📝 Application sensitivity to QoS parameters (10 MCQs)

📖 From Data Communication and Networks • 30. Quality of Service • 10 questions available

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What is Application sensitivity to QoS parameters?

Application sensitivity describes how differently various network applications react to changes in network conditions, where real-time audio and video are highly sensitive to delay and jitter but tolerant of some loss, while file transfers are sensitive to packet loss but tolerant of variable delays.

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📝 All Application sensitivity to QoS parameters MCQs

Q1. Given the Huffman code set {A:00, B:010, C:011, D:10, E:11}, which of the following statements correctly infers the consequence of the prefix property?

A.No code can be a suffix of another code.

B.No code can be a prefix of another code. ✅

C.All codes have the same length.

D.The code can be decoded only with a lookup table.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Because the code is prefix‑free, once the decoder reads a complete codeword it knows that no longer code can begin with those bits. Therefore the decoder can immediately identify each symbol without needing to look ahead or backtrack, which is the defining consequence of the prefix property.

Q2. Compare the average bit length per symbol for the Huffman code above with a fixed‑length code for five symbols. Which statement best evaluates the compression gain?

A.Both use 2 bits per symbol, so no gain.

B.Huffman uses 2 bits per symbol, fixed‑length 3 bits, saving 1 bit.

C.Huffman uses 2.4 bits per symbol on average, fixed‑length 3 bits, saving 0.6 bits. ✅

D.Huffman uses 3 bits per symbol, fixed‑length 2 bits, saving -1 bit.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A fixed‑length code for five symbols requires \ $\\lceil\\log_2 5\\rceil = 3\$ bits per symbol. The Huffman code uses lengths 2,3,3,2,2 bits, giving an average of \ $(2+3+3+2+2)/5 = 2.4\$ bits. The difference of 0.6 bits per symbol represents the compression gain, matching option C.

Q3. In Huffman coding, what term describes a code where no codeword is a prefix of any other codeword?

A.Self‑delimiting code

B.Instantaneous code ✅

C.Block code

D.Variable‑length code

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: An instantaneous code is one in which no codeword is a prefix of any other, allowing immediate symbol identification as each codeword is received. This property is exactly the definition of the prefix‑free condition used in Huffman coding.

Q4. If a message consisting of the sequence A‑B‑C‑D‑E is encoded using the given Huffman code, and the receiver decodes the bitstream 0010011, which inference about the original sequence is correct?

A.The sequence must be A‑B‑E.

B.The sequence could be A‑D‑C. ✅

C.The sequence is uniquely A‑D‑C‑E.

D.Decoding is ambiguous.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Parsing 0010011 using the given code: 00 → A, remaining 10011; 10 → D, remaining 011; 011 → C. The decoder thus recovers the sequence A‑D‑C. No other parsing is possible because the prefix property guarantees unique segmentation, so option B is correct.

Q5. Evaluate the impact on average code length if the probability of symbol A doubles while all other symbols keep the same relative probabilities. Which statement best reflects the change?

A.Average length decreases because A gets shorter code.

B.Average length stays the same because code lengths are fixed.

C.Average length increases because more frequent A uses longer code.

D.Average length may increase or decrease depending on re‑construction of the Huffman tree. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Doubling the probability of A makes it more likely to be assigned a shorter code, but the Huffman tree must be rebuilt to reflect the new frequencies. The average length can either decrease or increase depending on how the tree restructures, so the safest statement is that the effect is uncertain without recomputing the tree.

Q6. Apply the principle of entropy to estimate the theoretical minimum average bits per symbol for the five‑symbol source with probabilities {0.4,0.2,0.2,0.1,0.1}. Which value is closest to the entropy \ $H = -\\sum p_i \\log_2 p_i\$ ?

A.1.5 bits

B.1.85 bits

C.2.0 bits ✅

D.2.3 bits

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The entropy is \ $H = -\\sum p_i\\log_2 p_i = -(0.4\\log_2 0.4 + 2\\cdot0.2\\log_2 0.2 + 2\\cdot0.1\\log_2 0.1) \\approx 2.12\$ bits. Among the options, 2.0 bits is the closest approximation, making option C the best choice.

Q7. Given that no Huffman codeword is a prefix of another, what can be deduced about the decoding process when reading a bitstream sequentially?

A.The decoder must backtrack to resolve ambiguities.

B.The decoder can decide each symbol as soon as its codeword is recognized. ✅

C.Decoding requires knowledge of total message length.

D.Decoding is impossible without a separate delimiter.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because the code is prefix‑free, the decoder can stop as soon as it matches a complete codeword; there is no need for backtracking or extra delimiters. This immediate decision capability is a direct consequence of the prefix property, validating option B.

Q8. Contrast the compression ratio achieved by the Huffman code with that of a naive fixed‑length code for a large text where symbol frequencies follow a Zipf distribution. Which analysis is most accurate?

A.Huffman provides negligible improvement over fixed‑length.

B.Huffman can achieve up to 50 % reduction in average bits per symbol. ✅

C.Fixed‑length always outperforms Huffman for Zipf‑distributed data.

D.Both methods yield identical average lengths for Zipf distributions.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Zipf‑like frequencies give a few very common symbols and many rare ones. Huffman assigns short codes to the frequent symbols and longer codes to the rare ones, often cutting the average bits per symbol roughly in half compared with a uniform 3‑bit fixed length for five symbols. Hence option B accurately reflects the typical improvement.

Q9. Synthesize a method to extend the given Huffman code to support an additional symbol F with probability 0.05, while preserving the prefix property. Which of the following steps correctly describes the process?

A.Replace the longest codeword with a two‑bit extension for F.

B.Merge the two least probable symbols (F and the current least probable) and rebuild the tree. ✅

C.Append F with code 111, ensuring no existing code starts with 111.

D.Assign F the code 00, and shift all other codes one bit longer.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: To add a new symbol F with probability 0.05, the standard procedure is to combine F with the current least probable symbol (the one with the smallest weight) and rebuild the Huffman tree. This maintains the optimal prefix property while incorporating the new symbol, which corresponds to option B.

Q10. If the Huffman coding tree for the five symbols is altered by swapping the positions of symbols D and E in the tree, how does this affect the average code length and why?

A.Average length remains unchanged because swapping leaf nodes does not alter depths. ✅

B.Average length increases because E now has a longer code.

C.Average length decreases because D now has a shorter code.

D.The effect depends on the probabilities of D and E; swapping may increase or decrease average length.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Swapping two leaf nodes that reside at the same depth does not alter the depth of any symbol; therefore the total weighted path length, i.e., the average code length, stays the same. The probabilities of D and E are unchanged, so option A correctly describes the effect.

🔗 Related Topics (MCQs)

📝 Data flow characteristics in QoS MCQs 📝 QoS definitions in computer networks MCQs 📝 Flow classes in quality of service MCQs 📝 Flow control techniques for QoS MCQs 📝 Scheduling algorithms for QoS MCQs

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