26. Using
the generator (divisor) polynomial x

^{3}+ x + 1 for CRC, what frame will be transmitted for the data M = x^{7}+ x^{5}+ x^{3}+ 1?__Solution__

M(x)=10101001

C(x)=1011

r=3

T(x) = 10101001 111

**Note:**

__Cyclic Redundancy Check__**:**A major goal in designing error detection algorithms is to maximize the probability of detecting errors using only a small number of redundant bits.

Add k bits of redundant data to an n-bit
message

Represent n-bit message as n-1 degree
polynomial

__Algorithm for computing the checksum__
1. shift left r bits (append r zero bits to
low order end of the frame), i.e., M(x)x

^{r}
2. divide the bit string corresponding to
G(x) into (x

^{r})M(x).
3. subtract (or add) remainder of M(x)x

^{r}/ G(x) from M(x)x^{r}using XOR, call the result T(x). Transmit T(x).
27. Using
the generator (divisor) polynomial x

^{4}+ x + 1 for CRC, what frame will be transmitted for the data M = x^{7}+ x^{5}+ x^{3}+ 1?__Solution__

M(x)=10101001

C(x)=10011

r=4

T(x) = 10101001 1100

28. A
disadvantage of a broadcast subnet is the capacity wasted when multiple hosts
attempt to access the channel at the same time. As a simplistic example,
suppose that time is divided into discrete slots, with each of the n hosts
attempting to use the channel with probability p during each slot. What fractions
of the slots are wasted due to collisions?

__Solution__

The fraction of the slots are wasted due to
collisions is

1 - the fraction of the slots without
collisions.

Two situations are without collisions:

1) Successful transmission. One station
transmits and the rest of them idle. The probability is n x p x (1 - p)^(n -
1).

2) All stations are idle. The probability is
(1 - p)^n.

Hence, the fraction that are wasted is

1 - n x p x (1 - p)^(n - 1) - (1 - p)^n.

29. What
is the maximum data rate for a noiseless 1-MHz channel if QAM-256 is used?

__Solution__

Maximum data rate = 2 B log

_{2}V
= 2 x 1 x log

_{2}256
= 2 x 8 = 16 Mbps

30. If
a binary signal is sent over a 1-MHz channel whose signal-to-noise ratio is 20
dB, what is the maximum achievable data rate?

__Solution__

Signal-to-Noise ratio (S/N) = 20 dB

=>
10 log

_{10}(S/N)=20
=> log

_{10}S/N = 2,
=> S/N =10

^{2}= 100
Now, from Shannon's theorem we know,

Max. data rate = B log

_{2}(1 + S/N) bits/sec
= 1 x log

_{2}101 = 6.658 Mbps or about 7 Mbps**Note:**

The

**Shannon-Hartley theorem**states that the channel capacity is given by
C = B log

_{2}(1 + S/N)
where C is the capacity in bits per second, B
is the bandwidth of the channel in Hertz, and S/N is the signal-to-noise ratio.