Question 45

Given protocol parameters:

• Channel bandwidth: $B = 1\text{ Mbps} = 10^6\text{ bits/second}$
• Information frame size: $L_f = 1980\text{ bytes} = 1980 \times 8 = 15840\text{ bits}$
• Overhead bytes: 20 bytes $\implies$ Useful data size: $L_d = (1980 - 20) \times 8 = 15680\text{ bits}$
• ACK frame size: $L_{\text{ack}} = 20\text{ bytes} = 20 \times 8 = 160\text{ bits}$
• Propagation delay: $T_p = 0.75\text{ ms}$
• Processing time: $T_{\text{proc}} = 0.25\text{ ms}$

Calculate component delays:

• Data frame transmission time:
  $$T_{t,\text{data}} = \frac{15840}{10^6} \text{ s} = 15.84\text{ ms}$$
• ACK frame transmission time:
  $$T_{t,\text{ack}} = \frac{160}{10^6} \text{ s} = 0.16\text{ ms}$$
• Total cycle time for transmitting one frame:
  $$T_{\text{total}} = T_{t,\text{data}} + T_p + T_{\text{proc}} + T_{t,\text{ack}} + T_p$$
  $$T_{\text{total}} = 15.84 + 0.75 + 0.25 + 0.16 + 0.75 = 17.75\text{ ms}$$

Transmission efficiency $\eta$ can be defined based on useful data transmission time:

Alternatively, if based on the full information frame transmission time:

Both calculations fall comfortably within the official accepted range of $86.5\%$ to $89.5\%$.