Cn-ques1 ethernet and csma-cd

Question

Suppose two nodes, A and B, are attached to opposite ends of a 900 m cable, and that they each have one frame of 1000 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t=0. Suppose there are four repeaters between A and B, each inserting a 20 bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K=0 and B draws K=1 in the exponential backoff protocol. Ignore the jam signal.

Ques-1. What is the one-way propagation delay (including repeater delays) between A and B in seconds. Assume that the signal propagation speed is 2 * 108m/sec.

Ques-2. At what time (in seconds) is A's packet completely delivered at B.

Ques-3. Now suppose that only A has a packet to send and that the repeaters are replaced with bridges. Suppose that each bridge has a 20 bit processing delay in addition to a store-and-forward delay. At what time in seconds is A's packet delivered at B?

My work:- propagation tim = 4.5 *10-6 sec

Transmission time by A = 100  *10-6 sec.

For 2nd ques.I am not getting how to take collision and back off delay...

then how to solve for bridges.

Comments

But here You haven't considered repeater delays..So your propagation delay is wrong

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would you plz give me proper solution?
you may be right...

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900/(2 * 10^8)m/s + (4*20)/10000000) = 12.5 micro seconds

will be the answer of A]

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@khushtak see here same type of problem is solved

https://wetalldid.files.wordpress.com/2013/11/ece-374-computer-networks-and-internet-homework-4-solutions-spring-2013.pdf

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@Bikram sir, I think that the accepted answer is not entirely correct. You can see this answer https://gateoverflow.in/90981/csma-cd

After the collision is detected at 25us, A waits for k*51.2us according to back off algorithm (For A, k=0) and then starts sending. Now B will wait till 25+51.2us = 76.2us as k=1. So A cannot transmit the frame fully after 100us as it will take 100+25 = 125us and B will start transmitting at 76.2us. So, collision will occur before A completely transmits the frame.

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@ Xylene

Go to page number 4, problem 4 https://wetalldid.files.wordpress.com/2013/11/ece-374-computer-networks-and-internet-homework-4-solutions-spring-2013.pdf

and solve that way ..

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.......

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@ Deepakk Poonia (Dee)

can you brief more about the procedure

Answer 1

A]900/(2 * 10^8)m/s + (4*20)/10000000) = 12.5 micro seconds

B]frame transmission time is (1000bits/10*106bps)=100 μsec
At time t = 0 , both A and B transmit.
At time t = 12.5μ sec , A detects a collision.
At time t = 25μ sec last bit of B 's aborted transmission arrives at A .
At time t = 37.5μ sec first bit of A 's retransmission arrives at B .
At time t= 37.5 μsec + 100 μsec =137.5 μsec. A 's packet is completely delivered at B .

C]Each Bridge introduces additional 1000-bit store-and-forward delay and 20-bit processing delay. Total delay introduced is 4080-bit time or 408 µs. Transmission delay is 1000-bit time or 100 µs. Propagation delay is 4.5 µs. A’s packet reaches B at time 408 + 100 + 4.5 = 512.5µs 

Comments

why we are not adding interframe gap which is 96 bit times?

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at t=63.7 microseconds B has scheduled it retransmission because at t=12.5 microseconds B detects collision and it waits for 51.2 microseconds as k=1 for B, therefore collision will occur again. so how A's packet will be completely delivered to B?

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Recall, In CSMA, trasmitter senses the channel before initiating its transmission and refrains from sending its frames when channel is busy.