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Consider the following chain topology:
A ---- B ----- C ---- D ---- E

A is sending packets to E using a reliable transport protocol.
Each link above can transmit one packet per second. There are no queues or
other sources of delays at the nodes (except the transmission delay of course).

Consider a sliding window protocol running at the optimum window size.
Suppose nodes in the network get infected by a virus that causes them to
drop packets when odd sequence numbers. The sliding window protocol starts
numbering packets from sequence number 1. Assume that the sender uses a timeout
of 40 seconds. The receiver buffers out-of-order packets until it can deliver
them in order to the application. The number of packets in this buffer 35 seconds
after the sender starts sending the first packet?

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With a window size of 8, the sender sends out packets 1--8 in the first 8 seconds. But it gets back only 4 ACKs because packets 1,3,5,7 are dropped. Therefore, the sender transmits 4 more packets (9--12) in the next 8 seconds, 2 packets (13--14) in the next 8 seconds, and 1 (sequence number 15) packet in the next 8 seconds. Note that 32 seconds have elapsed so far. Now the sender gets no more ACKs because packet 15 is dropped, and it stalls till the first packet times out at time step 40. Therefore, at time 35, the sender would have transmitted 15 packets, 7 of which would have reached the receiver. But because all of these packets are out of order, the receiver's buffer would have 7 packets.

Source:

http://web.mit.edu/6.02/www/s2011/handouts/tutprobs/transport.html

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