Bit Stuffing Numerical Questions
- Given the data stream:
111110101110, what will be the transmitted data after bit stuffing (with the rule that after five consecutive 1’s, a 0 is inserted)?- A)
1111100101110 - B)
1111101011100 - C)
1111100111010 - D)
111110101110
1111100101110- Explanation: After five consecutive 1’s (
11111), a0is inserted. So, the transmitted data will be1111100101110.
- A)
- If the data stream is
111110000110, what will be the result of bit stuffing (with 5 consecutive 1’s rule)?- A)
1111100011010 - B)
1111100001100 - C)
111110001101 - D)
1111100001101
1111100011010- Explanation: After five consecutive 1’s, a 0 is inserted. So, the correct output is
1111100011010.
- A)
- For the input
101111110011, how many bits will be added using bit stuffing?- A) 1 bit
- B) 2 bits
- C) 3 bits
- D) 0 bits
- Explanation: There is only one occurrence of
11111, so one 0 bit will be added.
- What is the stuffed output for the data
1111001110111100(5 consecutive 1’s rule)?- A)
1111001011011011100 - B)
1111001111010111100 - C)
1111001110111011100 - D)
11110011101111000
1111001011011011100- Explanation: After five consecutive 1’s (
11111), a 0 bit is inserted.
- A)
- Given a data stream
111010101010111110, how will bit stuffing affect the transmission?- A) No bits will be added
- B) 1 bit will be added
- C) 2 bits will be added
- D) 3 bits will be added
- Explanation: The sequence
11111is found once, so one 0 bit is inserted after the fifth 1.
- What will be the result of bit stuffing for the sequence
111111010(5 consecutive 1’s rule)?- A)
1111110100 - B)
1111110010 - C)
1111111010 - D)
11111101010
1111110010- Explanation: After five consecutive 1’s, a
0is inserted, resulting in1111110010.
- A)
- For the bit stream
101011111111, how many extra bits will be added after performing bit stuffing?- A) 0 bits
- B) 1 bit
- C) 2 bits
- D) 3 bits
- Explanation: After five consecutive 1’s (
11111), one 0 bit will be added.
- For the data stream
111111011110, what will be the stuffed output?- A)
1111110101110 - B)
1111111011110 - C)
1111110111110 - D)
1111110111010
1111110101110- Explanation: A 0 is added after five consecutive 1’s.
- A)
- If the input data is
1110001111110, what is the stuffed output?- A)
11100011111101 - B)
11100011111010 - C)
11100011111011 - D)
11100011111100
11100011111101- Explanation: The bit pattern
11111is detected, and one 0 bit is added.
- A)
- What is the number of bits added to the sequence
1111101010111110after bit stuffing?- A) 1 bit
- B) 2 bits
- C) 3 bits
- D) 0 bits
- Explanation: The sequence
11111occurs once, so one 0 bit is inserted.
Byte Stuffing Numerical Questions
- If the data stream is
0x7E 0x45 0x7E, and the escape byte is0x7D, what will be the transmitted data after byte stuffing?- A)
0x7D 0x5E 0x45 0x7D 0x5E - B)
0x7E 0x45 0x7D 0x5E - C)
0x7D 0x7E 0x45 0x7D 0x7E - D)
0x7E 0x7E 0x45 0x7E
0x7D 0x5E 0x45 0x7D 0x5E- Explanation: The byte
0x7E(flag) is replaced by0x7D 0x5E.
- A)
- What will be the transmitted data if the input stream is
0x7D 0x7Eafter byte stuffing?- A)
0x7D 0x5E 0x7D 0x5E - B)
0x7D 0x7E 0x7D 0x5E - C)
0x7D 0x5E 0x7E - D)
0x7E 0x7D 0x7E
0x7D 0x5E 0x7D 0x5E- Explanation: Both occurrences of
0x7Dand0x7Eare escaped using the0x7Descape byte.
- A)
- If the data stream is
0x7E 0x40 0x7E, how will byte stuffing affect the transmission?- A)
0x7D 0x5E 0x40 0x7D 0x5E - B)
0x7E 0x40 0x7D 0x5E - C)
0x7D 0x5E 0x40 - D)
0x7D 0x5E 0x40 0x7D 0x5E
0x7D 0x5E 0x40 0x7D 0x5E- Explanation: Each occurrence of
0x7Eis replaced by0x7D 0x5E.
- A)
- What will the transmitted data be if the input stream is
0x7D 0x5E(no need to escape the control byte)?- A)
0x7D 0x5E - B)
0x7D 0x7E - C)
0x7E 0x5E - D)
0x7D 0x5E 0x7D 0x5E
0x7D 0x5E- Explanation: No byte stuffing is required as
0x5Eis not a control character.
- A)
- For a byte stream
0x7F 0x6F 0x7D, how will byte stuffing work?- A)
0x7D 0x5E 0x6F 0x7D 0x5E - B)
0x7D 0x5E 0x6F 0x7D 0x7F - C)
0x7F 0x6F 0x7D - D)
0x7D 0x5E 0x6F 0x7D
0x7D 0x5E 0x6F 0x7D- Explanation: The
0x7Dneeds to be escaped by inserting a0x7D 0x5E.
- A)
- If the input data is
0x7E 0x45 0x7D 0x6F 0x7E, what will be the transmitted data after byte stuffing?- A)
0x7D 0x5E 0x45 0x7D 0x5E 0x6F 0x7D 0x5E - B)
0x7D 0x5E 0x45 0x7D 0x5E 0x6F 0x7E - C)
0x7E 0x45 0x7D 0x5E 0x6F - D)
0x7D 0x5E 0x45 0x7D 0x7E
0x7D 0x5E 0x45 0x7D 0x5E 0x6F 0x7D 0x5E- Explanation: The bytes
0x7Eand0x7Dare escaped with0x7D 0x5E.
