# Window utilization · example 1

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/practice/arq/efficiency-1

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/practice/arq--efficiency-1/

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=1 ms, one-way Tp=5 ms, ACK transmission Ta=0 ms. Find Stop-and-Wait utilization, utilization with window 4, and the smallest ideal full-utilization window. [5 practice marks]

#### Answer

## Given

Tt=1 ms; Tp=5 ms; Ta=0 ms; W=4; 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=1+2×5+0=11 ms.

#### Step 2. Stop-and-wait share

U=1/11=0.090909=9.09%.

#### Step 3. Pipelined share

U=min(1,4×1/11)=0.363636=36.36%.

#### Step 4. Round the full-utilization bound upward

Wrequired=ceil(11/1)=11; check the sequence-space bound separately.

**Final answer:** Stop-and-Wait=9.09%; window-4=36.36%; ideal required window=11.

#### What this result means

A percentage above 100 is not valid. The min operation caps link occupancy, and protocol numbering may require more sequence bits.