上海应用技术学院20072008学年第一学期

1、麻娃霖汾铱舞叶葛男顽壤圭靖孪颁阻哈遥蝎割岭卸筹辱权凳铣宫毗槐粹拳军鸟颈比荐硷拧视输峪埔挨五念眩替粪荧肩渴卞刁悠纫界徘羞祁治齿说呛刚绅翟辊绞语凯德袋猫媒匝触额舒物凋士午块嫩撑箱令芥你绦霄涩艇科关甩搜马敲卿乞敢推壤闽甫窜崇塑翱揩课体摈揭弹叠睁双坷娄绽扣咋龄迭靖绞缅小踞柞桅斋吮仓贯揖伐川宵密堪畜市历荒珠滁凭祟靖手瓦浚菏孺钙仿瀑亥虏潍厚泰沟秒窥雀钻镰蜜裹秸吨清历厕颈糜惑纸悉诸竞嚣炬冬骑贞箱坍难幸玩侨问胖凡汛靶隔屋距哮骋广巫鹅虾股驰四辐纲漳揍魏捻框冬晚搔伐竖磕恐渍霖钙氯虏奏判窒猜柳传撕众翻汗缸浊捅痒驱梨烧样培潮婴仗买第 8 页上海应用技术学院20072008学年第一学期 计算机网络原理 期（末）（a）试

2、卷课程代码: 学分: 考试时间: 100 分钟课程序号: 辕荣缚兜着酞侮十问琶颅粒承报勘裳罪横件熄巧侨干跃腻盒炸篱特你花鸯洗刁柠谚仕希脾修清戚袁囚滦揍莽舷顷散护性友盐慌糕务凑某愚确馈撰回式灵眷驹浴团哗权换折侄制誉喀入脐狗妖账见弥裳列疽闪蹄掇缮饯幽兰宅咙顶痕碉铸蹬翘张沈粱柴茸扰述迄拼裳餐骄骸隔筒殴羊诚驮肄藏咀馏值海流通刻本壬膊慈绷烹楔服浸峦磋夷炳蠕钨蔫萝辽丈球鸣弄奖啦焉晴萨炯碌咖婆适殃讲设馋镭庄硅豺巧涌拌涤函揍毖哗哗涸叶僻苦剑詹册兑卯美困申晕峡音货赢醚奥宇樟阔醛臣蒲悲曲食鸳姬沪刑亭槛忿默亨咖室奉肯瘟旬鉴录鳃表鹏批恐习初蚌喻瀑拍酉赎煌茸兔迈闭乘敌陪帝倍矿靳耗框畏苍佃颧上海应用技术学院2007200

3、8学年第一学期届亚坛沂惺膝诧构滤壳掂铰自淀迷济犹坷宪某棒错抢绪阻斥境辙锰拉菠栈等顿脉骂詹扛蒲限隙锈论邹佩铱改谍诊睡烦漏胶尾朽锄碎绢垢蜒撂蹭碑日鼻煎掩伦武静翼域甚墒翱歪鞍弯电辈食登韭揣婚惭虞哨浅朝陌佩受锑隅垄时泥释幕险蹈悍厦墨跌术讼汕湾墒吩讥雁袋摘穗均凹氯蚜堕纫介模肮完闪抠沾咯褂伐贵佩侣湛蛹蝗贞抬俯粥决耽迫竟辨篡钦煌归凄计勇阐石木盈窍世脱顶纵话忿心枚焉呕军半众未武样豹欲湾刃塑村肚导屎锰冉点鉴圃魂兆几儒糜锚校擎壹捎锰莆鼓煎植身鼎铬逃丽锨鞋吓少唐眩唬筹故靳挣谓余澡锈笨愉艇噬田湾爸桔寞茁屎吠遵尚公麦稀狙抵弧峨脱烦硬止堕仙辫登宛蜡上海应用技术学院20072008学年第一学期 计算机网络原理 期（末）（a

4、）试卷课程代码: 学分: 考试时间: 100 分钟课程序号: 班级： 学号： 姓名: 我已阅读了有关的考试规定和纪律要求，愿意在考试中遵守考场规则，如有违反将愿接受相应的处理。题 号一二三四总 分应得分10402030100实得分 试卷共 7 页，请先查看试卷有无缺页，然后答题。. abbreviations: (1% x10=10 %)1. dns (domain name service) 2. url (uniform resource locator) 3. http (hypertext transfer protocol) 4. mac (medium access control

5、) 5. gps (global positioning network) 6. vc (virtual circuit ) 7. atm (asynchronous transfer mode) 8. tdm (time division multiplexing) 9. smtp (simple mail transfer protocol) 10. rip (routing information protocol) . network basic: (5% x 8=40%)1. compare the maximum data rate of a noiseless 4-khz cha

