GATE 2007

The distance between two stations M and N is L kilometers. All frames are K bits
long. The propagation delay per kilometer is t seconds. Let R bits/second be the
channel capacity. Assuming that processing delay is negligible, the minimum
number of bits for the sequence number field in a frame for maximum utilization,
when the sliding window protocol is used, is:
 A ) log { \((2LtR +2K) / {K }\)}

B) ​ ​​ ​log { ​​ \((2LtR/K)\) ​​ ​}

C) log { \((2LtR+K)/K\)}

D) log{ \((2LtR+K)/2K\)}

  • Explanation

    Ans : C.

    I dont how many of you know this that this problem is given in exercise of William Stallings Book Data and Network Communication. Chapter 6 Data link control. Any way that is not our concern.

    Bandwidth R bps , Frame size = K bits , PT = t sec,/ kilometer, distance L kilometer. So total PT ( not RTT) = L*t sec .Processing delay =0.Assuming W is window size

    Now, it says that We need to maximize utilization for sliding window protocol.

    Maximum utilization if=> useful time/ total time =1 . 

    it is possible when ,useful time = Total time.

    Useful time= W * {K/R } sec

    Total time = K/R + 2* L*t sec

    Now  =>  (W * {K/R } ) / ( K/R + 2* L*t ) = 1

    W * {K/R } = K/R + 2* L*t 

    W = K + 2LTR / K

    Suppose W= \(2^n\)where n = number of bits in sequence number

    \(2^n\)​ ​= K + 2LTR / K

    n= \(log (( K + 2LTR )/K)\)

    ans C.

    I hope it is clear.


abhishekeee2's picture

I think, you have calculated the Sender Window Size and considered the same as total Sequence number required. While, if sender window size is s then it will be (s+1) in case GBN and 2s in case of SR Protocol. So here minimum sequence number required will be in case of GBN that is (s+1).

So, Final Answer will be A.

vivek14's picture

Abhishek, I didn't exactly get your idea?

By the way, it does not mention GBN or SR. So I took generalized window size that is ,for n bit sequence number, window size 2n.

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