17,838 views
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

0 votes
0 votes

Some people having confusion with capacity: Capacity will always be in bits/bytes.
We can take the given value as BW.

max eff would be when we can send exactly 1+2a packets in one round trip.

ie (1+2LtR/k) pakets.

to send these many packets in sliding window(GBN/SW)
we need atleast  floor(log(1+2a)) +1 sequence number bits
ie  ceil(log(1+2a) sequence number bits  in GBN.

Ans=ceil(log((k+2Ltr)/k)) bits ie option(c)

–2 votes
–2 votes
Total time required for  K bit delivery = K/ R + 2* RTT = K/R+ 2* ( L* t)

                                                     = K/R+ 2Lt

optimal window size = 1+(2 * BW* prop delay)/Frame size = 1+ (2*R*Lt /K)

                              = (K+2RLt)/K

 For maximum utilization window size should be 2^n-1 = (K + 2RLt)/K

                                                                         2^n = (2K +2RLt)/K

                                                                            n = log (2K+2RLt)/K

                                                                        Answer option A
Answer:

Related questions

27 votes
27 votes
4 answers
9
Kathleen asked Sep 21, 2014
33,742 views
The message $11001001$ is to be transmitted using the CRC polynomial $x^3 +1$ to protect it from errors. The message that should be transmitted is:$11001001000$$110010010...
32 votes
32 votes
7 answers
10
16 votes
16 votes
3 answers
11
Kathleen asked Sep 21, 2014
11,622 views
In Ethernet when Manchester encoding is used, the bit rate is:Half the baud rateTwice the baud rateSame as the baud rateNone of the above
27 votes
27 votes
4 answers
12
Kathleen asked Sep 21, 2014
13,766 views
Match the following:$$\begin{array}{llll} \text{(P)} & \text{SMTP} &(1)& \text{Application layer} \\ \text{(Q)} & \text{BGP}& (2) & \text{Transport layer} \\ \text{(R)}...