2010計(jì)算機(jī)網(wǎng)絡(luò)原理A卷答案及評(píng)分標(biāo)準(zhǔn)-北交大(共4頁(yè))

1、精選優(yōu)質(zhì)文檔傾情為你奉上精選優(yōu)質(zhì)文檔傾情為你奉上專(zhuān)心專(zhuān)注專(zhuān)業(yè)專(zhuān)心專(zhuān)注專(zhuān)業(yè)精選優(yōu)質(zhì)文檔傾情為你奉上專(zhuān)心專(zhuān)注專(zhuān)業(yè)Questions(40)What are the five layers in the Internet protocol stack? What are the principal responsibilities of each of these layers? (5)答案：從上向下各層：應(yīng)用層：提供與用戶(hù)應(yīng)用有關(guān)功能，包括網(wǎng)絡(luò)瀏覽、電子郵件、不同類(lèi)文件系統(tǒng)的文件傳輸、虛擬終端軟件、過(guò)程作業(yè)輸入、目錄查詢(xún)和其他各種通用的和專(zhuān)用的功能等。（1分）傳輸層：完成端到端的可靠數(shù)據(jù)傳輸。（1

2、分）網(wǎng)絡(luò)層：關(guān)系到子網(wǎng)的運(yùn)行控制，其中一個(gè)關(guān)鍵問(wèn)題是確定分組從源端到目的端的路由選擇，以分組為單位進(jìn)行傳輸。（1分）鏈路層：主要任務(wù)是加強(qiáng)物理傳輸原始比特的功能，以幀為單位進(jìn)行傳輸。完成相鄰結(jié)點(diǎn)間的可靠數(shù)據(jù)傳輸。（1分）物理層：負(fù)責(zé)提供和維護(hù)物理線路，并檢測(cè)處理爭(zhēng)用沖突，提供端到端錯(cuò)誤恢復(fù)和流控制以比特為單位進(jìn)行傳輸。（1分）Why is it said that FTP sends control information out-of-band but HTTP sends control information “in-band”? (6)答：HTTP和FTP都是文件傳送協(xié)議，它們有許多共

3、同的特征（如都運(yùn)行在TCP之上）。不過(guò)這兩個(gè)應(yīng)用層協(xié)議最重要的差別是FTP使用兩個(gè)并行的TCP連接來(lái)傳送文節(jié)件，一個(gè)是控制連接，一個(gè)是數(shù)據(jù)連接。（2分）控制連接用于在客戶(hù)主機(jī)和服務(wù)器主機(jī)之間發(fā)送控制信息，例如用戶(hù)名和口令、改變遠(yuǎn)程目錄的命令、取來(lái)或放回文件的命令。數(shù)據(jù)連接用于真正發(fā)送文件。既然TCP使用一個(gè)獨(dú)立的控制連接，我們說(shuō)FTP在帶外(out-of-band)發(fā)送控制信息的。（2分）HTTP中，同一個(gè)TCP連接既用于承載請(qǐng)求和響應(yīng)頭部，也用于承裁所傳送的文件，因此我們說(shuō)HTTP在帶內(nèi)(in-band)發(fā)送控制信息。（2分）What is the difference between ne

4、twork architecture and application architecture? (6)答：網(wǎng)絡(luò)體系結(jié)構(gòu)是指以分層的方式來(lái)描述通信過(guò)程的組織體系。（例如五層網(wǎng)絡(luò)結(jié)構(gòu)）另一方面，應(yīng)用體系結(jié)構(gòu)是由應(yīng)用程序的研發(fā)者設(shè)計(jì)，并規(guī)定應(yīng)用程序的主要結(jié)構(gòu)（例如客戶(hù)機(jī)/服務(wù)器或P2P）從應(yīng)用程序研發(fā)者的角度看，網(wǎng)絡(luò)體系結(jié)構(gòu)是固定的，并為應(yīng)用程序提供了特定的服務(wù)集合。（評(píng)分標(biāo)準(zhǔn)：網(wǎng)絡(luò)體系結(jié)構(gòu)和應(yīng)用體系結(jié)構(gòu)各占3分）Suppose Host A sends two TCP segments back to back to Host B over a TCP connection. The first

5、 segment has sequence number 92; the second has sequence number 111. (6)a. How much data is in the first segment?b. Suppose that the first segment is lost but the second segment arrives at B. In the acknowledgment that Host B sends to Host A, what will be the acknowledgment number?答：a) 19 bytes b) a

