# Lecture Notes 12: Scheduling - Cont.

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1 Online Algorithms Professor: Yossi Azar Lecture Notes 12: Scheduling - Cont. Scribe:Inna Kalp 1 Introduction In this Lecture we discuss 2 scheduling models. We review the scheduling over time model that was introduced in the previous lecture and prove that SRPT algorithm is optimal with respect to this model. We also discuss another scheduling model where each job also has size, release time, deadline and value. In this model we ll present: EDF for unit size and unit value jobs EDF for jobs of different size Max value first for unit size jobs with arbitrary value 2 Reminder 2.1 Interval Scheduling In the previous lecture we completed analyzing the Interval Scheduling Problem (assuming the interval is of length T). Table 1: val =1 deterministic alg. random alg. no preemption T logt with preemption logt Const Table 2: val =length deterministic alg. random alg. no preemption T logt with preemption Const 2.2 Scheduling Over Time The Model: One (or more) processors Jobs are released over time. A job i has release time r i and size w i. No more than one job is executing in any given time. Preemption is allowed. Preemption in this model means that a processor can stop the execution of a job and continue executing it later. Each job has Completion Time c i, and Flow Time f i = c i r i. 12-1

2 We consider 2 possible goal functions: 1. Minimize maxc i : we saw a reduction from release time to batch - we collect several jobs and then process the batch, and so on. The reduction is 2 competitive. We also showed that greedy algorithm is 2 competitive for identical machines. 2. A more reasonable goal function is to minimize f i, meaning the total (or average) processing time. If all jobs arrive in time 0, we proved that SPT (=Shortest Process Time) algorithm is optimal. 3 Scheduling over time - General Case Consider the general case where job i arrives at time r i. Theorem: SPRT (Shortest Remaining Processing Time) is optimal in the general case. Proof: Assume by contradiction that the optimal schedule is not SRPT. Look at the first time t 0 that a job A should run according to SRPT, but instead, job B is executing in the optimal schedule. The sum of all the segments in which A or B execute after time t 0 equals to the sum of their remaining sizes from time t 0. Define t 2 as the time in which the first job from {A,B} ends its execution. Also define t 1 as the time in which the second job ends. If we ll change the execution order only in the segments in which jobs A or B execute after time t 0, the end of execution of all the other jobs will remain unchanged. Notice that our goal is to minimize f i = c i r i, and we can alternatively minimize c i. (The } {{ } constant two sums are equivalent for exact solution. This would not work if we were attempting to find an approximation algorithm). Therefore, the optimal schedule brings to a minimum C A + C B. In every schedule t 1 = max{c A, C B }. If we will execute job A Continuously (only in the slots that are originally of A or B), time t 2 will decrease, and by definition C A + C B will decrease. We decreased C i in contradiction to the assumption that the initial schedule is optimal. Figure 1: Upper figure - The slots of jobs A and B in the assumed optimal schedule. Lower figure: the schedule after the reordering of the execution in the slots of A and B. 12-2

4 figure 3: example - EDF can be preemptive Theorem 3: In the general case (jobs of different sizes), if there exists a schedule that executes all jobs, then EDF executes all jobs. Proof: Consider the first time such a schedule is not EDF. We have a job A that should execute according to EDF, but job B executes instead (d B > d A ). Let s reorder the slots where jobs A and B are executing. We ll execute job A in those slots until it finishes, and then execute job B. In any reordering of these slots, B will finish before d B, and in our reordering, A will finish before its original completion time and therefore before d A, so the schedule will remain feasible. We will repeat the reordering until we get EDF schedule, since with every iteration, the prefix of the schedule that equal to EDF increases. Note: If there is no schedule that can execute all jobs doesn t exist, EDF can be Unboundedly bad. Examples: 1. Define nɛn :1 x (n,0,n-1,1), n x (1,0,n,1): EDF will perform only 2 jobs - one job from the first group,and another one from the second group. OPT will perform n jobs from group 2. Ratio of ( n 2 ) = O(n). 2. Define M, nɛn, (n-1) x (1,0,n-1,1), n x (1,0,n, M): EDF will execute the n-1 jobs from the first group and one job from the second group. EDF = n-1+m. OPT will nm perform n jobs from the second group. OPT = nm. Ratio of ( n 1+M ) 4.3 Max value first for unit size jobs with arbitrary value Definition: MVF = Max Value First algorithm executes the job with the maximum value among the living jobs. Theorem: MVF is 2 competitive, when p i is 1 and,b i is arbitrary. Proof: Let s define RA =the group of jobs that OPT executed and MVF rejected. We will show that ( ) b i MV F = b i. After we ll prove (*), we can conclude : OP T (σ) b i + b i 2 b i = 2MV F (σ) and the competitve ratio follows. We still have to prove (*): Let s take. Suppose OPT executes at time t. There are several reasons why MVF doesn t execute i at time A: 1. MVF already executed it - can t be since. 2. i is not alive (t<r i or t>d i ) - can t be since OPT executes i at time t. 12-4

5 3. MVF executes in time t a job with an higher value. since option 3 is the only viable option, we found a one-to-one function between RA to A, such that every job in RA corresponds to a job in A with a higher value. Therefore, bi = b i MV F follows. Note: The competitive ratio is tight. Example: (1,0,1,1), (1,0,2,1+ε): MVF will execute the second job. OPT Will execute 2 jobs: it will execute the first job and then the second one. MVF = 1+ε. OPT = 1+1+ε. The ratio is

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