axτ, 1 We are now ready to lower bound the quantty n 3 We have the optal soluton to LP 2 j S βj + T ρ α T j β j 2T j β j γ ln1/δ + T ρ α T τ 4T axτ, 1γ ln1/δ Snce βj [0, γ] where the second nequalty s a wth probablty at least 1 δ nequalty, e, we apply Chernoff bounds for j S β j, along wth the observaton that each βj [0, γ] We now set T so that the quantty n 4 s at least T 2 Notng that τ γ lnn/δ, settng T = Θ wll ensure 2 ths nequalty and hence proves the lea Leas 11, 12 and 13 prove that the gap-verson of the xed coverng-packng proble can be solved n te γ logn/δ O, thus provng Theore 6 2 5 Onlne Algorths wth Stochastc Input In ths secton, we use the potental functon based algorth to solve the onlne verson of the resource allocaton proble ntroduced n Secton 22 The followng LP descrbes the resource allocaton proble 4 axze j,k w j,k x j,k st 5, j,k j, k a, j, kx j,k c x j,k 1, j, k, x j,k 0 Our algorth coputes ncreasngly better estates of the objectve value by coputng the optal soluton for the observed requests, and uses t to gude future allocatons Ths s slar to the algorth n [AWY09], except that we only need to estate the value of the optal soluton as aganst the entre soluton tself Through Leas 14 and 15, we show that our algorth acheves a copettve rato of 1 O thus provng Theore 2 We assue that the nuber of requests s known n advance Algorth 1 descrbes our algorth The frst requests are not served but used just for coputatonal purposes After these frst requests, the algorth proceeds n l stages, naely 0, 1,, l 1, where l s such that 2 l = 1 and s a postve nuber between 0 and 1 that the algorth desgner gets to choose In stage r the algorth serves t r = 2 r requests Note that the stage r conssts of all steps t t r, t r+1 ] Let W denote the optal soluton to the dstrbuton nstance of the proble Let X t be as defned n Secton 3 Let Y t be the aount of proft earned durng step t, e, Y t = w j,k, f n step t, request j was served usng opton k Instead of the usual S t, we now defne St r = t u=t r+1 Xu, whch s the su of Xu s tll t for u s belongng to stage r alone, e, u t r, t r+1 ] Slarly, V t r = t u=t Y u r+1 Let w ax = ax j,k w j,k The potental functon for constrant n step t when t t r + 1, t r+1 ] s defned by φ t = η c r1 + c r S t r γc 1 + tr+1 t cr, γ tr 1+cr η c r = 1 + c r γ, 12
Algorth 1 : Algorth for stochastc onlne resource allocaton 1: Intalze t 0 : t 0, 2: for r = 0 to l 1 do 3: for t = t r + 1 to t r+1 do 4: f t = t r + 1 then 5: If the ncong request s j, use the followng opton k: arg n k c r/γ 1 + cr γ 6: For all, S tr = Xt, and V t r = Y t 7: else 8: If the ncong request s j, use the followng opton k: 9: For all, S t 10: end f 11: end for 12: end for arg n k c r/γ 1 + cr γ φ nt a, j, k r or/w ax c 1 orzr w ax φ t 1 a, j, k or/w ax c 1 orzr r = St 1 r + X t, and, V t r = V t 1 r + Y t c r = Slarly, the potental functon for objectve at step t s, 4γ lnn + 1/δ t r w ax φ t obj = η obj r1 o r V t r wax 1 tr+1 t orzr, w ax tr Zr 1 or η obj r = 1 o r 2w ax lnn + 1/δ o r = t r Zr wax φ nt obj rw j,k φ t 1 obj w j,k When t = t r + 1, whch s a specal case arkng the begnnng of a new stage, the potental functon for constrant s defned by φ nt r = η c r 1 + tr cr γ and the potental functon for the objectve functon s gven by φ nt obj r = η obj r 1 tr orzr w ax Note that apart fro constants, the only dfference between o r and c r s that nstead of γ, o r has w ax /Zr The value Zr, as we defne below, gets progressvely updated, but wthn a sngle stage r reans the sae After stage r, the algorth coputes the optal objectve value e r to the followng nstance I r : the nstance I r has the t r requests of stage-r, and the capacty of resource s trc1+cr, e, the capacty of resources are scaled down accordng to the nuber of requests by a factor of tr, along wth a slght extra allowance by a factor of 1 + cr It uses e r to copute the value Zr + 1 to be used n the potental functon for objectve n stage r + 1 13
