on the value of preemption in schedulinggilsh/presentations/on_the... · 2021. 4. 20. · outline...

OutlineOverview + Uses

The ModelAlgorithm - Upper Bounds

Lower BoundsSummary

Open Problems

On the Value of Preemption in Scheduling

Yair Bartal1 Stefano Leonardi2 Gil Shallom1

Rene Sitters3

1Department of Computer Science, Hebrew University, Jerusalem, Israel

2Dipartimento di Informatica e Sistemistica,Universit di Roma La Sapienza, Rome, Italy

3Max-Planck-Insitut fur Informatik, Saarbrucken, Germany

TAU Algorithms Seminar, 2007

Bartal,Leonardi,Shallom,Sitters On the Value of Preemption in Scheduling



Lower BoundsSummary

Open Problems

Outline of Topics

Overview + Uses

The Model

Algorithm - Upper Bounds

Lower Bounds

Summary

Open Problems




Lower BoundsSummary

Open Problems

Overview

I Preemptive scheduling algorithms can achieve efficientperformance.

I Example: SRPT is optimal for minimizing total flow timeI Assumption:

I preemption is costlessI unbounded number of preemptions per job

I Real life: Preemption incurs a significant overhead.




Lower BoundsSummary

Open Problems

Focus

I A new model where preemption is costlyI New preemptive scheduling algorithm

I take the cost of preemption into account




Lower BoundsSummary

Open Problems

Competitive analysis

A common technique for measuring effectiveness of onlinealgorithms, by comparing the performance of an online algorithm,for any input sequence, to that of the optimal offline algorithm

ALG (I ) ≤ C · OPT (I ) + α




Lower BoundsSummary

Open Problems

Standard Scheduling Problems

I DefinitionsI Flowtime of Job j: Fj = Cj − rjI Total Flowtime:

∑j∈J Fj

I Goal Function: min∑

j∈J Fj




Lower BoundsSummary

Open Problems

SRPT

SRPT Algorithm (Shortest Remaining Processing Time)

I Optimal with respect to total flowtime in the single machinecase (in the standard model)




Lower BoundsSummary

Open Problems

The Standard Model Assumptions

I No info is available regarding future jobsI Online Algorithms (hard to plan)

I All job info is available on its arrivalI Relevant job info (job processing time)

I Preemption is costlessI Model does not take into account resources wasted due to

preemption




Lower BoundsSummary

Open Problems

The Model

I Standard ModelI Minimizing total flowtime∑

j∈J

Fj

I New ModelI Preemption cost is KI Minimize the sum of total flowtime and the total cost of

preemptions ∑j∈J

Fj + K ·M




Lower BoundsSummary

Open Problems

Model Boundaries

I Cost of preemption is K∑j∈J

Fj + K ·M

I K = 0 =⇒ standard model

I K is large =⇒ non-preemptive model




Lower BoundsSummary

Open Problems

WTP Algorithm (Wait to Preempt)

I When starting a new job, choose the job with the shortestremaining processing time. (Which job?)

I If there is an idle machine and a job is available, then startprocessing the shortest of the available jobs. (When toexecute?)

I Denote ε as the processing time of the shortest job

I Preempt a job as soon as it has been processed withoutinterruption for exactly

√Kε time units. That is, every job is

partitioned in parts of length√

Kε. (When to preempt?)




Lower BoundsSummary

Open Problems

WTP Algorithm (Wait to Preempt)

I One proof for single and parallel machines

I WTP competitive ratio:

(C (α+ 1) + α)

I SRPT competitive ratio:where C = 1 or C = O(log(min n

m , ρ))and α = minρ,

√κ , κ = K

ε

I ρ is the ratio between the longest and shortest job




Lower BoundsSummary

Open Problems

WTP Analysis

1. Total Flowtime ∑j∈J

Fj

2. Total cost of preemption

K ·M

∑j∈J

Fj + K ·M




Lower BoundsSummary

Open Problems

Total cost of preemption

I At mostpj√Kε

preemptions for job j

I Total cost WTPP(I ) ≤ K (P

pj )√K ε

=√κ∑

pj

OPT(I ) ≥∑

pj

I WTPP(I ) ≤√κ OPT(I ) ≤ α ·OPT(I )

α = minP,√κ




Lower BoundsSummary

Open Problems

Total Flowtime - Analysis Steps

I Jobs can arrive at any time, but WTP is not always allowed topreempt a job (unlike SRPT)

I For the analysis, We create a new schedule and shift the jobrelease times to the times when WTP is also allowed topreempt

