1 Identical Parallel Machines

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1 FB3: Matheatik/Inforatik Dr. Syaantak Das Winter 2017/18 Optiizing under Uncertainty Lecture Notes 3: Scheduling to Miniize Makespan In any standard scheduling proble, we are given a set of jobs J = {j 1, j 2, j n } and a set of achines M = {i 1, i 2, i }. A job j J requires p ij tie units to process on achine i M. Throughout this note, we shall assue that a job cannot be interrupted once it starts processing on a particular achine. We consider the online proble of scheduling a sequence of jobs (note that J here is a sequence rather than a set) on the achines M with the objective of iniizing the axiu copletion tie of a job. In other words, we want to iniize the tie point when the last job finishes processing. We observe here that copletion tie of the last job on any achine equal to the total load or the total processing volue of jobs that are assigned to this achine. We shall use this view of the objective function for the rest of the notes. So, the objective is to assign a job to a achine iediately upon arrival so as to iniize the axiu load on any achine. This is also known as the Online Load Balancing proble. 1 Identical Parallel Machines We first consider the case when all achines are identical. According to our notation, this eans, for any job j, p ij = p j for any achine i M. For a subset J J of jobs, we define the total volue V (J ) = j J p j. Define p ax = ax j J p j. The following is a natural greedy algorith for the proble. When job j arrives, pick the achine which has iniu load so far. Assign j to that achine. The algorith is also known as List Scheduling (LS). Theore 1. LS is (2 1 )-copetitive Proof. We first observe two siple lower bounds for the optiu solution for the entire sequence J. Let T denote the optiu solution. Then T p ax (1) T V (J) Now, consider the achine which has the axiu load in the schedule for LS - call the achine i and let j be the last job assigned to i and let t be the tie point when j arrived. Let load i,t be the total volue of jobs assigned to achine i before j arrives. Further, let (2) 1

2 J t be the set of all jobs that has arrived till t, including j. Since LS picks the least loaded achine when j arrives, load i,t V (Jt\{j }), by averaging arguent. Further, after j was assigned to i, the total load on i is load i,t + p j, Using the lower bounds 1 and 2, load i,t + p j V (J t \ {j }) + p j V (J t) + (1 1 )p j T + (1 1 )T (2 1 )T Theore 2. LS is Θ(2 1 )-copetitive. Proof. The above notation eans that the copetitive ratio for LS is tight. Theore 1 shows the upper bound. The lower bound is left as an exercise. 2 Restricted Assignents We consider a ore general scheduling odel defined as follows. For each job j, there is a subset of achines S j M such that for i S j, p ij = p j and p ij = for i / S j. Assue log() = log 2 (). Theore 3 (Azar et al., 1995). Any deterinistic online algorith is Ω(log )-copetitive. Proof. (Sketch) Consider a setting with achines (assue to be a power of 2). The idea would be to release 1 unit sized jobs, all at tie 0. However, the adversary would choose the sequence of release in such a way that any online algorith would incur a load of O(log ) on one of the achines while there exists an offline solution with 1 job assigned to each achine. Jobs are released in phases and we denote the set of jobs released in phase i by J i. Further, let M i denote the set of achines to which any job of J i can be dispatched (restricted assignents). For phase 1, we assue an arbitrary but fixed ordering 1, 2, of M 1. /2 jobs 1, 2, /2 are released such that the jth job can be dispatched to one of the two achines 2j 1 or 2j. Note that after this phase, one of the achines 2j 1 or 2j ust have unit load for all j. Now, let M 2 be a subset of M 1 such that all achines of M 2 has a load of 1. Let us create a fresh ordering only on M 2, 1, 2, /2. J 2 jobs are released such that jth job can be dispatched to 2j 1 or 2j according to the new ordering. Clearly, after phase 2, one of the achines 2j 1 or 2j will have a load of 2 for any j. It is not hard to see that we can continue this process for log phases and force at least one achine to incur a load of log. On the other hand, we can show as follows that offline OPT is 1. We build an offline schedule such that for any i, all achines of M i are epty before jobs J i are released. This is clearly true when J 0 are released. Assue the sae is true for phase i. Now, we ensure this is true for the phase i + 1. Consider any job j in J i. We dispatch it to the achine which is not included in M i+1. Together with the fact that all achines of M i were epty at the start of phase i, the above dispatch ensures that all achines of M i+1 are epty when jobs of J i+1 are released. See Figure 1 for an illustration with = 8. We prove that a greedy algorith siilar to LS achieves the upper bound of O(log ). When a job j arrives at tie t, assign it to the achine i S j which has the iniu load aong all achines in S j till tie t 1. 2

