Chapter 19

Chapter 19

19.1	Consider a TCP client application that writes a small application header (8 bytes) followed by a small request (12 bytes). It then waits for a reply from the server. What happens if the request is sent using two writes (8 bytes, then 12 bytes) versus a single write of 20 bytes?
A:	Two application writes, followed by a read, cause a delay because the Nagle algorithm will probably be invoked. The first segment (with 8 bytes of data) is sent and its ACK is waited for before sending the 12 bytes of data. If the server implements delayed ACKs, there can be a delay of up to 200 ms (plus the RTT) before this ACK is received.
19.2	In Figure 19.4 we are running tcpdump on the router sun . This means the data in the arrows from the right to the left still have to go through bsdi , and the data in the arrows from the left to the right have already come through bsdi . When we see a segment going to slip , followed by a segment coming from slip , the time differences between the two are: 34.8, 26.7, 30.1, 28.1, 29.9, and 35.3 ms. Given that there are two links between sun and slip (an Ethernet and a 9600 bits/sec CSLIP link), do these time differences make sense? ( Hint : Reread Section 2.10.)
A:	Assuming 5-byte CSLIP headers (IP and TCP) and 2 bytes of data, the RTT across the SLIP link for these segments is about 14.5 ms. We have to add to this the RTT across the Ethernet (normally 5 “10 ms), plus the routing time on sun and bsdi. So yes, the observed times do appear correct.
19.3	Compare the time required to send a special function key and have it acknowledged using the Nagle algorithm (Figure 19.6) and with the algorithm disabled (Figure 19.8).
A:	In Figure 19.6 the time difference between segments 6 and 9 is 533 ms. In Figure 19.8 the time difference between segments 8 and 12 is 272 ms. (We measured the time for the F2 key, not the F1 key, since the first echo of the F1 key was lost in the second figure.)