Minimizing Makespan and Preemption Costs ona System of Uniform

Machines

Hadas ShachnaiBell Labs and The Technion IIT

Tami TamirUniv. of Washington

Gerhard J. WoegingerUniv. of Twente

2

Generalized Multiprocessor Scheduling

We need to schedule n jobs on m uniform machines.

•Jj can be preempted at most aj times, aj 0.

•The machine Mi has the rate ui

1.

•The job Jj has processing time tj.

Non-preemptive

(aj=0).

Preemptive

(aj=).

The goal:

Minimum makespan

Generalizes the classic preemptive and non-preemptive scheduling problems

3

What is Known:

The non-preemptive scheduling problem (j, aj = 0)

is

strongly NP-hard (admits PTAS [HS88, ES99]).

The preemptive scheduling problem (j, aj=) can

be

solved optimally, using at most 2(m-1) preemptions

[GS78] (at most m-1 preemptions on identical machines).

Note: Similar solvability/approximability results in the identical and uniform machine environments.

Question: How many preemptions (in total, or per job) suffice in order to guarantee an optimal polynomial time algorithm?

4

We Investigate this Hardness Gap:

1. The GMS problem (generalized multiprocessor scheduling): Minimize the makespan, where we have job-wise or total bound on the number of preemptions throughout a feasible schedule.

2. The MPS problem (minimum preemptions scheduling): The only feasible schedules are preemptive schedules with smallest possible makespan.The goal is to find a feasible schedule that minimizes the overall number of preemptions.

5

Our Results• Hardness of GMS for j, aj=1 and for any total < 2(m-1).

• For MPS, matching upper and lower bound on the number of preemptions required by any optimal schedule. Jj can be processed simultaneously by j

machines (i.e., j, j 1).

• A PTAS for GMS instances with fixed number of machines. The scheme has linear running time, and can be applied to instances with release dates, unrelated machines, and arbitrary preemption costs.

• PTASs for arbitrary number of machines, and a bound on the total number of preemptions.

(A distinction between the uniform and identical machine environments)

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The Power of Unlimited Preemption

For a given instance I, with aj 0 and m uniform

machines, let w denote the minimum makespan of a schedule with preemptions.

Theorem 1: It is easy to compute w (a function of the tj’s and the ui’s).

Consider the LPT algorithm, that assigns jobs of I to the machines in order of non-increasing processing times. This is a feasible non-preemptive schedule of I.

Theorem 2: Any LPT schedule yields a 2-approximation for w.Holds also for parallelizable jobs (i.e., j, j 1).

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The Power of Unlimited Preemption

This 2 bound is tight already for identical jobs and identical machines:

Consider an instance with m machines and m+1 jobs,

j, aj 0, tj =t. i, ui =1.

preemptive

Makespan = t(1+1/m).

Non-preemptive

Makespan = 2t.

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A PTAS for GMS: Overview ‘Guess’ the minimum makespan, Topt, and the

maximum load on any machine, P, to within factor of (1+). Partition the jobs to the sets ‘big’, ‘small’ and ‘tiny’ (as in [SW-98] ).

• Find a feasible preemptive schedule of the big jobs in the interval [0, Topt (1+)].

• The small jobs have non-negligible processing times; however, the overall processing time of the set is small relative to an optimal solution. Schedule the small jobs non-preemptively.

•Add the tiny jobs greedily, with no preemptions.

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A PTAS for GMS (Cont.)

big

0 P

u1 u2 u3 u4 ; after scheduling B: C1 = C2 = C3 = C4 =

Topt

small

tiny

1

2

3

4

a1 = 3

a2 = 6

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Analysis of the Scheme The big jobs can be scheduled optimally with preemptions in [0, Topt (1+)] in polynomial time, by

taking

B= {Jj| tj > P}, (for some (0,1]).

Use a fixed number of scheduling points, and a fixed number of segment sizes.

Lemma: There exists (,m)(0,1], such that taking S= {Jj | ·P < tj P}, we get that tj P.

JjS

We may add all the small jobs at the end of the schedule on the fastest machine.

Let = (, m) be a parameter.

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Analysis of the Scheme (Cont.)

