Chapter 5:

5) The two-segment-lifetime timeout results from the need to purge old late 
   duplicates, and uncertainty of the sender of the last ACK as to whether
   it was received. For the first issue	we only need one connection
   endpoint in TIMEWAIT; for the second issue, a host in the LAST ACK state 
   expects to receive the last ACK, rather than send it.

12) (a) This is 125MB/sec; the sequence numbers wrap around when we send 
    2^32 B = 4GB.  This would take 4GB/(125MB/sec) = 32 seconds.

    (b) Incrementing every 32 ms, it would take about 32×4×10^9 ms, or
    about four years, for the timestamp field to wrap.

25) If every other packet is lost, we transmit each packet twice.

    (a) Let E => 1 be the value for EstimatedRTT, and T = 2 × E be the
    value for TimeOut.  We lose the first packet and back off TimeOut to 
    2×T. Then, when the packet arrives, we resume with EstimatedRTT = E, 
    TimeOut = T. In other words, TimeOut doesn’t change.

Chapter 6:

16) (a) In slow start, the size of the window doubles every RTT. At the end
    of the ith RTT, the window size is 2^i KB. It will take 10 RTTs before the 
    send window has reached 2^10 KB = 1MB.

    (b) After 10 RTTs, 1023KB = 1MB - 1KB has been transferred, and the
    window size is now 1MB. Since we have not yet reached the maximum
    capacity of the network, slow start continues to double the window each 
    RTT, so it takes 4 more RTTs to transfer the remaining 9MB (the amounts 
    transferred during each of these last 4 RTTs are 1MB, 2MB, 4MB, 1MB; 
    these are all well below the maximum capacity of the link in one RTT of 
    12.5MB). Therefore, the file is transferred in 14 RTTs.

    (c) It takes 1.4 seconds (14 RTTs) to send the file. The effective 
    throughput is (10MB/1.4s) = 7.1MBps = 57.1Mbps. This is only 5.7% of
    the available link bandwidth.

17) Let the sender window size be 1 packet initially. The sender sends an 
    entire windowful in one batch; for every ACK of such a windowful that the 
    sender receives, it increases its effective window (which is counted in 
    packets) by one. When there is a timeout, the effective window is cut into 
    half the number of packets.

    Now consider the situation when the indicated packets are lost. The window 
    size is initially 1; when we get the first ACK it increases to 2. At the 
    beginning of the second RTT we send packets 2 and 3. When we get their 
    ACKs we increase the window size to 3 and send packets 4, 5 and 6. When 
    these ACKs arrive the window size becomes 4.

    Now, at the beginning of the fourth RTT, we send packets 7, 8, 9, and 10;
    by hypothesis packet 9 is lost. So, at the end of the fourth RTT we have a 
    timeout and the window size is reduced to 4/2 = 2. Continuing, we have:

    RTT |  5     6     7     8     9
    ---------------------------------
    Sent| 9-10 11-13 14-17 18-22 23-28

    Again the congestion window increases up until packet 25 is lost, when it
    is halved, to 3, at the end of the ninth RTT. The plot should show the 
    window size vs. RTT and will basically have a sawtooth shape with peaks
    right before lost packets.

20) We may still lose a batch of packets, or else the window size is small 
    enough that three subsequent packets aren’t sent before the timeout. Fast 
    retransmit needs to receive three duplicate ACKs before it will retransmit 
    a packet. If so many packets are lost (or the window size is so small) 
    that not even three duplicate ACKs make it back to the sender, then the the
    mechanism cannot be activated, and a timeout will occur.

21) We will assume in this exercise and the following two that when TCP 
    encounters a timeout it reverts to stop-and-wait as the outstanding
    lost packets in the existing window get retransmitted one at a time,
    and that the slow start phase begins only when the existing window is
    fully acknowledged. In particular, once one timeout and retransmission 
    is pending, subsequent timeouts of later packets are suppressed or 
    ignored until the earlier acknowledgement is received. Such timeouts
    are still shown in the tables below, but no action is taken.

    We will let Data N denote the Nth packet; Ack N here denotes the 
    acknowledgement for data up through and including data N.

(a) Here is the table of events with TimeOut = 2 sec. There is no idle 
    time on the R-B link.

Time	 A recvs	A sends			R sends		cwnd size
-------------------------------------------------------------------------
0			Data0			Data0		1
1	 Ack0		Data1,2			Data1		2
2	 Ack1		Data3,4 (4 dropped)	Data2		3
3	 Ack2		Data5,6 (6 dropped)	Data3		4
4	 Ack3/timeout4	Data 4			Data5		1
5	 Ack3/timeout5&6			Data4		1
6	 Ack5		Data 6			Data6		1
-------------------------------------------------------------------------
7	Ack 6		Data7,8 (slow start)	Data7		2