計(jì)算機(jī)網(wǎng)絡(luò)試卷B理科班

1、2003年理科班計(jì)算機(jī)網(wǎng)絡(luò)試卷B 補(bǔ)考用一、填空題（每空0.5分，共10分）1網(wǎng)絡(luò)互連設(shè)備的作用是_1_。在不同層次上的網(wǎng)絡(luò)互連，名稱分別為數(shù)據(jù)鏈路層的互連設(shè)備叫_2_。網(wǎng)絡(luò)層的互連設(shè)備叫_3_；傳輸層的互連設(shè)備叫_4_；傳輸層上的設(shè)備叫_5_。網(wǎng)絡(luò)互連設(shè)備統(tǒng)稱為_6_。2TCP/IP協(xié)議參考模型由_7_、_8_、_9_、_10_。3在海明碼中，信息比特的位數(shù)與校驗(yàn)比特位數(shù)之間的關(guān)系為_11_。4IP地址是_12_比特的二進(jìn)制數(shù)。5互連網(wǎng)絡(luò)內(nèi)部的路由選擇分二個(gè)層次_13_和_14_。 6Internet上的每個(gè)接口必須有一個(gè)_15_。7距離矢量路由選擇算法存在收斂慢的問題，為了克服此問題，AR

2、PANET采用了_16_。8ARP協(xié)議用于 17 。RARP用于 18 9DNS是_19_的簡(jiǎn)稱。10TCP/IP網(wǎng)絡(luò)標(biāo)準(zhǔn)的傳輸服務(wù)原語(yǔ)是 20 。二、名詞解釋（5´2¢）1. 網(wǎng)橋2. 多路分解技術(shù)3. TCP協(xié)議與UDP協(xié)議4. 流量控制5. 差錯(cuò)控制Question 3: Quickies'' (20 points)a. What is meant by the term statistical multiplexing?b. Consider an http client that wants to retrieve a WWW document a

3、t a given URL. The IP address of the http server is initially unknown. The WWW object at the URL has one embedded GIF image that resides at the same server as the original object. . What transport and application layer protocols besides http are needed in this scenario? . Suppose that the time neede

4、d to contact and receive a reply from any server (for any protocol) is RTT. How many RTTs are needed from when the user first enters the URL until the complete document is displayed? Assume that non-persistent http is used. Consider the delays of all protocols in your answer, not just those of http.

5、 c. How are ports numbers used by UDP in demultiplexing incoming segments? d. Which protocol Go-Back-N or Selective-Repeat - makes more efficient use of network bandwidth? Why?二、請(qǐng)描述Ethernet的CSMA/CD協(xié)議的工作原理。（5分）三、請(qǐng)說明橋接器（Bridges）的工作原理，與交換機(jī)(Switches)比較各有什么優(yōu)缺點(diǎn)。（5分）四、傳輸層的數(shù)據(jù)可靠傳輸機(jī)制是采用那些措施保證的？（5分）五、（5分）請(qǐng)根據(jù)如圖

6、1所示的網(wǎng)絡(luò)，說明ARP協(xié)議的工作原理。 HOST1 HOST2 HOST3IP= MAC=08005A7A7ED1 IP= MAC=08005A7A7ED2 IP= MAC=08005A7A7ED1 圖1六、（10分）考慮建立一個(gè)CSMA/CD網(wǎng)，電纜長(zhǎng)1公里，不使用重發(fā)器，運(yùn)行速率為1Gbps。電纜中的信號(hào)速度是200000公里/秒。問最小幀長(zhǎng)度是多少？七、（10分）以太網(wǎng)幀必須至少64字節(jié)長(zhǎng)，才能保證在線纜的遠(yuǎn)端發(fā)生碰撞的情況下發(fā)送方仍然在發(fā)送?？焖僖蕴W(wǎng)同樣有一個(gè)64字節(jié)的最小幀長(zhǎng)規(guī)范，但位速率提高到了10倍。它是

7、如何使得最小幀長(zhǎng)規(guī)范能夠維持不變的？八、（10分）在下列情況下，計(jì)算傳送1000KB文件所需要的總時(shí)間，即從開始傳送時(shí)起直到文件的最后一位到達(dá)目的地為止的時(shí)間。假定往返時(shí)間RTT是100毫秒，一個(gè)分組是1KB（即1024字節(jié)）的數(shù)據(jù)，在開始傳送整個(gè)的文件數(shù)據(jù)之前進(jìn)行的起始握手過程需要2×RTT的時(shí)間。1 帶寬是1.5Mbps，數(shù)據(jù)分組可連續(xù)發(fā)送。(2¢)2 帶寬是1.5Mbps，但在結(jié)束發(fā)送每一個(gè)數(shù)據(jù)分組之后，必須等待一個(gè)RTT才能發(fā)送下一個(gè)數(shù)據(jù)分組。(3¢)3 帶寬是無限大的值，即我們?nèi)“l(fā)送時(shí)間為0，并且在等待每個(gè)RTT后可發(fā)送多達(dá)20個(gè)分組。(2¢)

8、4 帶寬是無限大的值，在緊接起始握手后我們可以發(fā)送一個(gè)分組，此后，在第一次等待RTT后可發(fā)送21個(gè)分組，在第二次等待RTT后可發(fā)送22個(gè)分組，在第n次等待RTT后可發(fā)送2n個(gè)分組。(3¢)Question 9: Addressing, and following the data (10 points)Consider the simple network shown below:AEWrite down an IP address for all interfaces at all hosts and routers in the network. The IP addresses

9、 for A and E are as given. You should assign IP addresses so that interfaces on the same network have the same network-part of their IP address. Indicate the number of bits in the network-part of this address.a) Choose physical addresses (LAN addresses) for only those interfaces on the path from A t

10、o E. Can these addresses be the same as in part a)? Why?b) Now focus on the actions taken at both the network and data link layers at sender A, the intervening router, and destination E in moving an IP datagram from A to E:1. How do A, E and the router determine the IP addresses needed for the IP da

11、tagram? 2. What, specifically, are the addresses in the IP datagram that flows from A to the router. What, specifically, are the addresses in the IP datagram that flows from the router to E.3. What are three other fields found in an IP datagram?4. How do A, E and the router determine the physical (LAN) addresses needed for the data link layer frame?d) Suppose that the router in the figure below is replaced by a bridge.1. How would the IP addresses change in this case?2. Ho