Hw01 hint

--- 27Visiting k of them incurs an RTT of D1 per DNS and visiting each of the remaining incurs an RTT of D22RTT0 is required to set up each of the m TCP connections and to request and receive each object8k = 2 three DNS server (a) no parallel TCP => ? RTT0 (b) five parallel TCP => ? RTT0

9(a) The time to transmit an object of size L over a link or rate R is L/R. The average time is the average size of the object divided by R (960,000 bits)/(16,000,000 bits/sec) = 0.06 sec (b)The traffic intensity on the access link is reduced by 60% since the 60% of the requests are satisfied within the institutional network. Thus the average access delay is (0.06 sec)/[1 (0.4)(0.9)] = 0.094 seconds10Note that each downloaded object can be completely put into one data packet. Let Tp denote the one-way propagation delay between the client and the server22calculating the minimum distribution time for client-server distribution: Dcs = max {NF/us, F/dmin} calculating the minimum distribution time for P2P distribution

F = 15 Gbits = 15 * 1024 Mbitsus = 30 Mbpsdmin = di = 2 Mbps

23(a) us/N dmin this rate is less than each of the clients download rate(b) us/N dmin the aggregate rate, N dmin, is less than the servers link rate us(c) DCS max {NF/us, F/dmin} + Suppose that us/N dmin + from (a) we have DCS NF/us----------------------------------------------------------------------------------------------------------------------- DCS = NF/us when us/N dmin DCS =F/dmin when us/N dmin 24(a) Define u = u1 + u2 + .. + uN. By assumption us = (us + u)/N

Let ri = ui/(N-1) and rN+1 = (us u/(N-1))/N

28Peer 6 would first send peer 15 a message, saying what will be peer 6s predecessor and successor?32UDPServer determines the client port number by unraveling the datagram it receives from the client.