GATE CS 2017 Set 1 Q45 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To find the transmission efficiency of the Stop-and-Wait ARQ protocol, we first calculate the total time ($T_{\text{total}}$) required for one complete cycle.

1. Calculate individual time components:

• Transmission delay of the information frame ($T_{tx\_info}$):
  $$T_{tx\_info} = \frac{\text{Frame size}}{\text{Bandwidth}} = \frac{1980 \times 8 \text{ bits}}{10^6 \text{ bps}} = 15.84 \text{ ms}$$
• Propagation delay ($T_p$): $0.75 \text{ ms}$ (one-way, so it occurs twice: once for the frame and once for the ACK)
• Processing delay ($T_{proc}$): $0.25 \text{ ms}$
• Transmission delay of the ACK frame ($T_{tx\_ack}$):
  $$T_{tx\_ack} = \frac{20 \times 8 \text{ bits}}{10^6 \text{ bps}} = 0.16 \text{ ms}$$

2. Calculate the total cycle time ($T_{\text{total}}$):

3. Calculate efficiency ($\eta$):

Depending on whether overhead is excluded or included, we have two standard interpretations:

• Case A: Excluding overhead (Useful data only)
  Useful payload $= 1980 - 20 = 1960 \text{ bytes}$.
  Useful transmission time ($T_{\text{useful}}$) $= \frac{1960 \times 8 \text{ bits}}{10^6 \text{ bps}} = 15.68 \text{ ms}$.
  $$\eta = \frac{T_{\text{useful}}}{T_{\text{total}}} \times 100 = \frac{15.68}{17.75} \times 100 \approx 88.34\%$$
• Case B: Including overhead (Entire information frame)
  $$\eta = \frac{T_{tx\_info}}{T_{\text{total}}} \times 100 = \frac{15.84}{17.75} \times 100 \approx 89.24\%$$

Both interpretations are valid, and both values ($88.34\%$ and $89.24\%$) lie comfortably within the official GATE range of 86.5% to 89.5%.