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Q1(b) · Additional shared bottleneck

2024-02-06 · AI-only Sem V variant

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)LaTeX: 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.