Lea 14 Wth probablty at least 1 2δ, t r W 1 c r e r t rw 1 + 2 c r Proof: Our goal s to establsh bounds on e r, whch s the optal soluton to the nstance I r For ths we consder the dstrbuton nstance of the followng nstance Îr: ths nstance s the sae as I r, except that each resource has a capacty of t r c / nstead of t r c 1 + c r/ We denote ths dstrbuton nstance of Îr by DÎr Let p r be the objectve value obtaned by the algorth Pure-rando on the nstance I r, e, the objectve value obtaned by followng the allocaton used by the optal soluton to DÎr Snce the resource usage and objectve value obtaned by the algorth Pure-rando can be seen as the su of d rando varables, we can apply Chernoff bounds on the Wth a probablty at least 1 δ, Pure-rando s usage of resource wll be wthn a factor of 1 + c r of trc and Pure-rando s objectve value wll fall short of t r W / whch s the optal objectve value for DÎr by a factor of 2w at ost 1 ax lnn+1/δ t rw Snce the nstance I r allows for ths extra resource usage by havng an addtonal capacty by a factor of 1 + c r as copared to Îr, the real excess resource usage occurs only wth probablty δ Thus, wth probablty 1 δ, resources are consued wthn capacty and the objectve value p r s at least p r t rw 2w ax lnn + 1/δ 1 t r W t rw 1 c r The soluton obtaned by Pure-rando s just a feasble soluton for I r, and thus p r s saller than the optal objectve value for I r, whch s e r Thus, wth probablty ore than 1 δ, we have, e r t rw 1 c r 6 We now get an upper-bound on e r To do ths, we consder the LP whch defnes e r, along wth ts dual The LP whch defnes e r s gven by: Consder the dual of LP 7 axze j I r w j,k x j,k 7, j I r,k a, j, kx j,k t rc 1 + c r j I r, x j,k 1 k j I r, k, x j,k 0 nze β j + t r1 + c r j I r α c 8 j I r, k, β j + a, j, kα w j,k, α 0 j I r, β j 0 The optal value of LPs 7 and 8 are the sae and equal to e r To upper bound e r, we now wrte down the LP for the dstrbuton nstance and ts dual The LP for dstrbuton nstance s gven by 14
axze j,k p j w j,k x j,k 9, j,k p j a, j, kx j,k c j, k x j,k 1 j, k, x j,k 0 and ts dual wrtten usng a dual ultpler of p j β j nstead of just β j s gven by nze j p j β j + α c 10 j, k, p j β j + p j a, j, kα p j w j,k, α 0 j, β j 0 Note that the set of constrants n LP 10 s a superset of the set of constrants n LP 8 Thus any feasble soluton to LP 10 s also feasble to LP 8, and n partcular, the optal soluton to LP 10 gven by βj s and α s s feasble for LP 8 So the optal value e r of LP 8 s upper-bounded by ts value at the soluton consttuted by the βj s and α sṫhe value of LP 8 at α, β s equal to βj + t r1 + c r j I r α c 11 To upper bound the expresson n 11, we apply Chernoff bounds to j I r βj Notce that by the constrants of LP 10, and by the fact that t s a nzaton LP, we get βj w ax Usng ths, we upper bound the expresson n 11 and hence e r as = βj + t r1 + c r α c j I r t r p j βj + 4t r p j βj w ax ln1/δ + t r1 + c r j j t rw + 2t rw c r α c where the frst nequalty s a wth probablty at least 1 δ nequalty, e, we apply Chernoff bounds for j I r βj along wth the observaton that βj w ax The second nequalty follows fro notng that j p jβj + α c = W Ths s because W s the optal value of the dstrbuton nstance and hence ts dual too Thus e r trw 1+2 cr wth probablty at least 1 δ Thus we have proved that the upper bound and the lower bound on e r hold wth probablty 1 δ each and hence together wth a probablty at least 1 2δ Ths proves the lea Usng these e r s, we defne our Zr + 1 as follows: Zr + 1 = e r t r 1 + 2 c r 15
Usng the bounds on e r n Lea 14, we note that Zr+1 W and that Zr+1 W 1 cr 1+2 cr W 1 3 c r Thus wth probablty at least 1 2 log1/δ, Zr satsfes these bounds for all r Gven the bounds on Zr, we use Lea 12 to see that wth a probablty at least 1 δ, the objectve value acheved n stage r s at least trzr 1 or, and the aount of resource consued n stage r s at ost trc 1 + cr Hence, these bounds are true for all r wth probablty at least 1 log1/δ, snce the total nuber of stages l = log1/ The total falure probablty s upper bounded by the su of the falure probablty durng estaton of Zr through e r, gven by 2 log1/δ and the falure probablty of our algorth n all stages together gven by log1/δ Thus, the total falure probablty s at ost 3 log1/δ Wth a probablty of 1 3 log1/δ, the algorth obtans an objectve value of at least l 1 r=0 and for each, the aount of resource consued s at ost l 1 r=0 t r Zr1 o r, t r c 1 + c r On settng γ = O 2 logn/, and δ = log1/, the above equatons can be splfed to the followng lea Lea 15 Wth probablty hgher than 1 O, the objectve value acheved at the copleton of the