727 views

Assume that X and Y are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both X and Y attempt to transmit a frame, they wait to get the control of channel using binary exponential algorithm. The probability that both were successfully allowed to send the frame on fifth round of the algorithm (assuming every time both X and Y will collide in back-off race till 4th round) is ________.

I think it will be 15/16.

After four collisions (Since it is given to assume four collision have happened I think we need not consider that probability), each host will choose a number between (0,$2^{4-1}$).

Probability of collision is 1/16 (Both choosing the same number).

Hence probability of success=15/16.
i think 31/32 will be the ans
Thanks for noticing, Updated it.
All the ans provided below are wrong, correct ans would be (1/2)*(1/4)*(1/8)*(1-1/16)=0.0146.

Here they are asking about the fifth iteration. That means we have to find the probability of success in 5th iteration not in 4th iteration.

For K=5, i.e. (0 to 25-1) = (0,31)PT.

So the probability of collision is 1/32 and hence the probability of success will be (1-1/32) = 31/32.

Collision occurred 4 times but successful in only 5th time na
When first collision occurred k=1 is used for second round

When  second collision occurred k=2 used for resolving the third round

When third collision occurred k=3 is used for resolving the 4th round

When 4th collision occurred k=4 is used for resolving the 5th round.

I think so.

probability that both were successfully allowed to send the frame on fifth round of the algorithm

Isn't this means to find the success in 5th round. Please explain.

After ith collision,sender choose a number between(0 to 2^i-1).

Therefore,after 4th collision sender choose between(0 to 2^4-1) i.e (0 to 15).

Collision occurs when both x and y choose the same number.

probability of collision is 1/16.

Therefore probability of success=1-11/16=15/16.

1 vote