Minimizing Total Completion Time Each job specified by procesing time (leng


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Minimizing Total Completion Time Each job specified by procesing time (leng

Austin Peay State University, US has reference to this Academic Journal, 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 Graham?s notation: Non preemptive version: 1|rj,pj|?Cj Preemptive version: 1|rj,pj,pmtn|?Cj Preemptive Scheduling Example: 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 alongside the shortest remaining processing time Theorem: Algorithm SRPT computes an optimal solution (that is, SRPT is 1-competitive) Example: Schedule B on previous slide is SRPT

 Chin, Taylor Austin Peay State University


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Proof: Exchange agument. Convert any schedule S into one satisfying the SRPT rule. Take any jobs i in addition to j in S. W.l.o.g. assume ri ? rj. Let T = set of time slots occupied by i in addition to j starting at rj Reschedule i in addition to j within T using SRPT: So ?Cj cannot increase !!! So we know: that one SPRT-reschedule does not increase ?Cj let k = minimum length job alongside rk = 0 SRPT will schedule one unit of k at time 0 start alongside any schedule Q repeat: apply SRPT-reschedule so that all pairs i,j until Q does not change after phase 1, one unit of k will be scheduled at time 0 in Q in addition to will remain there So Q in addition to 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 so that rj=1 in addition to apply the argument recursively so that the new instance Thus we convert Q into SRPT without increasing ?Cj So ?Cj(SRPT) ? ?Cj(Q) Non-Preemptive Scheduling Example:

Algorithm SPT: at each step, when no job is running, execute the job alongside shortest processing time Question: Is SPT?s schedule optimal? Intuition: If a big job is pending, an optimal schedule sometimes needs so that wait so that execute more small jobs that are about so that arrive Question: Is SPT competitive? No: Schedule B on previous slide is SPT Question: Is SPT competitive? ?Cj ? np ?Cj ? n(n+1) + (n+1+p) ? 2n(n+1)+p So in consideration of p = 2n(n+1) ratio SPT/Q ? nú2n(n+1) / 4n(n+1) = n/2 ?? Theorem: No online algorithm has competitive ratio < 2 ?Cj(A) ? n(2p-1) = 2pn+?(n) So in consideration of p = n2 ratio A/Q ? 2 - O(1/n) ? 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(Q) ? n(p-1+n)+(2p+n) = pn+?(n+p)

Lecture 3 Antibody-Antigen Reactions Binding of the epitope in the antigen binding site Antibody Avidity Effect of multivalent interactions Biological Consequences of Antibody Affinity/Avidity Cross-Reaction Antigen-Antibody Interactions Immune Precipitation Agglutination Enzyme (ELISA) Immunoassay Complement activation Overview of the Complement Cascade Effector Actions of Complement Terms so that remember

Intuition: To be competitive, an algorithm needs so that wait even if jobs are available Algorithm SSPT: When job j arrives, reset its release time so that r?j = rj+pj Apply SPT alongside respect so that 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 ú optimum(I) SSPT?s schedule in consideration of I = SRPT schedule in consideration of I? So ?Cj(SPT on I) = ?Cj(SRPT on I?) = optimum(I?) ? 2 ú optimum(I) Claim 1: optimum(I?) ? 2 ú optimum(I) Proof: Q = optimal schedule in consideration of I. Consider schedule Z in consideration of I? where Sj(Z) = 2Sj(Q)+pj Then Z is feasible in consideration of I? because jobs in Z do not overlap in addition to 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) Let T = schedule of SSPT on I Define I? by changing release times so that r?j = min { Sj(T) , 2rj+pj } Claim 2: T is also an SRPT schedule in consideration of I? Proof: Suppose T executes k in addition to h is pending Recall: T = schedule of SSPT on I I? obtained by changing rj so that r?j = min{Sj(T),2rj+pj} We need (SRPT of k) ? ph at any time t ? r?h. It?s enough so that show it in consideration of t=Sh(T) or t= 2rh+ph (since T runs k) T runs h after k so Sh(T) ? Ck(T) ? we are OK in consideration of t = Sh(T) At t=2rh+ph we need ph ? Ck(T)-(2rh+ph) This is equivalent so that 2(rh+ph) ? Sk(T)+pk True, because 2(rh+ph) ? 2Sk(T) ? Sk(T)+r?k ? Sk(T)+pk

Chin, Taylor General Manager

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