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Minimizing Total Completion Time

Each job specified by procesing time (length p j ) release time r j Goal: compute a schedule (preemptive or not) that maximizes the total completion time. all integer. Minimizing Total Completion Time. C j = completion time of j in a given schedule In Graham’s notation:

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Minimizing Total Completion Time

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  1. Each job specified by procesing time (length pj) release time rj Goal: compute a schedule (preemptive or not) that maximizes the total completion time all integer Minimizing Total Completion Time Cj = completion time of j in a given schedule In Graham’s notation: • Non preemptive version: 1|rj,pj|∑Cj • Preemptive version: 1|rj,pj,pmtn|∑Cj 1

  2. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 Schedule A ∑Cj = 4+7+8+9+11+12 = 51 Schedule B ∑Cj = 1+2+3+5+6+9+12 = 38 Preemptive Scheduling Example: 2

  3. 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) 3

  4. unit a+b in T unit a+b in T i i j j j i j i i i j i i j j rj rj ri ri at rj: b units of j and a≤b units of i unit ≥ a in T i i i i i i i i j j j j j j j unit a in T 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 !!! 4

  5. phase 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 rj=0 to rj=1 • and apply the argument recursively to the new instance Thus we convert Q into SRPT without increasing ∑Cj So ∑Cj(SRPT) ≤ ∑Cj(Q) 5

  6. 0 1 2 3 4 5 6 7 8 9 10 11 12 13 1 2 3 4 5 6 Schedule A ∑Cj = 4+7+8+9+11+12 = 51 Schedule B Schedule C ∑Cj = 4+5+6+7+9+11 = 42 ∑Cj = 2+3+5+6+10+13 = 39 Non-Preemptive Scheduling Example: 6

  7. Algorithm SPT: at each step, when no job is running, execute the job with shortest processing time Question: Is SPT’s 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? 7

  8. r1=0 p1=p n jobs with ri=1 and pi=1 SPT: 0 1 Q: Question: Is SPT competitive? ∑Cj ≥ np ∑Cj ≤ n(n+1) + (n+1+p) ≤ 2n(n+1)+p So for p = 2n(n+1) ratio SPT/Q ≥ n·2n(n+1) / 4n(n+1) = n/2  8

  9. r1=0 p1=p In A: Consider Q: Theorem: No online algorithm has competitive ratio < 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-1+n)+(2p+n) = pn+(n+p) So for p = n2 ratio A/Q ≥ 2 - O(1/n)  2 9

  10. 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 r’j = rj+pj • 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) • SSPT’s schedule for I = SRPT schedule for I” So ∑Cj(SPT on I) = ∑Cj(SRPT on I”) = optimum(I”) ≤ 2 · optimum(I) 10

  11. Let T = schedule of SSPT on I Define I” by changing release times to r”j = min { Sj(T) , 2rj+pj } 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 ≥ r”j • ∑Cj(Z) = 2∑Cj(Q) because • Cj(Z) = Sj(Z)+pj • = 2 Sj(Q)+pj+pj • = 2[Sj(Q)+pj] = 2Cj(Q) 11

  12. Case 1: h pending at Sk(T). Since T chose k, we have ph ≥ pk and k only gets shorter so T obeys the SRPT rule for k and h k h r’h = rh+ph Case 2: h was released after Sk(T). k h h r’h = rh+ph 2rh+ph k h h Recall: T = schedule of SSPT on I I” obtained by changing rj to r”j = min{Sj(T),2rj+pj} 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 ≥ r”h. It’s enough to show it for t=Sh(T) or t= 2rh+ph (since T runs k) • T runs h after k so Sh(T) ≥ Ck(T) we are OK for t = Sh(T) • At t=2rh+ph we need ph≥ Ck(T)-(2rh+ph) • This is equivalent to 2(rh+ph) ≥ Sk(T)+pk True, because 2(rh+ph) ≥ 2Sk(T) ≥ Sk(T)+r’k ≥ Sk(T)+pk 12

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