# Q1(b) · Additional shared bottleneck

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2024-feb-ai-reexam/q1b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2024-feb-ai-reexam--q1b/

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

### Q1(b). Four server/client paths share R=1 Gbps; Rs=400 Mbps, Rc=300 Mbps. Host 1 and Host 2 additionally share link Re before their separate Rc links. Assuming fair sharing on R, find the minimum Re so links Rc or Re do not bottleneck these two connections. [5 marks]

### Answer

### Given

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

### To find

Minimum extra shared link Re.

### 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**.

### Step 3. Carry both downstream flows

Each of Hosts1 and2 requires 250 Mbps. Re must carry their sum: **Re≥500 Mbps**. Rc=300 Mbps is already at least 250 Mbps.