Computer Networks · TEE · papers
Q1(a) · Shared bottleneck throughput
2023-01-27 · AI-only Sem V
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
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
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.