Networking Basics Questions and Answers

Q: The 1EEE802·4 token bus standard mainly designed by keeping motivations as :

The answer is All of the above.The IEEE 802.4 token bus (printed '1EEE802.4') was developed largely for factory automation. A station transmits only when it holds the token, which passes around a logical ring, so the worst-case waiting time is bounded: this supports real-time traffic and removes the probabilistic, collision-based access of CSMA/CD Ethernet.Physically it is a bus, a linear cable, which was considered more robust than a physical ring, where a single break in the cable stops the whole network. So all three motivations apply.

Q: Consider the stop and wait protocol, if transmission time is 'a' at the source and propagation delay is 'b' then after what time the sender can send the second packet ?Assume data packet and ACK packet of the same size with no pissy backing :

The sender can send the second packet after 2a + 2b.In stop-and-wait the sender must receive the acknowledgement before sending the next packet. With the ACK the same size as the data packet and no piggybacking:\[ \begin{aligned} T &= a\ (\text{send data}) + b\ (\text{data travels}) \\ &+ a\ (\text{send ACK}) + b\ (\text{ACK travels}) \\ &= 2a + 2b \end{aligned} \]a + 2b is the answer when the ACK's transmission time is negligible, which the question rules out by making the ACK as large as the data. 2b + a is the same mistake written differently, and (a + b)/2 has no meaning here.

Q: Suppose a channel has bandwidth B = 4 kHz, determine the channel capacity for each of the following signal to noise ratio.(i) 20 dB (ii) 30dB (iii) 40 dB

The capacities are 26.6 kbps, 39.8 kbps, 53.1 kbps.Shannon's formula is \(C = B\log_{2}(1 + \text{SNR})\), with SNR as a ratio, so first convert decibels: 20, 30 and 40 dB are ratios of 100, 1000 and 10000.\[ \begin{aligned} C_{1} &= 4000 \log_{2} 101 \approx 26.6\ \text{kbps} \\ C_{2} &= 4000 \log_{2} 1001 \approx 39.9\ \text{kbps} \\ C_{3} &= 4000 \log_{2} 10001 \approx 53.2\ \text{kbps} \end{aligned} \]Only option C has these values, in this order (the paper truncates the last two). Option A lists the same numbers in the wrong order, and B and D do not follow from the formula.

Q: A.................... is said to be strongly symmetric, if in the transmission matrix each row is a permutation of each column.

The definition is that of a strongly symmetric discrete memoryless channel.A discrete memoryless channel is described by its transition (channel) matrix of probabilities p(y | x). Information theory texts classify DMCs by the pattern of that matrix: a DMC is called strongly symmetric when its rows are permutations of each other and its columns are permutations of each other. For such channels the capacity is reached with equiprobable inputs.The binary symmetric channel is just one example of a strongly symmetric DMC, not the general class being defined. A binary asymmetric channel does not have this property, and a burst channel has memory, so it is not a DMC at all.

Q: Hamming distance between the codes 1110100 and 1111001 is :

The Hamming distance is 3.The Hamming distance is the number of positions in which two codewords differ. Comparing them bit by bit:\[ \begin{aligned} &1110100 \\ &1111001 \\ &0001101 \end{aligned} \]The last line, the bitwise XOR, has 1s in positions 4, 5 and 7, so the codes differ in 3 places. 0 would mean identical codes, and 1 or 2 undercount the differences.

Q: What is the dimension of the subspace spanned by 0111, 1010, 0011 and 1110 using generator matrix ?

The dimension is 3.Working over GF(2) (addition is XOR), check for linear dependence. The fourth vector is the sum of the other three:\[ \begin{aligned} 0111 + 1010 &= 1101 \\ 1101 + 0011 &= 1110 \end{aligned} \]So 1110 adds nothing new. The first three are independent: no single one is zero, no two are equal, and no two add to the third (0111 + 1010 = 1101, 0111 + 0011 = 0100, 1010 + 0011 = 1001). Three independent vectors span a subspace of dimension 3, with 8 codewords; dimensions 0, 1 and 2 are too small.

Q: A telephone line normally has a bandwidth of 3000 Hz assigned for data communications. The signal-to-noise ratio is usually 3162. Then the channel capacity for this channel is :

The channel capacity is 34,860 bps.By Shannon's formula for a noisy channel:\[ \begin{aligned} C &= B \log_{2}(1 + \text{SNR}) \\ &= 3000 \log_{2}(1 + 3162) \\ &= 3000 \times 11.62 \approx 34{,}860 \text{ bps} \end{aligned} \](More precise logarithms give about 34,880 bps; the textbook value, from Forouzan, rounds the logarithm to 11.62.) Option (C), 11.62, is the logarithm alone without multiplying by the bandwidth. The other values do not come from the formula.

