Department of Computer Science Stanley P. Y. Fung Online Preemptive Scheduling with Immediate Decision or Notification and Penalties
Motivation Preemption is not really “free” E.g. context switch in CPU scheduling Introduce preemption penalty (PP): each preemption introduces a penalty proportional to value of job Related but different scenario: we want commitment to process a job If a job cannot be processed, better to let the client know early Introduce non-completion penalty: if an accepted job is later not finished, gives a penalty
Immediate Decision/Notification How to model “commitment”: Immediate Notification (IMM NOTIF): Upon a job’s arrival, have to decide immediately whether to accept the job, or reject it right away If rejected, no value (customer goes away) If an accepted job is eventually not completed, incurs a penalty (cannot charge customer since work not done; in addition, customer is unhappy when he finds this out) Immediate Decision (IMM DEC): A stronger notion where in addition, an accepted job needs to be immediately allocated a time window for execution (“your delivery will take place on Friday morning”)
Problem Formulation Input: a set of jobs Parameters: release time r(j), deadline d(j), length p(j), weight w(j) Value of a job v(j) = p(j) x w(j) Unit length (UL) case: p(j) = 1 for all j Unit weight (UW) case: w(j) = 1 for all j Objective: Find a schedule with maximum total value of completed jobs, minus any penalties Penalty model: penalty = × v(j) Preemptive We mainly consider the restart model But also some on resume model (Doesn’t matter in IMM DEC)
Online Algorithms Jobs arrive online No information about future Decisions irrevocable Measure of performance: competitive ratio An online algorithm A is c-competitive if for any input instance I, OPT(I)/A(I) ≤ c Where OPT = offline optimal algorithm, A(I) = value of A on instance I
Previous and Our Results (PP) Upper boundLower bound UW (resume) No penalty4 [BKM+92]4 [Woeg 94, BKM+92] Penalty 4(1+ ) (resume and restart) 4(1+ ) UL (restart) No penalty 4.56 [ZFC+06], 4 for tight jobs 4 [Woeg 94] penalty [ZXDZ05] 4(1+ )
Previous and Our Results (PP) Upper boundLower bound Resume, k = max w / min w No penalty [KS95] [BKM+92] Penalty Restart, = max p / min p No penalty O( /log ) [Ting08] ( /log ) [ZFC+06, Ting08] Penalty (1+ ) + o( ) [ZXP09]
Previous and Our Results (IMM) Previous results: IMM DEC + UW: an upper bound from [TL09] No other prior result for preemptive case Non-preemptive case considered in various papers (usually UL, and parallel machines) Our results: IMM DEC (or IMM NOTIF + restart) UW: UL: Lower bound 4(1+ ) carries over to here Lower bound with laxity
Tight jobs: PP = IMM NOTIF = IMM DEC No algorithm gives competitive ratio better than 4(1+ ) Idea: (similar to [BKM+92, Woeg94]) If switch to short job, stop If not preempt from v i to v i+1, UW Lower Bound (PP/IMM) v0v0 v1v1 v2v2
Lemma: if < 4(1+ ), {v i } cannot be strictly increasing forever Let v n be the last term. Then If ONL is about to finish v n, release another job of value v n just before it finishes Lower Bound (cont’d) v n-1 vnvn vnvn
PP + UW: Algorithm 4(1+ )-competitive algorithm: Let = 2(1+ ) 2, = 2( – 1) Job arrival: add to pool When idle, start the pending job with earliest expiry time While j is running, if another job k reaches its expiry time: If v(k) > v(j) and j has preempted other jobs before, or If v(k) > v(j) and j has not preempted others before Then preempt j and start k; j never considered again Else continue with j (and k will expire)
Ideas of Proof Basic subschedules: sequence of preemption followed by one job completion v k-1 ≤ v k / , v k-2 ≤ v k-1 / , …, v 1 ≤ v 2 / Value obtained = v k – (v 1 +…+v k-1 ) Times in OPT when ONL is busy OK (1 to 1) Times in OPT when ONL is idle For each (part of) such job j executed by OPT, map it to some basic subschedule in ONL v1v1 vkvk
Ideas of Proof When OPT starts (part of) a job j while ONL idle: If ONL is idle at x(j) (expiry time of j), only possible if j started/completed before its expiry j is the first job in a basic subschedule map to it Otherwise, x(j) falls within some basic subschedule (v 1 …v k ) map to it can show total of v(j) < v k (or v 1 if k=1) Each basic subschedule is mapped by: OPT ONL v 1 +…+v k v1v1 vkvk v1v1 vkvk
IMM DEC + UW: Algorithm At any time, the algorithm maintains the provisional schedule of accepted jobs Execute according to provisional schedule When a new job j arrives: Identify a set of jobs { j i } in the provisional schedule so that their preemption create a long enough timeslot to fit the new job, and that v(j i ) < v(j) / If no such set exists, reject j Else preempt those j i, schedule j in the new empty interval (as early as possible)
Properties of Provisional Schedule Theorem: -competitive Simple but useful properties of provisional schedule: “No gap”, i.e. no idle period between planned jobs in it “monotone”, i.e. once a time t becomes busy, won’t become idle later on If a preemption happens, it preempts a suffix of the provisional schedule Only linear number of potential job sets to consider (prov. sch.)
