Minimizing Total Completion Time

Ecole Polytechnique, Nov 7, 20071

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

all integer

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: 0 1 2 3 4 5 6 7 8 9 10 11 12 13

12

3

4

56

Schedule A

∑Cj = 4+7+8+9+11+12 = 51

Schedule B

∑Cj = 1+2+3+5+6+9+12 = 38


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

Theorem: Algorithm SRPT computes an optimal solution (that is, SRPT is 1-competitive)

Example: Schedule B on previous slide is SRPT


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

i ji j j j ji i i i i i j j

rjriat rj: b units of jand a≤b units of i

unit a+b in T

i ji jj j ji iiiii j j

rjri

Reschedule i and j within T using SRPT:

unit ≥ a in T

unit a+b in T

unit a in T

So ∑Cj cannot increase !!!


So we know: that one SPRT-reschedule does not increase ∑Cj

let k = minimum length job with rk = 0

• SRPT will schedule one unit of k at time 0

start with any schedule Qrepeat: apply SRPT-reschedule to all pairs i,juntil Q does not change

phase

• 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)


Non-Preemptive Scheduling

Example: 0 1 2 3 4 5 6 7 8 9 10 11 12 13

12

3

4

56

Schedule A

∑Cj = 4+7+8+9+11+12 = 51

Schedule C

∑Cj = 2+3+5+6+10+13 = 39

Schedule B

∑Cj = 4+5+6+7+9+11 = 42


Algorithm SPT: at each step, when no job is running, execute the job with shortest processing timeQuestion: Is SPT’s schedule optimal?

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?

No: Schedule B on previous slide is SPT


Question: Is SPT competitive?

r1=0 p1=p

n jobs with ri=1 and pi=1

∑Cj ≥ np

∑Cj ≤ n(n+1) + (n+1+p) ≤ 2n(n+1)+p

SPT:

0

Q:

1

So for p = 2n(n+1)

ratio SPT/Q ≥ n·2n(n+1) / 4n(n+1) = n/2


Theorem: No online algorithm has competitive ratio < 2

r1=0 p1=p

∑Cj(A) ≥ n(2p-1) = 2pn+(n)

So for 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

Consider Q:∑Cj(Q) ≤ n(p-1+n)+(2p+n) = pn+(n+p)

In A:


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 timesTheorem: 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)


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)

Let T = schedule of SSPT on IDefine I” by changing release times 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

Recall: T = schedule of SSPT on I I” obtained by changing rj to r”j = min{Sj(T),2rj+pj}

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

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

h

k

Case 2: h was released after Sk(T). h

r’h = rh+ph

k h

h

r’h = rh+ph

k h

2rh+ph

True, because 2(rh+ph) ≥ 2Sk(T)

≥ Sk(T)+r’k ≥ Sk(T)+pk

Minimizing Total Completion Time

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