4. Suppose a 100-Mps point-to-point link is being set up between Earth
and a new lunar colony. The distance from the moon to earth is
approximately 385,000 km, and data travels over the link at the speed
of light-3 x 10^8 m/s.
a.) Calculate the minimum RTT for the link.
b.) Using the RTT as the delay, calculate the delay x bandwidth
product for the link.
c.) What is the significance of the delay x bandwidth product computed in (b)?
d) A camera on the lunar base takes pictures of Earth and saves them
in digital format to disk. Suppose Mission Control on Earth wishes to
download the most current image, which is 25 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?

Hi!!


a.) Calculate the minimum RTT for the link.

The minimum RTT is two times of the propagation delay on the link:
minimum RTT = 2 * Distance / Speed of propagation =
            = 2 * 385,000,000 m / (3*10^8 m/s) = 
            = 2.566 s

-----------------------------------------------------------

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

delay x bandwidth product = 2.566 s * 100 Mbps =
                          = 256.66 Mbits =
                          = 32.08 Mbytes

-------------------------------------------------------------

c.) What is the significance of the delay x bandwidth product computed in (b)?

The delay x bandwidth product computed in (b) measures the number of
bits that can be ?inflight?, i.e., the amount of data the sender can
spend up before receiving a response.
256.66 million bits corresponds to how many bits the sender can send
before the sender receives an acknowledgement that the receiver
receives the first bit. 128.33 million bits corresponds to how many
bits the sender can send before the receiver actually receives the
first bit.

------------------------------------------------------------

d) A camera on the lunar base takes pictures of Earth and saves them
in digital format to disk. Suppose Mission Control on Earth wishes to
download the most current image, which is 25 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?

File size = 25 MB = 25 * 1024 * 1024 bytes = 26,214,400 bytes =
                  = 26,214,400 * 8 b = 
                  = 209,715,200 bits

transmission time = File Size/Bandwidth =
                  = 209,715,200 bits / 100,000,000 bits/s =
                  = 2.097152 s 


It would then take a half of RTT for the Mission Control on Earth to
make the download request and another half of RTT for the propagation
delay for sending the data from the moon to the earth. Then the
minimum amount of time is the sum of the transmission time and the two
halves of RTT:

Minimum amount of time = transmission time + 2*(1/2)*RTT =
                       = 2.097152 s + RTT = 
                       = 2.097152 s + 2.566 s =
                       = 4.663152 s


--------------------------------------------------------------

I hope this helps you. Again feel free to request for a clarification
if you need it.


Best regards.
livioflores-ga

Hi livioflores-ga,

This is just to tell you that your answers are great. look forward to
working with you again.