N 10 100 1000 7680 51200 512000 300 Kbps u 700 Kbps 7680 512000 2 Mbps 7680 51200 512000 Peer to Peer N 10 100 1000 7680 25904 47559 300 Kbps u 700 Kbps 7680 15616 21525 2 Mbps 7680 7680 7680 Problem 23 Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of a rate of Note that this rate is less than each of the client’s download rate, since us/N. by assumption Thus each client can also receive at rate Since each client us/N dmin. us/N. ≤ receives at rate the time for each client to receive the entire file is F/( = us/N, us/N) NF/ us. Since all the clients receive the file in the overall distribution time is also NF/ us, NF/ us. Consider a distribution scheme in the server sends the file to each client, in parallel, at a rate of Note that the aggregate rate, is less than the server’s link rate since dmin. N dmin, us, by assumption Since each client receives at rate the time for each client us/N dmin. dmin, ≥ to receive the entire file is F/ Since all the clients receive the file in this time, the dmin. overall distribution time is also F/ dmin. From Section 2.6 we know that max {NF/us, (Equation 1) DCS F/dmin} ≥ Suppose that Then from Equation 1 we have But from (a) we us/N dmin. DCS NF/us . ≤ ≥ have Combining two gives: DCS NF/us . ≤ = when (Equation 2) DCS NF/us us/N dmin. ≤ We can similarly show that: =F/dmin when (Equation 3). DCS us/N dmin ≥ Combining Equation 2 and Equation 3 gives the desired result. Problem 24
N 10 100 1000 u 300 Kbps 7680 51200 512000 700 Kbps 7680 512000 2 Mbps 7680 51200 512000 Peer to Peer N 10 100 1000 u 300 Kbps 7680 25904 47559 700 Kbps 7680 15616 21525 2 Mbps 7680 7680 7680 Problem 23 Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of a rate of us/N. Note that this rate is less than each of the client’s download rate, since by assumption us/N ≤ dmin. Thus each client can also receive at rate us/N. Since each client receives at rate us/N, the time for each client to receive the entire file is F/( us/N) = NF/ us. Since all the clients receive the file in NF/ us, the overall distribution time is also NF/ us. Consider a distribution scheme in the server sends the file to each client, in parallel, at a rate of dmin. Note that the aggregate rate, N dmin, is less than the server’s link rate us, since by assumption us/N ≥ dmin. Since each client receives at rate dmin, the time for each client to receive the entire file is F/ dmin. Since all the clients receive the file in this time, the overall distribution time is also F/ dmin. From Section 2.6 we know that DCS ≥ max {NF/us, F/dmin} (Equation 1) Suppose that us/N ≤ dmin. Then from Equation 1 we have DCS ≥ NF/us . But from (a) we have DCS ≤ NF/us . Combining two gives: DCS = NF/us when us/N ≤ dmin. (Equation 2) We can similarly show that: DCS =F/dmin when us/N ≥ dmin (Equation 3). Combining Equation 2 and Equation 3 gives the desired result. Problem 24
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