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A $3000$km-long T1  trunk having Transmission time $1.544$ Mbps is used to transmit  $64$byte frames using Go back -N.

If the propagation speed is $6$ μsec/km, how many bits should the sequence numbers be?

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6 votes

Answer: 7 bits

For Go Back to N:

  • 'N' buffers are maintained on sender side and 
  • 1 buffer is maintained on receiver side

So, we need unique N + 1 sequence numbers

So the bits required would be  Ceil $log_{2}(N + 1)$ bits to represent these numbers.

Now let us find what the N should be.

  • Since the propagation delay is 6 μsec/km, which means for 1 Km it takes 6 μsec to travel from one end to another
  • So, for 3000 Km it would take 3000*6 μsec to travel from one end to another end, which is 18000 μsecs of total propagation delay.
  • So RTT would be $2*18000$

So, to achieve full capacity, sender must keep on transmitting for this time.

T1 trunk has a 1.544 Mbps transmission speed and is full duplex.

This means,

  • In 1 sec, $1.544 * 10^{6}$ bits can be loaded on link
  • Or $1,000,000$ μsec, $1.544 * 10^{6}$ bits can be loaded on link
  • Or 1 μsec, $\frac{1.544 * 10^{6}}{1000000}$ bits can be loaded on link
  • So, In $2*18,000$ μsec, $\frac{1.544 * 10^{6} * 2*18000}{1000000}$ bits can be loaded on link
  • Which comes to $55,584$ bits or $6948$ bytes

Frame size is given as 64 bytes. So $\frac{6948}{64}$ packets or 108 packets can be sent

So, our N is 108 packets, so $N + 1$ is $109$ packets

Bits needed for Ceil $log_{2}109$ which is bits

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Given, Propagation Speed is 6 microsec/Km. So for 3000 Km It will be  3000*6 = 18000 microsec.
Therefore the RTT = 2* 18000 microsec

Given, Bandwidth is 1.544Mbps means for 1 sec it has the data rate as 1.544*106
therefor the data rate for RTT will be (1.544 *106) * (2*18000*10-6) = 55584 bits = 6948 bytes.

Given, Frame size = 64bytes. Therefore the number of frames transmitted in RTT = 6948/64 = 108.56 $\cong$ 109.
Therefore N = 109

and in GBN #sequence bit = Ceil [log(N+1)] = Ceil[log109] = 7

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propagation time  tp= 18000 μsec= 18 msec

transmission speed of T1  trunk = 1.536 mbps

tx = 0.3

RTT = 36.3

in RTT  time sender can send upto 121 frames = (N)

in Go back N bits required would be ceil  log2(N+1) => 7 bits
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0 votes
Given,

         distance=6000 km , propagation delay=6 usec per km.

         so,total pd=6*1000*6 usec

         bandwith=1.544 Mbps

         packet size=64bytes

        Here,Go-Back-N is given so,

                          Sender window size is N and Receiver Window size is 1.

        Efficency =N/1+2a

       N=Sender window size

       a=pd/td

       for 100%utilization we must have N=1+2a

       And we know, Ws+Wr<=Sequence No.

       N=1+2a=218.

       sequence no bits >= ceil(log(N))=8.

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