End to End Throughput and Bottleneck Links

Tags: #◇CSNETWK #◇Computer Science #◇Computer Networks #◇Networking Protocols #◇Graph Theory Basics

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◇Overview

Definition: Throughput

◇Key Concepts

Topology Server Links ( Client Links ( Shared Middle Link ( Fair Sharing

◇Details / Theory

◇1. Concept Overview: Throughput & Bottlenecks

Definition: Throughput

Throughput is the actual rate at which data can be transferred from a sender to a receiver. It is determined by the "slowest" link in the path.

Definition: The Bottleneck Link

In any chain of links, the Bottleneck Link is the link with the minimum transmission rate. The entire path cannot be faster than this single link.

The Formula (Fair Share)

If $N$ connections share a common link with rate $R$, each connection gets a "fair share" of $R/N$.

$\text{Throughput} = \min(R_s, R_c, \frac{R}{N})$
$R_s$: Rate of Server Link $R_c$: Rate of Client Link $R$: Rate of Shared Link $N$: Number of concurrent flows

◇2. The Network Topology

Scenario Data
!◇Pasted image 20251121094535.png
Topology: 4 Servers sending to 4 Clients (4 pairs). Server Links ($R_s$): 10 Mbps each. Client Links ($R_c$): 90 Mbps each. Shared Middle Link ($R$): 400 Mbps total. Fair Sharing: The middle link divides its capacity equally among the 4 pairs.

◇3. Problem Analysis

Question 1

What is the maximum achievable end-end throughput (in Mbps) for each of four client-to-server pairs?

Analytical Breakdown:

1. Identify Constraints: We need to find the minimum speed along the path for one single pair.

2. Link 1 (Server Side): $R_s = 10$ Mbps.

3. Link 2 (Shared Middle): The total is 400 Mbps, but it is shared by 4 pairs.
   Per-pair capacity = $400 / 4 = 100$ Mbps.

4. Link 3 (Client Side): $R_c = 90$ Mbps.

5. Calculate Minimum:
   $\min(10, 100, 90)$

$10 < 90 < 100$

6. Conclusion: The server link is the slowest part of the chain.

Correct Answer: 10 Mbps

Question 2

Which link is the bottleneck link?

Analytical Breakdown:

1. Compare Capacities:
   $R_s = 10$ Mbps
   Shared Share ($R/4$) = 100 Mbps
   $R_c = 90$ Mbps

2. Identify Lowest: The value 10 is the lowest.

3. Map to Link: This value corresponds to $R_s$ (Server Link).

Correct Answer: Rs

Question 3

Assuming that the servers are sending at the maximum rate possible, what are the link utilizations for the server links (Rs)?

Analytical Breakdown:

1. Definition: $\text{Utilization} = \frac{\text{Actual Traffic Flow}}{\text{Link Capacity}}$
2. Actual Flow: The maximum throughput we calculated in Q1 is 10 Mbps.
3. Link Capacity: The capacity of $R_s$ is 10 Mbps.
4. Calculation: $10 / 10 = 1.0$ (or 100%).

Correct Answer: 1

Question 4

Assuming that the servers are sending at the maximum rate possible, what are the link utilizations for the client links (Rc)?

Analytical Breakdown:

1. Actual Flow: The traffic arriving at the client is limited by the bottleneck, so it is still arriving at 10 Mbps.
2. Link Capacity: The capacity of $R_c$ is 90 Mbps.
3. Calculation: $10 / 90 \approx 0.111...$

Correct Answer: 0.11

Question 5

Assuming that the servers are sending at the maximum rate possible, what is the link utilizations for the shared link (R)?

Analytical Breakdown:

1. Actual Flow: There are 4 pairs, each sending 10 Mbps.
   Total Flow = $4 \times 10 = 40$ Mbps.

2. Link Capacity: The total capacity of the shared link $R$ is 400 Mbps.

3. Calculation: $40 / 400 = 0.1$.

4. Alternative Method: Calculate per-pair utilization. Share = 10 Mbps. Capacity Share = 100 Mbps. $10/100 = 0.1$.

Correct Answer: 0.1

◇Summary of Logic (Cheat Sheet)

Critical Check:
If $R_s$ is small (e.g., dial-up), the server is the bottleneck. If $R$ is small (e.g., congested backbone), the core is the bottleneck. If $R_c$ is small (e.g., weak Wi-Fi), the client is the bottleneck.

◇Common Pitfalls

Forgetting that end-to-end throughput is determined by the bottleneck (slowest) link, not the fastest. Not dividing the shared backbone link $R$ by the number of active connections $N$ when multiple flows compete. Confusing link capacity with actual per-flow throughput — $\min(R_s, R_c, R/N)$ is the correct formula.

◇Summary

TL;DR — End-to-end throughput equals $\min(R_s, R_c, R/N)$ where $R_s$ is server link rate, $R_c$ is client link rate, and $R/N$ is the shared backbone divided among $N$ flows. The bottleneck link limits overall performance.

◇Links

Related: - ◇Pasted image 20251121094535.png Source / Reference:

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