GATE 2025 CS2 Q.36: solution and answer

GATE 2025 · CS2Q.36 · 2 marks · MCQ

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?

(A)
88.23
(B)
93.75
(C)
85.44
(D)
66.67

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Answer and solution

ANSWER

The channel utilization is about 88.23%, option (A).

Official answer: ASource: Official answer key, GATE 2025 CS2 (IIT Roorkee)

Transmission time of one frame: $T_{t} = \frac{3000 bits}{2000 bps} = 1.5 s$ .
Propagation delay: $T_{p} = 100 ms = 0.1 s$ . The acknowledgement is negligible in size, so its transmission time is $0$ , and processing times are negligible.
In Stop-and-Wait the sender sends one frame, then waits for the acknowledgement. One cycle takes $T_{t} + 2 T_{p} = 1.5 + 0.2 = 1.7 s$ .
Utilization $U = \frac{T _{t}}{T _{t} + 2 T _{p}} = \frac{1.5}{1.7} = 0.88235 \dots$
As a percentage this is about $88.2353%$ . The closest option is 88.23, which is option (A).

KEY IDEA

In Stop-and-Wait the channel is busy only for the frame's transmission time $T_{t}$ in each cycle of $T_{t} + 2 T_{p}$ (with a negligible acknowledgement), so $U = T_{t} / (T_{t} + 2 T_{p})$ .

WATCH OUT FOR

Using $T_{t} + T_{p}$ as the cycle time, or forgetting that the acknowledgement must travel back, which gives $1.5/1.6 = 93.75%$ (option B). Another slip is using the window-size formula $1/ (1 + 2 a)$ with $a$ computed from mismatched units.

From the GATE 2025 CS2 paper. Every question of this paper · Download the full paper · All 7 Performance questions, with notes

GATE 2025 CS2 Q.36: solution and answer

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