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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?

+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
- 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 $1,000,000$
**μ**sec, $1.544 * 10^{6}$ 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 *7 *bits

0

- 'N' buffers are maintained on receiver side

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

+1

@Arunav

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 = $7$

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 = $7$

0

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 ?

0

There may be some difference in the bits due to different formulaes.

Also, I didn't calculated upto the precision. It was approx answer .

Also, I didn't calculated upto the precision. It was approx answer .

0

@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?

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?

0

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

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

0

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

T1 trunk has a 1.544 Mbps means that the bandwidth of the link is 1.544Mbps

0

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.

0

@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.

0

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.

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*10^{6}

therefor the data rate for RTT will be (1.544 *10^{6}) * (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

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

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.

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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