I Now SRPT(I) and WTP(I) act the same on this new inputsequence

I We finally show that the flowtime incurred due to the shift isalso bounded




Lower BoundsSummary

Open Problems

Shifting jobs

1. Create a new schedule I ′ from II Take any optimal schedule for I and shift it forward in time

over αε time units. The resulting schedule is feasible for I ′

2. Result: WTPF (I ′) = SRPT (I ′)

3. Flowtime cost for waiting is r ′j − rj for job j




Lower BoundsSummary

Open Problems

Total Flowtime - The Analysis

WTPF (I ) = WTPF (I ′) +∑

j

(r ′j − rj) =

= SRPT(I ′) +∑

j

(r ′j − rj) ≤

≤ C ·OPT(I ′) +∑

j

(r ′j − rj)




Lower BoundsSummary

Open Problems

Total Flowtime - The Analysis

OPT(I ′) ≤ OPT(I ) +∑

j

(αε+ rj − r ′j )

WTPF (I ) ≤ C (OPT(I ) + nαε) + (C − 1)∑

j

(rj − r ′j )

≤ C (α+ 1)OPT(I ).




Lower BoundsSummary

Open Problems

Upper Bound Result

WTP = WTPP(I ) + WTPF (I ) ≤ (C (α+ 1) + α)OPT(I )




Lower BoundsSummary

Open Problems

WTP: m machines

Algorithm Wait To Preempt (WTP):

(i) If there is an idle machine and a job is available, then startprocessing the job with the shortest remaining processing timeamong the available jobs.

(ii) Every moment t that an additional m√

Kε units (summedover all machines) are processed do the following: If themaximum processing time over all jobs released so far is morethan

√Kε preempt all jobs; Otherwise, do nothing.




Lower BoundsSummary

Open Problems

WTP analysis (changes for m machines)

If Pmax ≤√

Kε then no job gets preempted. The preemption costWTPP(I ) is zero in this case. Otherwise we use the sum of allprocessing times as a lower bound for the optimum.




Lower BoundsSummary

Open Problems

WTP analysis (changes for m machines)

The number of preemptions is at mostmb∑

j pj/(m√

Kε)c ≤ OPT(I )/(√

Kε). The preemption cost is at

most K ·OPT(I )/(√

Kε) = OPT(I )√κ.

Further, the assumption Pmax >√

Kε equals ρ >√κ which

implies α =√κ. The preemption cost is at most αOPT(I ).




Lower BoundsSummary

Open Problems

Lemma: Schedule I’

LemmaSchedule σ satisfies the shortest remaining processing time rule forinstance I’.

We need to show:

At any moment t the SRPT-rule is satisfied for schedule σ with respect

to instance I ′




Lower BoundsSummary

Open Problems

Definition: Schedule I’

I Let σ be a schedule produced by WTP for instance I

I Let Sj be the start time of job j in σ

Define new release time r ′j = minSj , rj + αε to create instance I’




Lower BoundsSummary

Open Problems


Assume that at time t job j is processed and job k is available ininstance I ′ but is not processed. We have to show the followingproperty:

At time t the remaining processing time of job j is at most theremaining processing time of job k




Lower BoundsSummary

Open Problems


2 cases:

I α = ρ (i.e., ρ ≤√κ)

I α =√κ (i.e., ρ ≥

√κ)




Lower BoundsSummary

Open Problems

Lemma: Schedule I’: α = ρ

If α = ρ then Pmax = ερ ≤ ε√κ =

√Kε =⇒ no job gets

preempted since preemption is expensive =⇒ job k does not getpreempted whence it is not processed before time t.Assumption: job k is available in I ′ at time t but does not start attime t. Thus r ′k < Sk , the start time of job k. Now, using thedefinition of I ′, we must have r ′k = minSk , rk + αε = rk + αε.On the other hand, the start time of job j is at leastt − pj ≥ t − Pmax = t − ρε = t − αε ≥ r ′k − αε ≥ rk . Since thealgorithm started j while k was available we have pk ≥ pj which isobviously at least the remaining processing time of job j at time t.Hence, the property holds in this case.




Lower BoundsSummary

Open Problems

Lemma: Schedule I’: α =√

κ (i.e., ρ ≥√

κ)

If job k was processed before time t then it must have beenpreempted. Consider the last time before t that job k waspreempted. When a job gets preempted then all jobs getpreempted at that time. Therefore, the segment of job j that isprocessed at time t started after job k was preempted. Theproperty now follows directly from the algorithm (point (ii)).




