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56 votes
56 votes

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 the processing delay is negligible, the $\text{minimum}$ number of bits for the sequence number field in a frame for maximum utilization, when the $\text{sliding window protocol}$ is used, is:

  1. $\lceil \log_2 \frac{2LtR +2K}{K} \rceil$

  2. $\lceil \log_2 \frac{2LtR}{K} \rceil$

  3. $\lceil \log_2 \frac{2LtR +K}{K} \rceil$

  4. $\lceil \log_2 \frac{2LtR +2K}{2K} \rceil$

10 Answers

2 votes
2 votes
Check this...

Utilization = amount of data send / amount of data can be send

Utilization = (window size *frame size) / ((transmission time+RTT)*channel capacity)

Utilization = 1 when maximum

 window size *frame size = (transmission time+RTT)*channel capacity

W * K =  ( K/R + 2Lt ) * R

W = (K + 2LtR) / K

let  sequence number  field in frame contains n bits

Then  2^n = w

n = log w

n= log ( (K + 2LtR) / K)
1 votes
1 votes

Adding to the other brilliant answers here,

  • Here, "capacity" means "bandwidth". Look at the given units.
  • "minimum number of bits for the sequence number field" using sliding windows, would be for GBN and not SR, obviously.
  • 100% utilisation means The size of the sender window = 1 + 2a.

Hence, Option C

0 votes
0 votes
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= 2n, where n = number of bits in sequence number

2n​ ​= K + 2LTR / K

n= log((K+2LTR)/K)
Answer:

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