Question

a. Suppose the information content of a packet is the bit pattern 1110101010101111 and an even parity scheme is being used. What would the value of the checksum field (sequence of parity bits) be for the case of a two-dimensional parity scheme? Your answer should be such that a minimum-length checksum field is used.

b. With the same bit-pattern and your derived checksum, prove that the 2D scheme will not be able to detect all burst errors?

Answer 1

The packets length is 16 bit. so in 2D arrangement it can be written as:

16x1 matrix :total of 17 parity bits used, 16 for rows and 1 for column

8x2 matrix : total of 10 parity bits used, 8 for rows and 2 for column

4x4 matrix :total of 8 parity bits used, 4 for rows and 4 for column

other cases 2x8,1x16 are symmetric to above.

So in 4x4 arrangement we would need minimum number of parity bits.

In even parity scheme, bit 1 is inserted if column (or row) sum is odd (mod 2 arithmetic). and 0 is inserted if even.

so the arrangement is :

1 1 1 0 | 1

1 0 1 0 | 0

1 0 1 0 | 0

1 1 1 1 | 0

_ _ _ _

0 0 0 1

b. to prove that it can not detect all burst error, we will find such a burst error, consider the entries in location (3,2) (3,4) (4,2)(4,4).

Now suppose the burst error flips these bits. Nottice that the checksum still remains same and valid.

1 1 1 0 | 1

1 0 1 0 | 0

1 1 1 1 | 0

1 0 1 0 | 0

_ _ _ _

0 0 0 1

when sender recieves the packet and calculates the checksum it does not finds any error and accepts but infact it is errorred packet.