計(jì)網(wǎng)-第二章作業(yè)

1、Chapter 2注：括弧之中紅色標(biāo)注為與課本習(xí)題對(duì)應(yīng)的編號(hào)2.1，2.2節(jié)后作業(yè)：1. (R2)For a communication session between a pair of processes, which process is the client and which is the server?答：The process is always-on host called the server.The process which initiates the communication is the client. 2. (R3)What is the difference b

2、etween network architecture and application architecture?答：Network architecture refers to the organization of the communication process into layers.Application architecture, on the other hand, is designed by an application developer and dicates the broad structure of the application.3. (R4)What info

3、rmation is used by a process running on one host to identify a process running on another host?答：The IP address of the destination host and the port number of the destination socket.4. (R7)Suppose you wanted to do a transaction from a remote client to a server as fast as possible. Would you use UDP

4、or TCP? Why?答：We would use UDP. With UDP, the transaction can be completed in one roundtrip time the client sends the transaction request into a UDP socket, and the server sends the reply back to the clients UDP socket. With TCP, a minimum of two RTTs are needed one to set-up the TCP connection, and

5、 another for the client to send the request, and for the server to send back the reply.5. (R8)List the four broad classes of services that a transport protocol can provide. For each of the service classes, indicate if either UDP or TCP (or both) provides such a service.答：(1) Reliable Data Transfer:

6、TCP provides a reliable byte-stream between client and server but UDP does not.(2) Throughput: A guarantee that a certain value for throughput will be maintained; Neither.(3) Timing: A guarantee that data will be delivered within a specified amount of time; Neither.(4) Security: Neither.6. (R9)Why d

7、o HTTP, FTP, SMTP, and POP3 run on top of TCP rather than on UDP?答：The applications associated with those protocols require that all application data be received in the correct order and without gaps. TCP provides this service whereas UDP does not.7. (R11)What is meant by a handshaking protocol?答：A

8、protocol uses handshaking if the two communicating entities first exchange control packets before sending data to each other. SMTP uses handshaking at the application layer whereas HTTP does not.2.3，2.4節(jié)后作業(yè)8. (R15)Suppose Alice, with a Web-based e-mail account (such as Hotmail or gmail), sends a mes

9、sage to Bob, who accesses his mail from his mail server using POP3. Discuss how the message gets from Alices host to Bobs host. Be sure to list the series of application-layer protocols that are used to move the message between the two hosts.答：Message is sent from Alices host to her mail server over

10、 HTTP. Alices mail server then sends the message to Bobs mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3.9. (R17)From a users perspective, what is the difference between the download-and-delete mode and the download-and-keep mode in POP3?答：With downlo

11、ad and delete, after a user retrieves its messages from a POP server, the messages are deleted. This poses a problem for the nomadic user, who may want to access the messages from many different machines. In the download and keep configuration, messages are not deleted after the user retrieves the m

12、essages. This can also be inconvenient, as each time the user retrieves the stored messages from a new machine, all of non-deleted messages will be transferred to the new machine.10. (R18)Is it possible for an organizations Web server and mail server to have exactly the same alias for a hostname (fo

13、r example, foo. com)? What would be the type for the RR that contains the hostname of the mail server?答：Yes, an organizations mail server and Web server can have the same alias for a host name. The MX record is used to map the mail servers host name to its IP address.11. (R19)Why is it said that FTP

14、 sends control information out-of-band?答：FTP uses two parallel TCP connections, one connection for sending control information and another connection for actually transferring the file. Because the control information is not sent over the same connection that the file is sent over, FTP sends control

15、 information out of band.2.5，2.6節(jié)后作業(yè)12. (P1)True or false?a. A user requests a Web page that consists of some text and three images. For this page, the client will send one request message and receive four response messages.b. Two distinct Web pages (for example, /research.Html and www.mi

16、/students. html) can be sent over the same persistent connection.c. With nonpersistent connections between browser and origin server, it is possible for a single TCP segment to carry two distinct HTTP request messages.d. The Date: header in the HTTP response message indicates when the object in

17、 the response was last modified.答：a. Fb. Tc. Fd. F13. (P7)Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local host, so a DNS lookup is necessary to obtain the IP address. Suppose that n DNS servers are visited be

18、fore your host receives the IP address from DNS; the successive visits incur an RTT of . Further suppose that the Web page associated with the link contains exactly one object, consisting of a small amount of HTML text. Let denote the RTT between the local host and the server containing the object.

