exponential backoff

Question

Nodes A and B are connected with 100 Mbps ethernet segment with 6 microsec pop.delay between them.Suppose A,B send frames at t=0 and frames get collided.after first collision A draws k=0 and bdraws k=1.if jam signal is ignored and timeout is 1 RTT then at what time A's packet gets completely delivered to B...assume packet size 1000 bits.

• A. 28 microsec
• B. 16 microsec
• C. 22 microsec
• D. 38 microsec

Comments

What changes can we expect in the result if the jamming signal has not been ignored?

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@  dan31 

What changes can we expect in the result if the jamming signal has not been ignored?

then jamming signal transmission time also be added before sending packet again . 

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ok..Can you please confirm if we need to add jamming signal for min frame size calculation?

As there are some previous gate question, where it is included. But I don't think we need to add it.

ref- https://gateoverflow.in/1397/gate2005-74

Answer 1

at t=0 both A and B transmit and after 1RTT i.e  (TT+2PT) =10+12 =22 microsecond collision detected

A will again transmit while B wait for P.T i.e(6 microsecond) and A segment reach B after TT+PT=10+6=16

therefore after 22+16 =38 microsec A segment delivered to B

Comments

@kapil,sekhar,akhil

is the solution correct: as  thecollision detection time is 2*prop delay..why considering the transmisson delay

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@saurav04

Could you please let me know, when A will transmit while B wait for PT and A segment reach B after TT + PT, why you've added TT with PT here ?

Thank You!

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We don't need to calculate the TT for the 1st time. When 1st bit of both stations meet at the middle, collision will occur, so it is simply TP/2 when collision occurs and at TP both A & B aborts transmission.

Subtracting total 10 from ur result will be 28 microsec.

Answer 2

at t=0 both A and B transmit.

at t=3 collision occurs

at t=6 both station can detect collision hence abort transmission

at t=12 the last bit of B's transmission reaches A

at t=18 ist bit of A 's transmission reaches B

at t=28 last bit of A's transmission reaches A , hence at t=28 the frames is successfully sent to B

referring: https://gateoverflow.in/27094/cn-ques1-ethernet-and-csma-cd

Comments

I agree with this answer only!!

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This explanation holds greater sense. The point(reasoning at t = 12us) which includes the fact that the stations wait for the last garbled signal bit to arrive  is an important and essential  aspect to deliver error free frames!

Answer 3

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 

Answer 4

T = 0 : Start Transmission

T = TT + TP/2 : Collision

T = TT + TP : Detect Collision

T = TT + TP + TP : Clear Channel

Now, A draws K = 0, B draws K = 1

T = TT + TP + TP + TT + TP : A's Packet gets completely delivered to B

 

T = 2TT + 3TP

T = 2 * 10 + 3 * 6 = 38 Units