In a Time Division Multiplexing (TDM) medium access control bus LAN, each station is assigned one time slot per cycle for transmission. The length of each time slot is defined as the time to transmit 100 bits plus the end-to-end propagation delay.
We are given:
- Propagation speed = $2 \times 10^8$ m/s
- Length of LAN = 1 km = 1000 meters
- Bandwidth = 10 Mbps = $10 \times 10^6$ bps
- Required throughput per station = $\frac{2}{3}$ Mbps
Propagation delay ($t_p$) is the time for a signal to travel from one end of the LAN to the other:
$$
t_p = \frac{\text{distance}}{\text{propagation speed}} = \frac{1000}{2 \times 10^8} = 5 \times 10^{-6}~\text{seconds} = 5~\mu s
$$
Transmission time ($t_t$) for 100 bits at 10 Mbps:
$$
t_t = \frac{100}{10 \times 10^6} = 10^{-5}~\text{seconds} = 10~\mu s
$$
Each slot duration = transmission time + propagation delay:
$$
T_{\text{slot}} = t_t + t_p = 10~\mu s + 5~\mu s = 15~\mu s
$$
If there are $n$ stations, and each gets one slot per cycle, the total cycle time is:
$$
T_{\text{cycle}} = n \cdot T_{\text{slot}} = n \cdot 15~\mu s
$$
In TDM, each station transmits 100 bits every cycle. So, the throughput per station is:
$$
\text{Throughput}_{\text{per station}} = \frac{100~\text{bits}}{T_{\text{cycle}}} = \frac{100}{n \cdot 15 \times 10^{-6}}~\text{bps}
$$
Simplifying:
$$
= \frac{100 \times 10^6}{15n} = \frac{10^8}{15n}~\text{bps}
$$
We require this to be at least $\frac{2}{3}$ Mbps = $\frac{2}{3} \times 10^6$ bps:
$$
\frac{10^8}{15n} \geq \frac{2}{3} \times 10^6
$$
Multiplying both sides by $n$:
$$
\frac{10^8}{15} \geq \frac{2}{3} \times 10^6 \cdot n
$$
Dividing both sides by $\frac{2}{3} \times 10^6$:
$$
n \leq \frac{10^8}{15} \cdot \frac{3}{2 \times 10^6} = \frac{3 \times 10^8}{30 \times 10^6} = \frac{300}{30} = 10
$$
Thus, the maximum number of stations that can be allowed in the LAN so that each station achieves a throughput of at least $\frac{2}{3}$ Mbps is 10.
$$
\boxed{10}
$$
