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

@ 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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@ 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.