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By John G. Webster (Editor)

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Even though the work load of a display is distributed across the clusters with a round-robin assignment of blocks, a group might experience a higher work load as compared to other groups. For example, in Fig. 24, if the system services a new request for object X using group G4, then all servers in G4 become busy, while several other groups have two idle servers. This imbalance might result in a higher startup latency for future requests. For example, if another request for Z arrives, then it would incur a two time period startup latency because it must be assigned to G5 because G4 is already full.

V W Q S Heat decreases (a) knob = 0 Lemma 2: The optimal value for knob is C/Aug_Slice1. Proof: Let U be the total number of unique objects that are referenced over a period of time. We define Avg Slice1 = X Y U heat(x) ∗ size(Sx,1 ) x ... V W Q S Heat decreases heat(x) 1 = U U Avg Least = knob ∗ Avg Heat ∗ Avg Slice1 Avg Heat = ... x (b) knob = C/Avg_Slicel Figure 20. Status of the first slice of objects. The ideal case is when the LEAST of almost all the objects that constitute the database are disk resident (C is the total number of disk blocks): C ≈ U ∗ Avg Least ≈ U ∗ knob ∗ 1 ∗ Avg Slice1 U Solving for knob, we obtain: knob Ȃ C/Avg_Slice1.

23). If group G5 accesses cluster C2 during a time period, G5 would access C3 during the next time period. During a given time period, the requests occupying the slots of a group retrieve blocks that reside in the cluster that is being visited by that group. Therefore, if there are C clusters (or groups) in the system, and each cluster (or group) can support N simultaneous displays, then the maximum throughput of the system is m ϭ N ϫ C simultaneous displays. , playing musical chairs) with the C clusters using each for a Tp interval of time, and (2) at most, C Ϫ 1 failures might occur before a request can be activated (when the number of active displays is fewer than N ϫ C).

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