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Frames of $\text{1000 bits}$ are sent over a $10^6$ $\text{bps}$ duplex link between two hosts. The propagation time is $\text{25 ms}$. Frames are to be transmitted into this link to maximally pack them in transit (within the link).

What is the minimum number of bits $(I)$ that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.

  1. $I=2$
  2. $I=3$
  3. $I=4$
  4. $I=5$
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18 Answers

Best answer
137 votes
137 votes

Bandwidth won't be halved in full duplex.
http://superuser.com/questions/335979/does-1-gbit-s-port-in-full-duplex-mean-1-gbit-s-send-and-1-gbit-s-receive

Propagation time is given as $25\;\text{ms}$. 

Bandwidth $=10^6\;\text{bps}$. 

So, to fully utilize the channel, we must send $10^6$ bits into the channel in a second, which will be $1000$ frames per second as each frame is $1000$ bits.

Now, since the propagation time is $25\;\text{ms},$ to fully pack the link we need to send at least $1000\times 25\times 10^{-3}=25$ frames.

So, we need $\lceil\log_{2} 25\rceil=5\;\text{bits}.$

Correct Answer: $D$

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50 votes
50 votes

BW both the sides will be $10^6$​ ​bps for a full-duplex channel/wire/link.

We are actually only considering transit to one side while transmission of the packets to fully pack them in the link(106 bps). So, instead of the general approach of taking $1+2a$ packets, here we are taking $1+a$ packets since we are only considering the time of $1$ Tp, we need to keep the link full for $1$ Tp.

So, Min. no. of packets required $= 1 + a = 1 + \frac{25}{1} = 26$ packets.

Min. no. of bits required $= log_2​(26) = 5$ bits.

So, $5$ bits are enough to pack one side of the link while the frames are in transit.


Actually, both the hosts will send data packets and acknowledgments, but for both the sides, $5$ bits are enough individually to fully pack $10^6$​ bits from both the sides(full-duplex). Here, to keep things simple, I have only discussed considering one sender and one receiver only.  

For optimal case, when we also consider the time of the acknowledgment coming back, that time also packets can be kept on transmitting in the channel,

So, Min. no. of packets required $= 1 + 2a = 1 + 2*\frac{25}{1} = 51$ (doesn't matter if the ack. is piggybacked or not).

and Min. no. of bits req $= log_2​(51) = 6$ if we want this kind of link utilization.

Again, here also for simplicity, only considered one sender and one receiver.

But in reality both the hosts will act as source and destination simultaneously. Obviously, the acknowledgments will be piggybacked in such a practical case.

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23 votes
23 votes

When the link is duplex do we have to take Bandwidth  = B/2 ?

My solution:

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15 votes
15 votes

Answer (D).

Given, propagation time =25 ms.

Time required to transmit 1 frame = 1000 bits (frame size)/ 10^6 bps(bandwidth) = 1 ms

To fully utilize the channel, we need to transmit frames for propagation time(25 ms), 25 frames.

therefore, minimum no. of bits required for numbering 25 frames is = log2 (25) = 5 bits.

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