CE-413 Computer Networks & Data Communication-I
7th Semester Electronics Engineering
(All Sections)
Assignment 4
Date of Submission: 20thApril, 2009
1. Referring to the CRC-8 polynomial in table given below, answer the following questions:
CRC-8 |
x8 + x2 + x + 1 |
ATM header |
2. Given the dataword 1010011010 and the divisor 10111,
a. Show the generation of the codeword at the sender site (using binary division).
b. Show the checking of the codeword at the receiver site (assume no error)
3. Answer the following questions:
a. What is the polynomial representation of 101100?
b. What is the result of shifting 101100 three bits to the left?
c. Repeat part b by using polynomials.
d. What is the result of shifting 101100 four bits to the right?
e. Repeat part d by using polynomials.
4. A sender needs to send the two data items 0x3456, and 0xABCC. Answer the following:
a. Find the checksum at the sender site
b. Find the checksum at the receiver site if there is no error.
c. Find the checksum at the receiver site if the second data item is changed to 0xABCE
5. Assuming even parity, find the parity bit for each of the following data units.
6. A receiver receives the bit pattern 01101011. If the system is using even parity, is the pattern in error?
7. A system uses two-dimensional parity. Find the parity unit for the following two data units. Assume even parity.
10011001 01101111
8. Given a 10-bit sequence 1010011110 and a divisor of 1011, find the CRC. Check your answer.
9. Given a remainder 111, a data unit of 10110011, and a divisor 1001, is there an error in data unit?
10. Find the checksum of the following bit sequence. Assume 16-bit segment size.
11. For each data unit of the following sizes, find the redundancy bits needed to correct single-bit error.
12. Construct the Hamming code for the bit sequence 10011101.
13. Find the parity bits of the following bit pattern, using simple parity. Do the same for the two dimensional parity. Assume even parity.
0011101 1100111 1111111 0000000
14. Show that the Hamming code C(7,4) of Table 10.4 can correct one-bit errors but not more by testing the code in the following cases. The character in the burst error means no error; the character "E" means an error.
15. Draw the sender and receiver window for a system using Go-back-N ARQ given the following
16. Repeat exercise 15 from Selective repeat ARQ.
17. What does the number of RRJ frame means in Selective Repeat ARQ?
18. What does the number of RR/ack frame means in Selective Repeat ARQ?
19. A Go-Back-N ARQ uses window size of 15. How many bits are needed to define the sequence number?
20. A Selective Repeat ARQ is using 7-bits to represent sequence numbers. What is the size of the window?
21. Bit-stuff the following data:
0001111110111110011110011111001
22. Bit-stuff the following data:
000111111111111111111111111111110011111001
23. Prove that utilization of sliding window protocol is:
N >2a+1 for U=1
N < 2a+1 for U= N/2a+1
23. Why the size of sender and receiver window in case of Selective Reject ARQ is at most one half of 2^{k}. Justify your answer with the help of an example