Question

1. (10 points) Suppose the information content of a packet has the pattern as follows. An even parity is used. What are the values of the parity bits? Fill in the blanks. 1 1 1 0 0 1 1 0 1 0 0 1 1 1 0...

1. (10 points) Suppose the information content of a packet has the pattern as follows. An even parity is used. What are the values of the parity bits? Fill in the blanks.

1 1 1 0

0 1 1 0

1 0 0 1

1 1 0 1

2. (10 Points) Suppose there is a single bit error in the packet in problem 1 (see below). Show that two-dimensional parity checks can detect and correct it (by finding the row and column of the bit error). How about a double bit error?

1 1 1 0

0 1 1 0

1 0 1 1

1 1 0 1

3. (10 Points) Consider the 5-bit generator, G=10011, and suppose that D has the value 1010101010. What is the value of R? Show your steps.

4. (10 Points) Bonus: How many Multiple Access Protocols are we discussed in chapter 5? What are their pros and cons?

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Answer #1

1)

The parity bits are in bold and have been computed using even parity.

In even parity we add a 1 if the numbers of 1's is odd and we add a zero if the number of 1's is even

1 1 1 0 1
0 1 1 0 0
1 0 0 1 0
1 1 0 1 1
1 1 0 0 0

2)

The parity bits in the row and the column of the bit which is in error are changed. I have underlined the changed bits. So a single error can be detected and it can be corrected by just inverting the bit,

1 1 1 0 1
0 1 1 0 0
1 0 1 1 1
1 1 0 1 1
1 1 1 0 1

For the case of two errors, two errors are in italic and bold

1 1 1 0 1
0 1 1 0 0
1 0 1 0 0
1 1 0 1 1
1 1 1 1 0

The two errors are in row 3, column 3 and column 4.

The row parity is not changed but the column parity is changed.

So double errors can be detected but cannot be corrected.

3)

We append (n - k - 1) 0's to D and then divide D by G to get R, where

n is number of bits in D

k is number of bits in G

1011011100 10011 10101010100000 10011 0110010100000 10011 10100100000 10011 0111100000 10011 11010000 10011 1001000 10011 000

So R = 100

4)

Following Multiple Access Protocols are there :

  1. Aloha
  2. Slotted Aloha
  3. CSMA
  4. CSMA / CD
  5. CSMA / CA

Aloha pros:

It achieves high link utilization.

Aloha cons :

Collisions are also high as any station can send whenever it has a frame to send.

Slotted Aloha pros:

It achieves high link utilization and collisions are less than Aloha but a frame can be sent only at the start of a slot

Slotted Aloha cons :

. Link utilization decreases as stations have to wait for the next slot.

CSMA pros:

Channel is sensed before data is sent to avoid collisions

CSMA cons :

Distant stations can still collide they can't sense each others frame at the start of their respective transmissions.

CSMA / CD pros:

The process of sending and detecting collision is a continous. Distant collisions are also detected.

CSMA / CD cons :

It cannot be used in wireless networks. Works for wired networks only.

CSMA / CA pros:

As collisions cannot be detected in wireless networks, this protocol tries to avoid collisions

CSMA / CA cons :

Latency is high.

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