Minimizing%20Total%20Completion%20Time

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Minimizing Total Completion Time

• Each job specified by
• procesing time (length pj)
• release time rj
• Goal compute a schedule (preemptive or not) that
  maximizes the total completion time

• Cj completion time of j in a given schedule
• In Grahams notation
• Non preemptive version 1rj,pj?Cj
• Preemptive version 1rj,pj,pmtn?Cj

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Preemptive Scheduling Example
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• Online version
• jobs are revealed at their release time
• algorithm must execute a job without knowing
  future jobs
• Note different from list scheduling!
• (Decision time processing time)

Algorithm SRPT At each step schedule the job
with the shortest remaining processing time
Example Schedule B on previous slide is SRPT
Theorem Algorithm SRPT computes an optimal
solution (that is, SRPT is 1-competitive)
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Proof Exchange agument. Convert any schedule S
into one satisfying the SRPT rule. Take any jobs
i and j in S. W.l.o.g. assume ri rj. Let T
set of time slots occupied by i and j starting at
rj
Reschedule i and j within T using SRPT
So ?Cj cannot increase !!!
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So we know that one SPRT-reschedule does not
increase ?Cj
start with any schedule Q repeat apply
SRPT-reschedule to all pairs i,j until Q does not
change
let k minimum length job with rk 0

• SRPT will schedule one unit of k at time 0

• after phase 1, one unit of k will be scheduled
  at time 0 in Q
• and will remain there

• So Q and SRPT are the same at time unit 0 after
  phase 1

• We can modify the instance
• change pk pk-1
• change all release times rj0 to rj1
• and apply the argument recursively to the new
  instance

Thus we convert Q into SRPT without increasing
?Cj So ?Cj(SRPT) ?Cj(Q)
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Non-Preemptive Scheduling Example
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Algorithm SPT at each step, when no job is
running, execute the job with shortest processing
time
Question Is SPTs schedule optimal?
No Schedule B on previous slide is SPT
Intuition If a big job is pending, an optimal
schedule sometimes needs to wait to execute more
small jobs that are about to arrive
Question Is SPT competitive?
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Question Is SPT competitive?
?Cj np
?Cj n(n1) (n1p) 2n(n1)p
So for p 2n(n1) ratio SPT/Q n2n(n1) /
4n(n1) n/2 ??
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Theorem No online algorithm has competitive
ratio lt 2
Proof A online algorithm. At time 0 release
job 1
If A schedules 1 at time t p-2 then ratio
(2p-2)/p 2-2/p ? 2
Else, release n jobs at time p-1 of length 1
?Cj(A) n(2p-1) 2pn?(n)
?Cj(Q) n(p-1n)(2pn) pn?(np)
So for p n2 ratio A/Q 2 - O(1/n) ? 2
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Intuition To be competitive, an algorithm needs
to wait even if jobs are available

• Algorithm SSPT
• When job j arrives, reset its release time to
  rj rjpj
• Apply SPT with respect to these new release times

Theorem SSPT is 2-competitive.

• Proof idea I input instance. Define an
  SSPT-induced instance I such that
• optimum(I) 2 optimum(I)
• SSPTs schedule for I SRPT schedule for I

So ?Cj(SPT on I) ?Cj(SRPT on I)
optimum(I) 2 optimum(I)
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Let T schedule of SSPT on I Define I by
changing release times to rj min Sj(T) ,
2rjpj
Claim 1 optimum(I) 2 optimum(I)
Proof Q optimal schedule for I. Consider
schedule Z for I where Sj(Z) 2Sj(Q)pj

• Then
• Z is feasible for I because jobs in Z do not
  overlap and
• Sj(Z) 2Sj(Q)pj
• 2rj pj rj

• ?Cj(Z) 2?Cj(Q) because
• Cj(Z) Sj(Z)pj
• 2 Sj(Q)pjpj
• 2Sj(Q)pj 2Cj(Q)

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Recall T schedule of SSPT on I I
obtained by changing rj to rj minSj(T),2rjpj
Claim 2 T is also an SRPT schedule for I
Proof Suppose T executes k and h is pending
We need (SRPT of k) ph at any time t rh.
Its enough to show it for tSh(T) or t 2rhph
(since T runs k)

• T runs h after k so Sh(T) Ck(T) ? we are OK
  for t Sh(T)

• At t2rhph we need ph Ck(T)-(2rhph)
• This is equivalent to 2(rhph) Sk(T)pk

True, because 2(rhph) 2Sk(T)
Sk(T)rk Sk(T)pk