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.
 1197 reads
Similar content
Did not found what you are looking for, Ask your doubt or Help by your contribution
Responses
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.
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 2^{n}.