Bandwidth(B)=100 Mbps,Distance(d)=10000 km Packet size(M)=100000 bits, Prop speed (v)= 2 * 10^8 m/s
we know in ideal case to send 1 packet we need Tt time.So,in Tt +2*Tp time,we can send Tt + 2*tp/Tt=1+ 2a packets (a=Tp/Tt).
therefore (a)sender window size(Ws) =1+ 2a and since it is GBN receiver window size(Wr)=1 .
(b)We know Ws +Wr<=available sequence numbers. Number of bits in sequence number field = ceil (log (2+ 2a)).
(c)In TCP, the timeout value is not static and so we need to use algorithms like Jacobson Algorithm for changing the timerout timer value after each transmission cycle but in data link layer it is fixed and its value is 2*RTT=2*2Tp=4Tp.
Putting the values,we get Ws=40000 bits,Wr=1,No of bits in sequence number =16,Timer out timer value=0.20 sec.