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An upper-layer packet is split into 5 frames, each of which has an 90% chance 
of arriving undamaged. If no error control is done by the data link protocol, 
how many times must the message be sent on average to get the entire thing through?
A. 2.1
B. 1.9
C. 1.7
D. 2.4
asked in Computer Networks by Junior (569 points) | 74 views
C) 1.7 times
could you pls explain more yes answer is right

1 Answer

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What would be the probability that all reach safe?
$(\frac{9}{10})^{5}$ = 0.59049

Intuitively speaking, the probability of getting a head is $\frac{1}{2}$ then how many attempts would get you a head (provided once tail has come it wont appear again)?
$(\frac{1}{2})^{-1}$, right?

Here also, it would be $\frac{1}{0.59049}$ = 1.69350... ie roughly 1.7
answered by Active (3.3k points)
Also, hard and fast formula,
$\frac{1}{P}$ where P is the probablity of getting all the packets thru.

Vikas Verma

i am not getting your point 

"Intuitively speaking, the probability of getting a head is 1212 then how many attempts would get you a head (provided once tail has come it wont appear again)?"

what does is mean that no error handling is done by DLL aand what about remaining 10%


You can ignore the intuitive idea.

DLL is responsible for error detection and correction. If is does no error handling then compulsorily packets has to be resent.

and 10% of the time packets do no reach their destination because of many reasons involving, routing issues, congestion, noise in the channel, etc.
Don't take it personally but I would advise you to first clear your concepts before attempting questions.
yes I agree @vikas

but i was unable to understand last line of it seems difficult for me to solve this

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