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+11 votes

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 * 10^{8}m/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.

0

@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 ..

+9 votes

Best answer

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

0

i am not getting why we solve like this in ques 2

Somewhere i saw

Time of collision =512/10^{7}= 51.2 u sec added in ques 2. Dont know why

I already hv this solutions.. But i dont know why to do tht??

Reasons and logic behind it i want to know.

+2

In ques 2] we are asked when the packet will be completely transmitted to B.So here in question they told that 1 time collision occurs, so we consider the case 1 when both transmit frame at time t=0, so at 12.5 usec both will know about collisiion, so the bits will come back to A and B ,so A must wait till B's last bit reaches to A and then A will transmit its packet, so now waiting time of A is 12.5+12.5=25usec and propagation time of A=12.5usec and transmission time of A =100usec

so the total time to get delivered a packet from A to B will be=25+12.5+100=137.5 usec

In ques 3] the repeaters are replaced with bridges, so bridges have their store and forward delay of 1000 bit and 20 bit delay is their own as mentioned in ques 3] so total bit delay=4*(1000+20)=4080 and that is why the bit time will be (4080)/10000000=408 usec and the propagation delay of A will be(Not included bridges as they are already included)=4.5 usec and transmission time =100 usec so total time required to reach to B =408+100+4.5=512.5 usec

so the total time to get delivered a packet from A to B will be=25+12.5+100=137.5 usec

In ques 3] the repeaters are replaced with bridges, so bridges have their store and forward delay of 1000 bit and 20 bit delay is their own as mentioned in ques 3] so total bit delay=4*(1000+20)=4080 and that is why the bit time will be (4080)/10000000=408 usec and the propagation delay of A will be(Not included bridges as they are already included)=4.5 usec and transmission time =100 usec so total time required to reach to B =408+100+4.5=512.5 usec

0

thank you ... i ws thinking about B 's role ..which is clear now. B will send its packet too. meanwhile A have to wait... and Is there no use of k=0 and k=1 exponential back-off intervals ??

0

I want to clear your concept abut collision. Here A chooses k=0 slot and B chooses slot k=1, it means that after collision A can immediately send packet, but actually it is not correct coz after detecting collision, A must wait till the last aborted bit of B arrives, means that at the time of collision , the packets of A and B will colllide so those packets will come back to A and B, so that time when they come completely, A must wait till the last aborted bit of them come at A and B, nd the time taken to come those bits will be the 1 propagation delay.so A must wait 12.5+12.5 usec=25 usec and then it can transmit packets..

I hope now you are clear with your ans 2 douubt.

I hope now you are clear with your ans 2 douubt.

0

So from above comment, you just remember the concept of collision where always after collision, Station must wait 1 propagation delay time to get back the aborted bits.(The bits in the packet which were collided)

0

So the aborted bits of packets go back to their respective sender?

I didn't know that aborted packets went back to either end..I think they would be just dropped bcz of collision.

I think ,in addition, for ques 2)

at time t=0 both will send packet and collision occurs (**this will take 12.5 one propagation delay)**

{time for collision or slot time = 512/1o^{7} =51.2 u sec }

and both will back off with (**2 ^{k} *slot time)**

A use k=0 so it will back off for 2^{0} *51.2 u sec= **51.2 u sec **(so A will send packet after this time).

A will send packet in

= 100+12.5+51.2+ 12.5

=176.2 u sec

that's wht I think... correct me where I am wrong...

i used the link for above calculation of collision

http://www.erg.abdn.ac.uk/users/gorry/course/lan-pages/csma-cd.html

and if this is the right case..

what would happen in case of bridges ? how would the back off time be used with store and forward policy?

0

Here you are adding up the slot time or collision time but here the propagation delay we know so consider the case when both transmits frame at time t=0, so both station's packets will collide at half of the propagation delay time. so the stations will come to know about collision at the time of 12.5 usec which propagation delay. so A will then wait for the last aborted bit to arrive and that is why 12.5+12.5=25 usec will the waiting time of A and then A would send packet so propagation delay it will take so 25+12.5+100=137.5 will be total time taken by A to deliver the packet.. You are not wrong but here you are taking slot time 51.2 usec in consideration but here propagation delay is known so no need to take that slot time..When there is nothing given in the ques at that time we have to assume the slot time 51.2 usec .

+1

Just to simplify

A's propagation time = 12.5microsec

tx = 100microsec

In worst-case A's first bit about to reach B time lapsed till then is 12.5microsec .And at that point B transmits then and collision occurs then collision signal to propagated back to A it would require 12.5microsec .Then we apply collision algorithm exponential backoff protocol in which A wins as given K=0(for A) and k=1(for B).so then again A transmits the signal and further calculations =25 +12.5+100

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