Introduction
Fifteen GATE-style questions on Phase 1. Delay numericals carry most of the marks — write units on every line.
Practice questions
- MCQ · 1 mark
Process-to-process delivery is handled by:
- data link layer
- network layer
- transport layer
- session layer
- MSQ · 1 mark
Which statements about encapsulation are correct?
- The data link layer adds a header and a trailer.
- The transport PDU is called a segment.
- A router reads the transport header to forward.
- Network-layer addresses stay the same end to end.
- NAT · 1 mark
A 700-byte message gets headers of 20, 20 and 14 bytes and a 4-byte trailer. Fraction of the frame that is message data (3 decimals)?
Write your answer before opening the key.
- MCQ · 1 mark
Which device works only at the physical layer?
- Switch
- Repeater
- Router
- Bridge
- MSQ · 1 mark
Which are true of virtual-circuit switching?
- It has a setup phase.
- Bandwidth is reserved.
- Packets follow one path.
- Packets carry short labels.
- NAT · 2 marks
A 1.544 Mbps link carries 24 circuits. Time (s) to send 3,860,000 bits on one circuit after 0.5 s setup?
Write your answer before opening the key.
- NAT · 1 mark
Transmission delay (µs) of a 1500-byte packet on a 1 Gbps link?
Write your answer before opening the key.
- NAT · 2 marks
8 packets of 1000 bytes over 3 links of 2 Mbps, Tp = 3 ms per link, store-and-forward. Total in ms?
Write your answer before opening the key.
- NAT · 2 marks
4 packets of 2000 bits over links of 4 Mbps then 1 Mbps (Tp = 0). Arrival of the last packet in ms?
Write your answer before opening the key.
- MCQ · 1 mark
Propagation delay depends on:
- packet size
- bit rate
- distance and signal speed
- number of packets
- NAT · 2 marks
B = 1 Mbps, Tp = 250 ms, units 10,000 bits, send-and-wait. Utilisation (3 decimals)?
Write your answer before opening the key.
- NAT · 1 mark
Bandwidth-delay product in bits: B = 50 Mbps, Tp = 8 ms.
Write your answer before opening the key.
- MCQ · 2 marks
Which is the minimum unit size for 50 % send-and-wait utilisation on B = 8 Mbps, Tp = 5 ms?
- 10,000 bytes
- 5,000 bytes
- 80,000 bytes
- 40,000 bytes
- NAT · 2 marks
M = 6 × 10⁴ bits, packets of 1000 data + 20 header bits, 4 links of 1 Mbps, Tp = 0. Total in ms (3 decimals)?
Write your answer before opening the key.
- MCQ · 1 mark
With a fixed header per packet, the delay-minimising packet size on a multi-hop path is roughly:
- as large as possible
- as small as possible
- √(M·H/(h − 1))
- M / h
Answer key
Try the questions first. Open when you are ready to check.
- C — Transport = process to process.
- A: Hop to hop.
- B: Host to host.
- D: Dialogue control.
- ABD — Frames have trailers; segment = transport PDU; network addresses are end to end.
- C: Routers forward using the network header only.
- 0.923..0.924 — 700 / 758 = 0.9235.
- B — A repeater regenerates bits.
- A: Layer 2.
- C: Layer 3.
- D: Layer 2.
- ACD — Setup, fixed path, short labels, no bandwidth reservation.
- B: Only circuit switching reserves bandwidth.
- 60.5 — Circuit rate 64,333.3 bit/s → 60 s; + 0.5 s = 60.5 s.
- 12 — 12,000 / 10⁹ s = 12 µs.
- 49 — Tt = 4 ms; (8 + 2) × 4 + 9 = 49 ms.
- 8.5 — First: 0.5 + 2 = 2.5 ms; then 3 more × 2 ms (bottleneck) = 8.5 ms.
- C — Tp = d / v.
- A: That affects Tt.
- B: That affects Tt.
- D: Irrelevant to Tp.
- 0.019..0.020 — Tt = 10 ms; U = 10/510 = 0.0196.
- 400000 — 5 × 10⁷ × 0.008 = 4 × 10⁵.
- A — L ≥ 2·B·Tp = 80,000 bits = 10,000 bytes.
- B: Uses Tp instead of 2Tp.
- C: Left in bits but labelled bytes.
- D: Uses B·Tp in bits.
- 64.26 — (60 + 3) × 1.02 = 64.26 ms.
- C — Balance of pipelining against header overhead.
- A: Large packets lose pipelining.
- B: Tiny packets drown in headers.
- D: Not the optimum.