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G + n j( + 1) 1 G 1
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(10.29)
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G To obtain the probability of having m fully-occupied groups, we rst notice that there are m possible combinations of picking m fully-occupied groups among G groups. Given that there
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A Concurrent Push-Parallel Server Architecture
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are n active video sessions and m fully-occupied groups, the number of ways to distribute the remaining (n m /g) video sessions among the remaining (G m) groups with none of those groups fully occupied can be obtained from equation (10.29) as N (n m /g, G m, ( /g) 1). Hence the total number of ways for exactly m of the groups fully occupied is given by Nfull (n, m) = G m N (n m g , G m, g 1) (10.30)
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The probability of having m fully-occupied groups given n active video sessions can then be obtained from Pfull (n, m) = Nfull (n, m) N (n, G, ) (10.31)
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Knowing this, we can derive the average scheduling delay in the following way. Given m out of G groups are fully occupied, the probability for the assigned group to be available (not fully occupied) is given by V0 = G m G (10.32)
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Hence P0 = (1 V0 ) will be the probability of the assigned group being fully occupied. It can be shown that the probability for a client to wait k additional groups provided that the rst k assigned groups are all fully occupied is Vk = Pr{(k + 1)th group available |Pk } = G m , G k 1 k m (10.33)
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and the probability for the rst k groups all being fully occupied is
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Pk =
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m!(G k)! , G!(m k)!
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1 k m.
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(10.34)
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Hence, we can solve for the probability of a client having to wait k additional groups, denoted by Wk , from Wk = Pr{(k + 1)th group free |Pk }Pk = (G m)m!(G k 1)! , G!(m k)! 1 k m.
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(10.35) Therefore, given the number of groups that are fully occupied m, the average number of groups a client has to wait can be obtained from
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Wavg (m) =
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(10.36)
Similarly, given the number of active video sessions n, the average number of groups a client has to wait can be obtained from equations (10.31) and (10.36) as follows:
Mavg (n) =
Wavg ( j)Pfull (n, j)
(10.37)
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And the corresponding average scheduling delay given a system utilization of n is DS = Mavg (n)Q RV (10.38)
As the admission scheduler reduces the transmission jitter to equal to the clock jitter, the new pre ll delay can be obtained by replacing with in equation (10.23): DP = 2 + + f + f TE TF TF + f + (10.39)
10.5 Sub-Schedule Striping Scheme
The AGSS algorithm presented in the previous section substantially reduces the server buffer requirement as well as the scheduling delay. However, the client buffer requirement and, consequently, the pre ll delay are only slightly reduced as a side effect of the admission scheduler. In this section, we consider another modi cation to the concurrent-push algorithm that can substantially reduce the client buffer requirement and the pre ll delay. Speci cally, the analysis in Section 10.3 reveals that the main reason for the increase in client buffer requirement with the number of servers stems from the increase in the average lling time in equation (10.1). This suggests that we can reduce the buffer requirement by using smaller striping size Q. However, as the server retrieves data from the disk in units of Q bytes, reducing the striping size will adversely affect disk retrieval ef ciency. To solve this problem, we propose decoupling the transaction size for disk retrieval and transmission from the striping size sub-schedule striping (SSS). In particular, we maintain the disk transaction size at Q bytes but use a striping size (denoted by U ) inversely proportional to the number of servers in the system (Figure 10.6): U = Q/N S (10.40)