edited by
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48 votes
48 votes

A bit-stuffing based framing protocol uses an $\text{8-bit}$ delimiter pattern of $01111110.$ If the output bit-string after stuffing is $01111100101,$ then the input bit-string is:

  1. $0111110100$
  2. $0111110101$
  3. $0111111101$
  4. $0111111111$
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7 Answers

Best answer
41 votes
41 votes
$011111$ *one zero emitted here* $0101$

Correct Answer: $B$
edited by
32 votes
32 votes

here delimeter is "01111110" .....

so rule will be like this ...

at sender :- add 0 after each occurance of "011111" in input data..

at Reciever :- remove 0 for each occurance of "011111" in output.....

According to que Output string is "01111100101

so input will be "01111100101".. (i.e - 0111110101)

15 votes
15 votes

When delimiter pattern appears in data, we do stuffing so that data is not interpreted as a delimiter.

We usually stuff 0 and break the delimiter pattern present in data.

Now we have some data which has a delimiter pattern. (there could be any sequence of 0's and 1's before and after this delimiter pattern)

 ... 0 1 1 1 1 1 1

0

 ...

We can break the pattern by stuffing 0 after

  • 2 consecutive 1's
  • 3 consecutive 1's
  • 4 consecutive 1's
  • 5 consecutive 1's

We cleverly choose to stuff a 0  only after 5 consecutive 1's because by doing so we reduce the number of 0's to be stuffed and thereby also number of 0's un-stuffed later while reading this frame.

(Ex: If we stuff a 0 after every 2 consecutive 1's, we will have to stuff every time 2 consecutive 1's appear in the data. This would be too many times. And also while reading the frame, every 0 that occurs after 2 consecutive 1's has to be un-stuffed. This is not so clever)

4 votes
4 votes
B option
Answer:

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