Computer Science Study Materials for Competitive Exams

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Wednesday, 11 January 2017

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)xr
2. divide the bit string corresponding to G(x) into (xr)M(x).
3. subtract (or add) remainder of M(x)xr / G(x) from M(x)xr 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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