Chapter 21

21.1	In Figure 21.5 the first timeout was calculated as 6 seconds and the next as 24 seconds. If the ACK for the initial SYN had not arrived after the 24-second timeout expired , when would the next timeout occur?
A:	The next timeout is for 48 seconds: 0 + 4 —12. The factor of 4 is the next multiplier in the exponential backoff.
21.2	In the discussion following Figure 21.5 we said that the timeout intervals are calculated as 6, 24, and then 48 seconds, as we saw in Figure 4.5. But if we watch a TCP connection to a nonexistent host from an SVR4 system, the timeout intervals are 6, 12, 24, and 48 seconds. What's going on?
A:	It appears SVR4 still uses the factor 2 D instead of 4 D in the calculation of RTO.
21.3	Compare the performance of TCP's sliding window versus TFTP's stop-and-wait protocol as follows . In this chapter we transferred 32768 bytes in about 35 seconds (Figure 21.6) across a link with an RTT that averaged around 1.5 seconds (Figure 21.4). Calculate how long TFTP would take for the same transfer.
A:	The stop-and-wait protocol used by TFTP is limited to 512 bytes of data per round trip. 32768/512 — 1.5 is 96 seconds.
21.4	In Section 21.7 we said that the receipt of a duplicate ACK is caused by a segment being lost or reordered. In Section 21.5 we saw the generation of duplicate ACKs caused by a lost segment. Draw a picture showing that a reordering of segments also generates duplicate ACKs.
A:	Show four segments, numbered 1, 2, 3, and 4. Assume the order of receipt is 1, 3, 2, and 4. The ACKs generated by the receiver will be ACK 1 (a normal ACK), ACK 1 (a duplicate ACK when segment 3 is received out of order), ACK 3 when segment 2 is received (acknowledging both segments 2 and 3), and then ACK 4. Here one duplicate ACK is generated. If the order of receipt were 1, 3, 4, 2, two duplicate ACKs would be generated.
21.5	There is a noticeable blip in Figure 21.6 between times 28.8 and 29.8. Is this a retransmission?
A:	No, because the slope is still up and to the right, not downward.
21.6	In Section 21.6 we said that the 4.3BSD Tahoe release only performed slow start if the destination was on a different network. How do you think "different network" was determined? ( Hint : Look at Appendix E.)
A:	See Figure E.1.
21.7	In Section 20.2 we said that TCP normally ACKs every other segment. But in Figure 21.2 we see the receiver ACK every segment. Why?
A:	In Figure 21.2 the segments contain 256 bytes of data, which takes approximately 250 ms to transfer across the 9600 bits/sec CSLIP link between `slip` and `bsdi.` Assuming the data segments are not queued somewhere between `bsdi` and `vangogh,` they arrive at `vangogh` about 250 ms apart. Since this exceeds the 200-ms delayed ACK timer, each segment is acknowledged when the next delayed ACK timer expires .