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# hw1solutions - 15-441 Computer Networks Homework 1 Assigned...

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15-441: Computer Networks Homework 1 Assigned: Sep 9, 2008 Due: Sep 16, 2008 Lead TA: Vijay Panghal < [email protected] > October 8, 2008 1. Calculate the total time required to transfer a 1000-KByte file in the following cases, assuming a RTT of 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 RTT before sending the next The bandwidth is infinite, i.e., the transmit time is zero, but only up to 20 packets can be sent per RTT Solution: Assume we are talking about transfer time from the perspective of transmitting at the source. We also accepted solutions that included propagation time to the destination. t = t handshake + t propagation + t transmission (2 * RTT) + (0.5 * RTT) + (datasize / bandwidth) 2 * (0 . 1 s ) + 0 . 05 s + (1000 * 1024 * 8 bits ) 1 . 5 * 10 6 bps = 5 . 71 s t = t handshake + t transmission + t wait (2 * RTT) + (datasize / bandwidth) + (1000 - 1) * RTT 2 * (0 . 1 s ) + (1000 * 1024 * 8 bits ) 1 . 5 * 10 6 bps + 999 * (0 . 1 s ) = 105 . 6 s Because t transmission is zero: t = t handshake + t wait (2 * RTT) + (49.5) * RTT 2 * (0 . 1 s ) + 49 . 5 * (0 . 1 s ) = 5 . 15 s Common errors included not converting bytes to bits, forgetting the 2 * RTT in problems (b) and (c), and forgetting that 1 KByte = 2 10 bytes. 1

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2. Consider a Network with a ring topology, link bandwidths of 100 Mbps, and propagation speed 2 * 10 8 m/s. (a) What would be the circumference of the loop be to exactly contain one 250 byte packet, assuming nodes do not introduce delay? (b) What would be the circumference of the loop be to contain atleast one 250 byte packet, if there was a node every 100 m, and each node introduced 10 bits of delay (bit time cost in queue) ? Solution: Data in transit = Delay * Bandwidth Latency in traversing single bit around the ring = ( Ring Circumference 2 * 10 8 ) 250 * 8 = Ring Circumference 2 * 10 8 * (100 * 10 6 ) Ring Circumference = 4000 m
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hw1solutions - 15-441 Computer Networks Homework 1 Assigned...

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