Suppose a 128-Kbps point-to-point link is set up between Earth and a rover on Mars. The distance from Earth to Mars (when they are closest together) is approx- imately 55 Gm, and data travels over the link at the speed of light-3 × 108 m/s. (a) Calculate the minimum RTT for the link. (b) Calculate the delay x bandwidth product for the link. (c) A camera on the rover takes pictures of its surroundings and sends these to Earth. How quickly after a picture is taken can it reach Mission Control on Earth? Assume that each image is 5 Mb in size.

Please check my answer and correct the part which is need to be foxed.And provide justification for the answer. a) The transmission time of a frame can be calculated as follows: Propagation delay (Tp) = distance/speed = 5km / 200000 km/s = 0.0025ms Transmission time (Tt) = frame size/bandwidth =  10000 * 8bits / (1Mbps * 10 ^ 6) = 0.008sec = 8ms Total time for transmission =  2 * Tp + Tt(since the frame has to travel from A to D and then D to A) =  2 * 0.0025ms + 8ms = 0.005ms + 8ms = 8.005ms   The transmission time of the frame from A to D is 8.005 ms.   b)  The efficiency of the CSMA/CD protocol is given by the formula: Tt is the transmission time of a frame  C is the number of collisions, Tp is the propagation time of a signal from one end of the segment to the other.   Efficiency = Tt / (C * 2 * Tp + Tt + Tp)    Assuming that there are no other stations transmitting or attempting to transmit on the segment, the transmission from A to D will succeed without collisions. Therefore,…

Distinguish the fundamental differences between synchronous and isochronous connections, and give an example of each.

We need to use synchronous TDM and combine 20 digital sources, each of 100 Kbps. Each output slot carries 1 bit from each digital source, but one extra bit is added to each frame for synchronization. Answer the following questions: a. What is the size of an output frame in bits? b. What is the output frame rate?