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**Unformatted text preview: **15-441: Computer NetworksHomework 1Assigned: Sep 9, 2008Due: Sep 16, 2008Lead TA: Vijay Panghal<vpanghal@andrew.cmu.edu>October 8, 20081. Calculate the total time required to transfer a 1000-KByte file in the following cases, assuming a RTTof 100ms, a packet size of 1KByte. and an initial 2RTT of handshaking before the actual data is sent.The bandwidth is 1.5 Mbps and data packets can be sent continuously.The bandwidth is 1.5 Mbps, but after we finish sending each data packet we must wait one RTTbefore sending the nextThe bandwidth is infinite, i.e., the transmit time is zero, but only up to 20 packets can be sent perRTTSolution:Assume we are talking about transfer time from the perspective of transmitting at thesource. We also accepted solutions that included propagation time to the destination.t=thandshake+tpropagation+ttransmission(2 * RTT) + (0.5 * RTT) + (datasize / bandwidth)2*(0.1s) + 0.05s+(1000*1024*8bits)1.5*106bps= 5.71st=thandshake+ttransmission+twait(2 * RTT) + (datasize / bandwidth) + (1000 - 1) * RTT2*(0.1s) +(1000*1024*8bits)1.5*106bps+ 999*(0.1s)= 105.6sBecausettransmissionis zero:t=thandshake+twait(2 * RTT) + (49.5) * RTT2*(0.1s) + 49.5*(0.1s)= 5.15sCommon errors included not converting bytes to bits, forgetting the 2 * RTT in problems (b)and (c), and forgetting that 1 KByte = 210bytes.12. Consider a Network with a ring topology, link bandwidths of 100 Mbps, and propagation speed 2*108m/s.(a) What would be the circumference of the loop be to exactly contain one 250 byte packet, assumingnodes do not introduce delay?(b) What would be the circumference of the loop be to contain atleast one 250 byte packet, if there wasa node every 100 m, and each node introduced 10 bits of delay(bit time cost in queue)?Solution:Data in transit = Delay * BandwidthLatency in traversing single bit around the ring = (Ring Circumference2*108)250*8 =Ring Circumference2*108*(100*106)Ring Circumference= 4000mLets say Number of node in Ring topology are...

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