哈工大计算机网络2018实验四及报告

所需积分/C币:26 2019-01-20 09:53:12 1.41MB PDF
48
收藏 收藏
举报

哈尔滨工业大学16级的计算机网络实验一 截止到2019年初这是最新的版本!! 实验报告清晰明了 (已经过TA验收)可以直接上交 这一次实验是使用wireshark 所以只有报告就可以了,没有代码
算机网终 报告 藏为刚[irhηk·1.11v1.1114017483 r torr masTer.1.1 hk edit Vew Gc CaptUre Analze 5taistcs telephony cols Inderra s Hdp 0d题 盆阳改命冒业国国aQ 感区愚|国 ear ApFl Pratcc-Lengthinio d query Uxd5as aaaA Www. n1twr. edu Ci 81295 7( standard query 0x4973 A wo t 5-andard query 0x49/3 A w. hitwh, ecu. 1484511,3703C40172.17.20.214 484611 Aaa Www. nitwit edu cn 12 CET /_uload/=itc,1,cty lc/3/3. CcE HTTr/1. 1 3n37.8Ri4750719.217.776.1577.17.70.714HTTF35Fhttp:/1.17c0 057.8506500-19,217.225.15 72.17,20.214 58 ITT 2C0 oK -porrletplugs/simlenews/css/si np lene 4 461 GET /js/-FortletPlug>/datepicker c33//date 1.1 Fr ane 23L bytes on wire (0736 b ts), 467 bytes captured (3736 bit 0 ver sior4.5rc:1!.1/.20.4/2.1!,L.214).Dt:219.21/.22,1}(9,2/∠4 印 pertex Transter"oco1 Uer -AgenL: vuill/.0winduno NT 10. D; WON54 Appl i L/ 537. 36 (KHTML, like Gecko)chir ume/510. 2704 ITari/537,36\t\n cion/ htn Ixnl, application/xn1: 9-3, 9,image/ wcbp. * /*:g-08v\n C3:20343评2五到 6图 Ethernet(eth,1 ①你的浏览器运行的是TTP1.0,还是mTP1.1?你所访问的服务器所运行 HTTP协议的版本号是多少? 我的浏览器:HTTP1.1 服务器:HTTP1.1 ②你的浏览器向服务器指出它能接收何种语言版本的对象 zh-CN,简体中文 ③你的计算机的IP地址是多少?服务器http://www.hit.edu.cn的IP地址是 多少? 我的计算机:172.17.20.214 服务器:219.217.226.15 ④从服务器向你的浏览器返回的状态代码是多少? 200 (2)HTTP条件GET/response交互 藏叫太 w Wireshark1,1211y1.2114g:i4c8 c from mmaster. Fik Edit Vew Gc CcptLre Analyze Statistcs Telephony cols Interras HdlF d鼴盛|3回(白命与冒是回国QR及图|画区赐|国 Clear Apply Save rotocol Length inic 起速 Tify Http/l 20. 11.5921C30219.217.226.15 oE GET /_is/-Fortletplugs/simpleness css/si npleneN/. css lITTP/1 1 2.17.2C.214 HTTF DHTP/1.L∠CU3K ss/da-epicker.css Http/11 dynav. css Http/1 Htte 4oe Http/1.l 2c0 OxL/CSS) f Etherret II, 5rc: Lcf cHefe 29: f4: 9d [28: d2: 44: 29: f4: yd cst: tr-Linkt e5: 8d: b [08: 57: 00: e: d: b5) D Transmission Contrcl Protocol, Src 6)-y-Dct-Poat+00-(80)r 5e0--Ack. Len Gecko) chrone/510.2704. 84 Saf ari/537 3E\\n n+xm, applicat lon/xn1: Q=J 9, Image webp 3"3绀3非出另3罡毀∵:;"; duen cere Ou Keady to load or capure ①分析你的浏览器向服务器发出的第一个 Http Get请求的内容,在该请求报文 算机风 报告 中,是否有一行是: IF-MODIFIED-SINCE? 没有 ⑨分析服务器响应报文的内容,服务器是否明确返回了文件的内容?如何获 知? 服务器明确返回了内容 Http Status code(状态代码)为304时不明确返回文件 httpStatuscoDe(状态代码)为200时明确返回文件 分析你的浏览器向服务器发出的较晚的“ Http Get请求,在该请求报文中 是否有一行是: IF-MODIFIED-SINCE?