Question

(a) A data link with a 1 Gigabits/sec capacity is used to transmit packets made up of 1400 bytes of data and 100 bytes of protocol control information. Each packet is acknowledged by a short frame of size 150 bytes. The propagation speed in the transmission medium is 200,000 km/sec. Consider the buffering and processing time in the nodes to be negligible. A node A communicates with a node B, with a distance between nodes A and B of 16 km What is the Round Trip Time on this connection (measured from the start of the transmission of the message from the source to full reception of the acknowledgement at the source)? [7 marks] A Sliding Window flow control is used on this link. What should be the value of the Window size to allow a continuous use of the link (assuming no errors)? [3 marks] The data link protocol used on this connection allows for only 4 bits sequence numbers Should the flow control work in Selective Repeat or Go Back N mode of operation? [3 marks]

Answer 1

1. Round trip time will include propagation delay and transmission delay.

Since the distance between node A and node B is 16 KM, hence propagation time from node A to node B = 16/300000 = 5.33*10^-5 sec = 0.0533

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Hence round-trip propagation delay = 2*0.0533 = 0.1066

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Packet size = 1400+100 = 1500 bytes = 1500*8 bits = 12000 bits

Bandwidth = 1 Gbits/sec =

Transmission delay in sending packet = packet size/ bandwidth = 12000/ 10⁹ = 12

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Transmission delay in sending acknowledgement = acknowledge size/ bandwidth = 1200/ 10⁹ = 1.2  

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Hence Round trip time = Transmission delay in sending packet + Transmission delay in sending acknowledgement + round-trip propagation delay = 12 + 1.2 + 0.1066 = 13.3066  

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2. Suppose N packets can be transmitted for efficient utilization. This means

N*(transmission delay of sending packet) = Round trip time

=> N*12 = 13.3066

This given N to be roughly 1.

Hence the sequence number should be N+N = 2

For sequence number of 2, window size of 2 is sufficient.

3. Yes since sequence number is upto 2, hence 4 bit sequence number with total 2⁴ = 16 sequence number will be sufficient for both selective repeat and Go-back N.

Please comment for any clarification.