📝 IPSec AH and ESP Two Security Protocols (25 MCQs)

📖 From Data Communication and Networks • 32. Internet Security • 25 questions available

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What is IPSec AH and ESP Two Security Protocols?

Authentication Header (AH) provides data integrity and origin authentication without encryption, whereas Encapsulating Security Payload (ESP) provides confidentiality through encryption along with optional integrity and authentication services.

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📝 All IPSec AH and ESP Two Security Protocols MCQs

Q1. Why does a very small current interval in static arithmetic coding lead to many 0 bits in the middle of the code?

A.Because low‑precision arithmetic forces padding with zeros ✅

B.Because the interval size determines the symbol alphabet

C.Because high‑precision arithmetic eliminates zeros

D.Because a terminating symbol is required

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: When the interval becomes extremely small, the encoder must represent the value with high precision. To keep the interval within the limited word size, the algorithm inserts zero bits as padding, which appear in the middle of the final code. This padding is a direct consequence of insufficient precision.

Q2. Which statement correctly compares the number of bits transmitted per symbol by static and dynamic arithmetic coding?

A.Static coding always sends fewer bits than dynamic coding

B.Dynamic coding may emit bits after each symbol, reducing average bits ✅

C.Both send the same fixed number of bits per symbol

D.Static coding never requires a terminating symbol

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Dynamic arithmetic coding updates the interval after each input symbol and can output the most significant bits immediately, often before the entire message is processed. This incremental emission typically lowers the average bits per symbol compared with static coding, which must wait until the whole message is known before emitting any bits.

Q3. Why are the differences between consecutive samples generally smaller than the original sample values in predictive coding?

A.Because quantization noise is larger for differences

B.Because neighboring samples are highly correlated ✅

C.Because the encoder discards high‑frequency components

D.Because the sampling rate is increased

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Predictive coding exploits the natural correlation of adjacent samples; the original values differ only slightly from one another. By encoding the difference rather than the full value, the magnitude of the numbers to be represented shrinks, allowing fewer bits to describe the same information.

Q4. If PCM uses 3 bits per sample and delta modulation uses 1 bit per sample, what fraction of the original bits is transmitted with delta modulation?

A.01-Feb

B.01-Mar ✅

C.02-Mar

D.03-Apr

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: PCM transmits three bits for each sample, while delta modulation reduces that to a single bit. The reduction factor is therefore 1 divided by 3, meaning delta modulation sends one‑third of the bits required by PCM for the same number of samples.

Q5. Which of the following is NOT a drawback of static arithmetic coding?

A.It requires a terminating symbol for decoding

B.It can produce long runs of 0 bits when precision is low

C.It cannot encode symbols until the entire message is received

D.It always yields the highest possible compression ratio ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Static arithmetic coding does not guarantee the highest possible compression; its efficiency depends on symbol probabilities and model accuracy. The other three options describe genuine disadvantages: the need for a terminating symbol, potential long runs of zeros due to low precision, and the inability to output bits incrementally.

Q6. How does a terminating symbol assist the decoder in static arithmetic coding?

A.It signals the end of the interval narrowing process ✅

B.It increases the precision of the interval

C.It removes the need for high‑precision arithmetic

D.It allows the encoder to skip low‑probability symbols

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The terminating symbol marks the point at which the encoder stops refining the interval. The decoder, upon recognizing this symbol, knows that no further bits will be generated, allowing it to finalize the current interval and retrieve the original message without ambiguity.

Q7. Given $y_{n-1}=5$ and a transmitted increment bit $C_n=1$ with step size 1, what is the reconstructed value $y_n$ ?

A.6

B.4

C.5 ✅

D.7

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: In delta modulation, a transmitted '1' typically indicates that the reconstructed value should increase by one step. Starting from $y_{n-1}=5$ , adding the step size of 1 yields $y_n=6$ . The answer choice reflecting this calculation is 6.

Q8. Which statement best differentiates PCM from delta modulation regarding quantizer resolution?

A.PCM uses a variable‑step quantizer while delta modulation uses a fixed step

B.PCM quantizes each sample independently; delta modulation quantizes the difference using a single‑bit step

C.Both use the same number of bits per sample

D.Delta modulation requires higher resolution than PCM ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Delta modulation reduces the transmitted data to a single bit per sample, effectively using a binary step size. PCM, by contrast, may allocate multiple bits per sample, providing finer resolution. Therefore, delta modulation operates with a coarser quantizer compared to PCM’s higher‑resolution quantization.