如果有,在该首部行后面跟着的信息是 什么? 1484)1∈.6122310172.17.23.214 19,217.225.15 TTP 616 GET cs5/_system/ system. cs5 I TTP/1.1 1484116.6139980219.217.226.15 72.17.20.214 259 Http /1.1 304 Nul moDiied 1484216.6117120172.17.23.214 n9.217.226.15 Http 619 GET up load/site//'style/3/3 Http/1.1 148431∈.6152270172.17.2.214 ITTP 626 GET _upload/site/00/14/20/ style/5/5 1444l.6I7no071号.217.76.15 l77.l7.20.714 HI IF 757 HI IP/1. 1 304 NOT Modified 1484516.6186180172.17.2.214 19.217.226.15 Http 42 GET /js/_port letF lugs/sip enEws/css/ simp Enews,Css 148461C.6187860219.217.226.13 172.17.20.214 257 Http/1.1 304 Not noDified 1484716.62006301了217.2.214 219.217.226.15 Http 98get /js/jquery. min. js htTp/1.1 1484B16.6215570172.17.23.214 19. 217.226.15 Htti 611 Get /js/jqucry. sudy wp. visitcount. is Http/1.1 1484916.6219950219.217,226,15 172,17,20,214 Http 251 htTp /1,1 304 Nol moDiied 1433正.62p99929.22.1 1/2,14.20.∠14 Http 262http:/.1304NotModified 148116.623567Q172.17,20。21 219.217.225.15 Http 525 Clt upload/tp1/c930/48/temp latc48/stylc. ce5 Http/1 Connection: Keen-all\r n Cache-Control: max-age-O\r\n User-Agent: Mozilla/ 5.0 windows NT 10. 0: wOwG4) App lewebKi/537,30 (KHTML, like Gecko) Chrome/51 0. 2704.84 safar:/537 Accept: i/s\\n fcrcr:http://ncws.hit.cducn/xxyw/main,htmr\n Accept-Enccdi ng: gzi flate ch\\n Accept-Lancu age GA.3.125139911.1462348758;J5Es510v0DE130847045E306375E9A64E99FCrn Ur-dLh:"1e6062『-31-524dbg8599400"\r -Mocifiec-since: Thu, 19 NoV 2015 02: 34: 16 GMT\\n Mwullreuesl URT: hLLu/ems hiLe u is/iwer y such,visilcuullis1 IPrev request mframc-R28R 03 53 3R IC 4a Do 28 02 64 so as d ia on db do 0030 有 这个字段后面代表的是时间,即咨询服务器在这个时候之后是否有更新 ④服务器对较晚的ITTPGET请求的响应中的HTTP状态代码是多少?服务器是 否明确返回了文件的内容?请解释。 藏以太网 ires hark1.1211(v1.1-ug/4 c33c from master-1.12 ile Edit view Go Capture Analyze Statistics Te ephory Tools Internals Help ⊙A题回翼|0中申苍国国|成回画圖嗯| Fills:hlp Exutessiulm.Clear ApplySare Destinat on Protocol Length Info 1484416.6170000219.217,226.15172.17.20,214 257HTTP/1.1304 Jd ified 4845L6.6⊥86180⊥72.⊥7,20.214 15/porretwlugs/simpl eNews/cS/simplenewsCSs HIIP/1.1 14B46i6.618/860∠19.21.∠6.1 L/2.1/.20.214 Http 5/ Http/i Not Modified 14B716.6200630172.17,20.21 219.217.226.15HTp598GET/_1 lery min. js Http/1.1 219.217.226,15 Http 611 Get /js/iaucry.sus visitcount.js HTP/1.1 14B4916.6219950219.217,226.15172.17.20,214 261 HITTP/1.1 104 Not Modified 145016.6229990219.217,226.1 172.17.20.214 ITTP 62HTTP/1 1 lo4 Not Modified 4B5116.6235670172.17.20.214 219.217.226.15 Http 635 Get /_uplpad/tp1/03/3c/48/template48/style. c5s Http/1.1 l485216.6239990219.217,226。