Computer Networks · TEE · practice

Supporting HTTP timing · example 2

Unit 6 · HTTP, DNS, FTP and SMTP · optional practice

Optional extension · no matching calculation in the two supplied papers

Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.

Practice question. A client fetches 5 equal objects sequentially. Each object is 20000 bytes; link rate=2000000 bit/s and RTT=20 ms. Ignore DNS, TLS, loss, processing, headers, slow start and parallelism. Compare separate nonpersistent TCP connections with one new persistent connection without pipelining. [5 practice marks]

Answer

Given

n=5; RTT=20 ms; each payload=20000 B; rate=2000000 bit/s; ideal HTTP/1.x timing model.

To find

Total time under each stated connection policy.

Formula and why

T_{non}=n(2RTT+T_{object})LaTeX: T_{non}=n(2RTT+T_{object})
T_{persistent}=RTT+n(RTT+T_{object})LaTeX: T_{persistent}=RTT+n(RTT+T_{object})

Nonpersistent uses one setup RTT plus one request/first-response RTT per object. Persistent pays setup once, then one sequential request RTT per object.

Step-by-step solution

Step 1. Serialize one object

Tobject=20000×8/2000000 seconds = 80 ms.

Step 2. Separate connections

Tnon=5×(2×20+80)=600 ms.

Step 3. One persistent connection

Tpersistent=20+5×(20+80)=520 ms.

Step 4. Check the saving

600−520=80 ms = (5−1)×20; exactly the removed setup round trips.

Final answer: Nonpersistent=600 ms; persistent=520 ms.