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Consider a network connecting two systems located $8000$ $\text{Km}$ apart. The bandwidth of the network is $500 \times 10^6$ $\text{bits}$ per second. The propagation speed of the media is $4 \times 10^6$ $\text{meters}$ per second. It needs to design a Go-Back-$N$ sliding window protocol for this network. The average packet size is $10^7$ $\text{bits}$. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be ______.
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Best answer
56 votes
56 votes

Answer = 8 bits.

In order to achieve full utilization, sender has to keep on sending frames till
the acknowledgement arrives for the first frame.

Time taken for acknowledgement to arrive is $2$ times propagation delay $+$ transmission time for a frame.

One way propagation delay $=\dfrac{8000\times 10^3}{(4\times 10^6)}= 2s$

Time taken to transmit one frame $=\dfrac{10^7}{(500\times 10^6)}=0.02s$

 So, RTT $=2\times 2 + 0.02 = 4.02s$

No. of frames that can be transmitted in $4.02$ secs $=\dfrac{4.02}{0.02}=201$

Being Go-Back-N protocol this means $W_S = 201$ and $W_R=1.$ So, total number of sequence numbers required $ = 201+1=202.$

Hence, minimum number of bits required for sequence numbers till $202$ is $\lceil \log_2 202 \rceil = 8.$ 

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40 votes
40 votes

Isn't the minimum sequence number $\geq$ SWS+RWS.
So we are getting $201$ for the sender window and since we are using $\text{GBN}$
we should add $1$ as the RWS.

So total sequence number $\geq 201+1 =202.$

Answer will still remain 8 though

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22 votes
22 votes
Given $T_{processing}=0,Transmission_{ack}=0$

My total RTT here is $T_{data}+T_{prop}+T_{prop}=T_t+2*T_p$

My useful time will only be $N*T_{data}$ where $T_{data}$ will be transmission time for one frame.

so my efficiency

$\frac{N*T_{data}}{RTT}=\frac{T_{data}*N}{T_{data}+2*T_{prop}}=\frac{N}{1+2*\frac{T_p}{T_t}}=\frac{N}{1+2a}$

so, my efficiency would be 100% only if I send 1+2a packets into network in RTT time.

This 1+2a comes to be 201.

Now, this is the sender's window size. Receiver window size would be 1. So, the total you would need 202 distinct sequence numbers for frames and this requires $\lceil log_2202 \rceil = 8$ bits
10 votes
10 votes

to know  the minimum size in bits of the sequence number field when the The network is to be used to its full capacity we need to send until RTT time.


propagation speed is given by which we can find propagation time ...
to travel 4x10^6 meters its take 1 second
so for 8x10^6 meters it will take 2 second which is our propagation time

now our RTT will be  2*PT=2*2=4 second 

now we can find that in 4 second how much data we can transfer using Bandwidth. 
in 1 second we transferring 500x10^6 bits
so in 4 second  we will transfer 500x10^6 bits*4=200x10^7 bits 
now we can find the total no packets to be transfer (window size) =200x10^7 bits/10^7=200 packets
to represent 200 we need 8 bits which is our answer

 

Answer:

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