15 L7z.17.20,214 260 HI IP/L 1 bo4 NOT Modified 14'316.6/h6"01.1!,P)/14 /14./1/.77b.15 HttP h19 Get/_p1pad/tp1/0): )/4( /48/ template4x/c<s/minex.css HttP/1.1 148516.6259930219.217,226.1 2.17.20.214 60http:/1.1104noTModitied 14B5516.6273430219.217,226.1 172.17.20.214 Http 259HTTP/1.1304No 14B5916.7474040172.17,20214 10.32.125.40 IITTP 50 GET /code/iathis_utility.html IITTP/1 145016.7490710172,17,20。214 210,32,125:40 Gg Get /code/cs/jiathis_share, c55 Http/11 1486116.7706630172.17,20.214 219.217.226.1 Http em/sysTen edi LUr. LS> MiTP/1 f Ethernet Frc: Lcfcllefe 29: f4 Link_es: Bd: b5(03: 57.00:E5: 8d: b5) 团 Internet protocol version4:Src:172.17.20.214(172.17.20.214),Dst:219.217.226.15(219.217.226.15) o Transmission Control Protocol, 5rt Port: 3350 (335G),Dst Port: 80(80),5eq: 1376, Ack: 99302, Len: 557 A Hyper Lext TI dris[er PrULuLUl B Gft /js/Iquery. sudy wf. Visitcnunt. is Http/1. 1\r\N Host: news. ht. edu. cn\r\n Connection keep-alive\r\ 请求响应中的HTP状态代码为304 不会明确返回文件,因为根据之前HTP的GET请求中 IF-MODIFIED-SINCE 算机风 报告 字段内的时间服务器判断结果为 Not Modified,于是客户端可以使用木地这个 没有过期的缓存文件 (三)TCP分析 报文捕获: *NLAN 文科编辑E)瓶图^跳转(G)龚获O)分析(A统计)电话Y)无线Y工具T帮助〔H ■⑧日为巴只乡香业国当照 区表达式+ Protocl Length I 33.053279 192.168.199.228 128.119.245.1 6614498+83[SY]Seq=0Win=8192Ler=MSs=1458kS=25 43.399538 192,168.199.228 128.119.245.12 661449983[sY]seq-0win-8192Ler-0Nss-1453ks-25 53.b/b4b 6b8→149 [SYN, ACK Seq=b Ack=l Win=29208 Len=U MS 63,67653 192,168.199228 128,119.245,12 414499+Ba LACK] Sey=_ Ack=1 Win=16384 Len=B 73.677726 192.168.199.228 128.119.245.1 TCP 1514 [TCP segment cf a reassembled PDU] 83.677751 99.228 TCP 15-4 [TCP segment cf a reassembled PDU] 93.677758 192.168.199.228 128.119.245.12 1514 TCP segment of a reassembled PDU] 103.677788 192,168.199.228 128.119.215.1 15-4 TCP segment cf a reassembled PDu] 8.119.245.12 l68.199.278 548→14499[ACK]seq=1Ak=1461w1r=?1?8Pn= 144,845516192.168.199.228 128. 119. 245.12 TCP 1514 TCP segment cf a reassembled PDU 15A.13512 192.168.199.2 128.119.4.12 15_ [ICP segment ct a reassembled PUU] 128,119.245.12 488→14499[ACK]5eq=1Ack=2921Wr=35972Len=0 174,84573 128,119.245.12 192,168.199.228 TCP 64 80>14402 ACK Seq-1 Ack-5341 Win-40960 Len-d 184,943859 192.168.199.228 128.119.