- A)
- For the input data
0x7A 0x7E 0x7D, how many escape bytes will be added?- A) 1 byte
- B) 2 bytes
- C) 3 bytes
- D) 4 bytes
- Explanation:
0x7Eand0x7Deach require an escape byte.
- What is the byte-stuffed version of
0x40 0x7D 0x45?- A)
0x40 0x7D 0x5E 0x45 - B)
0x40 0x7D 0x45 0x7D - C)
0x40 0x7D 0x5E 0x45 0x7D 0x5E - D)
0x40 0x7E 0x7D
0x40 0x7D 0x5E 0x45- Explanation: The
0x7Dis escaped as0x7D 0x5E.
- A)
- What will be the transmitted data if the stream is
0x7E 0x4F 0x4D 0x7D?- A)
0x7D 0x5E 0x4F 0x4D 0x7D 0x5E - B)
0x7E 0x4F 0x4D - C)
0x7D 0x5E 0x4F 0x4D - D)
0x7E 0x4F 0x4D 0x7D
0x7D 0x5E 0x4F 0x4D 0x7D 0x5E- Explanation: Both
0x7Eand0x7Dare escaped by0x7D 0x5E.
- A)
- If the byte stream is
0x7F 0x7D 0x5E, how will byte stuffing work?- A)
0x7D 0x5E 0x7D 0x5E - B)
0x7F 0x7D 0x5E - C)
0x7D 0x5E 0x7D 0x7E - D)
0x7D 0x5E 0x7F
0x7D 0x5E 0x7D 0x5E- Explanation: Both occurrences of
0x7Dand0x7Erequire byte stuffing.
- A)
Explanation of Each Numerical Question:
Bit Stuffing Questions
- Data:
111110101110- Rule: If there are five consecutive
1s, insert a0. - Solution:
11111is found in the beginning, so we insert a0after the fifth1. The stuffed output is1111100101110.
- Rule: If there are five consecutive
- Data:
111110000110- Solution: The sequence
11111is found, so we insert a0. The stuffed output is1111100011010.
- Solution: The sequence
- Data:
111110000110- Answer: One
0is inserted after the sequence11111→ Result:1111100011010.
- Answer: One
- Data:
1111001110111100- Solution: The first
11111pattern is found in the middle, so a0is inserted → Result:1111001011011011100.
- Solution: The first
- Data:
111010101010111110- Solution: Only one sequence of
11111occurs, so one0is inserted → Result:1111100101110.
- Solution: Only one sequence of
Byte Stuffing Questions
- Data:
0x7E 0x45 0x7E- Rule: If
0x7Eis found, escape it with0x7D 0x5E. - Solution: The byte
0x7Eappears twice and is replaced with0x7D 0x5E. The final output is0x7D 0x5E 0x45 0x7D 0x5E.
- Rule: If
- Data:
0x7D 0x7E- Solution: Both
0x7Dand0x7Eneed to be escaped. Thus,0x7Dis replaced by0x7D 0x5E, and0x7Eis replaced by0x7D 0x5E. The output is0x7D 0x5E 0x7D 0x5E.
- Solution: Both
- Data:
0x7E 0x40 0x7E- Solution:
0x7Eneeds to be escaped, but0x40is data, so no change. The result is0x7D 0x5E 0x40 0x7D 0x5E.
- Solution:
- Data:
0x7D 0x5E- Solution: No byte stuffing is needed because
0x7Dis a special escape character, but0x5Eis not a special control byte.
- Solution: No byte stuffing is needed because
- Data:
0x7F 0x6F 0x7D- Solution:
0x7Dis escaped using0x7D 0x5E. So the result is0x7D 0x5E 0x6F 0x7D 0x5E.
- Solution:
- Data:
0x7E 0x45 0x7D 0x6F 0x7E- Solution: Both
0x7Eand0x7Dare special bytes and need to be escaped. The final output is0x7D 0x5E 0x45 0x7D 0x5E 0x6F 0x7D 0x5E.
- Solution: Both
- Data:
0x7A 0x7E 0x7D- Solution: Both
0x7Eand0x7Dare control bytes, and thus each is replaced with0x7D 0x5Efor0x7Eand0x7D 0x5Dfor0x7D. The result is0x7D 0x5E 0x7D 0x5E.
- Solution: Both
- Data:
0x40 0x7D 0x45- Solution: The
0x7Dbyte needs to be escaped as0x7D 0x5E, but the other bytes are data. The output is0x40 0x7D 0x5E 0x45.
- Solution: The
- Data:
0x7E 0x4F 0x4D 0x7D- Solution:
0x7Eand0x7Dneed to be escaped. The result is0x7D 0x5E 0x4F 0x4D 0x7D 0x5E.
- Solution:
- Data:
0x7F 0x7D 0x5E- Solution: The
0x7Dis escaped as0x7D 0x5E. The final output is0x7D 0x5E 0x7D 0x5E.
- Solution: The
Conclusion
- Bit Stuffing ensures that reserved bit sequences (like flags) don’t accidentally appear in the data stream. It is applied by inserting a
0after five consecutive1s. - Byte Stuffing ensures that special control bytes, like
0x7E, do not appear in the data stream. When these special bytes are encountered, they are replaced by an escape sequence.
Both methods are essential for data communication to avoid misinterpretation of data, especially in protocols that use specific byte sequences for control purposes.
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