Ideas of Proof Classify jobs in OPT into four sets: C: Completed in both OPT and ONL R: Rejected by ONL (upon arrival) D: Discarded by ONL (preempted before getting started) A: Aborted by ONL (preempted while execution) Jobs in R: Cannot start in idle time in ONL If start in a busy period with completed jobs j 1,…,j m, then v(j) < v(j i ) OPT A j1j1 j2j2 < (v(j 1 )+v(j 2 ))
Suppose an R-job j starts at t when ONL is idle S t [t] is empty. By monotone property, S r(j) [t] is also empty By no-gap property, S r(j) [u] is all empty for u > t Hence j could not be rejected Suppose j starts at t when ONL is busy Rejected due to a set X of jobs in S r(j) [t] s.t. v(j) < v(X) Jobs in X may be preempted but replaced by other (higher- value) jobs in the same busy period (no-gap) Eventually goes to one of j i ’s Bounding R jobs t j j1j1 j2j2 < (v(j 1 )+v(j 2 ))
Ideas of Proof Consider one busy period Let V(C) = V Jobs in D/A: Can prove v(D) + v(A) V/( – 1) Jobs in R: v(R) L + V where L = length of b.p. v(C) + v(A) (1+1/( – 1))V OPT v(C) + v(A) + v(D) + v(R) ONL v(C) – (v(D)+v(A)) Competitive ratio = OPT/ONL, choose a optimised
IMM DEC + UL Same algorithm Provisional schedule different properties: Each preemption preempts either one job, or two including the currently executing one Efficient identification of candidate preemption sets Theorem: -competitive Proof: Similar ideas for C, A, D Different bound for R
Proof for UL Case Each job in R is “blocked” by at most 2 jobs in provisional schedule (i.e. jobs in C/A/D) Total value (value of R-job)/ Distribute the values to them Each job in C/A/D “blocks” at most two jobs in R (i.e. in OPT) V(R) 2 ( v(C)+v(A)+v(D) ) Prov.sch. OPT v1v1 < (v 1 +v 2 ) v2v2 v1v1 v2v2 v1v1 v1v1 v1v1
Discrete time steps, p(j) = 1 = time step size No issue of preemption Without immediate decision/notification: unit job scheduling problem Motivation: buffer management in QoS switches Previous results: Deterministic UB 1.828, LB Randomized UB 1.582, LB 1.25 Various restricted cases, adversary models etc Unit Jobs
Algorithm: Keep a provisional schedule S When each job j arrives: Find the earliest slot u before d(j) in S such that w(S t [u]) < w(j)/ for some . Evict this job (call it j’) and j takes its place. If no such slot, reject j Repeat above for the evicted job j’ (but with different condition w(S t [u]) < w(j’)), the subsequently evicted j’’, … -competitive Lower bound? Unit Jobs + IMM NOTIF
Algorithm: Keep a provisional schedule S When a new job j arrives, find any slot u such that w(S t [u]) < w(j)/ for some Replace S t [u] with j 4(1+ )-competitive ( ) lower bounds Unit Jobs + IMM DEC
Future Work PP + UL: gap still exists E.g. between (4, 4.56) for restart without penalty Better bounds for IMM DEC + UL (=UW?) Upper/lower bounds to separate the competitive ratio of IMM DEC / IMM NOTIF / no-commitment Algorithms here works for IMM NOTIF, but trivially Unit jobs lower bounds IMM DEC but limited re-scheduling / wider interval / …