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A simple network consist of two nodes X and Y. Each node has a single frame of F = 1500 bits (including all header and preambles) to send to each other. Both nodes attempt to transmit at ‘t = 0’. In case that a node executes the exponential Back-off algorithm. In the algorithm, a node experiencing the nth collision in a row for given frame, randomly chooses a value of k from {1, . . . 2m} where m is min (n, 10). The node then wait 8k μs before attempting to retransmit the frame. It is also given that the propagation speed of the signal over the network is 2 × 108 m/s and a data rate of 150 Mbps.

Assume that collision occur at time t = 0, and that the node X draws k = 1 and Y draws k = 2. The time taken by X’s packet (from t = 0) to be completely delivered toY is ________ μs.

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at t=0 collision occur . and X draws k=1 and Y draws
k=2 . (but question doesn't tell what node X and Y draw
K value in sub-sequent  collisions , it say only k value can be random)

best case :  
  1st collsion : X draws k=1 and Y draws k=2 (given)
  after 8 microsecond , node x can transmit the frame.
  but it can transmit and send last bit of frame upto
  16 microsecond. becuse after that , node y again try to
  transmit the frame and collision happen . so in
  16-8-4(propagation time )=4 microsec, node x can transmit
  only 4*150 mbps=600 bit .
  
  2nd collision: now let x draw =1 and y draw 3 or 4 becuse at
2nd tym collision , for node Y , n=2  so k ranges {1,2,3,4}
 . let assume 3 . so after another 8 microsec i.e after t=24 microsec
  ,node x again start transmission . and node y start transmission again
  after 24 microsec . so in 24-8-4=12 microsec , node x can transmit
  12*150=1800 bit . but only 900 bits frame is left . so transmission time
  will be= 6 microsec

  so total time taken by node X to completely deliver the frame to node Y
  is= 8+4+8+6+2*4=34 microsec

worst case :
   in the worst  case , we can assume everytime when collision happen after
t=0 , node x draws k value 1 and node y draws k value 2 . then total time taken by
node x to complete deliver the frame to node y =8+4+8+4+8+2+3*4=46 microsecond
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according to data given in the question

Tt = size/B.W

=150/150x10^6 = 10 micro sec

Tp= distance/speed

=800/2x10^6= 4 micro sec

now time taken by X to transfer data from X to Y will be

1. Tt and Tp to load and send data (loading time Tt and sending time Tp) at time t=0 collision  also got detected

2. It will wait 1 round trip time i.e 2Tp for retransmission of frame.

3. Now X=1 and X is ready to send the data it will only require Tp time to do so because data has already been loaded i.e time Tt is not required

now adding all we get

Tt + Tp + 2Tp +Tp = 10+4+8+4

Ans is 26 micro sec

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