Computer Networks problems and solutions for UGC NET/GATE exam - Set 6

26.       Using the generator (divisor) polynomial x³ + x + 1 for CRC, what frame will be transmitted for the data M = x⁷ + x⁵ + x³ + 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 x⁴ + x + 1 for CRC, what frame will be transmitted for the data M = x⁷ + x⁵ + x³ + 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 log₂ V

= 2 x 1 x log₂ 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 log₁₀(S/N)=20

=> log₁₀ S/N = 2,

=> S/N =10² = 100

Now, from Shannon's theorem we know,

Max. data rate = B log₂(1 + S/N) bits/sec

= 1 x log₂ 101 = 6.658 Mbps or about 7 Mbps

Note:

The Shannon-Hartley theorem states that the channel capacity is given by

C = B log₂(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.

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