What is the minimum timeout duration that should be set for this rdt scheme over this network?

Programming by Geoffrey

 

Problem 1 (19pts)

The diagram below shows the timing progress of a packet and ACK exchange to transfer 9 packets (each of size 2kB) between a sender and a receiver using a Go­Back-3 rdt, over a network with throughput 10Mbps, loss-error probability of 5% and a round-trip duration of 2ms.

 

012345678
012 345678
012 345678

012345578
012345578
012345578

012345678
012345678

012345678 012345678 012345678 012345678 012345678

012345678 012345678 012345678 012345678 012345678 012345678 012345678 012345678

sender

Pkt Pkt

kt

rAt 0 tim Peout

pkt pkt Pkt pkt pkt

okt 2 tirneout

Pkt pkt pkt Pkt Pkt

pkt 4 t..Meout

Pkt
Pkt

Pkt
Pkt
Pkt

receiver

Ack
Ack
Ack

Ack

Ack
Ack

Ack Ack

Ack Ack Ack

Ack Ack

 

  1. For the shown diagram: (13 pts)
  2. Enumerate all the sent packets and ACKs in the diagram in accordance to the employed rdt scheme. (7.75 in total, 0.25pt for each packet/ack)Please explain what the receiver does with each received packet on the right

    of the diagram (Example: packet delivered or buffered or discarded).

    1. Calculate the minimum number of sequence numbering bits that should be used in the transport layer header of this rdt scheme. (lpt)

    All steps and mathematical formulae must be written.

    1. Is the selection of the window size suitable with respect to the round-trip time of the network? Justify your answer. (1.5pts)

    All steps and mathematical formulae must be written.

    1. What is the minimum timeout duration that should be set for this rdt scheme over this network? (lpt)

    All steps and mathematical formulae must be written.

    1. What is the range of round-trip times on this network that justifies the use of a Stop-and-Wait rdt instead of the employed Go-Back-3 scheme? (2.5pts)

    All steps and mathematical formulae must be written.

    Highlight in each step the packets that are in the sender window on the left of the diagram. (5.25pt in total, 0.25pt for each packet)