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Analyzing Network Costs and Latencies: A Comprehensive Overview

This document provides a detailed analysis of network costs, performance metrics, and latency measurements across various configurations. It explores different bandwidth scenarios, evaluating their impact on data transmission times and costs. The content outlines the relationship between various data rates such as 100Kbps and 1Mbps and the associated latencies, offering insights into optimizing network performance. Additionally, it discusses next steps for improving network efficiency and achieving cost-effective solutions in data transmission.

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Analyzing Network Costs and Latencies: A Comprehensive Overview

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  1. Dst Dst Dst Dst Cost Cost Cost Cost Next Next Next Next B A A A 1 2 B A B B B C 2 2 C B B A D C C D 2 3 D E C E E D E 2 1 E F E F F F 2 1 F E E D F

  2. 5 1 2 B A 2 C 1 2 E G 1 1 6 D 3 F Step Set S D(A), p(A) D(B), p(B) D(D), p(D) D(E), p(E) D(F), p(F) D(G), F(G) 0 C 5, C 2, C ∞ ∞ 6, C 1, G 1 CG 3, G 2 BCG 3 ABCG 5, A 4 ABCDG 6, D 5 ABCDEG 6 ABCDEFG

  3. 5 1 2 B A 2 C 1 2 E G 1 1 6 D 3 F Step Set S D(A), p(A) D(B), p(B) D(D), p(D) D(E), p(E) D(F), p(F) D(G), F(G) 0 C 5, C 2, C ∞ ∞ 6, C 1, G 1 2 3 4 5 6

  4. Dst Dst Cost Cost Next Next A B 2 1 A B B C 2 2 B C 1 2 B D D 3 2 E D A E E 2 1 F E 2 C F F 1 2 E F 2 E 1 2 1 D 1 Dst Cost Next F A 4 C/F Dst Cost Next B 4 C/F A 3 E C 2 C/F B 3 C E 2 F C 1 D F 1 F D 1 D E 1 E

  5. A 100Kbps, 2ms 1Mbps C D 200Kbps, 1ms B I (ms) I (ms) 5 5 4 4 T -1.2 3 3 2 2 1 1 0 0 T (ms) 5 5 1 2 3 1 4 2 3.2 4 -1 -1 -1.2 -2 -2 T (ms)

  6. a (ms) 9 8 7 6 5 4 3 2 1 0 1 2 3 4 5 6 7 8 9 t (ms) -1 -2 -3 time = 0 600 bits 100Kbps, 2ms A 500Kbps C D time = T 1000 bits 0 200Kbps, 1ms T B 2 t+1 6 T+5 t+6 t+6+d 8 t+8 t+8+d t+9.2 t+9.2+d 0 0 2 2 t t t+1 t+1 6 6 8 t+5 8 t+5 8+d 8+d 9.2+d t+6 9.2+d t+6 11.2+d t+9.2+d

  7. 600 bits time = 0 100Kbps, 2ms A 500Kbps D C time = T 1000 bits 200Kbps, 1ms B (a) T+6 <= 8 (b) 8 < T+6 <= 9.2 (c) 9.2 < T+6 0 0 0 T 2 T+1 2 2 T T T+1 T+1 6 6 6 T+5 T+6 T+5 8 8 T+6+d 8 8+d T+5 8+d T+8 T+8+d T+6 9.2+d 9.2+d T+9.2 T+9.2+d T+6 9.2 11.2+d T+8+d I = T+8+d – (T+9.2+d) = -1.2ms I = 11.2+d – (9.2+d) = 2ms I = T+8+d – (9.2+d) = T-1.2 ms

  8. 600 bits time = 0 100Kbps, 2ms A 500Kbps D C time = T 1000 bits 200Kbps, 1ms B (b) (c) (a) T+6 <= 8 A B C A B C A B C 0 T 2 T+1 6 T+5 T+6 8 time

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