谁知道"An error occurred in the secure channel support" 造成这个错误的原因

chithany 2006-07-18 04:46:58

今天
突然在自己的event log中看到这样的一个错误。

SiteName : Basket

ASPCode :

Number : -2147012739

Source :

File : /Basket/include/CbstdCMCTranslateXml.asp

Line : 149

Description : An error occurred in the secure channel support

source:

Line : 149
是一个httprequest 对象的 send 方法
哪位高手有碰过这样的问题

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chithany 2006-07-18

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方法绝对没有问题。
因为已经运行了一年，这个log是第一次出现

这两天接到通知，说公司的一个网站访问不了，经访问发现页面提示如下错误：代码如下: msxml3.dll 错误 ‘80072ee7’ The server name or address could not be resolved 看提示应该是域名或地址无法解析，百度一番都是什么在hosts文件中加解析或者是DNS设置。我从我的实际情况来说，不是这个问题，因为之前这个网站所在的服务器是一直没法访问外网的，都是可以正常访问，只是这两天无法访问而已。经过查看提示文件，发现该文件是global.asa文件，而且修改时间还是最近两天的，而我们又没有更新网站。所以，第一念头就是“是不是被挂木马了？”

让Win7的WinHTTP支持TLS1.2，修复WinHttp.WinHttpRequest或msxml3.dll出现安全频道支持出错问题

Computer Networking: A Top-Down Approach, 6th Edition Solutions to Review Questions and Problems Version Date: May 2012 This document contains the solutions to review questions and problems for the 5th edition of Computer Networking: A Top-Down Approach by Jim Kurose and Keith Ross. These solutions are being made available to instructors ONLY. Please do NOT copy or distribute this document to others (even other instructors). Please do not post any solutions on a publicly-available Web site. We’ll be happy to provide a copy (up-to-date) of this solution manual ourselves to anyone who asks. Acknowledgments: Over the years, several students and colleagues have helped us prepare this solutions manual. Special thanks goes to HongGang Zhang, Rakesh Kumar, Prithula Dhungel, and Vijay Annapureddy. Also thanks to all the readers who have made suggestions and corrected errors. All material © copyright 1996-2012 by J.F. Kurose and K.W. Ross. All rights reserved Chapter 1 Review Questions There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, PDAs, Internet-connected game consoles, etc. From Wikipedia: Diplomatic protocol is commonly described as a set of international courtesy rules. These well-established and time-honored rules have made it easier for nations and people to live and work together. Part of protocol has always been the acknowledgment of the hierarchical standing of all present. Protocol rules are based on the principles of civility. Standards are important for protocols so that people can create networking systems and products that interoperate. 1. Dial-up modem over telephone line: home; 2. DSL over telephone line: home or small office; 3. Cable to HFC: home; 4. 100 Mbps switched Ethernet: enterprise; 5. Wifi (802.11): home and enterprise: 6. 3G and 4G: wide-area wireless. HFC bandwidth is shared among the users. On the downstream channel, all packets emanate from a single source, namely, the head end. Thus, there are no collisions in the downstream channel. In most American cities, the current possibilities include: dial-up; DSL; cable modem; fiber-to-the-home. 7. Ethernet LANs have transmission rates of 10 Mbps, 100 Mbps, 1 Gbps and 10 Gbps. 8. Today, Ethernet most commonly runs over twisted-pair copper wire. It also can run over fibers optic links. 9. Dial up modems: up to 56 Kbps, bandwidth is dedicated; ADSL: up to 24 Mbps downstream and 2.5 Mbps upstream, bandwidth is dedicated; HFC, rates up to 42.8 Mbps and upstream rates of up to 30.7 Mbps, bandwidth is shared. FTTH: 2-10Mbps upload; 10-20 Mbps download; bandwidth is not shared. 10. There are two popular wireless Internet access technologies today: Wifi (802.11) In a wireless LAN, wireless users transmit/receive packets to/from an base station (i.e., wireless access point) within a radius of few tens of meters. The base station is typically connected to the wired Internet and thus serves to connect wireless users to the wired network. 3G and 4G wide-area wireless access networks. In these systems, packets are transmitted over the same wireless infrastructure used for cellular telephony, with the base station thus being managed by a telecommunications provider. This provides wireless access to users within a radius of tens of kilometers of the base station. 11. At time t0 the sending host begins to transmit. At time t1 = L/R1, the sending host completes transmission and the entire packet is received at the router (no propagation delay). Because the router has the entire packet at time t1, it can begin to transmit the packet to the receiving host at time t1. At time t2 = t1 + L/R2, the router completes transmission and the entire packet is received at the receiving host (again, no propagation delay). Thus, the end-to-end delay is L/R1 + L/R2. 