Computer Networks · TEE · practice

Window utilization · example 3

Unit 3 · Flow control and ARQ · 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. On an ideal error-free link, Tt=4 ms, one-way Tp=8 ms, ACK transmission Ta=1 ms. Find Stop-and-Wait utilization, utilization with window 6, and the smallest ideal full-utilization window. [5 practice marks]

Answer

Given

Tt=4 ms; Tp=8 ms; Ta=1 ms; W=6; no processing/queueing delay.

To find

Two utilizations and the required window.

Formula and why

C=T_t+2T_p+T_a,\quad U_1=T_t/C,\quad U_W=\min(1,WT_t/C)

Both propagation directions matter. All durations already share the same unit.

Step-by-step solution

Step 1. Compute a complete cycle

C=4+2×8+1=21 ms.

Step 2. Stop-and-wait share

U=4/21=0.190476=19.05%.

Step 3. Pipelined share

U=min(1,6×4/21)=1.000000=100.00%.

Step 4. Round the full-utilization bound upward

Wrequired=ceil(21/4)=6; check the sequence-space bound separately.

Final answer: Stop-and-Wait=19.05%; window-6=100.00%; ideal required window=6.