Question

Let host A send a large file to host B over a pipelined transport protocol with a fixed sliding window (e.g., GBN or SR). How big does the sender’s window need to be to ensure maximum use of the network path capacity between A and B if the path has the following idealistic characteristics:

(i) the round-trip delay between the end hosts (measured for a very small packet so that it is not affected by the transmission delay) is 10 msec, (ii) traffic traverses 10 routers along the path from A to B and the same 10 routers on the return path from B to A, (iii) the bandwidth of the each link in the path is 1 Megabyte per second (1 Megabyte = 106bytes), (iv) Segment size is 1000 bytes while the acknowledgement size is negligible (i.e., the transmission delay of the ack packets can be assumed to be 0) and (v) there is no interfering traffic and the processing delay at both end hosts and routers is negligible.

Answer 1

solution:

Here the entire propogation delay is 10msce. Propogation delay at the router is even considered here.

Transmission delay of data packet = packet size/bandwidth

= 1000B/10^6 B = 1 msec

Here we have 10 routers between HostA and HostB. So total transmission delay is going to be 11msec.

Thus total time to send a packet from HostA to HostB = 11+10=21msec.

Now to achieve maximum use of network path, we must send packets untill acknowledgement of first packet is received.

Here acknowledgement of first packet will come after 21msec.

HostA can transmit one packet at 1msec of time. So in 21msec, it will send 21 packets.

Thus there sender window size has to 21.

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