GATE CS 2026 Set 1 Q35 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To achieve $100\%$ link utilization in a sliding window protocol, the sender must be able to transmit frames continuously without waiting for acknowledgements. This requires the window size $W$ to be at least the total number of frames that can be transmitted during one round-trip cycle.

1. Calculate Transmission Delay ($T_t$):

$$T_t = \frac{\text{Frame Size}}{\text{Bandwidth}} = \frac{1000 \text{ bits}}{100 \times 10^3 \text{ bps}} = 0.01 \text{ seconds} = 10 \text{ ms}$$

2. Propagation Delay ($T_p$):

$$T_p = 100 \text{ ms}$$

3. Calculate Round-Trip Time (RTT):
Since the processing time and acknowledgement transmission time are zero, the total time to complete one cycle (from sending the first bit of a frame to receiving its acknowledgement) is: $$T_{\text{total}} = T_t + 2 T_p = 10 \text{ ms} + 2 \times 100 \text{ ms} = 210 \text{ ms}$$

4. Calculate Minimum Window Size ($W$):
For $100\%$ utilization, the window size $W$ must satisfy: $$W \ge \frac{T_{\text{total}}}{T_t} = \frac{210 \text{ ms}}{10 \text{ ms}} = 21$$

Thus, the minimum size of $W$ required is 21.