Bit Stuffing and Byte Stuffing Numerical Questions

Bit Stuffing Numerical Questions

  1. 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
    Answer: A) 1111100101110
    • Explanation: After five consecutive 1’s (11111), a 0 is inserted. So, the transmitted data will be 1111100101110.
  2. 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
    Answer: A) 1111100011010
    • Explanation: After five consecutive 1’s, a 0 is inserted. So, the correct output is 1111100011010.
  3. 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
    Answer: A) 1 bit
    • Explanation: There is only one occurrence of 11111, so one 0 bit will be added.
  4. What is the stuffed output for the data 1111001110111100 (5 consecutive 1’s rule)?
    • A) 1111001011011011100
    • B) 1111001111010111100
    • C) 1111001110111011100
    • D) 11110011101111000
    Answer: A) 1111001011011011100
    • Explanation: After five consecutive 1’s (11111), a 0 bit is inserted.
  5. 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
    Answer: B) 1 bit will be added
    • Explanation: The sequence 11111 is found once, so one 0 bit is inserted after the fifth 1.
  6. 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
    Answer: B) 1111110010
    • Explanation: After five consecutive 1’s, a 0 is inserted, resulting in 1111110010.
  7. 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
    Answer: B) 1 bit
    • Explanation: After five consecutive 1’s (11111), one 0 bit will be added.
  8. For the data stream 111111011110, what will be the stuffed output?
    • A) 1111110101110
    • B) 1111111011110
    • C) 1111110111110
    • D) 1111110111010
    Answer: A) 1111110101110
    • Explanation: A 0 is added after five consecutive 1’s.
  9. If the input data is 1110001111110, what is the stuffed output?
    • A) 11100011111101
    • B) 11100011111010
    • C) 11100011111011
    • D) 11100011111100
    Answer: A) 11100011111101
    • Explanation: The bit pattern 11111 is detected, and one 0 bit is added.
  10. What is the number of bits added to the sequence 1111101010111110 after bit stuffing?
    • A) 1 bit
    • B) 2 bits
    • C) 3 bits
    • D) 0 bits
    Answer: A) 1 bit
    • Explanation: The sequence 11111 occurs once, so one 0 bit is inserted.

Byte Stuffing Numerical Questions

  1. If the data stream is 0x7E 0x45 0x7E, and the escape byte is 0x7D, 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
    Answer: A) 0x7D 0x5E 0x45 0x7D 0x5E
    • Explanation: The byte 0x7E (flag) is replaced by 0x7D 0x5E.
  2. What will be the transmitted data if the input stream is 0x7D 0x7E after byte stuffing?
    • A) 0x7D 0x5E 0x7D 0x5E
    • B) 0x7D 0x7E 0x7D 0x5E
    • C) 0x7D 0x5E 0x7E
    • D) 0x7E 0x7D 0x7E
    Answer: A) 0x7D 0x5E 0x7D 0x5E
    • Explanation: Both occurrences of 0x7D and 0x7E are escaped using the 0x7D escape byte.
  3. 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
    Answer: A) 0x7D 0x5E 0x40 0x7D 0x5E
    • Explanation: Each occurrence of 0x7E is replaced by 0x7D 0x5E.
  4. 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
    Answer: A) 0x7D 0x5E
    • Explanation: No byte stuffing is required as 0x5E is not a control character.
  5. 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
    Answer: D) 0x7D 0x5E 0x6F 0x7D
    • Explanation: The 0x7D needs to be escaped by inserting a 0x7D 0x5E.
  6. 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
    Answer: A) 0x7D 0x5E 0x45 0x7D 0x5E 0x6F 0x7D 0x5E
    • Explanation: The bytes 0x7E and 0x7D are escaped with 0x7D 0x5E.
  7. 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
    Answer: B) 2 bytes
    • Explanation: 0x7E and 0x7D each require an escape byte.
  8. 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
    Answer: A) 0x40 0x7D 0x5E 0x45
    • Explanation: The 0x7D is escaped as 0x7D 0x5E.
  9. 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
    Answer: A) 0x7D 0x5E 0x4F 0x4D 0x7D 0x5E
    • Explanation: Both 0x7E and 0x7D are escaped by 0x7D 0x5E.
  10. 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
    Answer: A) 0x7D 0x5E 0x7D 0x5E
    • Explanation: Both occurrences of 0x7D and 0x7E require byte stuffing.