Q: Which provision can resolve/overcome the shortcomings associated with duplication or failure condition of Stop-and-Wait Automatic Repeat Request protocol especially due to loss of data frames or non-reception of acknowledgement ?

The provision is a sequence number in the header of the message.In Stop-and-Wait ARQ, if an acknowledgement is lost, the sender times out and retransmits the frame, and the receiver would accept the same frame twice. Numbering frames, modulo 2 being enough (0 and 1), lets the receiver recognise a duplicate and discard it while still acknowledging it. Acknowledgements are numbered too, so the sender knows which frame they refer to.A checksum detects corrupted frames, but it cannot tell a correct duplicate from a new frame. So the duplication problem is solved by sequence numbers, not by the checksum, and the options including or relying on the checksum are wrong.

Q: Binary cyclic code :(i) Detects all single errors.(ii) Detects all odd number of error bits.(iii) Detects all burst errors of 16 or fewer bits.

The answer is (i), (ii) and (iii).A cyclic redundancy check with a well-chosen generator polynomial detects all single-bit errors (the generator has more than one term), all errors with an odd number of bits (the generator has \(x + 1\) as a factor), and all burst errors of length up to the degree of the generator. For a 16-bit CRC such as CRC-16 or CRC-CCITT, that is every burst of 16 or fewer bits.These three properties are the standard textbook summary of CRC error detection, so options naming only one or two of them are incomplete.

11.

The 1EEE802·4 token bus standard mainly designed by keeping motivations as :

A.To support real time traffic
B.It is less vulnerable compared to token ring
C.To remove probabilistic nature of CSMA/CD
D.All of the above

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The answer is All of the above.

The IEEE 802.4 token bus (printed '1EEE802.4') was developed largely for factory automation. A station transmits only when it holds the token, which passes around a logical ring, so the worst-case waiting time is bounded: this supports real-time traffic and removes the probabilistic, collision-based access of CSMA/CD Ethernet.

Physically it is a bus, a linear cable, which was considered more robust than a physical ring, where a single break in the cable stops the whole network. So all three motivations apply.

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12.

Consider the stop and wait protocol, if transmission time is 'a' at the source and propagation delay is 'b' then after what time the sender can send the second packet ?

Assume data packet and ACK packet of the same size with no pissy backing :

A.2a + 2b
B.(a + b)/2
C.2b + a
D.a + 2b

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The sender can send the second packet after 2a + 2b.

In stop-and-wait the sender must receive the acknowledgement before sending the next packet. With the ACK the same size as the data packet and no piggybacking:

\[ \begin{aligned} T &= a\ (\text{send data}) + b\ (\text{data travels}) \\ &+ a\ (\text{send ACK}) + b\ (\text{ACK travels}) \\ &= 2a + 2b \end{aligned} \]

a + 2b is the answer when the ACK's transmission time is negligible, which the question rules out by making the ACK as large as the data. 2b + a is the same mistake written differently, and (a + b)/2 has no meaning here.

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13.

Suppose a channel has bandwidth B = 4 kHz, determine the channel capacity for each of the following signal to noise ratio.

(i) 20 dB (ii) 30dB (iii) 40 dB

A.39.8 kbps, 53.1 kbps, 26.6 kbps
B.40.8 kbps, 56.1 kbps, 22.2 kbps
C.26.6 kbps, 39.8 kbps, 53.1 kbps
D.56 kbps, 29.1 kbps, 54.1 kbps

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The capacities are 26.6 kbps, 39.8 kbps, 53.1 kbps.

Shannon's formula is \(C = B\log_{2}(1 + \text{SNR})\), with SNR as a ratio, so first convert decibels: 20, 30 and 40 dB are ratios of 100, 1000 and 10000.

\[ \begin{aligned} C_{1} &= 4000 \log_{2} 101 \approx 26.6\ \text{kbps} \\ C_{2} &= 4000 \log_{2} 1001 \approx 39.9\ \text{kbps} \\ C_{3} &= 4000 \log_{2} 10001 \approx 53.2\ \text{kbps} \end{aligned} \]

Only option C has these values, in this order (the paper truncates the last two). Option A lists the same numbers in the wrong order, and B and D do not follow from the formula.

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14.

A.................... is said to be strongly symmetric, if in the transmission matrix each row is a permutation of each column.

A.Binary symmetric channel
B.Discrete memoryless channel
C.Binary assymmetric channel
D.Burst channel

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The definition is that of a strongly symmetric discrete memoryless channel.

A discrete memoryless channel is described by its transition (channel) matrix of probabilities p(y | x). Information theory texts classify DMCs by the pattern of that matrix: a DMC is called strongly symmetric when its rows are permutations of each other and its columns are permutations of each other. For such channels the capacity is reached with equiprobable inputs.

