This problem is about the TCP silly window syndrome. (a) Fill in the blank: Silly windows can happen when the sender only has very small data packets to send or when the receiver advertises a very small, but nonzero . (b) What is the official name of the solution to this problem? (c) Why is it important to some video games to disable this feature of TCP? (d) All the headers of a TCP/IP packet total 64 bytes in length. Suppose one end of a connection is constantly sending packets that have only 1 byte of actual data (in addition to the header). What percentage of the connections bandwidth is being used for actual data? (e) Compare with the percentage when the sender is sending full segments of 1460 bytes of data

(a) Suppose TCP slow start is active. A sender has sent a total of 8 packets and has received 8 ACKs in reply. What is the maximum number of segments the sender is allowed to send? (b) In slow start the number of segments a receiver is allowed to send effectively doubles every round trip time, going 1, 2, 4, 8, . . . . However, this applies only if the receiver sends an individual ACK for each segment they receiver. TCP has a feature where sometimes if a receiver gets multiple segments at once, they will only ACK every second segment instead of every segment. In this case, instead of 1, 2, 4, 8, what will the first four values be? Assume that if a receiver gets n segments, they will send back dn/2e ACKs.

7. If you send a random TCP packet to www.google.com at port 80 with the ACK bit set and you havent already established a connection via the three-way handshake, Googles servers will likely send you back a packet with what TCP flag (besides ACK) set?

(a) TCP sequence numbers can range from 0 up to what value? (b) Suppose you are on a 1 Tbps (terabits per second) connection, maxing out the connection with data. How long will it take until the TCP sequence numbers wrap around? (c) Explain why this extremely quick wraparound is a problem. (d) The solution to this fast wraparound is something called PAWS. Explain how PAWS uses a particular TCP option to mitigate the problem of quick sequence wraparound

8. In class, the way I describe the TCP congestion avoidance algorithm was incomplete. There are actually two possibilities: If a packet is lost, then what TCP does depends on whether the loss was detected by a duplicate ACK or a timeout. If the loss is detected by a duplicate ACK, then things work the way I described them in class. However, if the loss is detected by a timeout, then slow start is begun all over again, and a threshold (called the slow start threshold) is set at half the last congestion window. Once slow start gets back up to that threshold, then slow start stops and additive increase is begun. The graph below shows a trace of the congestion window size over time. Use it to answer the following questions. (a) During what time periods is TCP slow start active? (b) During what time periods is additive increase in effect? (c) There are three times that a packet was lost. What times are they? (d) Of those three times, which ones correspond to losses detected by duplicate ACKs, and which ones correspond to losses detected by timeouts? (e) Notice that packet losses dont occur at the exact same window size. Explain why this is. (f) The dashed line indicates what? (See the paragraph at the start of this problem.)

Congestion Window Size 8