Question

An 8-bit byte with binary value 11001101 is to be encoded using an even-parity Hamming code.
What is the binary value after encoding?

A bit stream 10101010 is transmitted using the standard CRC method. The divisor is 1011.
Show the actual bit string transmitted. Suppose the second bit from the left is inverted during
transmission. Show that how this error is detected by the receiver?

What is the maximum size of the sender window and receiver windows for each of the following
protocols when the number of bits in the sequence field is 8?
c. Go-Back-N ARQ
d. Selective-Repeat ARQ

Compare and contrast the Go-Back-N ARQ Protocol with Selective-Repeat ARQ describing
the process of each protocol with a suitable example.

Answer 1

1) Binary string=11001101

step 1: finding all the bits in positions of power of 2: i.e. 1,2,4,8

These will respectively be parity bits P1, P2, P3, P4

Bits at locations not in the power of 2 i.e. 3,5,6,7 can be represented as:

3= 1+2

5=1+4

6=2+4

7=1+2+4

each of these is represented as a sum of powers of 2.

To find P1: notice position of ' 1' in 3, 5 and 7. We see bit 1 occuring over there. Counting the number of times 1 appears=3, which is odd. P1=1

To find P2: notice position of '2' in 3,6 and 7. '2' appears only in 3 and 7 at lthe second position. Hence the number of times it occurs=2. which is even. Thus P2=0

To find P3: notice position of '4' in 5, 6 and 7. '4' occurs only in 7 at the third position. hence the number of times it occurs=1. Which is odd. Thus P3=1

To find P4: notice position of '8' in above. As it doesn't occur at all, count is considered to be=0. Which is even. Thus P4=0

After encoding, we get

10011100

2) Part A: CRC actual bit transmission

2 B) Second bit from the left is inverted

3) c) Sender window size Go- back N= 2 to the power of n -1= 2 to the power of 8 -1= 64-1=63

Receiver window size=1

3) d) Sender window size Selective repeat = at most half of 2 to the power of n= 2 to the power of 8/2=64/2=32

Receiver window size=32