∠45.12 I CP 1514 [ICP segment ct a reassembled PLU] 194,915893 192.168.199.228 128.119.245.12 1514 [TCP segmenL uf d reassembly PDU] Frame 3: 56 by les un wire (528 biLs), 66 by Les Captured (528 bils) ur inlerfdLt a Ethernet II, Src: IntelCon_c7: 02: 4c (60: 67: 23: C7: 02: Lc),Dst: Hiwifi Cc: be: 56 (d4: cc: 07: 0c: bc: 55) nternet Protocol version 4. Src:192.68.193.778. st:128.111.745.17 Transmission Cortrol Protocol, Src Port: 14493(144S8,Dst Port: 80(E0),5eq:a,Len:0 000d4ee76eb5E6726c7324cE8450 0818E343e664地e4ca8口7487.fg.@ 0029千58c38a23358sedd7bb33ece90882.8..p 08382E88d16328829405b4219383989191 949492 ①向gaia.cs. umass.edu服务器传送文件的客户端主机的IP地址和TCP端 口号是多少? 192.168.199.228 14498 ②Gaia.cs.umas.edu服务器的IP地址是多少?对这一连接,它用来发送和 接收TCP报文的端口号是多少? l28.119.245.12 ③客户服务器之间用于初始化TCP连接的 TCP SYN报文段的序号( sequence number)是多少?在该报文段中,是用什么来标示该报文段是SYN报文段的? ITCP Segment Len: 9] Sequence number: 0 (relative sequence number) Acknowledgment number: a Header Leng Lh: 32 by Les gS: 0x0a2 (SYN) 0e0..- Reserved: Not set Nonce: Nct set 0.,...=Congestion Window Reduced (CWR): Not set ,8. ECN-Echo: Not set Urgent: Not set Acknow lodgment: Not set Fush: Not sct 月.= eset:Nct 如图,初始化TCP连接的 TCP SYN报文段的序号是0 计算机风实验报告 通过 Flags标志位,将其中的SYN位置为1,表示该报文段是SYN报文段 ④服务器向客户端发送的 SYNACK报文段序号是多少?该报文段中, Acknowledgement字段的值是多少?Gaia.cs. umass. edu服务器是如何决定 此值的?在该报文段中,是用什么来标示该报文段是 SYNACK报文段的? [TCP Segment Len: 0] Sequence number: 0 (relative sequence number) Acknowledgment number: 1 (relative ack number) Header Length: 32 bytes ˇ Flags:x812(SY,ACK) 000= Reserved Not set Nonce: not set l■■■ Congestion Window Reduced (CWR): Not set :日 ECN-Echo Not set …∴.0.∴...= Urgent: Not set .1 Acknowledgment: Set I1■■■■1■ 9 ■■■ Push: Not set Reset: ot set 1.= Syn: Set 0= Fin: Not set TCPF1ags:*x事不率A*S重 如上图,服务器端向客户端发送的报文段序号为0; 服务器发的 acknowledgment number字段是根据上一次客户端发给服务器的 seq+1得到的; 通过 Flags标志位中的SY位和ACK位都是1来确定该报文段是一个 SYNACK 报文段的 ⑤你能从捕获的数据包中分析出tcp三次握手过程吗? 1460 W5=255 SACK FERM1 ∈3.676693102.168.199.22828.119.24512 TCP 54 14492>80 [ACK Seq-l Ack-1 Win-16384 Len-C 66 14499+8E SYN Sey=0 Win=8192 Len=0 MSS=1460 WS=256 SACK PERM=1 6680+14499 [SYN, ACK] Seq=0 Ack-1 Win-29200 Len-0 MSS-146B SACK_ PERM-1 NS=128 541449988[ACK]seq=1Ack=1win=16384Len=0 首先客户端向服务器发送seq=0的建立连接的请求 然后服务器向客户端返回seq=0,ack=0+1=1的响应 ⑥包含 Http Pc0sT命令的TCP报文段的序号是多少 2915.664115 192.168.199.228 128.119.245.12 1135 PosT /w1resHark-labs/lab3-1-reply. htm Http/1.1(teXt/plaln) 128.119.245.12 192168:199.228 66[TCP Dup ACK 244it1] 80+ 14499 [ACK] Sc 2511575718419.168.199223 111221.29.254 TI Sv1. 