GATE CSE 2014 Set 1 | Question: 28

Question

Consider a selective repeat sliding window protocol that uses a frame size of $1$ $\text{KB}$ to send data on a $1.5$ $\text{Mbps}$ link with a one-way latency of $50$ $\text{msec}$. To achieve a link utilization of $60\%$, the minimum number of bits required to represent the sequence number field is ________.

Comments

My assumption on reading

to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%.

Bandwidth that can be used = 60% of 1.5Mbps

Instead, efficiency is 60%. How did we interpret this?

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@Mk Utkarsh True. This question checks patience level if solved from GO pdf!!

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GO to the Root of Concept: Video Explanation with timestamp:Sliding Window PYQs| GATE 2009 Q57, 58| 2006 Q46, 64| 2003, 2014, 2015, 2004, 2016, 2007| With NOTES

Answer 1

Sender window size is  e*[1+2a] where e is the efficiency

Divided by 2 as window size of sender + window size of receiver is less then available sequence number

after putting all the values of a = Tp/Tt and e=0.6

we got answer to be log[2e(1+2a)] = 5 as our bits for sequence number

Answer 2

For 100% efficiency we--------- sent (1+2a) Frames
For 60% efficiency we----------- sent ? Frames
= 60/100(1+2a) , where a = Propogation delay(Tp)/Transmission delay(Tt)
= 11.85
We are using the Selective repeat sliding window protocol. Therefore, Sender window size(Ws) = Reciever window size(Wr).
We know, Available sequences number >= Ws+Wr.
Available sequences number = 11.85 +11.85 = 23.70
Therefore, sequence number bits required = ceil(Log 23.70) = 4 bits.