Computer Networks Module 2
Switching, performance and the physical layer interview questions
Interview questions on switching, performance and the physical layer: warm-ups, then what each harder question tests and how a good answer starts.
Warm-up questions
Short questions with short answers, each answered in full.
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Fresher
What is the difference between a virtual circuit and a datagram network?
In datagram IP, each packet carries its destination address and routers use destination routes. IP forwarding needs no row for each conversation, although usable routes and link state are needed. A virtual circuit establishes logical-path state before data uses it. Switches look up an incoming key in the applicable local label space and select an outgoing port and label; the carried packet can retain its original addresses. A label path can aggregate flows. Reservation and ordering are separate service properties: the fixed serial model keeps order, while recovery needs restored state or a usable protection path. Both designs share packet links; circuit switching instead assigns a recurring fixed channel for the call.
From the lesson Circuit, packet and virtual-circuit switching
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Fresher
What is the bandwidth-delay product, and why does it matter?
The path rate times RTT is the modeled amount of unacknowledged data needed to sustain that rate. A window smaller than this budget can leave the path idle, with a window-only throughput ceiling of window/RTT. At 100 Mb/s and 50 ms, the budget is 5,000,000 bits, or 625,000 bytes. TCP's unscaled 65,535-byte advertised window allows about 10.5 Mb/s over that RTT; window scaling allows larger advertisements. Other limits still matter: congestion control, available data, endpoint work and competing traffic. The product sizes in-flight data; it does not require a router to maintain a standing queue of that size.
From the lesson Delay, throughput and bandwidth-delay product
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Fresher
What is the difference between baud rate and bit rate?
Baud counts symbols sent per second. Identical adjacent symbols need not create a signal transition. Bit rate counts bits per second; an M-valued symbol alphabet represents log₂M encoded bits per symbol before additional overhead. PAM4 represents two encoded bits per symbol, while 256-QAM represents eight. Manchester instead uses two half-bit signal elements per data bit, so the stated 10 Mb/s example has 20 million elements a second. Bandwidth in hertz is a third quantity. The 2B Nyquist signaling bound here belongs to an ideal real low-pass, zero-inter-symbol-interference model; noise and a practical code impose separate limits.
From the lesson Signals, line coding and channel capacity
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Fresher
What is the difference between multimode and single-mode fibre?
Multimode fibre supports several spatial modes, so differing paths spread a pulse. The cited 10GBASE-SR optics reach 300 m on OM3 and 400 m on OM4. Single-mode fibre supports one spatial mode at its operating wavelength; mode field diameter measures the optical field and is distinct from core diameter. The specified 10GBASE-LR and ER options reach 10 km and 40 km on compatible single-mode fibre. Loss and chromatic dispersion still constrain the link. Choose fibre and optics from their compatibility, reach and loss-budget specifications rather than assuming that any equal-rate optical parts work together.
From the lesson Transmission media: copper, fibre and radio
Questions with model answers
What each question tests, the first step of a good answer and the follow-ups to expect.
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Fresher to mid
A 50 MB backup crosses five store-and-forward hops of 1 Gb/s as 1,500-byte packets with 52 bytes of headers each. Estimate when its last byte arrives, compare sending it as one 50 MB message, and say what else you would check.
- What it tests
- Transmission time per hop, the store-and-forward pipeline, header overhead, and judging which delays matter at this size.
- First step
- Each packet carries 1,448 bytes of the file, so 50,000,000 / 1,448 rounds up to 34,531 packets. A full packet needs 12,000 bits / 1 Gb/s = 12 microseconds on each hop, and P packets over N hops take about (N + P - 1) packet times.
- Follow-ups
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- One of the five links is 100 Mb/s. Now what?
- Does 200 km of fibre on the path matter?
- Would 9,000-byte jumbo frames help?
From the lesson Circuit, packet and virtual-circuit switching
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
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Mid to senior
A file copy between Mumbai and Frankfurt reaches about 9 Mb/s on a 1 Gb/s path. RTT is 120 ms and no packet loss was observed. What could limit the copy, how would you distinguish the causes, and what evidence would justify a change?
- What it tests
- Window-limited throughput, the bandwidth-delay product, TCP window scaling and receive buffers, and telling a window limit from a loss limit.
- First step
- If the flow were window-limited, 9 Mb/s × 0.12 s implies about 135 KB in flight, against a 15 MB bandwidth-delay product. That is a hypothesis: inspect actual in-flight data, advertised and congestion windows, application pacing and endpoint work before treating the observation as a window diagnosis.
- Follow-ups
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- How large must the receive buffer be?
- What if 0.1 % of packets were lost?
