New Online Algorithms for Story Scheduling in Web Advertising

Transcription

1 New Online Algorithms for Story Scheduling in Web Advertising Susanne Albers TU Munich Achim Paßen HU Berlin

2 Online advertising Worldwide online ad spending 2012/13: $ 100 billion Expected to surpass print ad spending soon Display advertising: images, videos, animations Content shown depending on browsing history of user

3 Story boarding

4 Story boarding Maintain ad position of a web site during browsing session of a user. Position depicts image sequences of advertisers. Advertiser pay unit shown. Depending on history/state user becomes interesting for advertisers. Maximize revenue of session.

5 Model Session time is slotted Time t: user continues surfing with probability β 0 < β < 1 I = J 1,...,J N J i = (a i,l i,v i ) a i = arrival time l i = length v i = per-unit value Preemption: Job J i may be scheduled for < l i time units max t 0β t v(t) v(t) = per-unit value of job scheduled at time t Dasgupta, Ghosh, Nazerzadeh, Raghavan SODA 09

6 Competitive analysis A: Online algorithm OPT: Offline algorithm A(I) OPT(I) A is c-competitive if for all I A(I) 1 c OPT(I).

7 Previous results Lower bound: c 2 for general β c β +β 2 Upper bound: c = 7 ALG: t v i of current J i loss in delaying J k with v k > v i for 1 time unit Dasgupta, Ghosh, Nazerzadeh, Raghavan SODA 09

8 Previous results Lower bound: c 2 for general β c β +β 2 Upper bound: c = 7 Jobs to be scheduled immediately upon arrival c = Ω( logµ/loglogµ) µ = max{l max /l min, v max /v min } Total value of v i gained only if entire J i is shown c = O(log(v max /v min )) Dasgupta, Ghosh, Nazerzadeh, Raghavan SODA 09

9 Our contribution Upper bound: c = 4/(2 β) Upper bound: c = 1+Φ 2.62 Φ = Golden Ratio Problem extension: Web page with m ad positions, where stories can be shown simultaneously; job migration not allowed max m β t v(t,j) t 0 j=1 v(t,j) = per-unit value of job scheduled on ad position j at time t Upper bound: /(2 2) c 1/(3 2 2) 5.828

10 Algorithmic approach Time: P 1 P 2 P 3 0 k 1 k 2k 1 2k 3k 1 Phases P 1,P 2,P 3,... of k consecutive time steps. Scheduling decisions are made at the beginning of the phase. Jobs arriving during the phase are ignored.

11 Simple algorithm Time: Q n : P n (n 1)k nk 1 ALG1: Phase P n : Q n = {unscheduled jobs J i with a i (n 1)k} Schedule jobs of Q n in order of non-increasing per-unit value Preempt job at the end of P n Thm: c = 1/(β k 1 (1 β k )) c = 4/(2 β) for k = log β 2 c = 1/(1 β) for k = 1

12 Refined algorithm Time: Q n : P n (n 1)k nk 1 ALG2: Phase P n : J n = remainder of last job in P n 1 Q n = {J n + unscheduled jobs J i with a i (n 1)k} Schedule jobs of Q n in order of non-increasing per-unit value If J n contained in schedule, move it to the front Thm: c = 1/β k 1 max{1/β k 1,1/(1 β 2k ),β 3k /(1 β k )} c = 1+Φ 2.618, where Φ = (1+ 5)/2 for k = 1 2 log β(1+φ) +1

13 Refined algorithm Time: Q n : P n (n 1)k nk 1 ALG2: Phase P n : J n = remainder of last job in P n 1 Q n = {J n + unscheduled jobs J i with a i (n 1)k} Schedule jobs of Q n in order of non-increasing per-unit value If J n contained in schedule, move it to the front Thm: c = 1/β k 1 max{1/β k 1,1/(1 β 2k ),β 3k /(1 β k )} c = 1+Φ 2.618, where Φ = (1+ 5)/2 for k = 1 2 log β(1+φ) +1

