Computer Networks · TEE · papers
Q1(b) · Shared bottleneck throughput
2023-11-22 · AI-only Sem V variant; incomplete recovered scan
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
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.
Total delivered rate=4×250=1000 Mbps.