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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

Best answer
75 votes
75 votes

Answer: C

We can send $\dfrac{\text{RTT}}{\text{Transmission Time}}$ number of packets

for maximum utilization of the channel, as in this time, we get the first $\text{ACK}$ back and till that time, we can continue sending packets.

So, $\dfrac{\text{Transmission Time + 2$\times $Propagation Time}}{\text{Transmission Time}}$ number of packets should be sent.

Therefore, bits required for the sequence number field:

$\left \lceil \log_2 \left(\frac{\dfrac{K}{R}+\large 2Lt}{\dfrac{K}{R}}\right) \right\rceil = \left\lceil \log_2\left(\dfrac{K+2LtR}{{K}}\right) \right\rceil$

Edit : here it is asked for general sliding window protocol not GBN nor SR .

In general sliding window protocol:

Sequence number bit $=\log \text{(sender window size)}$

edited by
88 votes
88 votes

for maximum utilization $\eta = 1$,

$\begin{align*} \eta &= SWS \times \frac{\text{Transmission Time}}{\text{Transmission Time + 2$\times$Propagation Time}} \\ \\ 1 &= SWS \times \frac{\text{Transmission Time}}{\text{Transmission Time + 2$\times$Propagation Time}} \\ \end{align*}$

so, $\dfrac{\text{Transmission Time + 2$\times$Propagation Time}}{\text{Transmission Time}} =\text{SWS}$

gives the number of packets that are sent. It means that it is the Sender's Window Size.

But we also know that,

$\text{available sequence numbers} \ \geq \text{ SWS + RWS}$                                                         

so, to get the minimum number of bits needed to represent the sequence numbers - we should consider what protocols are in use.

if GBN ARQ: 

$\begin{align*} \substack{\large\text{# bits required to rep.} \\ \large\text{ sequence numbers}}
&= \lceil \log_2{(SWS + RWS)} \rceil \\ &= \left \lceil \log_2{\left(\frac{\frac{K}{R} + 2Lt}
{\dfrac{K}{R}}+1 \right)} \right \rceil \\ &= \left \lceil \log_2{\left(\frac{2K + 2LtR}{K}\right)}
 \right \rceil \end{align*}$

if Selective Repeat ARQ :

$\begin{align*} \substack{\large\text{# bits required to rep.} \\ \large\text{ sequence numbers}}
&= \lceil \log_2{(2\times SWS)} \rceil \\ &= \left \lceil \log_2{\left(2\times \frac{\frac{K}{R} + 2Lt}{\frac{K}{R}}\right)}
\right \rceil \\ &= \left \lceil \log_2{\left(\frac{2K + 4LtR}{K}\right)} \right \rceil \end{align*}$

edited by
7 votes
7 votes

$Answer$ C

$a = \frac{Tp}{Tt}$

$\text{For maximum utilisation,}$

$\frac{W_{s}}{1+2a} = 100 $%$ $

$log(1+2a)$ is the answer 

$Tt = \frac{K}{R}$

$Tp = t\times L$

$log(1+2a) = log(1+2\frac{t\times L}{\frac{K}{R}})$

$log(1+2a) = log(\frac{2LtR+K}{K})$

 

 

edited by
4 votes
4 votes
Answer : (C)

Ws(window size of sender)=1+2a  where a=Tp/Tt

 minimum sequence number possible= 1+2a

min. num of bit for sequence number = ⌈log(1+2a)⌉ ************************** Equation 1

Tp/Tt=Lt/(K/R)

1+2a=(K+2LtR)/K

put value of 1+2a in eq 1

answer is C.
Answer:

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