The binary symmetric channel is just one example of a strongly symmetric DMC, not the general class being defined. A binary asymmetric channel does not have this property, and a burst channel has memory, so it is not a DMC at all.

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15.

Hamming distance between the codes 1110100 and 1111001 is :

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The Hamming distance is 3.

The Hamming distance is the number of positions in which two codewords differ. Comparing them bit by bit:

\[ \begin{aligned} &1110100 \\ &1111001 \\ &0001101 \end{aligned} \]

The last line, the bitwise XOR, has 1s in positions 4, 5 and 7, so the codes differ in 3 places. 0 would mean identical codes, and 1 or 2 undercount the differences.

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16.

What is the dimension of the subspace spanned by 0111, 1010, 0011 and 1110 using generator matrix ?

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The dimension is 3.

Working over GF(2) (addition is XOR), check for linear dependence. The fourth vector is the sum of the other three:

\[ \begin{aligned} 0111 + 1010 &= 1101 \\ 1101 + 0011 &= 1110 \end{aligned} \]

So 1110 adds nothing new. The first three are independent: no single one is zero, no two are equal, and no two add to the third (0111 + 1010 = 1101, 0111 + 0011 = 0100, 1010 + 0011 = 1001). Three independent vectors span a subspace of dimension 3, with 8 codewords; dimensions 0, 1 and 2 are too small.

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17.

A telephone line normally has a bandwidth of 3000 Hz assigned for data communications. The signal-to-noise ratio is usually 3162. Then the channel capacity for this channel is :

A.34,860 bps
B.6,262 bps
C.11.62 bps
D.3,854 bps

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The channel capacity is 34,860 bps.

By Shannon's formula for a noisy channel:

\[ \begin{aligned} C &= B \log_{2}(1 + \text{SNR}) \\ &= 3000 \log_{2}(1 + 3162) \\ &= 3000 \times 11.62 \approx 34{,}860 \text{ bps} \end{aligned} \]

(More precise logarithms give about 34,880 bps; the textbook value, from Forouzan, rounds the logarithm to 11.62.) Option (C), 11.62, is the logarithm alone without multiplying by the bandwidth. The other values do not come from the formula.

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18.

Which of the following CRC generator guarantee the detection of a single bit errors ?

A.\(X^{4} + X^{2}\)
B.1
C.\(X^{3} + X\)
D.\(X^{2} + 1\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The generator that guarantees detecting a single-bit error is \(X^{2} + 1\).

A single-bit error at position i is the error polynomial \(x^{i}\). It goes undetected only if the generator divides \(x^{i}\). A generator with at least two terms, one of them the constant 1, cannot divide any \(x^{i}\), so it catches every single-bit error.

\(X^{2} + 1\) has two terms, one of them 1, so it qualifies. The generator 1 divides every polynomial, so every remainder is zero and no error at all is detected. \(X^{4} + X^{2} = X^{2}(X^{2} + 1)\) and \(X^{3} + X = X(X^{2} + 1)\) have no constant term, so they fail the condition and the guarantee does not hold for them.

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19.

Which provision can resolve/overcome the shortcomings associated with duplication or failure condition of Stop-and-Wait Automatic Repeat Request protocol especially due to loss of data frames or non-reception of acknowledgement ?

A.Provision of sequence number in the header of message
B.Provision of checksum computation
C.Both (A) and (B)
D.None of the above

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The provision is a sequence number in the header of the message.

In Stop-and-Wait ARQ, if an acknowledgement is lost, the sender times out and retransmits the frame, and the receiver would accept the same frame twice. Numbering frames, modulo 2 being enough (0 and 1), lets the receiver recognise a duplicate and discard it while still acknowledging it. Acknowledgements are numbered too, so the sender knows which frame they refer to.

A checksum detects corrupted frames, but it cannot tell a correct duplicate from a new frame. So the duplication problem is solved by sequence numbers, not by the checksum, and the options including or relying on the checksum are wrong.

Discuss in Forum

20.

Binary cyclic code :

(i) Detects all single errors.
(ii) Detects all odd number of error bits.
(iii) Detects all burst errors of 16 or fewer bits.

A.Only (i)
B.Only (ii)
C.(ii) and (iii)
D.(i), (ii) and (iii)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The answer is (i), (ii) and (iii).

A cyclic redundancy check with a well-chosen generator polynomial detects all single-bit errors (the generator has more than one term), all errors with an odd number of bits (the generator has \(x + 1\) as a factor), and all burst errors of length up to the degree of the generator. For a 16-bit CRC such as CRC-16 or CRC-CCITT, that is every burst of 16 or fewer bits.

These three properties are the standard textbook summary of CRC error detection, so options naming only one or two of them are incomplete.

Discuss in Forum

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