2 875 Application Dat 192.168.199.228 TLSVI, 2 779 Application Data 5 bytes on wire (9080 bits), 11]5 bytes captured (9080 bits)on inter face E 上 thernet⊥, Src: IntelLor c/: 02: 4c(bU: 6/: 20: c/: 02: 4c Ust: Hiwiti de: be: 5b(d4: ee: 0/: ke: be: 56) nLer neL Protocol version 4. sr.c 192.168-199228,DL:128.119.245.12 Transmission Control Protocol, Src Port: 1499(1/99), Dst Port: 80(B0),Seq: 151791, Ack: 1, Len: 108_ Source port: 14499 Lon pcrt: 0 [Stream index: 1] LTCP equence number: 151791 (relative sequence number) 2872re1 (relative ark numher h: 26 byt 20f538a3ee5e4 ds e0 af ea g962215018..8.P4,,b!P Sea=151791 算机网终实验报告 ⑦如果将包含HTPP0ST命令的T¢P报文段看作是TcP连接上的第一个报 文段,那么该TCP连接上的第六个报文段的序号是多少?是何时发送的?该 报文段所对应的ACK是何时接收的? 24915.684115 2.168.199.228 Http 1135 PosT wireshark labs/lab3 1 reply htm htTp/1.1(TeXt/plcin) 25715.981201128.119.245.12192.168.199228HTTP 833 Http/1. 1 200 ok (text/htm1) 31524.812194e8::b41b:574c:4fd…fe89::1d96:332a:d72HTTx,37 PoSt de427db--842-4595-a151-a35f6e9d2dad/HT/1.1 31924.813189fe88::41b:574c:4fd,fe86::1d66:532a:d72. Http/x837 POST d8427dbc-1842-4595-a151-a25f6e9d2dad/HTP/1.1 2324.828989192.1∈8.199.18239.255.255.250 179mseArch*http:/1.1 P8::1:3a:77.8::h1h:574(:4千.HTpX985HTTP1.1 32924.836721e8e::1ce:32a:d72fe89:b41b:574c:45d.HTTx.985HTp1.129 3454.852818 19.1F8.199778 12.168.199.-8B Http/x..787 Post d9477dhr-d847-4595-a151-a25f6e9d7dad/ htTp/'l.'1 34924854772192,168.199228192,168,199.-89HTPX787PoST!d9427db=-#842-4525-a151-a35f692dad/HP/1,1 bH||P/1.12 35624.865516 192.168.199.18E 192.168.199.228 TTPX945|TP/1.1269 Transmission Control Protocol, Src Port: 14499(14499,Dst Port: Ba (80),Seg: 151791, Ack: 1, Len: 1081 ˇ[109 Rcasscmblcd TCP Segments(152871 bytes):#7(1460),#8(1459),#9(1459),#13(1450),#14(1460),#15(1460),#18(14∈e),t19(1460),t20(1460 IFrame: 8, payload: 1450-2919(1460 bytes)I IFrame;9,pala;:2920-437)(146htes)1 IFrame: 10,payload: 4380 5839(1460 bytes] Frame:14,pay12a:554-7299(1460btes iFrame:15,pal:7399-8759(1460 bytes) 「 Frame:18,pay13小:8768-18219(146htes)l 「「 rame:19,pay1d:193229-11679(1460btes) 第六个报文段Seg=7301在httppost发送之前,tcp连接建立之后发送。 144.845516 192.168.199.228 128. 119. 