Q9. For a signal where most sample differences are zero, which coding method generally yields the fewest bits?

A.Static arithmetic coding

B.Dynamic arithmetic coding

C.Delta modulation ✅

D.PCM

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: When the differences between consecutive samples are frequently zero, delta modulation can represent each sample with a single bit, often the same bit repeatedly. This redundancy leads to extremely low bit rates compared with methods that encode full sample values or probabilities, making delta modulation the most efficient in such scenarios.

Q10. How does dynamic arithmetic coding emit bits immediately after each symbol is read?

A.By postponing interval updates until the end of the message

B.By fixing the interval width after the first symbol

C.By outputting the most significant bits of the current interval as soon as they become certain ✅

D.By using a separate terminating symbol for each symbol

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Dynamic arithmetic coding maintains a narrowing interval as symbols arrive. Whenever the most significant bits of the interval become invariant regardless of future symbols, those bits are flushed to the output stream. This incremental emission enables real‑time bit generation after each symbol is processed.

Q11. If a video compression algorithm applies predictive coding across frames, what is the primary effect on required bandwidth compared to encoding each frame independently?

A.Bandwidth increases due to added motion vectors

B.Bandwidth remains unchanged

C.Bandwidth decreases because inter‑frame redundancies are exploited ✅

D.Bandwidth fluctuates unpredictably

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Predictive coding across frames captures temporal redundancy by transmitting only the differences (residuals) and motion information. Since many pixels remain similar from one frame to the next, the residual data is usually far smaller than a full frame, leading to a reduction in the overall bandwidth needed for transmission.

Q12. If PCM needs 3 bits per sample for a maximum value of 7, and dynamic arithmetic coding averages 1.8 bits per symbol, what is the compression ratio?

A.0.6 ✅

B.0.3

C.0.9

D.0.45

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Compression ratio is calculated as (compressed size) / (original size). Here, 1.8 bits / 3 bits = 0.6, meaning the dynamic arithmetic coding representation uses 60 % of the bits required by PCM, indicating a 40 % reduction in size.

Q13. How does lossy compression exploit the human visual system's inability to detect small distortions?

A.By increasing quantization steps only in high‑frequency components ✅

B.By removing all color information

C.By discarding every other pixel

D.By converting images to grayscale only

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The human eye is less sensitive to fine detail in high‑frequency components. Lossy algorithms therefore apply coarser quantization to those components, removing subtle variations that are unlikely to be perceived, while preserving low‑frequency information that carries most of the visual content.

Q14. Which statement about error accumulation in delta modulation is true?

A.Error never accumulates because each bit is independent

B.Error can build up if the step size is too large relative to signal changes ✅

C.Error is eliminated by the terminating symbol

D.Error is corrected automatically by the decoder

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Delta modulation updates the reconstructed signal by a fixed step. If the step size exceeds the actual change in the original signal, the reconstructed value will overshoot, and subsequent corrections may not fully compensate, causing the error to accumulate over time.

Q15. Why is a terminating symbol still required in dynamic arithmetic coding even though bits are emitted incrementally?

A.Because it marks the end of the message to resolve the final interval ✅

B.Because it resets the encoder for the next message

C.Because it indicates a change in probability model

D.Because it doubles the compression efficiency

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Even though bits are flushed as soon as they become certain, the encoder continues to refine the interval until the entire message is processed. The terminating symbol signals that no further refinement will occur, allowing the decoder to finalize the interval and recover the exact message.

Q16. What advantage does a hybrid approach that combines predictive coding with delta modulation provide?

A.It eliminates the need for any quantization

B.It allows both large and small changes to be represented efficiently ✅

C.It doubles the transmission speed regardless of signal type

D.It removes the requirement for a terminating symbol

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Predictive coding reduces the magnitude of values by encoding differences, while delta modulation conveys those differences with a single bit per sample. The combination means that when differences are small, delta modulation transmits minimal bits, and when larger changes occur, predictive coding still keeps the values compact, yielding overall higher efficiency.

Q17. In which situation is sacrificing accuracy justified when choosing between lossless and lossy compression?

