A scale free network is heterogeneous as the degrees of the nodes have a

A scale free network is heterogeneous as the degrees

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A scale-free network is heterogeneous, as the degrees of the nodes, have a negative expo- nential distribution: p ( k ) = 1 ζ ( γ, k min ) k - γ . (21) In this expression k min is the smallest degree of any vertex and for the applications we discuss in this paper k min = 1; ζ is the Hurvitz zeta function. p ( k ) is the probability that a vertex has degree k , in other words it is connected to other k vertices. Thus, the number of nodes with a high degree of connectivity is relatively small as we can see from Table 6. Table 6: A power-law distribution with degree γ = 2 . 5; the probability, p ( k ), and N k , the number of vertices with degree k , for a network with a the total number of vertices N = 10 8 . k p ( k ) N k k p ( k ) N k 1 0.745 74 . 5 × 10 6 6 0.009 0 . 9 × 10 6 2 0.131 13 . 1 × 10 6 7 0.006 0 . 6 × 10 6 3 0.049 4 . 9 × 10 6 8 0.004 0 . 4 × 10 6 4 0.023 2 . 3 × 10 6 9 0.003 0 . 3 × 10 6 5 0.013 1 . 3 × 10 6 10 0.002 0 . 2 × 10 6 An organization where these nodes are selected as controllers has several advantages: There is a natural way to choose controllers, c i is a controller if deg( c i ) > κ , with κ a relatively large number, e.g. κ = 10. There is a natural way to cluster the nodes around the controllers; a node q joins the cluster built around the controller c i at the minimum distance d ( q, c i ). Thus, the average distance between the controller and the servers in the cluster is minimal. A number of studies have shown that scale-free networks have remarkable properties such as: robustness against random failures [12], favorable scaling [2, 3], resilience to congestion [32], tolerance to attacks [80], small diameter [18], and small average path length [11]. Problem 7. Use the start-time fair queuing (SFQ) scheduling algorithm to compute the virtual startup and the virtual finish time for two threads a and b with weights w a = 1 and w b = 5 when the time quantum is q = 15 and thread b blocks at time t = 24 and wakes up at time t = 60 . Plot the virtual time of the scheduler function of the real time. 51
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As in Problem 6, we consider two threads with the weights w a = 1 and w b = 5 and the time quantum is q = 15, and thread b blocks at time t = 24 and wakes up at time t = 60. Initially S 0 a = 0, S 0 b = 0, v a (0) = 0, and v b (0) = 0. The scheduling decisions are made as follows: 1. t=0 : we have a tie, S 0 a = S 0 b and arbitrarily thread b is chosen to run first; the virtual finish time of thread b is F 0 b = S 0 b + q/w b = 0 + 15 / 5 = 3 . (22) 2. t=3 : both threads are runnable and thread b was in service, thus, v (3) = S 0 b = 0; then S 1 b = max[ v (3) , F 0 b ] = max(0 , 3) = 3 . (23) But S 0 a < S 1 b thus thread a is selected to run. Its virtual finish time is F 0 a = S 0 a + q/w a = 0 + 15 / 1 = 15 . (24) 3. t=18 : both threads are runnable and thread a was in service at this time thus, v (18) = S 0 a = 0 (25) and S 1 a = max[ v (18) , F 0 a ] = max[0 , 15] = 15 . (26) As S 1 b = 3 < 12, thread b is selected to run; the virtual finish time of thread b is now F 1 b = S 1 b + q/w b = 3 + 15 / 5 = 6 . (27) 4. t=21 : both threads are runnable and thread b was in service at this time, thus, v (21) = S 1 b = 3 (28) and S 2 b = max[ v (21) , F 1 b ] = max[3 , 6] = 6 . (29) As S 2 b < S 1 a = 15, thread b is selected to run again; its virtual finish time is F 2 b = S 2 b + q/w b = 6 + 15 / 5 = 9 . (30) 5. t=24 : Thread b was in service at this time, thus, v (24) = S 2 b = 6 (31) S 3 b = max[ v
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