Compare and contrast questions require you to explain the similarities and diffe

Question

Compare and contrast questions require you to explain the similarities and differences to get credit for your answer): 1.What is the definition of a linear block code? 2. (2 parts )In a codeword, we add two redundant bits to each 8-bit word. Find the number of a. valid codewords & b. invalid codewords 3. If we want to be able to detect 2-bit errors, what should be the minimum Hamming distance? 4. In CRC, if the dataword is 5 bits and the codeword is 8 bits, how many 0s need to be added to the dataword to make the dividend? What is the size of the remainder? What is the size of the divisor? 5. In CRC, which of the following generators (divisors) guarantees the detection of an odd number of errors? a. 10111 b. 101101 c. 111 6. Assume we are sending data items of 16-bit length. If two data items are swapped during transmission, can the traditional checksum detect this error? Explain your answer. Chapter 11 7. Explain why flags are needed when we use variable-size frames. 8. Compare and contrast byte-oriented and bit-oriented protocols. 9. In a byte-oriented protocol, should we first unstuff the extra bytes and then remove the flags or reverse the process? 10. Compare and contrast flow control and error control. 11. Compare and contrast HDLC with PPP.

Explanation / Answer

Solution:

1)

A linear block code is error correction code for which the linear combination of code words is also a code word. So if from a k-dimensional subspace of vector space is forms 2k code words of all n-tupples over the field GF(2), then that block code of length l with 2k code words is called a linear code.

2)

So we have 8-bit word which means that number of valid code word= 2^8

and after 2 redundant bits added, the total number of code word= 2^10.

Number of invalid code word= 2^10-2^8= 1024-256= 768 invalid codes.

3)

hamming distance= s+1, and here s= 2

minimum hamming distance would be= 2+1= 3

4)

here

k=5, n=8

code word size and dividend size both are the same which is 8 bits

now we need 3 bits (0's) in data word

Reminder size r= n-k= 3 bits

Divisor size= r+1= 4 bits.

11)

  HDLC generally operates at data link layer and also used for synchronous PPP connections.

HDLC functionality depends on the following:
• HDLC is a bit-oriented protocol.
• On transmitting end, it receives data from the application layer and delivers it to the receiving end on the other side of the application.
• On receiving end, it accepts data and transmits it to the high-level application layer.
• Both sides of the modules exchange control information, encoded into a frame.

PPP Protocol:

Point to point protocol is a byte-oriented protocol.
• It is operative at layer-2 and layer-3. PPP generally uses HDLC frame format at layer-2 and uses IPv4 and IPv6 formats at layer-3.

Other than IP, PPP supports other network layer protocols too. This is a main advantage of PPP.
• PPP also have a built-in security mechanism such as PAP(Password Authentication Protocol) and CHAP(Challange handshake authentication protocol).
• The crc-16 field is included in the frame which helps in error reduction.
• PPP uses same start and end flags used in HDLC frame which raises transparency problem.

I hope this helps. :)

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