Lower BoundsSummary

Open Problems

Lemma: Schedule I’: α =√

κ (i.e., ρ ≥√

κ)

If job k is not processed before time t then we have r ′k < Sk , sinceit is available in I ′ at time t but does not start at t. By definitionof I ′ we must have r ′k = rk + αε, implying rk ≤ r ′k − αε ≤ t − αε.If the segment of job j (that is processed at time t) started laterthan rk then the claim obviously holds since the algorithm started(or resumed) the processing of j while job k was available. Soassume the segment started before time rk . We show that thisgives a contradiction. Clearly, there can be no idle time between rkand t since the algorithm has the choice to start job k. Since thereis no idle time, at least m(t − rk) ≥ mαε = m

√Kε units are

processed in the interval [rk , t] and it will contain at least onepreemption point. At this point all jobs get preempted since job jis processed and pj ≥ t − rk ≥ αε =

√κε =

√Kε




Lower BoundsSummary

Open Problems

Lower Bounds

What we are going to see:

I Lower bound 1:Idle Time Not Allowed Model

I Lower bound 2: Idle Time Allowed Model

I Inapproximability




Lower BoundsSummary

Open Problems

Idle Time Not Allowed Model

I We show a lower bound on the competitive ratio achievable byany online algorithm due to the online nature of the problem.

I The algorithm has to choose an action with information onlyregarding the past and therefore is bound to make mistakes.

I We construct an example which makes the algorithm performat its worst.

Ω(α), where α = minP,√κ




Lower BoundsSummary

Open Problems


0 x 2x 3x

N processes

4x

A

B

(a) The job release sequence in the first case

N processes

0 x 2x 3x 4x

A

B

(b) The job release sequence in the second case

Figure: Idle Time Not Allowed Release Sequences




Lower BoundsSummary

Open Problems


0 x 2x 3x 4x

N processes

N processes

AlgA

OptAB

B

(a) The algorithm chooses to executethe longer process first

0 x 2x 3x 4x

N processes

Alg

OptA

N processes

B

AB

(b) The algorithm chooses to executethe shorter process first

Figure: Idle Time Not Allowed Execution Sequences




Lower BoundsSummary

Open Problems


Either many machines choose to:

I Execute Job A first and incur a significant cost

I Execute Job B and incur a significant cost

I Preempt and incur a significant cost due to preemption

The lower bound result:





Lower BoundsSummary

Open Problems

Idle Time Not Allowed Model: proof

I Set x = min√

K ε, ερ2

I At time zero, release m jobs of type A and m jobs of type B

I No Idle Time =⇒ Start time of a job is a multiple of x

I Job starts at time t ≥ 3x ⇐⇒ Machine idle before time 3x=⇒ All jobs start at time t ≤ 3x




Lower BoundsSummary

Open Problems


I Let m1,m2,m3 be the expected number of jobs of type B thatstarts respectively, at time 0, x and 2x . We must havemk ≥ m/3 for some k ∈ 1, 2, 3

I Adversary releases N ·m small jobs of size ε each, beginningat time kx

A batch of m jobs is released at time kx + iε, for 0 ≤ i ≤ N − 1 where N = xε




Lower BoundsSummary

Open Problems


x 2 x 3 x 4 x0

AAA

BBBB

A B

A

Figure: The algorithm’s schedule if no small jobs would be released. Attime 2x there are still three B jobs with a remaining processing time of x .The adversary releases many small jobs (dotted in figure) at time 2x .




Lower BoundsSummary

Open Problems

Idle Time Not Allowed Model: proof: comparison toadversary

The adversary does not preempt any job and executes all smalljobs at their release date

I If k = 1 then the adversary completes all A-jobs at time x andall B-jobs at time 4x . The total flow time is at most 6xm inthis case.

I Similarly, the total flow time is 7xm for k = 2

I 5xm for k = 3

shortly explain the cost calculation due to which jobs are waiting




Lower BoundsSummary

Open Problems


0 x 2x 3x 4x

N processes

N processes

AlgA

OptAB

B

(a) The algorithm chooses to executethe longer process first

0 x 2x 3x 4x

N processes

Alg

OptA

N processes

B

AB

(b) The algorithm chooses to executethe shorter process first

Figure: Idle Time Not Allowed Execution Sequences




Lower BoundsSummary

Open Problems

Idle Time Not Allowed Model: proof: RandomizedAlgorithm

Consider a schedule σ given by the randomized algorithm A. Let Sbe the set of B-jobs that start processing at time (k − 1)x . Attime kx the remaining processing time of these jobs is x .Let b be the number of jobs that are preempted.




