GATE 2023 CS Q.17: solution and answer

Official answer: BSource: Official answer key, GATE 2023 CS (IIT Kanpur)

1. Stop-and-Wait sends one frame and then waits for its acknowledgement, so the link carries data only during the transmission time $T_t$ in each cycle of length $T_t+2T_p$.

2. Utilization $U=\frac{T_t}{T_t+2T_p}=\frac{1}{1+2a}$, where $a=\frac{T_p}{T_t}$.

3. Here $T_t=\frac{L}{B}$ (frame size over transmission rate) and $T_p=\frac{d}{v}$ (link length over propagation speed), so $a=\frac{dB}{vL}$.

4. $U$ falls as $a$ rises. $a$ grows with a longer link ($d$ larger) and with a higher transmission rate ($B$ larger), since the frame size $L$ and speed $v$ do not change between the options.

5. So $U$ is lowest when the link is longer and the transmission rate is higher, which is option (B).

6. Check the other options. In (A) and (D) one factor raises $a$ and the other lowers it, so $a$ is in between. In (C) both factors lower $a$, so (C) has the highest utilization.