19、Assuming zero transmission time of the object, how much time elapses from when the client clicks on the link until the client receives the object?答：The time to get IP address is RTT1+RTT2+RTTn, once the IP address is known, the response time is 2RTT0+RTT1+RTTn.14. (P8)Referring to Problem P7, suppos

20、e the HTML file references three very small objects on the same server. Neglecting transmission times, how much time elapses witha. Non-persistent HTTP with no parallel TCP connections?b. Non-persistent HTTP with parallel connections?c. Persistent HTTP (with pipelining)?答：a. RTT1+RTT2+RTTn+2RTT0+32R

21、TT0=8RTT0+RTT1+RTTn.b. RTT1+RTT2+RTTn+2RTT0+2RTT0=4RTT0+RTT1+RTTn.c. RTT1+RTT2+RTTn+2RTT0+RTT0=3RTT0+RTT1+RTTn.2.7，2.8節(jié)后作業(yè)15. (P9)Consider Figure2.12, for which there is an institutional network connected to the Internet. Suppose that the average object size is 900,000 bits and that the average requ

22、est rate from the institutions browsers to the origin servers is 10 requests per second. Also suppose that the amount of time it takes from when the routers on the Internet side of the access link forwards an HTTP request until it receives the response is two seconds on average (see Section 2.2.5).

23、Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use/(1-), where is the average time required to send an object over the access link andis the arr

24、ival rate of objects to the access link.a. Find the total average response time.b. Now suppose a cache is installed in institution LAN. Suppose the hit rate is 0.6. Find the total response time. 答：a. The time to transmit an object of size L over a link or rate R is L/R.The average time is the averag

25、e size of the object divided by R : =( bits)/(bits/sec)=0.06secThe traffic intensity on the link is ( 10 requests/sec)(0.06msec/request )=0.6The average access delay is ( 0.06sec )/( 1-0.6 )=0.15secThe total average response time is therefore 0.6sec+2sec=2.15sec.b. The average access delay is ( 0.06

26、sec )/( 1-0.60.4 )=0.0456secWith cache and the hit rate is 0.6 , when miss the cache the average response time is 0.0456sec+2sec=2.0456secThe average response time is (1-0.6)2.0456sec=0.832sec16. (P20)In our coverage of an overlay network using query flooding in Section 2.6, we describe in some deta

27、il how a new peer joins the overlay network. In this problem we want to explore what happens when a peer leaves the overlay network. Suppose every participating peer maintains TCP connections to at least four distinct peers at all times. Suppose Peer X, which has five TCP connections to other peers,

28、 wants to leave.a. First consider the case of a graceful departure, that is, Peer X explicitly close its application, thereby graceful closing its five TCP connections. What actions would each of the five formerly connected peers take?b. Now suppose that Peer X abruptly disconnects from the Internet

29、 without notifying its five neighbors that is closing the TCP connections. What would happen?答：a. Each of the five Gnutella clients immediately learns that it has one less neighbor. The one of these five clients called A. Suppose A has only three neighbors after X drops out . Then A needs to establi

30、sh a TCP connection with another peer . A should have a fresh list of active peers; he sequentially contacts peers on this list until one accepts his TCP connection attempt.b. We still use A. A doesnt immediately know that X has departed. A will only learn about Xs departure when it attempts to send

31、 a message (query or ping) to X. When A attempts to send a message, As TCP will make several unsuccessful attempts to send the message to B. As TCP will then inform the Gnutella client that X is down. A will try to establish a TCP connection with a new peer to rebuild a fifth connection.17. (P21)Con

32、sider query flooding, as discussed in Section 2.6. Suppose that each peer is connected to at most N neighbors in the overlay network. Also suppose that the node-count field is initially set to K. Suppose Alice makes a query. Find an upper bound on the number of query messages that are sent into the overlay network.答：Th