245.12 TCP 1514 [TCP segment of a reassembled PoU] 154.0451121619221212112170154 TCP segment of a reassemb1edU Y Trarsmission Control Protocol, Src Fort: 14499(14499), Dst Port: 80(80), Seq: 7301, Ack: 1, Len: 1460 ounce po叶:14499 DesLindliu fur L: 50 Stream index: 11 LTCP Segment Len:14」 quence nuimber: 7301 r≌1〓 tve seru [ Next sequence number: 8761 (relative sequence number)] Acknowledgment number: 1 (relative ack number) Header Length: 20 bytes Flags: 0x010(ACK) Window size value: 64 25斗49b2 92.162.199,22 12.11y,245,12 154|s∈ gment ot a reassembled P儿 744.41△481 1?8.119?45.1 9?168.199.228 5480+1499[a〔K]5eq=1ak=7301wr=43984len=8 254.414492 128.119.245,12 192.168.199.228 TCP 54890→14499ACK1Seq=1Ack=8761Wir=46723Len=0 264.414549 192.168.1 128.119.245.12 1514 ITCP segment of c reassembled PJU 274.41457492.168.199.228128.119.245.12 15-4 [TCP segment of a reassembled PoU] Transmission Control Protocol, Src Fort: 80(80), Dst Port: 14499(14499), Seq: 1, Ack: 8761, Len: 0 Destination port: 14499 .TCP Segment Len: el Sequence rumber: 1 (relative sequence number) Acknowledgment number: 8761 (relative ack nunber) Feaden Lergth: zB bytes Oxe10 (A 对应的ack即为服务器返回的第六个ack ⑧前六个TCP报文段的长度各是多少? 算机网终实验报告 ,1 Ler 24915.604115192.168.19.228128.119.245.12HTP1135P0ST/wireshark-las/1a3-1-ely.htmhttp:/1.1(text/plan) 25715.981201128.119.245.12 2.168.199 Http 833 Http/1.1 200 ok (text/htm1) 31524.812194 8::t41b:574c:4f…fe86::1d66:532a:d72HTTP!X 27PoST!d8427dbc-4842-4595-a151-a35f6e9d2dad/HTTP/1.1 1924.813199 e8::b41b:574c:4fd…fe89::1d66:332a:d72…HTpX,937poST/de427dbc-d842-4535-a151-a35f6e9d2dad/HT/1.1 234.8288192.1F8.199.18E 79.255.755.)5 SSDP 179 M-sfarch & Http/1.1 32624.831663e8::1cee:32a:d72fe8::b41b:574c:4 fda Http,985HTTp/1.129 24 34524.852018192.168.199.228192.1608,19.8TTP/X.737P0sTd427dbc-B42-4535-a151-a35f6e9d2dad/ITP/1.1 a924.84//2 192.168.199.222 192.168.199.88 HI IP/X. /8/ PoSI /d042dbc-d842-4595-albl-a05tbe9d2dad/ HI IP/1. 1 35224.86374 192.18.199.186 192.168,199.228TP/x.945|TP1.1290 35524.865516192.1E8.19918E 92.168.199.228 Http/X.. 945 Http/1.1 200 Transmission Cortrol Protocol, Src Frame: 7, payload: 3 460 bytes 「 Frame:3,pal3:1458-2919(1460htes) IFrame: 9, payload: 2920 4379(1460 bytes) fRame: 10, payload: 4380-5839_(1460 bytes] IFrame: 14, payload: 5940-7299(1460 bytes)1 「 Frame:15,pay1n3小:73-8759(146htes Iframe:18,pala:8760-10219(1460btes) -Frame; 12.payload: 12220-11679(1460 bytes ⑨在整个跟踪过程中,接收端公示的最小的可用缓存空间是多少?限制发送端 的传输以后,接收端的缓存是否仍然不够用? 如图,接收端公示的最小的可用缓存空问是29200,该窗凵大小会一直增加, 所以不会出现接收端的缓存是否仍然不够用的情况。 