Resource augmentation and on-line scheduling on multiprocessors - PowerPoint PPT Presentation

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Resource augmentation and on-line scheduling on multiprocessors

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Optimal time-critical scheduling via resource ... (Ik 1 Qk 1) is Iq for some q k. By IH, EDF on m (1 - 2/m)-speed procs meets all deadlines for Iq ... – PowerPoint PPT presentation

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Title: Resource augmentation and on-line scheduling on multiprocessors


1
Resource augmentation and on-line scheduling on
multiprocessors
  • Phillips, Stein, Torng, and Wein. Optimal
    time-critical scheduling via resource
    augmentation. STOC (1997). Algorithmica (to
    appear).

2
Background on-line algorithms
  • Optimization problems given problem instance I,
    algorithm A obtains a value valA(I) -- goal is to
    maximize this value
  • On-line algorithms vs an optimal off-line/
    clairvoyant algorithm (OPT)
  • Competitive ratio of on-line algorithm A
  • min all I ( valA(I)/ valOPT(I) )
  • Goal Design an on-line algorithm with largest
    competitive ratio

3
Background hard-real-time scheduling
  • The on-line problem
  • Instance I J1, J2, ..., Jn of jobs
  • Each job Jj (rj, pj, dj)
  • arrives at instant ri
  • needs to execute for pi units...
  • by a deadline at instant di
  • Job Ji is revealed at instant ri
  • Difficult to formulate as an optimization problem
    -- all deadlines must be met!
  • In uniprocessor systems, we dodged this issue
  • EDF/ LL are optimal algorithms (always meet all
    deadlines)
  • EDF/ LL are on-line algorithms...
  • ... with competitive ratio one

4
Hard-real-time scheduling multiprocessors
  • No optimal (in the EDF/LL sense) on-line
    algorithm exists
  • Must still meet all deadlines...So, give the
    on-line algorithm extra resources (more/ faster
    processors)
  • This paper asks how much extra resources do EDF/
    LL need, in order to meet all deadlines for sets
    of jobs known to be feasible on m processors?
  • The answers
  • EDF/ LL meet all deadlines if processors are (2
    - 1/m) times as fast
  • No on-line algorithm can meet all deadlines if
    processors are lt 1.2 times as fast
  • EDF cannot always meet all deadlines if
    processors are (2 - 1/m - ?) times as
    fast, for any ? gt 0

5
Why we care
  • Our (RTS) task systems
  • usually pre-specified (e.g., periodic tasks/
    sporadic tasks)
  • on-lineness usually not an issue
  • exception overload scheduling (later)
  • Well do feasibility analysis (does a schedule
    exist?)
  • If feasible, well use the results in this paper
  • choose an algorithm (usually, EDF)
  • overallocate resources as mandated by these
    results
  • sleep well, knowing that the system performs as
    expected
  • Why choose feasibility analysis (versus
    schedulability analysis with chosen algorithm)?
  • provably competitive performance translates to
    approximation guarantees

6
Model and definitions
  • Instance I J1, J2, ..., Jn of jobs
  • Each job Jj (rj, pj, dj)
  • arrives at instant ri
  • needs to execute for pi units...
  • by a deadline at instant di
  • If I is feasible on m processors, an s-speed
    on-line algorithm will meet all deadlines on m
    processors each s times as fast
  • (Thus, EDF is a (2 - 1/m)-speed algorithm)

7
Digression An example of how wed use these
results
8
Scheduling periodic tasks - taxonomy
Periodic task system ? ?1, ?2,..., ?n ?i
(Ti, Ci),
RM EDF
LL/ Pfair
9
Remember this? (last class)
  • RM-US(1/4)
  • all tasks ?i with (Ti/ Ci gt 1/4) have highest
    priorities
  • for the remaining tasks, rate-monotonic
    priorities
  • Lemma Any task system satisfying
  • (SUM ?j ? j ?? Ci /Ti) ? m/4 and
  • (ALL ?j ? j ?? Ci /Ti) ? 1/4
  • is successfully scheduled using RM-US(1/4)
  • Theorem Any task system satisfying
  • (SUM ?j ? j ?? Ci /Ti) ? m/4
  • is successfully scheduled using RM-US(1/4)

10
A new (job-level static priority) scheduling
algorithm
  • EDF-US(1/2)
  • If Ci/Ti ? 0.5, then jobs of ?i get EDF
    priority
  • If Ci/Ti gt 0.5, then jobs of ?i get highest
    priority
  • (EDF implementation set deadline to -?)
  • Lemma Any task system satisfying
  • (SUM ?j ? j ?? Ci /Ti) ? m/2 and
  • (ALL ?j ? j ?? Ci /Ti) ? 1/2
  • is successfully scheduled using EDF-US(1/2)
  • Theorem Any task system satisfying
  • (SUM ?j ? j ?? Ci /Ti) ? m/2
  • is successfully scheduled using EDF-US(1/2)

11
Scheduling periodic tasks w/ migration
RM-US(1/4) EDF-US(1/4)
Pfair
25 50
100
12
Back to the results in this paper...(faster
processors)
13
The big insight
  • Definitions
  • A(j,t) denotes amount of execution of job j by
    Algorithm A until time t
  • A(I,t) SUM j ?I A(j,t)
  • The crucial question Let A be any busy
    (work-conserving) scheduling algorithm executing
    on m processors of speed ? ? 1. What is the
    smallest ? such that at all times t, A(I, t) ?
    A(I,t) for any other algorithm A executing on m
    speed-1 processors?
  • Lemma 2.6 ? turns out to be (2 - 1/m)
  • Use Lemma 2.6, and an individual algorithms
    scheduling rules, to draw conclusions regarding
    these algorithms

14
The oh-so-important lemma 2.6
Lemma Let I be an input instance, t ? 0 any
time-instant. For any busy algorithm A using
(2-1/m)-speed machines, A(I,t) ? A(I, t) for any
algorithm A using 1-speed machines.
  • Proof by contradiction
  • Suppose there are time instants at which this is
    not true
  • Let ? i ? t ? A(I,t) lt A(I,t) and A(i,t)
    lt A(i,t)
  • Let j be the job with the earliest release time
    rj in ?
  • Let to be the earliest time instant at which
  • A(I,to) lt A(I,to) Eq (1)
  • A(j,to) lt A(j,to) Eq (2)

15
EDF is a (2 - 1/m)-speed algorithm
Instance I J1, J2, ..., Jn job Jj (rj, pj,
dj) is feasible on m procs Wlog, assume that di ?
di1 for all i Let Ik J1, J2, ..., Jk Proof
Induction on k Base EDF on m (1 - 2/m)-speed
procs meets all deadlines for I1, .., Im IH EDF
on m (1 - 2/m)-speed procs meets all deadlines
for I1, .., Ik Were considering Ik1.
  • Let Qk1 ? Ik1 denote the jobs in Ik1 with
    deadlines at dk1
  • (Ik1 \ Qk1) is Iq for some q ? k
  • By IH, EDF on m (1 - 2/m)-speed procs meets all
    deadlines for Iq
  • BY definition of EDF, EDF(Ik1) is identical to
    EDF(Iq) on jobs of Iq -- thus, all deadlines in
    Iq are met in EDF(Ik1)
  • By Lemma 2.6, EDF(Ik1,dk1) ? OPT(Ik1, dk1)
  • Since OPT meets all deadlines at dk1, so must
    EDF on m (1 - 2/m)-speed procs
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