What this topic covers
13 subtopics- Delays10
- Performance7
- Devices4
- Encapsulation4
- OSI model4
- Physical layer4
- Switching3
- Addressing2
- Architecture2
- TCP/IP model2
- Network types1
- Topology1
Show all 13 subtopics
- Transmission modes1
- Beginner 15
- Intermediate 15
- Advanced 15
About this topic
Every networks course starts with the layered models – OSI's seven layers and the four-layer TCP/IP stack – and the delays a packet suffers on its way: transmission (pushing bits onto the link), propagation (travelling down it), queueing and processing. Interviews ask which device works at which layer; GATE asks you to compute delays and capacities. Every numerical answer here was computed. 1 Mbps = 10⁶ bits/s and 1 KB = 1000 bytes.
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Sample questions
Try each one first, then open the answer for the explanation and why the other options are wrong.
Which is the correct order of the OSI layers from bottom (1) to top (7)?
Show answer and explanation
Explanation
A common memory aid from the bottom: "Please Do Not Throw Sausage Pizza Away".
Why the other options are wrong
- ANetwork is below Transport; Session below Presentation.
- CData link sits directly above Physical.
- DPhysical is layer 1.
A 1500-byte packet crosses 3 links (two routers in between), each 10 Mbps with 2 ms propagation. Routers store and forward; ignore processing and queueing. What is the end-to-end delay in ms? (one decimal)
The answer is a number – work it out, then open the solution.
Show answer and explanation
Explanation
Each link adds its transmission (12,000 bits / 10⁷ = 1.2 ms) and propagation (2 ms): 3 × 3.2 = 9.6 ms. A store-and-forward router must receive the whole packet before sending it on.
P = 20 packets of 1000 bytes cross N = 4 store-and-forward links of 2 Mbps each; ignore propagation and queueing. How many ms until the last packet arrives?
The answer is a number – work it out, then open the solution.
Show answer and explanation
Explanation
One transmission takes 8000 / 2 × 10⁶ = 4 ms. The first packet needs N × 4 = 16 ms; the remaining P − 1 = 19 arrive every 4 ms: (N + P − 1) × 4 = 23 × 4 = 92 ms.