8 10 pts Youve recently been hired by NASAJPL to work on dat

8 (10 pts) You\'ve recently been hired by NASA/JPL to work on data communications to the International Space Station (SS). As part of your new job, you\'ve been asked to work on the link layer protocol for a wireless data connection with a bandwidth of 100 Mbps to the ISS. Assuming the ISS is 250 km from the surface of the earth and the speed of light is c 3.0 x 10 m/s, answer the following (a) What is the bandwidth-delay product for communication between Earth and the ISS? (b) If your protocol will be the Go-Back-N sliding window protocol, what size window do you need to use in order to achieve a transmission speed of 100 Mbps? Assume the frame size is 4096 bytes, the window is always full, there are no losses, and the ACK traffic is negligible. (c) What size frame do you need in order to achieve an actual data transfer speed of 100 Mbps, assuming 128 bytes of overhead, and the window size is 256?

Solution

a.) Propogation Delay (T_p) = distance (in meter)/ velocity of light = (250 * 10³) / (3* 10⁸) = 8.33* 10^-4 sec

Now, bandwidth-delay product = 2*100 * 10^{6 *} 8.33 * 10^-4 = 2*8.33* 10⁴ = (500/3) kbits

b.) Transmission delay (T_t) = (frame size)/bandwidth = (4096 * 8 )/(100 * 10⁶⁾ sec

Let a = T_p/T_t = {(250 * 10³) / (3* 10⁸) } /{(4096 * 8 )/(100 * 10⁶) } = 2.54

For maximum utiliation, window size = 2^(1+2a) = 2^(1 + 2*2.5) = 2^6 =64

c.) For window size = 2^(1 + 2a) = 256

which gives a = 3.5.

Let we have x bytes of frame size (including overhead). Now,

a = T_p/T_t =   {(250 * 10³) / (3* 10⁸) } /{(x * 8 )/(100 * 10⁶) } = 3.5

which gives x = 2976 bytes

Note: actual answer may differ a bit but method will be the same. Please cross check the answer once. Hope it helps.