Explanation of Each Numerical Question:

Bit Stuffing Questions

  1. Data: 111110101110
    • Rule: If there are five consecutive 1s, insert a 0.
    • Solution: 11111 is found in the beginning, so we insert a 0 after the fifth 1. The stuffed output is 1111100101110.
  2. Data: 111110000110
    • Solution: The sequence 11111 is found, so we insert a 0. The stuffed output is 1111100011010.
  3. Data: 111110000110
    • Answer: One 0 is inserted after the sequence 11111 → Result: 1111100011010.
  4. Data: 1111001110111100
    • Solution: The first 11111 pattern is found in the middle, so a 0 is inserted → Result: 1111001011011011100.
  5. Data: 111010101010111110
    • Solution: Only one sequence of 11111 occurs, so one 0 is inserted → Result: 1111100101110.

Byte Stuffing Questions

  1. Data: 0x7E 0x45 0x7E
    • Rule: If 0x7E is found, escape it with 0x7D 0x5E.
    • Solution: The byte 0x7E appears twice and is replaced with 0x7D 0x5E. The final output is 0x7D 0x5E 0x45 0x7D 0x5E.
  2. Data: 0x7D 0x7E
    • Solution: Both 0x7D and 0x7E need to be escaped. Thus, 0x7D is replaced by 0x7D 0x5E, and 0x7E is replaced by 0x7D 0x5E. The output is 0x7D 0x5E 0x7D 0x5E.
  3. Data: 0x7E 0x40 0x7E
    • Solution: 0x7E needs to be escaped, but 0x40 is data, so no change. The result is 0x7D 0x5E 0x40 0x7D 0x5E.
  4. Data: 0x7D 0x5E
    • Solution: No byte stuffing is needed because 0x7D is a special escape character, but 0x5E is not a special control byte.
  5. Data: 0x7F 0x6F 0x7D
    • Solution: 0x7D is escaped using 0x7D 0x5E. So the result is 0x7D 0x5E 0x6F 0x7D 0x5E.
  6. Data: 0x7E 0x45 0x7D 0x6F 0x7E
    • Solution: Both 0x7E and 0x7D are special bytes and need to be escaped. The final output is 0x7D 0x5E 0x45 0x7D 0x5E 0x6F 0x7D 0x5E.
  7. Data: 0x7A 0x7E 0x7D
    • Solution: Both 0x7E and 0x7D are control bytes, and thus each is replaced with 0x7D 0x5E for 0x7E and 0x7D 0x5D for 0x7D. The result is 0x7D 0x5E 0x7D 0x5E.
  8. Data: 0x40 0x7D 0x45
    • Solution: The 0x7D byte needs to be escaped as 0x7D 0x5E, but the other bytes are data. The output is 0x40 0x7D 0x5E 0x45.
  9. Data: 0x7E 0x4F 0x4D 0x7D
    • Solution: 0x7E and 0x7D need to be escaped. The result is 0x7D 0x5E 0x4F 0x4D 0x7D 0x5E.
  10. Data: 0x7F 0x7D 0x5E
    • Solution: The 0x7D is escaped as 0x7D 0x5E. The final output is 0x7D 0x5E 0x7D 0x5E.

Conclusion

  • Bit Stuffing ensures that reserved bit sequences (like flags) don’t accidentally appear in the data stream. It is applied by inserting a 0 after five consecutive 1s.
  • 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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