📝 Chord finger table explained (10 MCQs)

📖 From Data Communication and Networks • 29. Peer to Peer Paradigm • 10 questions available

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What is Chord finger table explained?

A Chord finger table is a routing index maintained by each node that stores references to other nodes at exponentially increasing distances around the ring, allowing messages to be forwarded quickly toward their destination in O(log N) hops.

Easy

Medium

Hard

📝 All Chord finger table explained MCQs

Q1. What is the typical bit depth for a standard color image as mentioned in the example?

A.8

B.16

C.24 ✅

D.32

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The example specifies that a standard color image uses 24 bits per pixel, allocating 8 bits for each of the three primary colors (red, green, blue). This 24‑bit depth is common for true‑color images, allowing 16.7 million possible colors.

Q2. If the transmission rate is increased from 100 kbps to 200 kbps, what is the new transmission time for the black‑and‑white image (1280 × 720, 1 bit depth)?

A.4.5 seconds ✅

B.9 seconds

C.12 seconds

D.18 seconds

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The original time is calculated as $\frac{1280\times720\times1}{100\,000}\approx9$  seconds. Doubling the rate halves the time, giving $\frac{9}{2}=4.5$  seconds. Therefore the black‑and‑white image would be transmitted in roughly 4.5 seconds.

Q3. Which statement correctly compares the transmission times for gray (8 bit) and color (24 bit) images?

A.Gray takes longer than color

B.Gray takes one‑third of the time of color ✅

C.Gray takes the same time as color

D.Gray takes half the time of color

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A gray image uses one‑third the bits per pixel of a 24‑bit color image (8 vs 24). Since transmission time is proportional to total bits, the gray image requires roughly one‑third the time of the color image.

Q4. Suppose the image resolution is doubled to 2560 × 1440 while keeping the bit depth at 24 bits and transmission rate at 100 kbps. Approximately how many times longer will the transmission time be compared to the original color image?

A.2×

B.4× ✅

C.8×

D.16×

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Doubling each dimension quadruples the pixel count: $(2560\times1440)/(1280\times720)=4$ . Transmission time scales linearly with pixel count, so the time becomes about four times the original 215 seconds, i.e., roughly 860 seconds.

Q5. If a lossless compression algorithm reduces the size of the 24‑bit color image by 40%, what is the new approximate transmission time?

A.129 seconds ✅

B.86 seconds

C.215 seconds

D.1290 seconds

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A 40 % reduction leaves 60 % of the original size. Multiplying the original 215 seconds by 0.60 yields $215\times0.6\approx129$  seconds, which is the new transmission time after compression.

Q6. An application requires that the transmission time for a 1280 × 720, 8‑bit gray image be under 30 seconds. What minimum transmission rate (in kbps) must be provided?

A.100 kbps

B.200 kbps

C.250 kbps ✅

D.300 kbps

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The gray image contains $1280\times720\times8=7,372,800$  bits (≈7,372.8 kb). To finish in 30 seconds, the rate must be at least $7,372.8/30\approx245.8$  kbps, so a 250 kbps link satisfies the requirement.

Q7. If both the bit depth is reduced from 24 bits to 8 bits and the transmission rate is increased from 100 kbps to 150 kbps, what is the factor by which the transmission time for the image changes compared to the original color scenario?

A.Reduced to 1/4.5 of original ✅

B.Reduced to 1/2 of original

C.Increased by 1.5 times

D.Unchanged

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Reducing bits from 24 to 8 cuts size to one‑third. Raising the rate by 1.5 (150/100) further speeds transmission. The combined factor is $(1/3)/1.5 = 1/4.5$ , meaning the new time is about 22 % of the original.

Q8. Consider three scenarios for the same 1280 × 720 image: (i) black‑and‑white, (ii) gray with 40 % compression, (iii) color with 30 % compression. If the transmission rate is 100 kbps, rank the scenarios from shortest to longest transmission time.

A.i < ii < iii ✅

B.ii < i < iii

C.iii < ii < i

D.i < iii < ii

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Compute times: black‑and‑white ≈9 s; gray compressed = $74\times0.6≈44.4$  s; color compressed = $215\times0.7≈150.5$  s. Ordering from smallest to largest gives i < ii < iii.

Q9. A streaming service wants to support simultaneous transmission of 10 color videos (each 1280 × 720, 24‑bit) without buffering. Using the original transmission time of 215 seconds per video, what minimum aggregate bandwidth (in Mbps) is required to deliver all streams continuously?

A.0.5 Mbps

B.1 Mbps ✅

C.1.5 Mbps

D.2 Mbps

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Each video contains $1280\times720\times24=22,118,400$  bits. Dividing by 215 s gives ≈0.103 Mbps per stream. Ten streams need $10\times0.103≈1.03$  Mbps, so an aggregate bandwidth of about 1 Mbps is required.

Q10. If an image adds an alpha channel (8 bits) to the original 24‑bit color image, how does the transmission time change when using the same 100 kbps link?

A.Increase by 33 % ✅

B.Decrease by 25 %

C.No change

D.Increase by 50 %

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Adding an 8‑bit alpha raises bits per pixel from 24 to 32, a factor of $32/24=4/3$ ≈ 1.33. Thus the transmission time grows by roughly 33 %, from 215 seconds to about 286 seconds.

🔗 Related Topics (MCQs)

📝 Peer to peer architecture introduction MCQs 📝 P2P networks in computer networks MCQs 📝 Distributed Hash Table DHT in P2P MCQs 📝 Chord protocol in peer to peer networks MCQs 📝 Chord identifier space and hashing MCQs

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