GATE CS 2021 Set 1 Q49 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To find the minimum sender window size $W$ for a link utilization of $50\%$, we first calculate the transmission delays and the total round-trip cycle time.

1. Transmission Delays:

• Transmission delay of the data frame ($T_t$):
  $$T_t = \frac{\text{Frame Size}}{\text{Bandwidth}} = \frac{2000 \text{ bits}}{10^6 \text{ bps}} = 2 \text{ ms}$$
• Transmission delay of the acknowledgement frame ($T_{ack}$):
  $$T_{ack} = \frac{\text{ACK Size}}{\text{Bandwidth}} = \frac{10 \text{ bits}}{10^6 \text{ bps}} = 0.01 \text{ ms}$$

2. Total Cycle Time ($T_c$):

The total time elapsed from the start of sending a frame to receiving its acknowledgement is: $$T_c = T_t + T_p + T_{ack} + T_p$$ Given the one-way propagation delay $T_p = 100 \text{ ms}$: $$T_c = 2 \text{ ms} + 100 \text{ ms} + 0.01 \text{ ms} + 100 \text{ ms} = 202.01 \text{ ms}$$

3. Window Size ($W$):

The link utilization (efficiency $\eta$) is given by: $$\eta = \frac{W \times T_t}{T_c}$$ Setting $\eta = 50\% = 0.5$: $$0.5 = \frac{W \times 2}{202.01} \implies W = \frac{101.005}{2} = 50.5025$$

Rounding to the nearest integer, we get $W = 51$ (which lies within the accepted range of $50$ to $52$).