14 Algorithm for m ad positions Q n : Pos. 1 Pos. 2 (n 1)k nk 1 ALG3: Phase P n : Q n = {unscheduled jobs J i with a i (n 1)k} For t = (n 1)k,...,nk 1, schedule m jobs of highest per-unit value Preempt jobs at the end of P n Implementation: Units of a job are placed on same ad position

15 Algorithm for m ad positions Q n : Pos. 1 Pos. 2 (n 1)k nk 1 ALG3: Phase P n : Q n = {unscheduled jobs J i with a i (n 1)k} For t = (n 1)k,...,nk 1, schedule m jobs of highest per-unit value Preempt jobs at the end of P n Thm: c = 1/β k 1 (1+1/(1 β k )) c = (1+1/(1 β(2 2)))/(2 2) for k = log β (2 2) /(2 2) c 1/(3 2 2) 5.828

16 Simple algorithm Time: Q n : P n (n 1)k nk 1 ALG1: Phase P n : Q n = {unscheduled jobs J i with a i (n 1)k} Schedule jobs of Q n in order of non-increasing per-unit value Jobs of same value sorted in order of increasing arrival time Preempt job at the end of P n

17 Analysis simple algorithm I = J 1,...,J N J i = (a i,l i,v i ) k-quantized input I k = J 1,...,J N J i = (a i,l i,v i ) a i = k a i/k P n (n 1)k nk 1

18 Analysis simple algorithm Lemma: OPT(I k ) β k 1 OPT(I) Proof: Shift optimal schedule for I by k 1 time units to the right. Observation: ALG1(I k ) = ALG1(I)

19 Relaxed offline algorithm CHOP : Optimal algorithm that may resume preempted jobs Always schedule job with highest per-unit value. Jobs with same per-unit value are scheduled in the same order as in ALG1. Observation: CHOP(I k ) OPT(I k )

20 Timing property S = ALG1 s schedule for I k S = CHOP s schedule for I k t S (i) = start time of J i in S t S (i) = start time of J i in S Lemma: t S (i) t S (i)

21 t S (i) t S (i) Let J i be first job in S with t S (i ) > t S (i ) = t t J i CHOP ALG1: J i v i v i At time t CHOP has finished J i v i > v i : v i = v i : CHOP schedules jobs of value v i in the same order as ALG1 t S (i) t l i t S (i) t l i +1

22 Phase analysis P n Time: (n 1)k nk 1 J n = {jobs scheduled by ALG1 in P n } ALG1(P n ) = value achieved for J n ALG1(I k ) = n ALG1(P n ) CHOP(P n ) = value achieved for J n Lemma: CHOP(I k ) = n CHOP(P n ) Proof: For any J i scheduled by CHOP, {jobs scheduled by CHOP} n t S (i) t S (i) J n

23 Phase analysis P n Time: (n 1)k nk 1 Lemma: CHOP(P n ) ALG1(P n )/(1 β k ) Proof: J = last job of P n with value v CHOP(P n ) ALG1(P n )+extra value in scheduling preempted portion of J ALG1(P n ) = nk 1 t=(n 1)k β t v = (β (n 1)k β nk )/(1 β) v extra value = β t v = β nk /(1 β) v t nk CHOP(P n )/ALG1(P n ) 1+β nk /(β (n 1)k β nk ) = 1/(1 β k )

24 Wrapping up CHOP(I k ) ALG1(I k )/(1 β k ) β k 1 OPT(I) CHOP(I k ) OPT(I) ALG1(I)/(β k 1 (1 β k ))

25 Analysis refined algorithm Take care of delays if last job of previous phase is continued. Loss of preempted jobs have to amortized over several phases; segments of up to three phases.

26 Analysis m ad positions S value is at least as high as that of optimal schedule S schedules up to 2m jobs at any time t: each J i scheduled in S but not in S can be mapped (a) to a unit v i v i scheduled in S or (b) to a job preempted by S at time t < t

27 Open problems Tight bounds for deterministic algorithms Design randomized algorithms