Professor: Ian Foster TAs: Xuehai Zhang, Yong Zhao. Winter Quarter.

Transcription

1 Professor: Ian oster Ts: Xuehai Zhang, Yong Zhao Winter Quarter

2 alculate the total time required to transfer a 1 KB file (RTT=1 ms, packet size = 1 KB, handshaking = x RTT) (a)bandwidth=1.5 Mbps, data packets sent continuously initial RTT s (ms) + 1KB/1.5Mbps (transmit) + RTT/ (propagation) H.5 + 8Mbit/1.5Mbps =.5+5. sec=5.58 sec. If we pay more careful attention to when a mega is 1 6 versus, we get 8,19, bits/1,5, bits/sec = 5.46 sec, for a total delay of 5.71 sec.

3 (b) bandwidth=1.5 Mbps, wait one RTT before sending the next packet To the above we add the time for 999 RTTs (the number of RTTs between when packet 1 arrives and packet 1 arrives), for a total of = (c) bandwidth = ( transmit time = ), packets per RTT This is 49.5 RTTs, plus the initial, for 5.15 seconds. (d) bandwidth =, send n-1 packets at n-th RTT. t n RTTs past the initial handshaking we have sent n = n+1-1 packets. t n = 9 we have thus been able to send all 1, packets; the last batch arrives.5 RTT later. Total time is +9.5 RTTs, or 1.15 sec.

4 onsider a LN with a maximum distance of km. t what bandwidth would propagation delay (at a speed of 1 x 1 8 m/s) equal transmit delay for 1-byte packets? What about 51-byte packets? Propagation delay is 1 m/( 1 8 m/sec) = sec = 1ìs. 1 bytes/1ìs is 1 bytes/ìs, or 1 MB/sec, or 8 Mbit/sec. or 51-byte packets, this rises to 49.6 Mbit/sec. 4

5 What properties of postal addresses would be likely to be shared by a network address scheme? What differences might you expect to find? What properties of telephone numbering might be shared by a network addressing scheme? 5

6 Postal ddresses Postal addresses are strongly hierarchical (with a geographical hierarchy, which network addressing may or may not use). ddresses also provide embedded routing information. Unlike typical network addresses, postal addresses are long and of variable length and contain a certain amount of redundant information. This last attribute makes them more tolerant of minor errors and inconsistencies. 6

7 Telephone Numbers Telephone numbers are more similar to network addresses (although phone numbers are nowadays apparently more like network host names than addresses): they are geographically) hierarchical, fixed-length, administratively assigned, and in more-or-less one-to-one correspondence with nodes. 7

8 One property of addresses is that they are unique; if two nodes had the same address it would be impossible to distinguish between them. What other properties might be useful for network addresses to have? an you think of any situations in which network ( or postal or telephone) addresses might not be uique? 8

9 One might want addresses to serve as locators, providing hints as to how data should be routed. One approach for this is to make addresses hierarchical. nother property might be administratively assigned, versus, say, the factory-assigned addresses used by thernet. Other address attributes that might be relevant are fixed-length v. variable-length, and absolute v. relative (like file names). If you phone a toll-free number for a large retailer, any of dozens of phones may answer. rguably, then, all these phones have the same non-unique address. more traditional application for nonunique addresses might be for reaching any of several equivalent servers (or routers). 9

10 alculate the effective bandwidth for the following case. or (a) and (b) assume there is a steady supply of data to send; for (c) simply calculate the average over 1 hours (a) 1-Mbps thernet through store-and-forward switches The effective bandwidth is 1 Mbps; the sender can send data steadily at this rate and the switches simply stream it along the pipeline. We are assuming here that no Ks are sent, and that the switches can keep up and can buffer at least one packet. (b) sender wait for a 5-byte ack packet The data packet takes.4 ms as in 15(b) above to be delivered; the 4 bit Ks take 4ìs/link for a total of 4 4ìs + 4 1ìs = ìsec =. ms, for a total RTT of.4 ms. 5 bits in.4 ms is about. Mbps, or 8 KB/sec. (c) Overnight (1-hour) shipping of 1 compact disks (65 MB each) bytes / 1 hours = bytes/(1 6 sec) H 1.5 MByte/sec = 1 Mbit/sec. 1

11 alculate the bandwidth delay product for the following links. Use one-way delay, measured from first bit sent to first bit received. (a) 1-Mbps thernet with a delay of 1 ìs bits/sec 1-6 sec = 1 bits = 1.5 bytes. (b) 1-Mbps with 1 store-and-forward switch, packet size 5 bits, 1 ìs per link delay The first-bit delay is 5 ìs through the store-andforward switch, as in 15(a). 1 7 bits/sec sec = 5 bits. lternatively, each link can hold 1 bits and the switch can hold 5 bits. 11

12 (c) 1.5-Mbps T1 link, with a transcontinental one-way delay of 5 ms bits/sec sec = 75, bits = 975 bytes (d) 1.5-Mbps T1 link through a satellite in geosynchronous orbit, 5,9 km high. The only delay is speed-of-light propagation delay. This was intended to be through a satellite, ie between two ground stations, not to a satellite; this ground-toground interpretation makes the total one-way travel distance 5,9, meters. With a propagation speed of c = 1 8 meters/sec, the one-way propagation delay is thus 5,9,/c =.4 sec. Bandwidth delay is thus bits/sec.4 sec = 6, bits H 45 KBytes 1

