# Supporting numerical: stop-and-wait utilization

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/practice/arq/efficiency

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/practice/arq--efficiency/

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. Derive ideal Stop-and-Wait utilization and the window needed to keep a link busy. [5 practice marks]

#### Answer

Assume an error-free link, constant data-frame transmission time Tt, one-way propagation Tp, ACK transmission Ta, negligible processing and no queueing. One stop-and-wait cycle takes Tt+2Tp+Ta. Useful data occupies the forward link for Tt of that cycle.



$$
U_{SW}=\frac{T_t}{T_t+2T_p+T_a}
$$





$$
U_W\approx\min\left(1,\frac{WT_t}{T_t+2T_p+T_a}\right)
$$





$$
W_{\mathrm{required}}=\left\lceil\frac{T_t+2T_p+T_a}{T_t}\right\rceil
$$



These are ideal teaching bounds for equal frames and a sufficiently fast independent ACK path. Loss, ACK scheduling, headers and a limited sequence space reduce actual performance. A calculated window must also satisfy the selected ARQ protocol’s sequence-number bound.

#### Practical example

If sending a frame takes 1 ms but the round trip adds 10 ms, Stop-and-Wait occupies only 1/11 of the forward link. A window of at least 11 can hide that waiting time under the stated ideal model.