Solution#1(3.12)
To operate efficiently, the window size must be large enough to allow the transmitter to keep transmitting until the first ack has been received. The propogation time is (6)(3000)µsec = 18 ms. At T1 speed, which is 1.544 Mbps, a 64-byte frame takes 0.3 msec (approx). Therefore the first frame fully arrives 18.3 msec after the transmission was started. The ack takes another 18 mses to get back, plus a small (negligible) time for the ack to arrive fully. In all, this time is 36.3 msec. The transmitter must have enough window space to keep going for 36.3 msec. A frame takes 0.3 msec, so it takes 121 frames to keep fill the pipe. Seven bit sequence numbers are needed.
Solution#2(3.22)
Let t=0 denote the start of transmission. At t=1 msec, the first frame has been fully transmitted. At t=271 msec the first frame has fully arrived. At t=272 msec, the frame acknowleding a first one has been fully sent. At t=542 msec, the ack-bearing frame has fully arrived. Thus the cycle is 542 msec. A total of k frames are sent in 542 msec, for an efficiency of k/542 Hence
(a) k=1, efficiency = 1/542 = 0.18%
(b) k=7, efficiency = 7/542 = 1.29%
(c) k=4, efficiency = 4/542 = 0.74%
Solution#3
See the code in the book for Protocol 3.
Solution#4