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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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5 votes
Can we not use the formula for Utilization here it is fully utilized so 1=N/1+2a that way N=201 no of bits required in go back n for window size is 2^k-1=201..k-8 bits
4 votes
4 votes
Given, L =  10^7 bits , B = 500 * 10^ 6  SO , Tt = L / B = 0.02 sec

d = 8000 km ,V = 4*10^6 m/sec SO, Tp = d / v = 2 sec

a = Tp / Tt = 2 / 0.02 = 100

The network is to be used to its full capacity so η = 100 %

Go-back-N protocol, So η = w/(1+2a) where w = 2^n-1
w = 1+2*a

⇒ 2^n - 1 = 201
⇒ 2^n = 202 ⇒ n = 8
2 votes
2 votes

If nothing is given , we can safely assume that the channel is full duplex.

What is Channel Capacity? → The amount of data that the channel can contain before the 1st bit reaches the receiver.

Channel capacity is also known as Bandwidth-Delay Product . For a half-duplex channel it is $B.W * T_p$.

$T_p = 2s$ and $B.W = 500*10^6 bps$ , so the Channel capacity for half-duplex channel is $10^9 \ bits$ , and that for a full duplex channel it will be $2*10^9 \ bits$ . Size of one packet = $10^7 \ bits$ , so the maximum number of packets that the channel can hold is $200$.

Now , squence number $>=$ $SWS+RWS >=200+1 = 201$ . So the minimum number of bits required = $log_2(201) = 8$ . Thus $8$ is the answer. 

0 votes
0 votes
For 100% efficiency we --------- sent (1+2a) Frames.
Than to sent (1+2a) frames ------ (1+2a) sequence numbers required.
Now try to calculate (1+2a) , where a =  Propagation delay(Tp)/Transmission delay(Tt).
And Tt = packet size / bandwidth = 10^6 / 500 * 10^6 = 1/50 sec.
Tp = distance / speed = 8000 km / 4*10^6 m = 2 sec.
Therefore, (1+2a) = 201 after calculation.

We know, Available sequences number >= Ws+Wr.
In GBN, Sender window size(Ws) = (1+2a) and Receiver window size(Wr) = 1.
So total sequence numbers required, Ws+Wr = 201 + 1 = 202.
So bits required to represent this many sequence numbers = ceil(Log 202) = 8 bits.
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