The tiny jobs contribute at most P processing units to the maximally loaded machine, since the number of tiny jobs that extend the schedule is bounded by the number of ‘holes’ on any machine. We take

T= {Jj| tj P/m}.

Theorem: The above scheme yields a (1+ )-approximation

to the minimum makespan, in O( ) steps.

3/12

)αε

1(n

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Minimizing the Number of Preemptions

We count the number of segments, Ns(I), generated for an instance I.

Ns(I) = #preemptions(I) + n.

1. Lower Bound:

m,b, there exists an instance, I, in which j, j b, and in any optimal schedule of I, Ns(I) m+n+m/b-22. Upper Bound:

We present an algorithm that produces for any m,b, and any instance I in which j, j b, an optimal schedule with Ns(I) m+n+m/b-2

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Upper Bound Proof (Algorithm)Assume that j, j =1.

Step 1: calculate w (the optimal makespan).

Step 2: schedule the jobs one after the other, each job is scheduled on one DPS.

A DPS (Disjoint Processor System): a union of disjoint idle-segments of r machines with non-decreasing rates, such that the union of the idle segments is the interval [0,w].

0 w

An idle machine forms a DPS

(r=1)

0 w

Two machines that form a DPS

(r=2)

idle

idle

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1/2

Upper Bound Proof (Algorithm)

An Example: 3 machines, with rates 2,3,5

3 jobs, with lengths 4,3,3.In this case, w=tj/ui = 1.

Initially, each machine forms a DPS.

532

M3

M2

M1

M2,3

M1253

The longest job (t=4) is scheduled for 1/2 time unit on each of the machines M2, M3.

Available DPSs

The remainders of M2 and M3 form a new DPS.

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2/31/2

Upper Bound Proof (Algorithm)

532

M3

M2

M1

M2,3

M1253

Available DPSs

The next (t=3) is scheduled on a DPS consisting of M2,3 and M1. We need to solve one equation in order to find the time 2/3 (3 = 3*1/2 + 5*1/6 + 2*1/3).

We are left with one DPS, consisting of the remaining idle segments of M1,M3. The last job is scheduled on this DPS (3 = 2*2/3 + 5*1/3).

M3

M2

M1

M1,3 52

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Algorithm Analysis

• Using amortized analysis, we show that the total number of segments allocated by the algorithm is at most n+2(m-1).

•Since each job is scheduled on a single DPS, the requirement j, j =1 is preserved.

•For arbitrary values of j’s the algorithm schedules each job on at most j consecutive DPSs, and by similar arguments we have Ns(I) n+m+m/b-2.

•For the special case of j, j =1, our algorithm and its analysis are simpler than the known algorithm [GS-78].

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Related Work

• Preemptive scheduling (aj= ,j) .

(Horvath, Lam and Sethi, 1977).• Non-preemptive scheduling (aj=0, j).

LPT is 2-optimal (Gonzalez, Ibarra and Sahni, 1977)

PTASs (Hochbaum and Shmoys, 1987; Hochbaum and Shmoys, 1988; Epstein and Sgall, 1999)

• MPS for non-parallelizable jobs

(Gonzalez and Sahni, 1978)

• A wide literature on scheduling parallelizable jobs

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What is Known: The non-preemptive scheduling problem (j, aj =

0) is strongly NP-hard (admits PTAS [Hochbaum and Shmoys, 1987, 1988; Epstein and Sgall, 1999]).

The preemptive scheduling problem (j, aj=) can

be solved optimally [Horvath, Lam and Sethi, 1977], using at most 2(m-1) preemptions [Gonzalez and Sahni 1978] (at most m-1 preemptions on identical machines [McNaughton 1959]).Note: Similar solvability/approximability results in the identical and uniform machine environments. Question: How many preemptions (in total, or per job) suffice in order to guarantee an optimal polynomial time algorithm?

19

The Power of Unlimited Preemption

This 2 bound is tight already for identical jobs and identical machines:

Consider an instance with m machines and m+1 jobs,

j, aj 0, tj =t. i, ui =1.

Preemptive, Cmax=t(1+1/m).

Non-preemptive, Cmax= 2t.

•This result extends the result known for non-preemptive scheduling [Gonzalez, Ibarra and Sahni, 1977].