📝 IntServ service classes guaranteed controlled load (12 MCQs)

📖 From Data Communication and Networks • 30. Quality of Service • 12 questions available

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What is IntServ service classes guaranteed controlled load?

Guaranteed and Controlled Load are the two primary IntServ service classes, where Guaranteed provides firm mathematical bounds on delay and loss for hard real-time applications, and Controlled Load offers best-effort-like service with high assurance of low loss and delay under varying loads.

Easy

Medium

Hard

📝 All IntServ service classes guaranteed controlled load MCQs

Q1. If a DCT transformation removes redundant data, what will happen to the last three values of the p matrix after the transformation?

A.They become zero

B.They remain unchanged ✅

C.They are amplified

D.They are replaced by random noise

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: In a DCT the redundancy is captured by the DC component, while the AC components that are linearly dependent on the DC remain unchanged. Consequently the three values that were redundant do not change after transformation, illustrating the redundancy‑removal property.

Q2. Which matrix is applied twice to perform a two‑dimensional DCT on a square image block?

A.The identity matrix

B.The Fourier matrix

C.The T matrix ✅

D.The rotation matrix

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The two‑dimensional DCT is achieved by applying the same orthogonal DCT matrix on rows and then on columns. This matrix, commonly denoted as T, is used twice—once on the left and once on the right—producing the transformed block.

Q3. Given source data $p = (7,8,9,10)$ and the transformed vector $M = (17, -2.22, 0.00, -0.16)$ , which statement correctly explains why the first component of $M$ changed compared to a previous source vector with a different average?

A.The first component depends only on the variance of p

B.The first component reflects the overall average of p ✅

C.The first component is unaffected by any change in p

D.The first component is determined by the smallest value in p

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The DC term of a DCT encodes the mean value of the input data. When the average of the source vector changes, the DC component (the first element of M) changes proportionally, while the AC components remain stable because they capture relative variations.

Q4. How does the computational complexity of a two‑dimensional DCT compare to applying a one‑dimensional DCT separately on rows and columns of an $N \times N$ block?

A.Both have the same O(

N^2

) complexity

B.Two‑dimensional DCT is O(

N^3

) while separate 1‑D DCTs are O(

N^2

)

C.Two‑dimensional DCT can be performed with two O(

N^2 \log N

) operations, matching the separate approach ✅

D.Two‑dimensional DCT requires more memory but identical arithmetic count

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: A two‑dimensional DCT can be decomposed into a row‑wise 1‑D DCT followed by a column‑wise 1‑D DCT. Each pass costs O( $N^2 \log N$ ) operations, so the total cost is essentially two such passes, giving the same asymptotic complexity as performing the two 1‑D transforms separately.

Q5. When performing the inverse transformation, why must the T matrix be applied in reverse order compared to the forward DCT?

A.Because matrix multiplication is commutative

B.To undo the column‑wise transformation before the row‑wise one

C.To increase compression ratio

D.To reverse the row‑wise transformation before the column‑wise one ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: The forward DCT first transforms rows then columns ( $M = T\,p\,T^{\top}$ ). To retrieve the original data, the column transformation must be undone first, followed by the row transformation, which is the reverse order of the forward process.

Q6. Assume the 2×2 DCT matrix $T = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1\\ 1 & -1\end{bmatrix}$ . If the source block is $\begin{bmatrix}4 & 2\\ 6 & 0\end{bmatrix}$ , what is the element at position (1,2) of the transformed matrix $M = T \, p \, T^{\top}$ ?

A.-2

B.0

C.2

D.4 ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Carrying out the multiplication yields $M = \begin{bmatrix}6 & 4\\ 0 & -2\end{bmatrix}$ . The element in the first row, second column is therefore 4. This demonstrates how the DCT concentrates energy in the DC term while redistributing the remaining information into the AC coefficients.

Q7. If every element of the source vector $p$ is multiplied by a constant $k$ , how does the transformed vector $M$ produced by a DCT change?

A.All components of

M

are multiplied by

k

✅

B.Only the DC component is multiplied by

k

; AC components remain unchanged

C.The AC components are multiplied by

k

while the DC component stays the same

D.The transformed vector is unchanged

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The DCT is a linear transform; scaling the input by $k$ scales every output component by the same factor. Both the DC term (which reflects the mean) and the AC terms (which reflect variations) are proportionally affected, preserving the relative structure of the spectrum.

Q8. Suppose the correlation between the four elements of $p$ is altered, breaking the original linear relationship. Which of the following is most likely to occur in the transformed vector $M$ ?

A.The redundant three values will become identical to the first value

B.The previously invariant three values will now vary significantly ✅

C.All values of

M

will become zero

D.Only the DC component will change, the AC components will stay constant

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When the elements of $p$ lose their linear dependence, the AC coefficients that previously remained constant begin to reflect the new variations. Consequently the three values that were previously invariant will now exhibit noticeable changes, indicating loss of redundancy.

Q9. To recover the original source matrix $p$ from a transformed matrix $M$ using the inverse DCT, which sequence of operations must be performed?

A.Apply the T matrix once, then multiply by the transpose of T

B.Multiply by

T^{\top}

on the left and by

T

on the right ✅

C.Apply the inverse of T on both sides in the same order as the forward transform

D.Multiply by

T

on the left and by

T^{\top}

on the right

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The inverse DCT uses the same orthogonal matrix because $T^{-1}=T^{\top}$ . To undo the forward operation $M = T\,p\,T^{\top}$ , one premultiplies by $T^{\top}$ and post‑multiplies by $T$ , yielding $p = T^{\top} M T$ .

Q10. In the p matrix $(7,8,9,10)$ , which element(s) can be considered redundant based on their linear dependence, and why?

A.Only the first element, because it has the highest magnitude

B.The last three elements, because they can be expressed as linear combinations of the first ✅

C.None, all elements are independent

D.All four, because the DCT removes redundancy completely

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The DCT reveals that the three later values are highly correlated with the first; they can be approximated as scaled or shifted versions of the first element. This linear dependence means they carry little new information and are therefore redundant.

Q11. If the forward DCT uses the operation $M = T \, p \, T^{\top}$ , what would be the result of mistakenly applying the matrices as $M = T^{\top} \, p \, T$ instead?

A.The same transformed matrix, because

T

is symmetric ✅

B.A transposed version of the correct matrix

C.An entirely different matrix that destroys the frequency representation

D.A matrix scaled by a factor of

-1

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For the DCT matrix, $T = T^{\top}$ holds, making it symmetric and orthogonal. Swapping the order of $T$ and $T^{\top}$ therefore yields the identical product, so the mistaken ordering does not alter the resulting transformed matrix.

Q12. When the very small positive or negative values in the p matrix are ignored before transformation, what is the most plausible effect on the reconstructed data after applying the inverse DCT?

A.The reconstruction error becomes negligible because those values contribute little energy ✅

B.The error grows linearly with the number of ignored values

C.The reconstructed data becomes completely unrelated to the original

D.The DC component compensates for the loss, eliminating any error

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Neglecting tiny coefficients discards only a minimal amount of signal energy. Since the DCT concentrates most energy in the larger coefficients, the inverse transform can still approximate the original data closely, resulting in a very small reconstruction error that is often imperceptible.

🔗 Related Topics (MCQs)

📝 Data flow characteristics in QoS MCQs 📝 QoS definitions in computer networks MCQs 📝 Application sensitivity to QoS parameters MCQs 📝 Flow classes in quality of service MCQs 📝 Flow control techniques for QoS MCQs

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