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Nodes A and B are connected by a 100MbPs ethernet segment with 6μ μ s propagation delay between them.Suppose A and B send frames at t=0 and frames get collided.After first collision A draws k=0 and B draws k=1(exponential backoff algorithm ) jam signal is ignored and time out time is one RTT then at what time A's packet gets completely delivered to B?

Assume packet size is of 1000bits

asked in Computer Networks by (301 points) | 125 views
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22 microseconds
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32 micro second.
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How 32
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Actually as per given info A can't transmit the frame to B at k=0 because as soon as A's frame reaches B, B again start transmitting frame because it chose k = 1*PT. AND THUS transmission is not possible in the second round also. They have to go for 3rd round.
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can anyone verify what is answer for this question??? i am getting 22 .
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@Shubhanshu

Actually as per given info A can't transmit the frame to B at k=0 because as soon as A's frame reaches B, B again start transmitting frame because it chose k = 1*PT

I couldn't understand this point..Can you please explain a bit more?

The question says that the time out period is 1 RTT..here time out means what? Is it the wait backoff time? If yes then why are you taking k=1*PT? Shouldn't it be 1*RTT?

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Shivangi Parashar 2   can you provide the solution that's is given ???

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i think 32 is the right answer

since A and B are transmitting the data at same time and collision will occur in mid .when A will get first byte of B then A will assume that collision is occurred and stop transmit the data till total time is data transmission time + propagation time =10+6=16 micro seconds.now Backoff algo is applied since for A ,K=0 means waiting time for A is zero means A will send the data immediately and B will wait for 1 Tp ,so again 16 micro seconds required to retransmit the data .total time will be 32 micro seconds.

4 Answers

+1 vote

38-microsecond total time 

First If A is starting doing the transmission of data that is taking 10 microsecond and after that collision take place so that to detect collision its take 2*pt time it means that 12 microsecond and after using the back off algorithm A is win than A will transmits the data within taking 16 microsecond (10 +6) and total time is 22(10 +12)+16= 38 microseconds 

answered by Active (1.2k points)
edited by
0 votes
Both the station start at the time t=0

At t=3us collision occurs than 3us again will required to detect the collision so at t=6us collision will be detected

Now 1tp that is 6us will require to clear the link as collision has been occurred so we must clear the link before starting so now at t=12us link will be ready to again transfer.

Now 10us is time to transmit the frame so at t=22us at will finish transmitting

Further tp=6us will require it to propagate so toata time taken to deliver the packet is 22+6us=28us.
answered by (155 points)
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Now 1tp that is 6us will require to clear the link as collision has been occurred so we must clear the link before starting so now at t=12us link will be ready to again transfer.

Clearing the link means?

0 votes

 

ans ) 38 microsec

answered by Loyal (9.9k points)
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in last line it should be 22+16=38.

what will be changes if there is modification that K=1 and B choose k=0.

is it 22+12+16???
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@magma why would both retransmit the frames after 1RTT? A should transmit the frame immediately after detecting collision(since jam signal is ignored here) and B should wait for 1RTT before transmitting..isnt it so?
0 votes
Ans should be 32 micro sec

Both the station start transmitting at same time and they will collide at the middle.

so time after which collision will be detected=10+(2*3)=16μsec

time taken By A's packet to reach B=10+6=16μsec

Total time=32 μsec
answered by (149 points)

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