# Q1(a) · Shared bottleneck throughput

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-jan-ai-reexam/q1a

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-jan-ai-reexam--q1a/

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

### Q1(a). Ten server/client paths share middle link R=200 Mbps. Each server link Rs=25 Mbps and each client link Rc=50 Mbps. Assuming fair sharing and maximum server sending rates, find each pair’s maximum end-to-end throughput in Mbps. [5 marks]

### Answer

### Given

10 flows, server links 25 Mbps, client links 50 Mbps, shared link 200 Mbps.

### To find

Maximum throughput of each flow.

### Formula and why



$$
T=\min(R_s,R_c,R/n)
$$



A flow cannot exceed any link on its path; fair sharing divides the common capacity among n active flows.

### Solution

### Step 1. Convert and divide the shared capacity

200 Mbps ÷ 10 = 20 Mbps per flow.

### Step 2. Take the smallest per-flow capacity

T=min(25,50,20)=**20 Mbps**.

Total delivered rate=10×20=200 Mbps.