Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of $3 \times 10^8$ m/s. The time taken (in milliseconds, *rounded off to two decimal places*) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is _________.

To completely receive the packet, the total time required is the sum of the transmission time ($T_t$) and the propagation time ($T_p$):

1. Transmission Delay ($T_t$):

   $$T_t = \frac{\text{Packet Size}}{\text{Bandwidth}} = \frac{1000 \times 8\text{ bits}}{100 \times 10^6\text{ bps}} = \frac{8000}{10^8}\text{ s} = 8 \times 10^{-5}\text{ s} = 0.08\text{ ms}$$

2. Propagation Delay ($T_p$):

   $$T_p = \frac{\text{Distance}}{\text{Propagation Speed}} = \frac{2100 \times 10^3\text{ m}}{3 \times 10^8\text{ m/s}} = \frac{2.1 \times 10^6}{3 \times 10^8}\text{ s} = 7 \times 10^{-3}\text{ s} = 7.00\text{ ms}$$

3. Total Time:

   $$\text{Total Time} = 0.08\text{ ms} + 7.00\text{ ms} = 7.08\text{ ms}$$