I have a few doubts in solving this question.
B = 1 Mbps = 10^6 b/s
d = 3*10^4 km = 3*10^7 m
L = 1 kB = 2^13 b
v = 3*10^8 m/s
Tt = transmission time = L/B = (2^13/10^6) s
Tp = propagation time = d/v = 0.1 s
1 < WS <= (1 +2a) ( For sliding window, WS > 1)
For max efficiency, WS = (1 +2a) = 1 + 2*Tp/Tt = 25.4
Doubt 1: Is Bandwidth = 2^20 or 10^6?
Doubt 2: 1. If (1+2a) is not a natural number, do we take floor or ceil?
Doubt 3: For finding minimum sequence number, do we assume that efficiency should be high? Because if efficiency can be low, then minimum will be (2+1)=3 (WS=2 because in sliding window WS>1).
Thanks.