Question

25. Suppose you are designing a sliding window protocol for a 1-Mbps point-to-point link to the stationary satellite evolving around the Earth at an altitude of 3 x 104 km. Assuming that each frame carries 1 KB of data, what is the minimum number of bits you need for the sequence number in the following cases? Assume the speed of light is 3 x 108 m/s. (a) RWS-1 (b) RWS-SWS

Answer 1

Given, L = 3 times 10^4 km = 3 times 10^7 mk S = 1 KB B = 1 Mbps Using Bandwidth - delay product, N _f = B _fps times R _TT = 125 pps times 0.2 sec {always B _fps = B/S} N _f = 25(pallets) B _fps = 1 KB/1 Mbps = 125 pps {always Latency of link is around 100 ms ie, in one way} Given, RWS = 1: Thus, the sequence required for this is the number space be 2b. Therefore, the required number of hits be 26. Given, RWS = SWS: since, SWS has a large value. Therefore, sequence number can be cover around twice that of the SWS. So, the number space can be around 25 times 2 = 50 Thus, it requires 6 units which are needed. Note: - T _delay = L/C = 3 times 10^7 m/3 times 10^8 m = 0.1 S (or) = 100 ms R_TT = 2 times 0.1 =