41.309586192.163.139.228 128.119.245.1 TCP 5612499÷80L5N5eq-win-6192Len-6N55-1466 53.676468128.119.245.12192.168.199.228TCF 56 80+14499 [SYN, ACK Seq=e Ack=l Win=29200 Ler=0 192.168.199.228 5414499>80 ACKI Seg-1 Ack-l Win-16384 Len-a 73.677726 192.168,199.228 128.149.245.12 TCP 1514 TCP segment of a reassembled PDUI 83.67775192.168.199.228 128.119.245.1 1514 [TCP segment cf a reassembled PDU 9.6///曷292.168,19,228 18。119。21,12 51 [ICP segment ct a reassembled PDUI 13.677788 199.228 128.119.245.12 1514 [TCP segment cf a reassembled PDU] 134.45373 128.119.245.12 192.168.199.228 5488+14499[aK]saq1ack=1461win=32128Ln= 144.945516 19.168.199.278 128.119.245.12 1514 [TCP segment cf a reassembled Pnl! 192.168,199.228 28.119 245 12 TCF 1514 [TCP segmenL uf d Reassembled PDUI ITCP Segment Len: 0] Sequence number: g (relative sequence number Acknowledgment Crclativc ack number) Header Length 32 bytes ⊥ags:x12(SYN,K) Window size va_ue: 29202 [Calculated window size: 2920a] Checksum: Dxb44a [validation disabled] Urgent pointer: 0 ⑩在跟踪文件中是否有重传的报文段?进行判断的依据是什么? 没有出现重传,因为客户端发送的报文序列号没有出现重复。 Transmission Control Protocol, Src Port: 14499(14499), Dst Pc [109 Reassembled TCP Segments(152871 bytes): # 7(1460),#8(146 IFrame: 7, payload: 0-1459(1460 bytes)l Frame: 8, payload: 1460-2919(1460 bytes) Frame:9 payload: 2920-4379(1460 bytes)l Frame: 10, payload: 4380-5839(1460 bytes) Frame: 14, payload: 5840-7299(1460 bytes)1 ①TP连接的 throughput是多少?请写出你的计算过程 由图可知,发送数据总的长度为 发送时间间隔约为 因此吞吐量为 (四)IP分 A.对捕获的数据包进行分析 1.在你的捕获窗口中,应该能看到由你的主机发出的一系列 ICMP Echo Request 包和中闫路由器返回的一系列 I CMP TTL- exceeded消息。选择第一个你的主 计算机区络。实验报告 机发出的 ICMP Echo Request消息,在 packet details窗口展开数据包的 Internet protocol部分 ①你主机的IP地址是什么? 015.9320660172.17.150.171 219.217.226.1 70 Echo(pind 77115.9331630172.17.150.254 172.17.150.171 ICMP 70 Time-to-l 172.17.150.171 ②在IP数据包头中,上层协议( upper layer)字段的值是什么? F Time to live: 1 Protocol: ICMP (1) B Header checksum: oxb7e5 [validation disabled] [Good: Falsel [Bad; False] source:172,17,150,171(172.17,150,171) Destinat ion:219.217.226.15(219217.226.15) [Source GeoIP: unknown] Destination GeoIP: unknown] a Internet control Message protocol 0000024dcb81a1d78843cb12cboo80o4500.s..,.X,<,,,E 020e20f0800389800 02cc304550696e67 8 030506c6f7474657250726f332e34312e30 Plotter ro3.41.6 0703045506 pOPIn 01 ③IP头有多少字节?该IP数据包的净载为多少字节?