algorth s at least W 1 O and no resource s consued ore than ts capacty Thus Lea 15 proves that, our algorth acheves a copettve rato of 1 O for γ = O 2 proves Theore 2 6 Adwords n d settng logn/ and hence In ths secton, we gve a sple proof of Theore 5: greedy algorth acheves a copettve rato of 1 1/e n the adwords proble, where the pressons coe fro an adversaral stochastc nput odel As before, we llustrate our proofs for the d odel wth unknown dstrbuton below We now brefly descrbe the adwords settng Settng There are a total of n advertsers, and queres arrve onlne, fro soe pool of queres Let the unknown nuber of queres that arrve be The queres that appear each day are drawn d fro soe unknown dstrbuton Advertser bds an aount b j on query j Advertser has a budget B denotng the axu aount of oney that can be spent on a gven day The bd aounts b j are revealed onlne as the queres arrve The objectve s to axze the su of the bd aounts successfully allocated, subject to budget constrants Whenever a query j arrves, wth a bd aount b j > reanng budget of, we are stll allowed to allot that query to advertser, but we only earn a revenue of the reanng budget of, and not the total value b j Goel and Mehta [GM08] prove that the greedy algorth gves a 1 1/e approxaton to the adwords proble when the queres arrve n a rando perutaton or n d, but under an assupton whch alost gets down to bds beng uch saller than budgets We gve a uch spler proof for a 1 1/e approxaton by greedy algorth for the d unknown dstrbutons case, and our proof works rrespectve of the the relaton between the sze of the bds and the budgets nvolved Let p j be the probablty of query j appearng n any gven presson Let y j = p j Let x j denote the offlne fractonal optal soluton for the dstrbuton nstance Let w t denote the aount of oney spent by advertser at te step t, e, for the t-th query n the greedy algorth to be descrbed below Let f 0 = j b jx j y j Let f t = f 0 t r=1 w r Let ft = n =1 f t Note that f 0 s the aount spent by n the offlne fractonal optal soluton to the dstrbuton nstance Consder the greedy algorth whch allocates the query j arrvng at te t to the advertser who has the axu effectve bd for that query, e, argax n{b j, B t 1 r=1 w r} We prove that ths algorth obtans a revenue 16
of 1 1/e,j b jx j y j and thus gves the desred 1 1/e copettve rato aganst the fractonal optal soluton to the dstrbuton nstance The proof s slar to the proof we presented n Lea 8 for the resource allocaton proble Consder a hypothetcal algorth that allocates queres to advertsers accordng to the x j s We prove that ths hypothetcal algorth obtans an expected revenue of 1 1/e,j b jx j y j, and argue that the greedy algorth only perfors better Let w h t and f h t denote the quanttes analogous to w t and f t for the hypothetcal algorth, wth the ntal value f h 0 = f 0 = j b jx j y j Let f h t = n =1 f h t Let EXCEED t denote the set of all j such that b j s strctly greater than the reanng budget at the begnnng of te step t, naely b j > B t 1 r=1 wh r Lea 16 E[w h t f h t 1] f h t 1 Proof: The expected aount aount of oney spent at te step t, s gven by If j EXCEED t E[w h t f h t 1] = x j y j 1, then by 12, j EXCEED t t 1 B w h xj y j r + x j y j b j 12 r=1 j / EXCEED t E[w h t f h t 1] B t 1 r=1 wh r f h0 t 1 r=1 wh r = f h t 1 Suppose on the other hand x j y j < 1 We can wrte E[w ht f h t 1] as Snce b j B, and j EXCEED t E[w h t f h t 1] = f h 0 j EXCEED t j EXCEED t x j y j < 1, 13 can be splfed to t 1 b j B w h xj y j r 13 r=1 E[w h t f h t 1] > f h0 = f h t 1 t 1 r=1 wh r Lea 17 The hypothetcal algorth satsfes the followng: E[f h t f h t 1] f h t 11 1/ Proof: Fro the defnton of f h t, we have f h t = f h t 1 w h t E[f h t f h t 1] = f h t 1 E[w h t f h t 1] f h t 11 1, where the nequalty s due to Lea 16 Sung over all gves the Lea Lea 18 E[GREEDY] 1 1/e,j b jx j y j Proof: Lea 17 proves that for the hypothetcal algorth, the value of the dfference f h t 1 E[f h t f h t 1], whch s the expected aount spent at te t by all the advertsers together, condtoned on f h t 1, s at least f h t 1 But by defnton, condtoned on the aount of oney spent n frst t 1 steps, the greedy algorth earns the axu revenue at te step t Thus, for the greedy algorth too, the stateent of the lea 17 ust hold, naely, 17