3 Phase 3 Phase 2 Phase 1 ALG OPT Figure 1: An exaple proving a lower bound of Ω(log ) for any deterinistic online algorith for load balancing with restricted assignents. Here = 8. The arrows indicate the restricted assignent rules. Theore 4. The odified LS algorith is O(log )-copetitive Proof. Let T be the optial solution for a sequence of jobs σ = {j 1, j 2, j 3, j n }. We assue that job j r arrives at tie t r for r = 1, 2, n. Define load i,r to be the load on achine i at tie t r, for i = 1, 2,, r = 1, 2, n. Let ax i load i,tn = T be the axiu load on any achine after the greedy algorith has assigned all jobs. Let α = T. Note that, with T the above definition, T αt + T and hence the copetitive ratio is α + 1. We now prove that α = O(log ). We define subsets of achines according to the load on the in the LS algorith. Let us define M α s = {i : load i,tn (α s)t } for s = 0, 1, 2, α. Observe that M α M α 1 M α 2 M 0 = M. For any job j r that arrived at t r, let i r be the achine on which it was assigned by the greedy algorith. Define level(j) = load ir,tr + 1. Note a subtlety here. We assue that load ir,tr was the load on achine i r just before j r was assigned to it while level(j r ) is the load of i r after j r was assigned to it. Define a set of jobs J α s = {j : level(j) (α s)t + 1} for s = 0, 1, 2, α. In the illustration, the shaded area represents the set of jobs J α s. Consider a set J α s for any s {0, 1, α} and a job j r J α s. Due to the greedy nature of the algorith, we can clai that for any achine i S j, load i,tr level(j r ) 1 (α s)t. Since load on a achine never decreases throughout the algorith, we have load i,tn (α s)t, that is i M α s. Hence, in any feasible schedule, the set of jobs J α s can only be scheduled on the set of achines M α s. Fro this, we can write the following lower bound on T, using siple averaging arguent 3

4 M α s M α 1 Load M α αt (α 1)T (α s + 1)T (α s)t (α s 1)T 3T 2T T Figure 2: An exaple proving a lower bound of Ω(log ) for any deterinistic online algorith for load balancing with restricted assignents. Here = 8. The arrows indicate the restricted assignent rules. T J α s M α s Now let us estiate J α s. As entioned earlier, we need to find the area of the shaded portion of the illustration. It is not hard to see that this area is at least the following quantity. J α s ( M α s+1 + M α s+2 + M α )T Note that the above su is a lower bound on J α s since we ignore the jobs that have levels strictly between (α s)t and (α s + 1)T and the jobs that have level strictly above αt (indicated with red in the illustration). But this is sufficient for our purpose. Plugging in the above lower bound for J α s in 3, we get (3) T ( M α s+1 + M α s+2 + M α )T M α s (4) = M α s M α s+1 + M α s+2 + M α (5) 4

5 We prove later that the above recurrence solves to M α s 2 s 1 M α for any s = 2, 3, α (6) Now plugging in s = α in the above inequality, we get M 0 2 α 2 M α 2α 4 Thus we have 2 α 4 M 0 = 4 which gives us α = O(log ) as desired. Finally we prove that the recurrence relation (6) is correct. α s = M α s. Recall fro (5) To siplify notations, let α s α s+1 + α s α Substituting the above inequality for α s+1 on the RHS, α s 2( α s+2 + α s α ) Siilarly, substituting α s+2 and so on, we get α s 2 s 1 ( α s+s + α s+s α ), s = 1, 2, s Plugging in s = s, we then have as desired. α s 2 s 1 α 5