6、ck number = 92（評(píng)分標(biāo)準(zhǔn)：a，b各占3分）What are the two most important network-layer functions in a datagram network? What is the difference between them? (6)答：數(shù)據(jù)報(bào)網(wǎng)絡(luò)中網(wǎng)絡(luò)層兩個(gè)最重要的功能是轉(zhuǎn)發(fā)和選路。（2分）轉(zhuǎn)發(fā)（Forwarding）特指在一個(gè)路由器內(nèi)將數(shù)據(jù)從輸入端口移動(dòng)到合適的輸出端口（2分），而選路（Routing）特指確定從數(shù)據(jù)源端到目的端的整條傳輸路徑（2分）。If all the links in the Internet were to

7、 provide reliable delivery service, would the TCP reliable delivery service be redundant? Why or why not? (5)答：不多余（1分）。雖然每條鏈路都能保證數(shù)據(jù)包在端到端的傳輸中不發(fā)生差錯(cuò)，但它不能保證IP 數(shù)據(jù)包是按照正確的順序到達(dá)最終的目的地。IP 數(shù)據(jù)包可以使用不同的路由通過(guò)網(wǎng)絡(luò)，到達(dá)接收端的順序會(huì)不一致，因此，TCP 需要用來(lái)使字節(jié)流按正確的序號(hào)到達(dá)接收端（2分）。路由器也可能由于緩存溢出而丟棄IP數(shù)據(jù)包（2分）。In Cyclic Redundancy Check (CRC), co

8、nsider the 4-bit generator G is 1001, and suppose that D has the value . What is the value R? Show your calculation steps. (6)答：本題中G=1001，D=，需要求R，先將D補(bǔ)全3個(gè)0（G的位數(shù)-1），為（2分）.再對(duì)其用G進(jìn)行二進(jìn)制除法，得到的余數(shù)即為R=111。（2分）其具體計(jì)算過(guò)程如下：計(jì)算過(guò)程（2分）驗(yàn)算過(guò)程（不算分）三. Consider transferring an enormous file of L bytes from Host A to Host B

9、. Assume an MSS of 1,460 bytes. (10)What is the maximum value of L such that TCP sequence numbers are not exhausted? Recall that the TCP sequence number fields has 4 bytes.For the L you obtain in (a), find how long it takes to transmit the file. Assume that a total of 66 bytes of transport, network,

10、 and data-link header are added to each segment before the resulting packet is sent our over a 100 Mbps link. Ignore flow control and congestion control so A can pump out the segments back to back and continuously.一共有 232 = 4,294,967,296個(gè)可能的序列號(hào)。（2分）a) 由于序列號(hào)不隨著TCP報(bào)文段個(gè)數(shù)增長(zhǎng)而增長(zhǎng)。而是根據(jù)發(fā)送數(shù)據(jù)的字節(jié)數(shù)而增長(zhǎng)。因此，MSS的大小與

11、可以從A發(fā)送到B的文件的大小無(wú)關(guān)。文件大小可以簡(jiǎn)單估算為： 232 4.19 Gbytes 。（3分）b) TCP報(bào)文段的個(gè)數(shù)為= 2,941,758。66字節(jié)的首部加到每個(gè)報(bào)文段將帶來(lái)194,156,028字節(jié)的首部。要傳輸?shù)淖止?jié)總數(shù)為：(232+ 194,156,028) 8 = 3,591107 bits。（3分）因此將花費(fèi) 359 seconds = 6 minutes 來(lái)傳輸文件在一個(gè) 100Mbps 鏈路上。（2分）（思路正確，計(jì)算錯(cuò)誤給1半分）四. Consider a router that interconnects four subnets: N1, N2, N3 and

12、N4. Suppose all of the interfaces in each of these four subnets are required to have the prefix 4/26. Also suppose the N1 is required to support up to 24 interfaces, N2 is required to support up to 12 interfaces, and N3 and N4 are each required to support up to 6 interfaces. Provide four network add

13、resses of the form a.b.c.d/x that satisfy these constraints. (10)答：N1: 4/27 （3分）N2: 6/28 （3分）N3: 12/29 （2分）N4: 20/29 （2分）五. (10) Suppose CSMA/CD protocol is used in a LAN. The distance between hosts A and B is 2 km, propagation speed is km/s, transmission rate of the link is 10 Mbps. Please answer t

14、he following questions:(1) Suppose collision occurred when the two hosts send data. How long is the time from the beginning of data transmission to collision is detected?(2) What is the smallest frame from A to B?（1）2/=10-5(s)(2)2*10-5*10*106=200(bit)六. Suppose within your Web browser you click on a

15、 link to obtain a Web page. The IP address for the associated URL is cached in your local host. Further suppose that the Web page associated with the link contains a small amount of HTML text, and the HTML file references five very small objects on the same server. Let RTT2 denote the RTT between the local host and the server containing the object. Neglecting transmission times, how much time elapses with: (10)a. Non-persistent HTTP with no parallel TCP connec