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87 votes
87 votes

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

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

Suppose that the sliding window protocol is used with the sender window size of $2^{I}$, where $I$ is the numbers of bits as mentioned earlier and acknowledgements are always piggy backed. After sending $2^I$ frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time)

  1. $16ms$
  2. $18ms$
  3. $20ms$
  4. $22ms$
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4 Comments

Since propagation time is 25ms, but we are still sending 32 packets, assuming half duplex connection, clearly the receiver can start transmitting piggybacked acknowledgement, only after 32ms. The receiver has to acknowledge all the 32 packets it received, and thus it will need 32transmissiins and one propagation.

Therefore wont the wait time be imcreased to 32+25=57ms?
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RTT value includes transmission time of acknowledgement so to get waiting time exclude transmission time from it
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edited by
efficiency – 32/52 = 61 %
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7 Answers

122 votes
122 votes
Best answer

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 $25$ ms.
Bandwidth $= 10^6$ 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 $25$ ms, to fully pack the link we need to send at least $1000 \times 25  \times 10^{-3}= 25$ frames. So, we need $⌈\log_2  25⌉ = 5$ bits.

$I = 5$, so $2^I = 32$ frames are sent.

Now, we need to get RTT (which is the time between which a frame is sent and its ACK is received), to determine the waiting time.
Transmission time (for a frame of size $1000$ bits) $= 1000/10^6 = 1$ ms.
So, transmission time for $32$ frames $= 32$ ms.
RTT $=$ Propagation time for frame $+$ Transmission time for frame $+$ Propagation time for ACK $+$ Transmission time for ACK

$ = 25$ ms $+ 1$ ms $+ 25$ ms $+ 1$ ms (ACK is piggy backed and assuming frame size for piggy backing is also $1000$ bits)
$= 52$ ms
So, waiting time $= 52 - 32 = 20$ ms. (For the $32$ ms, the sender was transmitting and not waiting)
Correct Answer: $C$

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4 Comments

@ but the sender is sending till 32nd second.

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I am not going by any definition. Just proceeding in a logical manner.

At time $t=0$ the sender starts sending the first packet. Now since the sender window size is 32. At time $t=32$ the sender shall finish transmitting the $32nd$ packet.

The first packet which was transmitted, its last bit at the beginning of the sender at time $t=1$. It’s last bit shall reach the receiver at $t=1+25\text{ (propagation delay)}=26$. Now at $t=26$ the receiver starts transmitting the piggybacked acknowledgement. So the last bit of this acknowledgement shall be at the beginning the receiver at $t=27$. This last bit shall reach the sender at $t=27+25\text{ (propagation delay)} = 52$

So at time $t=52$ the sender can start sending a new packet again.

So waiting time = $52-32=20$
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@Arjun Sir, @Ayush Upadhyaya Sir, Beautiful explanation! & Thanks a ton to the professor who set the such concept clearing question

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18 votes
18 votes

Given, propagation time =25 ms.

Time required to transmit 1 frame = 1000 bits (frame size)/ 106 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 = log 2(25) = 5 bits.

We need to calculate minimum waiting time after sending 2 frames (32 frames) to initiate transmission of next frame, for that we need to understand for what time sender is in busy mode and for what time sender is in waiting mode.

Sender is in busy mode during transmission of all the 32 frames, that means, for 32 ms.

after transmission sender is in waiting mode which is

= ( time required to get the ack for 1st frame ) - ( time for which sender is busy)

= (1+25+1+25) - (32) = 20 ms.

1+25+1+25 = transmission time for 1st frame + propagation time for frame + transmission time for ack as it is piggy backed + propagation time for ack

 

4 Comments

In question it is mentioned that the ACKs are piggybacked, so the RTT will be 50 instead of 52.

Hence the answer will be 50-32=18.
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what about frame transmission time...??
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So why sender will wait after 32 ms? He can go on sending more data without waiting any further?

Please clarify
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In sliding window protocol sender sends the number of frames== window size before ACK. So once 32 frames are sent. Sender will wait for the acknowledgment. After it receives ACK then only it will start sending the next frames.
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14 votes
14 votes

This might help.

3 Comments

But the question says that acknowledgement is piggybacked.
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if ack is piggy backed so why everyone is not considering this fact?
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We are actually using that fact. Piggybacking implies that Transmission delay of ACK is not to be ignored.
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3 votes
3 votes
Size of sliding window = 2^5 = 32
Transmission time for a frame = 1ms
Total time taken for 32 frames = 32ms
The sender cannot receive acknoledgement before round trip time which is 50ms
After sending 32 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 50 – 32 = 18
1------------25------32(last frame)----------50(first ack receive) so waiting time is (50-32) = 18ms(using pipelining)
Please any one correct me if I am wrong..
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1 comment

Refer first answer .

Hope it will make you clear.
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Answer:

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