network delay throughput and end to end performance short notes

Topic Revision Summary

Topic Revision Summary Store-and-Forward: Router must receive entire packet before transmitting. Delay per hop = \frac{L}{R} . Bottleneck Bandwidth: R_{b o ttl e n ec k} = min (R_{1}, R_{2}, \dots, R_{N}) . Determines max throughput. Single Packet Time: T = \sum_{i = 1}^{N} \frac{L}{R _{i}} (ignoring prop/proc delay). Multi-Packet Time (Pipelining): T_{t o t a l} = First Packet i = 1 \sum N \frac{L}{R _{i}} + Remaining Packets (P - 1) \frac{L}{R _{b o ttl e n ec k}} Unit Trap: Ensure consistency. 1 Byte = 8 bits . 1 Mbps = 1 0^{6} bps . Key Insight: The first packet clears the path; subsequent packets fill the bottleneck.

One-Page Revision — Transmission and Propagation Delay

Revision Sheet Two Formulas d_{trans} = \frac{L}{R} d_{prop} = \frac{d}{s} Total Delay d_{total} = d_{trans} + d_{prop} Critical size: L_{c} = \frac{R \cdot d}{s} Unit Checklist: bytes \to bits (\times 8), Mbps \to bps (\times 1 0^{6}), km \to m (\times 1 0^{3}). Wording Decoder: "Completely receive" = sum. "First bit" = propagation only. "Equal delays" = solve for L . Dominance Rule: Long link + small packet \to propagation wins. Short link + large packet \to transmission wins.

Network Delay, Throughput and End-to-End Performance Short Notes for GATE CS

Network Delay, Throughput and End-to-End Performance short notes for GATE CS: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Topic Revision Summary

Store-and-Forward: Router must receive entire packet before transmitting. Delay per hop = $\frac{L}{R}$ .
Bottleneck Bandwidth: $R_{b o ttl e n ec k} = min (R_{1}, R_{2}, \dots, R_{N})$ . Determines max throughput.
Single Packet Time: $T = \sum_{i = 1}^{N} \frac{L}{R _{i}}$ (ignoring prop/proc delay).
Multi-Packet Time (Pipelining): $T_{t o t a l} = First Packet i = 1 \sum N \frac{L}{R _{i}} + Remaining Packets (P - 1) \frac{L}{R _{b o ttl e n ec k}}$
Unit Trap: Ensure consistency. $1 Byte = 8 bits$ . $1 Mbps = 1 0^{6} bps$ .
Key Insight: The first packet clears the path; subsequent packets fill the bottleneck.

One-Page Revision — Transmission and Propagation Delay

Revision Sheet

Two Formulas

d_{trans} = \frac{L}{R}

d_{prop} = \frac{d}{s}

Total Delay

d_{total} = d_{trans} + d_{prop}

Critical size:

L_{c} = \frac{R \cdot d}{s}

Unit Checklist: bytes → bits (

\times 8

), Mbps → bps (

\times 1 0^{6}

), km → m (

\times 1 0^{3}

Wording Decoder: "Completely receive" = sum. "First bit" = propagation only. "Equal delays" = solve for

L

Dominance Rule: Long link + small packet → propagation wins. Short link + large packet → transmission wins.