13 or the network given in igure 4.9, give global distance-vector tables like those of Tables 4.5 and 4.8 when (a) ach node knows only the distances to its immediate neighbors. (b) ach node has reported the info it had in the preceding step to its immediate neighbors. (c) step (b) happens a second time. 1

14 6 8 B 1 14

15 B B istance to Reach Node Information Stored at Node B B istance to Reach Node Information Stored at Node B B istance to Reach Node Information Stored at Node

16 or the network give in figure 4.9, show how the link-state algorithm builds the routing table for node. onfirmed Tentative 1. (,,-). (,,-) (,8,) (,,). (,,-) (,8,) (,,) (B,4,) (,,) 4. (,,-) (,6,) (,,) (B,4,) (,,) (,9,) 5. (,,-) (,6,) (,,) (,9,) (,,) (B,4,) 6. previous + (,6,) 7. previous + (,9,) 16

17 or the network in igure 4.9, suppose the forwarding tables are all established as in exercise 1 and then the - link fails. Give (a) the tables of, B, and after and have reported the news. (b) the tables of and after their next mutual exchange. (c) the table of after exchange with it. 17

18 B Nexthop cost est nexthop cost dest B Nexthop cost dest B Nexthop cost dest (a) B

19 B Nexthop cost dest B Nexthop cost dest B Nexthop cost dest (b) (c)

20 Routing table: SubnetNumbers (default) SubnetMask NextHop Interface Interface 1 R R R4

21 The router can deliver packets directly over interfaces and 1, or it can forward packets to router R, R, or R4. escribe what the router does with a packet addressed to each of the following destinations: (a) pplying the subnet mask , we get Use interface as the next hop. (b) pplying subnet mask , we get Use R as the next hop. (c) ll subnet masks give as the subnet number. Since there is no match, use the default entry. Next hop is R4. (d) Next hop is R. (e) None of the subnet number entries match, hence use default router R4. 1

22 onsider the simple network, in which and B exchange distance-vector routing information. ll link have cost 1. Suppose the - link fails. x

23 (a) give a sequence of routing table updates that leads to a routing loop between and B necessary and sufficient condition for the routing loop to form is that B reports to the networks B believes it can currently reach, after discovers the problem with the link, but before has communicated to B that no longer can reach. (b) stimate the probability of the scenario in (a), assuming and B send out routing updates at random times, each at the same average rate. t the instant that discovers the failure, there is a 5% chance that the next report will be B s and a 5% chance that the next report will be s. If it is s, the loop will not form; if it is B s, it will. (c) stimate the probability of a loop forming if broadcasts an updated report within 1 second of discovering the - failure, and B broadcasts every 6 seconds uniformly. t the instant discovers the failure, let t be the time until B s next broadcast. t is equally likely to occur anywhere in the interval d t d 6. The event of a loop forming is the same as the event that B broadcasts first, which is the event that t < 1. sec; the probability of this is 1/6.

24 The sequence number field in the TP header is bit long, which is big enough to cover over 4 billion bytes of data. ven if this many bytes were never transferred over a single connection, why might the sequence number still wrap around from 1 to? The sequence number doesn t always begin at for a transfer, but is randomly or clock generated. 4

25 esign a reliable byte-stream protocol that uses a sliding window (like TP). This protocol will run over a 1-Mbps network (RTT=1 ms, MSL=6 s) (a) How many bits would you include in the dvertisedwindow and SequenceNum fields in the header The advertised window should be large enough to keep the pipe full; delay (RTT) bandwidth here is 1 ms 1 Mbps = 1 Mb = 1.5 MB of data. This requires 1 bits ( 1 =,97,15) for the dvertisedwindow field. The sequence number field must not wrap around in the maximum segment lifetime. In 6 seconds, 75 MB can be transmitted. bits allows a sequence space of 14 MB, and so will not wrap in 6 seconds. (If the maximum segment lifetime were not an issue, the sequence number field would still need to be large enough to support twice the maximum window size; see inite Sequence Numbers and Sliding Window in Section.5.) 5

26 (b) How would you determine the numbers given above, and which values might be less certain? The bandwidth is straightforward from the hardware; the RTT is also a precise measurement but will be affected by any future change in the size of the network. The MSL is perhaps the least certain value, depending as it does on such things as the size and complexity of the network, and on how long it takes routing loops to be resolved. 6

27 Suppose a host wants to establish the reliability of a link by sending packets an measuring the percentage that are received; routers, for example, do this. xplain the difficulty doing this over a TP connection. The problem is that there is no way to determine whether a packet arrived on the first attempt or whether it was lost and retransmitted. Having the receiver echo back immediately and measuring the elapsed times would help; many Berkeley-derived implementations measure timeouts with a.5 sec granularity and round-trip times for a single link without loss would generally be one to two orders of magnitude smaller. But verifying that one had such an implementation is itself rather difficult. 7

28 Suppose TP operates over a 1-Gbps link. (a) How long would it take for the TP sequence numbers to wrap around completely? This is 15MB/sec; the sequence numbers wrap around when we send B = 4GB. This would take 4GB/(15MB/sec) = seconds. (b) Suppose an added -bit timestamp field increments 1 times during the wraparound time you found above. How long would it take for the timestamp to wrap around? Incrementing every ms, it would take about ms, or about four years, for the timestamp field to wrap. 8