# Q1(b) · Shared bottleneck throughput

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-ai-partial/q1b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-ai-partial--q1b/

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

### Q1(b). Four server/client paths share a link R = 1 Gbps. Each server link Rs = 400 Mbps and client link Rc = 300 Mbps. Assuming fair sharing and maximum sending rates, find the maximum throughput of each connection. [5 marks]

### Answer

### Given

4 flows, server links 400 Mbps, client links 300 Mbps, shared link 1000 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

1000 Mbps ÷ 4 = 250 Mbps per flow.

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

T=min(400,300,250)=**250 Mbps**.

Total delivered rate=4×250=1000 Mbps.