Lower BoundsSummary

Open Problems


Since at least |S | − b machines are not available for processing thesmall jobs, the total flow time of small jobs is more than(|S | − b)Nx/2.

The total cost is at least bK + (|S | − b)Nx/2 =bK + (|S | − b)x2/(2ε) ≥ bK + (|S | − b)K/2 ≥ |S |K/2




Lower BoundsSummary

Open Problems


Since E |S | ≥ m/3 the expected cost is at least mK/6. Weconclude that the expected competitive ratio is at leastmK/(42mx). Substituting K ≥ x2/ε yields the lower boundΩ(x/ε) = Ω(min

√κ, ρ)




Lower BoundsSummary

Open Problems

Idle Time Allowed Model

I The algorithm is allowed not to execute a job even if an idlemachine is available

I Even in this case the lower bound is the same:

The lower bound result:





Lower BoundsSummary

Open Problems


First we prove the deterministic case and then sketch how toextend it for randomized algorithms. Let A be any deterministicalgorithm and define N = 1

2 min √κ, ρ) and let x = εN. We

define an instance with mN large jobs of length 2x and a numberof batches of small jobs. A set of m large jobs is released at eachtime point 0, 4x , 8x , . . . , (N − 1)4x . We refer to the interval[(i − 1)4x , i4x [ as segment i .




Lower BoundsSummary

Open Problems


Depending on the algorithm the adversary may decide to release aset of mN2 small jobs, of length ε each, shortly after a set of largejobs. Consider the situation at time point t = (i − 1)4x , i.e., atthe beginning of segment i , for some (i ∈ 1, . . . ,N). Let σ bethe schedule that A constructs if no more jobs are released aftertime t. Let S be the set of large jobs in σ that start processing insegment i and are scheduled without preemption. If |S | ≥ m/2 wesay that the segment is of type 1. If |S | < m/2 we say thesegment is of type 2.




Lower BoundsSummary

Open Problems


If |S | ≥ m/2 then there is a j ∈ 1, 2, 3, 4 such that at least|S |/4 ≥ m/8 jobs from S start processing in interval[t + (j − 1)x , t + jx ]. The adversary will release mN small jobsstarting from time t + ix . More precisely, m jobs of length ε eachare released at every time point t + ix + kε for allk ∈ 0, . . . ,N − 1.If |S | < m/2 then no small jobs are released after the large jobs.




Lower BoundsSummary

Open Problems


We construct a schedule with small cost in order to lower boundthe optimal schedule. All small jobs are processed withoutpreemption and start at their release time. The flow time of onesmall job is ε and the number of small jobs is at most mN2 givinga total flow time of at most mN2ε. Next we add the large jobswithout preemption to the schedule. We can do this such that theflow time of any large job is at most 5x . (In any interval[(i − 1)4x + x , i4x + x ] there is at most one batch of small jobswhich block this interval for a time x.) The total flow time of thisschedule is mN2ε+ mN5x = 6mNx . Denote this value by z∗.Notice that the contribution of the large jobs is at most 5 timesthe contribution of the small jobs.




Lower BoundsSummary

Open Problems


In order to be better than Ω(N) competitive the algorithm mustcomplete almost all large jobs before time 4Nx . More precisely, ifαmN large jobs are completed after time 4Nx (for some α < 1),then the average flow time for these jobs is at least 4Nxα/2. Thetotal flow time for these jobs will be at least αmN times 4Nxα/2is 2α2mN2x = z∗Nα2/3.




Lower BoundsSummary

Open Problems


Similarly, in order to be better than Ω(N) competitive thealgorithm must schedule almost all large jobs without preemption.More precisely, if αmN large jobs are preempted (α < 1), then thetotal cost is at leastαmNK = αmNεκ ≥ αmNε4N2 = 4αmN2x = 2

3αz∗N.




Lower BoundsSummary

Open Problems


We conclude that in order to be better than Ω(N) competitive thealgorithm must process almost all large jobs without preemptionand before time 4xN. Let U be the set of all large jobs with thisproperty. By definition at most m/2 jobs from U start in anysegment of type 2. In general, at most 4m jobs from U can start ina segment since no more jobs fit in. Let N1 be the number ofsegments of type 1. Then (N − N1)m/2 + N14m ≥ |U|, implyingN1 ≥ (2|U|/m − N)/7. For example, if |U| ≥ (17/20)mN thenN1 ≥ N/10. Hence, we now assume that at least ten percent ofthe segments is of type 1.




Lower BoundsSummary

Open Problems


Since we assumed that at least N/10 segments are of type 1, thetotal cost is at least N/10 times 3z∗/16 is 3z∗N/160. Weconclude that there is a constant c such that the cost of anyschedule is at least cNz∗.