并解释你是怎样确定该 IP数据包的净载大小的? Ether net II, src: sony- b1:2c, bc (28: 84:3C, D1:2C: b0), bst, JuriperN-bE:1a:1d(00: 24: dc:b8=1a:1d. Internet protocol version4,src:172.17.150.171(172,37.150.171),Dst:213.217.226.15(219.217 version: 4 Header Length: 20 bytes a Differenti ated services Fiele: Oxo (DSCP 0x00: Default: ECN: 0x00: Not-ECT (Not ECN-Capable 0000 00.. m Differentiated servces codepoi nt: Default (Gxoo) 00- Explicit congestion Notification: Not-ECT (NGt ECN-capable Trans port)(0x00) Total Length: 56 Identifica ion: 0x013a (314) B Flags: 0xoo 0.+, itit- Reserved bit: Not set .O,,,t,,- Don'r fragment; Not set More fragments: Not set Fragment off set: o 日 Time to live:1 E [Expert Info (Note/ sequence):"Time To Live" only 1] Protocol: ICMP (1) a Header checksum: oxb7e5 [validation disabled [Good: False [Bad: False] Source;:172.17.150,171(172.17,150.171) Desti nation:219.217226.15(219217.226,15 [Source GeoIP: Unknown] [Destination GeoIp: unknown] Internet Control Message Protocol IP头有20字节。 IP包的净载为 Total length- Header Length-56B-20B-36B ④该IP数据包分片了吗?解释你是如何确定该P数据包是否进行了分片 dentification: 0x01 3a (314) 日F1ag5:0x00 o,,.- reserved bit: NOT SeT Don't fragment: Not set 0.∴∴,= More fragments: Not set Fragment offset: 0 日 Time to live:1 计算机网终 报告 没有,分片位移为0, More fragments为0表示后面无分段。 2.单击 Source列按钮,这样将对捕获的数据包按源IP地址排序。选择第一个 你的主机发出的 ICMP Echo Request消息,在 packet details窗口展廾数据 包的 Internet Protocol部分。在“ listing of captured packets”窗口, 你会看到许多后续的ICMP消息(或许还有你主机上运行的其他协议的数据包) ①你主机发出的一系列ICMP消息中IP数据报中哪些字段总是发生改变? ID、TTL、 Header checksum这三个字段总在变化。 ②哪些字段必须保持常量?哪些字段必须改变?为什么? 必须改变 ID鉴别码,用于区分不同的数据包; TTL来自于 traceroute的要求,用来测试路径上的路由信息; Header checksum首部校验和,前面的字段改变,该值也必须跟着改变 必须保持常量: 除以上(ID,TTL, Header Checksum)外的字段保持常量。 ③描述你看到的IP数据包 Identification字段值的形式。 16位,在某一范围内是+1递增的。 3.找到由最近的路由器(第一跳)返回给你主机的 I CMPTime-to-live exceeded消息 思考下列问题: ① identification字段和TTL字段的值是什么?最近的路由器(第一跳)返 回给你主机的 ICMP Time-to-1 ive exceeded消息中这些值是否保持不变?为 什么? 不变,IP是无连接服务,相同的标识是为了分段后组装成同一段,给同 个主机返回的ICM,标识不代表序号,TL消息是相同的,因此 Identification不变;因为是第一跳路由器发回的数据报,故TIL是最大值减 1,总是等于254 4.单击Time列按钮,这样将对捕获的数据包按时间排序。我到在将包大小改 为2000字节后你的主机发送的第一个 ICmP Echo Request消息。

...展开详情
试读 14P 哈工大计算机网络2018实验四及报告
立即下载 低至0.43元/次 身份认证VIP会员低至7折
一个资源只可评论一次,评论内容不能少于5个字
您会向同学/朋友/同事推荐我们的CSDN下载吗?
谢谢参与!您的真实评价是我们改进的动力~
  • 分享精英

    成功上传11个资源即可获取
关注 私信 TA的资源
上传资源赚积分or赚钱
    最新推荐
    哈工大计算机网络2018实验四及报告 26积分/C币 立即下载
    1/14
    哈工大计算机网络2018实验四及报告第1页
    哈工大计算机网络2018实验四及报告第2页
    哈工大计算机网络2018实验四及报告第3页

    试读结束, 可继续读2页

    26积分/C币 立即下载 >