Question 36

In a CSMA/CD Ethernet network, to ensure collision detection before transmission finishes, the packet transmission time $T_{tx}$ must satisfy:

where $T_p$ is the one-way propagation delay.

1. Calculate the transmission time $T_{tx}$:

   • Frame size $L = 10{,}000\text{ bits}$
   • Bandwidth $B = 500\text{ Mbps} = 500 \times 10^6\text{ bits/second}$
   $$T_{tx} = \dfrac{L}{B} = \dfrac{10{,}000}{500 \times 10^6} = \dfrac{10^4}{5 \times 10^8} = 2 \times 10^{-5}\text{ s} = 20\ \mu\text{s}$$

2. Propagation delay for cable length $d$ and signal speed $v = 200{,}000\text{ km/s}$:

   $$T_p = \dfrac{d}{v}$$

3. Setting $2 \cdot T_p \le T_{tx}$:

   $$2 \cdot \dfrac{d}{v} \le T_{tx} \implies d \le \dfrac{v \cdot T_{tx}}{2}$$
   $$d \le \dfrac{200{,}000\text{ km/s} \times 2 \times 10^{-5}\text{ s}}{2} = \dfrac{4\text{ km}}{2} = 2\text{ km}$$

Thus, the maximum length of the cable is 2 km.