- Would a 10 Gb/s link make the copy faster?
From the lesson Delay, throughput and bandwidth-delay product
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
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Mid to senior
An office's 20 Mb/s uplink runs at about 95 % utilisation all day. The throughput graphs look healthy, but video calls stutter. Explain why with numbers, and say what you would change.
- What it tests
- Queueing delay against traffic intensity, standing queues and bufferbloat, and the remedies: active queue management, prioritisation, shaping and capacity.
- First step
- A 1,500-byte packet takes 0.6 ms at 20 Mb/s. In the ideal Poisson-arrival, fixed-size M/D/1 model, 95% load gives an average wait of 0.95 × 0.6 ms / (2 × 0.05) = 5.7 ms. Actual arrival patterns and standing queues can differ; measure latency idle and under load.
- Follow-ups
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- Why does the link look fine on a throughput graph?
- What does active queue management change?
- Would doubling the uplink fix it?
From the lesson Delay, throughput and bandwidth-delay product
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
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Fresher to mid
A laptop shows an 80 MHz Wi-Fi channel at a 30 dB signal-to-noise ratio, yet a speed test reaches 400 Mb/s. Is that reasonable? Use Shannon's limit to explain the gap and say what would raise the rate.
- What it tests
- Applying Shannon's formula with the decibel conversion, knowing why real links run below it, and naming the levers: channel width, spatial streams, signal quality and contention.
- First step
- Convert 30 dB to S/N = 1,000. The ideal Gaussian-channel bound is 80 MHz × log₂1,001, about 797 Mb/s for the stated single-stream model. A 400 Mb/s speed test is below that bound; identify the actual stream, coding, airtime and path constraints rather than assuming that a fixed half has been lost.
- Follow-ups
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- The laptop uses two spatial streams. Does that change the answer?
- Why not simply use a 320 MHz channel?
- The ratio drops to 20 dB in the next room. What happens?
From the lesson Signals, line coding and channel capacity
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
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Mid to senior
1000BASE-X uses 8B/10B coding, while 10GBASE-R uses 64B/66B. Why the change in these binary serial-line interfaces, and what does 64B/66B give up to get its lower overhead?
- What it tests
- Line-code overhead and symbol rate, how block codes guarantee transitions and control codes, and the trade-offs of scrambling.
- First step
- For the compared binary serial-line model at 10 Gb/s of data, 8B/10B needs 12.5 Gbaud and 64B/66B needs 10.3125 Gbaud. The former rate is about 21% higher. Compare the required signaling rates and component demands; actual cost and reach also depend on the PHY and channel.
- Follow-ups
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- How does 64B/66B keep the receiver's clock in step?
- What does its 2-bit header carry?
- Why do faster lanes move to PAM4?
From the lesson Signals, line coding and channel capacity
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
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Fresher to mid
A user says file copies are slow. Their desk port has linked at 100 Mb/s instead of 1 Gb/s, and the port's CRC error counter rises during copies. How do you find the cause, step by step?
- What it tests
- Reading link state and counters, telling cabling faults from duplex and configuration faults, and proving a fix with measurements.
- First step
- Check both ends first: the negotiated speed and duplex on the switch port and on the computer (ethtool shows them on Linux), and whether either end is forced. Then read the counters twice, a minute apart, and note which grow: CRC errors, late collisions, link changes.
- Follow-ups
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- Both ends report 100 Mb/s full duplex. What next?
- The switch port is fixed at 100 Mb/s full duplex and the computer auto-negotiates. What happens?
- How do you prove the fix worked?
From the lesson Transmission media: copper, fibre and radio
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
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Fresher to mid
Two buildings 2 km apart need a 10 Gb/s link. Which medium and optics would you choose, how would you check the link budget, and what could make the link fail months later?
- What it tests
- Matching the medium and optics to the distance, working a loss budget with margin, and knowing how fibre runs fail.
- First step
- The named 10GBASE-T copper channel reaches up to 100 m, and the cited 10GBASE-SR optics reach 300 m on OM3 or 400 m on OM4. Compare compatible single-mode fibre and 10GBASE-LR optics, specified for 10 km at 1310 nm, then calculate the loss budget using the actual parts.
- Follow-ups
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- Why not use 10GBASE-ER optics for extra margin?
- The run turns out to be 12 km. What changes?
- Could a point-to-point radio link replace the fibre?
From the lesson Transmission media: copper, fibre and radio
Model answer, Pro.In the answer:
- Model answer
- Follow-up answers
- Common mistakes
- Rubric
- Variants
- about 3 minutes
All Computer Networks interview questions · The Computer Networks track