VKSF341
Exercise 3
Student Name
1. Compute the VRC for the following binary streams – using even parity
a. 10011101
X 1
b. 11110110
X 0
c. 00001111
X 0
d. 11001111
X 0
2. Compute the LRC for the following bit stream, assume a chunk of size 8. obtain
the row-wise and column-wise bits to be added in each case.
a. 10101011, 00011101, 11110000, 10101110
10101011 1
00011101 0
11110000 0
10101110 1
______________
11101000 0
b. 00000000, 11111111, 10101010, 01010101
00000000 0
11111111 0
10101010 0
01010101 0
____________
00000000 0
c. 10011101, 11000110, 11110101, 01111010
10011101 1
11000110 0
11110101 0
01111010 1
_______________
11010100 0
d. 11100110, 10111110, 00011110, 11110010
11100110 1
10111110 0
00011110 0
11110010 1
___________
10110100 0
3. Compute the CRC bits to be added to the following. Use the polynomial x3+ x2+1.
How many bits will be there in the CRC?
3 bits, as the divisor has 4 bits
a. 1110101101
1101 ) 1110101101 , 000 (
1101
1110
1101
1111
1101
1001
1101
1000
1101
1010
1101
1110
1101
11
- hence CRC of 3 bits is 011
b. 1011110110
1101 ) 1011110110, 000 (
1101
1101
1101
1011
1101
1100
1101
1000
1101
101
- hence CRC is 101
4. Compute the checksum for the following. Assume a chunk size of 8. Show that
the checksum computed is correct.
a. 10101011, 00011101,
10101011
00011101 - 1’s complement addition
11001000 - discard any carry over
complement this – 00110111 – this is the checksum
to check if it is correct add the checksum with the first 2 lines of bits – you
should get all 1’s
10101011
00011101
00110111
11111111 – this provides a check
b. 11110000, 10101110
11110000
10101110
10011110 – discard any carry over
complement this – 01100001 – this is the checksum
to check if it is correct add the checksum with the first 2 lines of bits – you
should get all 1’s
11110000
10101110
01100001
11111111 – this provides the check
5. a. Construct the Hamming Code sequence for 0001001 using even parity.
d
d
d
R8
d
D
d
R4
d
R2
R1
0
0
0
0
1
0
0
1
1
0
0
b. Identify the parity bit values at the receiving end for the above data stream with
an error in bit position 6.
Received after 6th bit flipped
00001101100, computing the redundancy bits
r8 = 0, r4 = 1, r2 = 1, r0= 1, which indicated bit flip in position 6
6. a. Construct the Hamming Code sequence for 0101101 using even parity.
D
d
d
R8
d
d
d
R4
d
R2
R1
0
1
0
1
1
1
0
0
1
0
0
b. Identify the parity value for the above data stream with an error in bit position
11.
Received bit stream after 11th bit flipped
11011100100 - computing the redundancy bits
r8 = 1 r4 = 0 r2 = 1, r0= 1, which indicated bit flip in position 11
7. In the Sliding Window scheme shown below, the window size is 7 and the
systems use a mod 16 counter. What is the window size at the check points
Sending
window = 7
Station 1
Station 2
time
Data frame, 0
Receiving
window = 7
Data frame, 1
Data frame, 2
Check point B
Sending Window =4
Check point A
Receiving Window =4
Ack, 1
Check point C
Receiving Window =5
Check point D
Sending Window =5
Data frame, 3
Data frame, 4
Data frame, 5
Check point F
Sending Window =2
Check point E
Receiving Window =2
Ack, 3
Check point H
Sending Window =4
Check point G
Sending Window =4
What is the maximum frame number that can be used in this scheme? 15
8. In the Sliding Window scheme shown below, the window size is 4 and the
systems use a mod 8 counter. What is the window size at the check points
Sending
window = 4
Station 1
Station 2
time
Data frame, 0
Receiving
window = 4
Data frame, 1
Data frame, 2
Check point B
Sending Window =1
Check point A
Receiving Window =1
Ack, 1
Check point C
Receiving Window =2
Check point D
Sending Window =2
Data frame, 3
Check point F
Sending Window =1
Check point E
Receiving Window =1
Ack, 4
Check point G
Sending Window =4
Check point H
Sending Window =4
Data frame, 4
Data frame, 5
Data frame, 6
Check point J
Sending Window =0
Data frame, 7
Check point I
Sending Window =0
9. The following is a sliding window implementation in HDLC, where the window
size is 4 and a mod 8 counter is used. Identify the frame number values in the
blanks provided
0, 1, 2, 3, 4, 5, 6, 7
N(s) 0
N(s) 1
Station 1
0
N(r)
0 N(r)
N(s) 2
0 N(r)
Station 2
0, 1, 2, 3
data, 0
data , 1
data, 2
N(s) 0
2 N(r)
N(s) 1
3
data, 0
N(s) 3
1 N(r)
data , 3
N(r)
data, 1
N(s) 2
data, 2
N(s) 4
3
N(s) 5
3
N(r)
data , 4
N(r)
data , 5
RNR , (? 6
RR , (?
N(s) 6
3
N(r)
Data frame, 6
N(s) 7
3 N(r)
Data frame, 7
6
)
)
4
N(r)