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exponential backoff

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4 4 votes

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.

  1. 28 microsec
  2. 16 microsec
  3. 22 microsec
  4. 38 microsec

9 Answers

Best answer
4 4 votes
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
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7 7 votes

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

1 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 

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1 1 vote

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

1 1 vote
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
1 1 vote

answer is 34 micro sec I added it wrong 

since  it took 6 micro sec for collision detection and then A will send the data 

now before sending it will wait 12 micro sec to avoid the collision 

then A takes 10 micro sec for transmission delay 

and 6 micro sec for propagation delay 

all add up and give 34 micro sec 

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Two nodes (A and B) are on the same network, and distanced by 500-meter. The two nodes are connected via 3 switches, each with a delay of 5 µs. Suppose that both of A and B transmit 1500 bytes at the same exact time and their frames collide. A chooses to retransmit immediately after jam signal is received, and B chooses to retransmit after 50 µs of receiving the jam signal. Assume the transmission speed is 10 Mbps, the channel is CSMA/CD with backoff intervals that are multiples of, the jam signal takes 10 µs to transmit, and the inter-frame gap is 9.6 µs. Find the time required to deliver A’s packet to B