edited by
22,181 views
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. $0111111111$
edited by

7 Answers

3 votes
3 votes

Answer B.

8-bit delimiter pattern is 01111110. 

The output bit-string after stuffing is 01111100101.

The above highlighted bit is stuffed bit.  
So input bit-string must be 0111110101. 
1 votes
1 votes

$\text{Bit-Stuffing}:$ Bits stuffing is the insertion of noninformation bits into data and it is used for synchronization purpose example

 

$A. 0111110100 \rightarrow 011111\color{red}{0}0100$ 

$B. 0111110101 \rightarrow 011111\color{red}{0}0101$ 

$C. 0111111101 \rightarrow 011111\color{red}{0}1101$

$C. 0111111111 \rightarrow 011111\color{red}{0}1111$

So Option $B$

–1 votes
–1 votes
I think end delimiter pattern should be 0111110 (five 1's).
Answer:

Related questions

50 votes
50 votes
6 answers
3
go_editor asked Sep 28, 2014
17,088 views
An IP router implementing Classless Inter-domain Routing (CIDR) receives a packet with address $131.23.151.76$. The router's routing table has the following entries:$$\be...