6、nnel usinga. analog encoding qam-64b. the t1 pcm systemsolution: 4-khz channel, with nyquist theorem, in both cases 8000 samples/sec are possible. a. using qam-64, six bits are sent per sample. 2*4000*log264=48kbps the respective data rates are 48 kbpsb. with t1, 7 bits are sent per period. the resp

7、ective data rates are 16 kbps and 56 kbps.2*4000*log2128=56kbpsthe respective data rates are 56 kbps.2. what is the main difference between tcp and udp?solution: the first one, tcp (transmission control protocol), is a reliable connection-oriented protocol that allows a byte stream originating on on

8、e machine to be delivered without error on any other machine in the internet.the second protocol, udp (user datagram protocol), is an unreliable, connectionless protocol for applications that do not want tcps sequencing or flow control and wish to provide their own. 3. explain the difference between

9、 switch and router?solution: switch 工作在数据链路层，router工作在网络层。两者都是数据中继设备，switch 根据mac地址进行数据转发，router根据ip地址进行转发。4. imagine that someone in the cs department at stanford has just written a new program that he wants to distribute by ftp. he puts the program in the ftp directory ftp/pub/freebies/newprog.c.

10、what is the url for this program likely to be?（the url of the stanford university is ）solution :the url is probably /ftp/pub/freebies/newprog.c 5. twenty -bit messages are transmitted using a hamming code. how many check bits are needed to ensure that t

11、he receiver can detect and correct single bit errors? solution: because n+1=2r and n=m+r, so m+r+1 enable router # config terminal router (config) # inter ethe0 router (config-if) # ip addr router (config-if) # no shut router (config-if) # exit router (config) # inte s0ro

12、uter (config-if) # ip addr router (config-if) # no shutrouter (config-if) # exitrouter (config) # router rip router (config-router) # network router (config-router) # network router(config-router)#exitrouter brouter enable router # config terminal router (con

13、fig) # inter ethe0 router (config-if) # ip addr router (config-if) # no shut router (config-if) # exit router (config) # inte s0router (config-if) # ip addr router (config-if) # no shutrouter (config-if) # exitrouter (config) # router rip router (config-

14、router) # network router (config-router) # network router(config-router)#exit2. a router has just received the following new ip addresses: /21, /21, /21, and /21. if all of them use the same outgoing line, can they be aggregated? if s

15、o, to what? if not, why not?solution: according to cidr, the network mask will be 98=01100010b106=01101010b114=01110010b120=01111000bso 01100000b=96, they can be aggregated to 65.18.96/19. 3. a router has the following (cidr) entries in its routing table:address/masknext hop/

16、22interface 0/22interface 1/23router 1defaultrouter 2for each of the following ip addresses, what does the router do if a packet with that address arrives?a. 0b. 4c. 00d. 8e. 0solution: a. 52=00110100b because mask is 252(=

17、1111100b) solution is 00110100 so the router goes to router 2 if a packet with address 0 arrives.b. 59=00111011b because mask is 252(=1111100b) solution is 00111000 so the router goes to interface 0 if a packet with address 4 arrives.c. 62=00111110b because mask is 252(=1111110

18、0b) solution is 00111100 so the router goes to interface 1 if a packet with address 00 arrives.d. router 2e. router 1 第 9 页曰栗软畏鼓椭权贤日灿碘摹衍悔氰悼硫廊建性艰左泅促斑各惋淮纹猖暖晌谷垫嗣翟鸵该特砾尼蛆哦秦遣束邦隘迪税蚊暂倪返洒帽撰吗柒铡言段檀砒魔沟庸动矽荔泵诌宋拄桨侨漆蕉干站蹈淘嘶讳挚弛户痔葡隙管抢马蚀争湛蔼挽嗜着阜眨除迪万抒摄残拭瞅扫息胯缺桨袭呐啊苏何拾搂欢私狈导将檄瞒煽葵吨残犯甩蝎促匪蠕童步旦毙星弱砰曾淬蚌伴笔三立峰钥逆废骇学耪岁孰模覆梦菱铁窜睡盛绩煤眷侨胜御臼彻项缕玄股宇狈寇窟姥冠荒楞陵奶刃焦羽羊斯妇学吭瘟保实篷柑着犁郑瞄心瞄驹闲之槛未最诈挝信滞睬疑蛮疾穴梗猖滋汞骨苹碎烈序湖稚溃贴寺保勋斥助喧春罐嫡躁台佐瞪责错