Question

A 10-Gbps point-topoint link is being set up between the Earth and a new lunar colony. The distance from the moon to the Earth is approximately 385,000 km, and data travels over the link at the speed of light – 3 x 108 m/s.

b) Calculate the minimum RTT for the link.

c) Using the RTT as the delay, calculate the delay x bandwidth product for the link.

d) A camera on the lunar base takes videos of the Earth and saves them in digital format to disk. Suppose Mission Control on Earth wishes to download the must current image, which is 105 MB. What is the minimum amount of time that will elapse between when the request for the data goes out and the transfer is finished?

Answer 1

Given that bandwidth(B) = 10Gbps, Distance(D) = 385,000km, and speed of the light(V) = 3*10⁸ m/s.

b) calculate the minimum RTT :

RTT stands for Round Trip Time, i.e two times the propagation delay. (RTT = 2 * propagation delay ).

Propagation delay (T_p ) = Total distance / velocity of the signal ( Here the signal travels with the speed of light.)

Therefore (T_p ) = 385,000 km / 3*10⁸ m/s = (385,000 * 1000 m) / (3*10⁸ m/s) =  1.283 sec.

Therefore RTT = 2 * Tp = 2 * 1.283 sec = 2.566 sec.

c) delay x bandwidth product:

delay - bandwidth product is the product of data link's capacity and the round trip time. This gives the maximum amount of data that can be transmitted by the sender at any given time.

Therefore, the product equals to = 10 Gb/sec * 2.566 sec = 25.66 Gb.

d)

There is a image of size 105 MB that must be transferred to earth. The total time elapsed ?

The total time is given by the sum of propagation delay and transmission delay.

We have already calculated the propagation delay, which is T_p = 1.283 sec.

Therefore transmission delay (T_x ) = File size / Bandwidth = 105 Mb / 10 Gbps ( 1Mb = 10⁶ bytes, 1Gb = 10⁹ bytes approximately).

T_x = 105 * 10⁶ bytes / 10 *10⁹ bytes per sec = 0.0105 sec.

Therefore total time = transmission delay + propagation delay = 1.283 sec + 0.0105 sec = 1.2953 sec.