Lower BoundsSummary

Open Problems


The lower bound extends easily to randomized algorithms. We useYao’s min-max principle, i.e., again we consider a deterministalgorithm but now we randomize the input. The set of large jobsremains the same but now a batch of small jobs is released forevery segment i at time ti , where ti is uniformly taken from the set(i − 1)4x + jx | j = 1, 2, 3, 4. The optimal schedule remains atmost z∗. Again we can argue that, to be better thanΩ(N)-competitive, the algorithm must process almost all jobswithout preemption and before time 4xN. But in that case theexpected flow time of the small jobs will be Ω(N)z∗




Lower BoundsSummary

Open Problems

Inapproximability

I Using Woeginger et al. Ω(√

n) lower bound on theapproximation ratio of polynomial-time approximationalgorithms for the problem of minimizing flow time in anon-preemptive environment

I We get an inapproximability result (in our model) of:

Ω(min

κ

14−δ,P

13−δ)




Lower BoundsSummary

Open Problems

Inapproximability

Kellerer et al. [?] showed a lower bound of Ω(√

n) on theapproximation ratio of polynomial-time approximation algorithmsfor the problem of minimizing flow time in a single-machinenon-preemptive environment. In our case preemption may be usedbut incurs a cost K . Therefore, we would use their proof in orderto achieve an inapproximability result which is a function of theparameters of our problem, that is, ρ and κ.




Lower BoundsSummary

Open Problems

Inapproximability: The NP-hard problem

Problem: Numerical three-dimensional matching (N3DM).Instance: Positive integers ai , bi and ci , 1 ≤ i ≤ ` , with∑`

i=1(ai + bi + ci ) = `D.Question: Do there exist permutations π, ψ such thatai + bπ(i) + cψ(i) = D holds for all i?




Lower BoundsSummary

Open Problems

Inapproximability

Given an arbitrary N3DM instance and some real number0 < γ < 1

2 , Kellerer et al. define the numbers

n =⌈(20`)

4γ D

2γ

⌉, r =

⌈2Dn

1−γ2

⌉, g = 100r`2.




Lower BoundsSummary

Open Problems

Inapproximability: Reduction to scheduling problem

The number of jobs in their instance is n. Further, the smallest jobhas processing time 1/(rg) and the largest has processing time atmost 8r + D < 9r . They show that the total flow time is smallerthan 200r`2 if there is a solution to the N3DM problem and atleast 100r2`2 if no such solution exists. This results in a lowerbound of 1

2 r on the approximation ratio achievable.

1

2r =

1

2

⌈2Dn

1−γ2

⌉=⇒ Ω(

√n) lower bound




Lower BoundsSummary

Open Problems

Inapproximability

In our model, preemption is allowed but if K ≥ 100r2`2, then theoptimal solution is non-preemptive. Thus, the lower bound of 1

2 ron the approximability applies directly to our problem if we setK = 100r2`2.




Lower BoundsSummary

Open Problems

Inapproximability

We would like to express this lower bound in terms of ρ and κ:

ρ =Pmax

ε<

9r

1/(rg)= 900r3`2,

κ =K

ε=

K

1/(rg)= 10000r4`4.




Lower BoundsSummary

Open Problems

Inapproximability

We simplify the parameters of the N3DM problem and use

` ≤ r2γ

1−γ in order to get:

ρ = O(r3+ 2γ1−γ ) =⇒ r = Ω

(ρ

1

3+2γ

1−γ

),

κ = O(r4+ 4γ1−γ ) =⇒ r = Ω

(κ

1

4+4γ

1−γ

).




Lower BoundsSummary

Open Problems

Inapproximability

We now recall that 0 < γ < 12 can be chosen arbitrarily. Given any

δ > 0, we choose a small enough γ in order to maintain thefollowing relations:

1

2r = Ω

(ρ

13−δ)

and1

2r = Ω

(κ

14−δ).




Lower BoundsSummary

Open Problems

Summary

I More realistic scheduling model

I Efficient online algorithms for this setting

I Lower bounds for the online settings

I Inapproximability bound




Lower BoundsSummary

Open Problems

Open Problems

I Minimizing completion time, weighted flow time, or stretch

I Cost of preemption to be dependent on the jobs involved, ormore generally on the state of the system

I Semi-clairvoyance with costly preemption




Lower BoundsSummary

Open Problems

The end

Questions?


on the value of preemption in schedulinggilsh/presentations/on_the... · 2021. 4. 20. · outline...

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