GATE CSE 2014 Set 3 | Question: 24

Question

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:

• A. $0111110100$
• B. $0111110101$
• C. $0111111101$
• D. $0111111111$

Comments

Delimiter pattern=$01111110$.From this pattern it is clear that we are going to insert $0$ in input string.It means no. of $1$ in input string and output string(after stuffing 0) should be same.So we can directly say ans $b$

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Similar question: https://gateoverflow.in/3724/gate2004-it-80

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Why we are not stuffing a ‘1’ after 6 consecutive 1?

Answer 1

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. 

Comments

you've highlighted the wrong zero

Answer 2

$\text{Bit-Stuffing}:$ 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 $B$

Answer 3

I think end delimiter pattern should be 0111110 (five 1's).