TCP has a 32-bit sequence number field and 16-bit advertised window field. Assume that RTT is 512 (2 9 ) ms, transmission speed is 1 Gbps (2 30 bps) and each segment transmitted is 1B byte. Note that Since not two identical sequence numbers can be unacknowledged in the pipe, half of the sequence numbers can used (2 31). (a) How long does it take for the sequence numbers to warp around? 5 pts (b) Now, instead of sending 1 B segment, let’s send a 16 B segment. How long does it take for the sequence numbers to warp around? 5 pts (c) What is the drawback in using large segments? 5 pts (d) What is the maximum achievable throughput?

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Explanation:

Usable sequence numbers = 2^31

Transmission speed = 2^30 bps

(a)Wrap around time = (2^31 * 8 * 1)/(2^30) = 16 seconds

(b)segment size = 16 Bytes

Wrap around time = (2^31 * 8 * 16)/(2^30) = 256 seconds

(c)if we use large segment size ,which increases size of MTU in data link layer .Maximum transmission unit is the maximum size of a packet or frame that can flow across the network, without being fragmented,which lead to overhead of fragmentation at network layer.

(d)Maximum network throughput equals the TCP window size divided by the round-trip time of communications data packets.

Round trip time = 2^9 ms

window size = 2^16

Throughput = (2^16)/(2^9) = 2^7 KBps = 128 KBps = 1Mbps

answer(igmp);

transport layer protocol

explanation: