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+5 votes

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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What about same question in SWP?

4 Answers

+5 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

edited by
Thanks for nice explanation
  • 'N' buffers are maintained on receiver side

Does it ? Then, how many buffers protocol $6$ maintains ?

My bad.. that is SR

No, Bits required will be $7$ only . There is some calculation mistake .
Why would they be 7, When only N + 1 packets are there ?

In $1$ sec, $1.5$Mbits can be loaded and so in $1$ RTT how much can we load (Acc. to bandwidth delay product)

RTT = $36300$ usec hence, in this much time we can send $56048$ bits (Something )

Hence, Total packets => $110$ (something)

Therefore, total sequence numbers = $111$ and bits req. are = $7$
Thank you Sir, I had taken prop delay time instead of RTT which should be the case. However, my answer is still 109 packets, bits are correct though, am I missing something ?
There may be some difference in the bits due to different formulaes.

Also, I didn't calculated upto the precision. It was approx answer .
@Arunav :- As you have calculated n= 108 ,but then number of frames that a sender can send are 108+1=109.

Then you should put 109 in N+1 as N=109.Although it wont change the answer here but it might change [email protected] Can you please verify?

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

where is the source? I am not getting in the question. !

Tanenbaum.Its mentioned in question title:p Just kidding.

T1 trunk has a 1.544 Mbps means that the bandwidth of the link is 1.544Mbps
cool! it must be mentioned in the question(Transmission time/Bandwidth )After reading the question , i shall not see the tanenbaum  exercise :) rather i would try to solve it by seeing the data provided.

@Arunav Khare.....Perhaps I have a trivial doubt, but if the Bandwidth is 1.544Mbps, then when we'll change this to bits, won't it be 1.544*2^20 bits? Kindly clarify.

Okay, I've cleared this doubt....seems like I was getting confused!....If anyone else faces a similar confusion, then here's what it's about....

When we talk about data say, 1M then, it's 2^20 as data is always in the form of bits.

Whereas, quantities such as bandwidth, which is measured as frequency, 1M would mean 10^6.
+2 votes

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

0 votes
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
0 votes

         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


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

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


       sequence no bits >= ceil(log(N))=8.
Why are they saying propagation speed and then proving unit as microsecond per km?

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