GATE CS 2025 Set 2 Q.36: Data Link Layer: Framing, Error Detection and Sliding Window

Suppose we are transmitting frames between two nodes using Stop-and-Wait protocol. The frame size is 3000 bits. The transmission rate of the channel is 2000 bps (bits/second) and the propagation delay between the two nodes is 100 milliseconds. Assume that the processing times at the source and destination are negligible. Also, assume that the size of the acknowledgement packet is negligible. Which ONE of the following most accurately gives the channel utilization for the above scenario in percentage?

88.23

93.75

85.44

66.67

Official answer and solution

Official answer: A
From: Official GATE 2025 answer key

Solution

Compute frame transmission time $T_t = L / R$ and substitute into the Stop-and-Wait utilization formula $\eta = \frac{T_t}{T_t + 2T_p}$ .

Frame transmission time ( $T_t$ ): $T_t = \frac{\text{Frame size}}{\text{Transmission rate}} = \frac{3000\text{ bits}}{2000\text{ bps}} = 1.5\text{ s} = 1500\text{ ms}$
One-way propagation delay ( $T_p$ ): $T_p = 100\text{ ms} = 0.1\text{ s}$ .
Channel utilization: $\eta = \frac{T_t}{T_t + 2T_p} \times 100\% = \frac{1500}{1500 + 2(100)} \times 100\% = \frac{1500}{1700} \times 100\% \approx 88.235\%$

Final answer

Option (A)

Common trap

Using $T_p$ instead of $2T_p$ in the denominator (which gives $1500/1600 = 93.75\%$ , Option B).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 2 (CS2)
Question: 36 of 65
Section: Computer Science and Information Technology

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