A.When transmitting mission‑critical text documents

B.When storing archival legal records

C.When streaming video to a mobile device ✅

D.When preserving medical imaging for diagnosis

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Streaming video to a mobile device often requires real‑time delivery over limited bandwidth. In such cases, a modest loss of visual fidelity (acceptable to human viewers) can be exchanged for higher compression, making lossy compression a practical choice.

Q18. Given $x_n=10$ , $y_{n-1}=9$ , and a transmitted increment bit $C_n=1$ with step size 2, what is the reconstructed $y_n$ ?

A.11 ✅

B.13

C.9

D.12

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: In delta modulation, a transmitted '1' indicates an increase by the step size. Adding the step size of 2 to the previous reconstruction $y_{n-1}=9$ yields $y_n=11$ . This matches option A.

Q19. How does the need for high‑precision arithmetic in static coding affect system latency?

A.It reduces latency because calculations are simpler

B.It increases latency due to extra processing for precision handling ✅

C.It has no impact on latency

D.It makes latency unpredictable

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Static arithmetic coding must maintain a very narrow interval, requiring high‑precision arithmetic operations. These operations are computationally intensive and often need multiple word‑size manipulations, which adds processing time and therefore increases overall latency before bits can be emitted.

Q20. If dynamic arithmetic coding keeps an interval width of at most $\frac{1}{2^k}$ after processing k symbols, what is the maximum number of bits that can be emitted after reading the next symbol?

A.0

B.1 ✅

C.k+1

D.2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When the interval width is bounded by $\frac{1}{2^k}$ , the most significant bit of the interval becomes fixed after the next symbol, allowing at most one additional bit to be emitted. Hence, the maximum number of new bits after reading the (k+1)‑th symbol is one.

Q21. For a slowly varying signal, which coding scheme typically yields the lowest entropy rate and why?

A.PCM, because it uses fixed‑length codes

B.Delta modulation, because it transmits only single‑bit changes

C.Dynamic arithmetic coding, because it adapts probabilities to the signal ✅

D.All produce identical entropy rates

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Dynamic arithmetic coding continuously updates symbol probabilities based on observed data, allowing it to assign shorter codes to more probable symbols. In a slowly varying signal, the probability model quickly converges, resulting in a lower entropy rate than the fixed‑length PCM or the binary step of delta modulation.

Q22. Designing a hybrid scheme that switches between dynamic arithmetic coding and delta modulation, which decision rule best determines the switch point?

A.Switch when the variance of recent samples exceeds a fixed threshold ✅

B.Switch after a fixed number of symbols regardless of content

C.Switch based on the presence of a terminating symbol

D.Switch when the interval width falls below

\frac{1}{2^{10}}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: A variance‑based threshold directly reflects the signal's local activity. High variance indicates that differences are large, favoring dynamic arithmetic coding, whereas low variance suggests small differences, where delta modulation excels. Therefore, using variance as the switch criterion balances compression efficiency across varying signal characteristics.

Q23. If a terminating symbol is omitted in static arithmetic coding, which decoding error is most likely to occur?

A.Undetected overflow error

B.Incorrect symbol ordering

C.Ambiguous final interval leading to wrong message reconstruction ✅

D.Loss of synchronization between encoder and decoder

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Without a terminating symbol, the decoder cannot determine where the encoded interval should stop narrowing. This ambiguity leaves the final interval open to multiple possible interpretations, often resulting in the reconstruction of an incorrect message at the end of the stream.

Q24. In a bandwidth‑limited but latency‑critical environment, which coding method offers the best trade‑off between compression efficiency and real‑time delivery?

A.Static arithmetic coding

B.Delta modulation ✅

C.Dynamic arithmetic coding

D.PCM

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Delta modulation emits a single bit per sample immediately after each symbol, minimizing both the amount of data transmitted and the delay before bits become available. While it may not achieve the highest compression ratios, its low latency makes it the most suitable choice when timely delivery is paramount.

Q25. What is the primary purpose of a terminating symbol in static arithmetic coding?

A.To increase the compression ratio

B.To indicate the end of the encoded message and finalize the interval ✅

C.To reduce the number of bits per symbol

D.To enable dynamic updating of symbol probabilities

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The terminating symbol signals to the decoder that no further symbols will follow, allowing it to close the current interval and uniquely determine the original message. This explicit marker is essential because static arithmetic coding does not emit bits incrementally and therefore needs a clear endpoint.

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