12. A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth. FDM requires sophisticated analog hardware to shift signal into appropriate frequency bands. 13. a) 2 users can be supported because each user requires half of the link bandwidth. b) Since each user requires 1Mbps when transmitting, if two or fewer users transmit simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link. c) Probability that a given user is transmitting = 0.2 d) Probability that all three users are transmitting simultaneously = = (0.2)3 = 0.008. Since the queue grows when all the users are transmitting, the fraction of time during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008. 14. If the two ISPs do not peer with each other, then when they send traffic to each other they have to send the traffic through a provider ISP (intermediary), to which they have to pay for carrying the traffic. By peering with each other directly, the two ISPs can reduce their payments to their provider ISPs. An Internet Exchange Points (IXP) (typically in a standalone building with its own switches) is a meeting point where multiple ISPs can connect and/or peer together. An ISP earns its money by charging each of the the ISPs that connect to the IXP a relatively small fee, which may depend on the amount of traffic sent to or received from the IXP. 15. Google's private network connects together all its data centers, big and small. Traffic between the Google data centers passes over its private network rather than over the public Internet. Many of these data centers are located in, or close to, lower tier ISPs. Therefore, when Google delivers content to a user, it often can bypass higher tier ISPs. What motivates content providers to create these networks? First, the content provider has more control over the user experience, since it has to use few intermediary ISPs. Second, it can save money by sending less traffic into provider networks. Third, if ISPs decide to charge more money to highly profitable content providers (in countries where net neutrality doesn't apply), the content providers can avoid these extra payments. 16. The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable. 17. a) 1000 km, 1 Mbps, 100 bytes b) 100 km, 1 Mbps, 100 bytes 18. 10msec; d/s; no; no 19. a) 500 kbps b) 64 seconds c) 100kbps; 320 seconds 20. End system A breaks the large file into chunks. It adds header to each chunk, thereby generating multiple packets from the file. The header in each packet includes the IP address of the destination (end system B). The packet switch uses the destination IP address in the packet to determine the outgoing link. Asking which road to take is analogous to a packet asking which outgoing link it should be forwarded on, given the packet’s destination address. 21. The maximum emission rate is 500 packets/sec and the maximum transmission rate is 350 packets/sec. The corresponding traffic intensity is 500/350 =1.43 > 1. Loss will eventually occur for each experiment; but the time when loss first occurs will be different from one experiment to the next due to the randomness in the emission process. 22. Five generic tasks are error control, flow control, segmentation and reassembly, multiplexing, and connection setup. Yes, these tasks can be duplicated at different layers. For example, error control is often provided at more than one layer. 23. The five layers in the Internet protocol stack are – from top to bottom – the application layer, the transport layer, the network layer, the link layer, and the physical layer. The principal responsibilities are outlined in Section 1.5.1. 24. Application-layer message: data which an application wants to send and passed onto the transport layer; transport-layer segment: generated by the transport layer and encapsulates application-layer message with transport layer header; network-layer datagram: encapsulates transport-layer segment with a network-layer header; link-layer frame: encapsulates network-layer datagram with a link-layer header. 25. Routers process network, link and physical layers (layers 1 through 3). (This is a little bit of a white lie, as modern routers sometimes act as firewalls or caching components, and process Transport layer as well.) Link layer switches process link and physical layers (layers 1 through2). Hosts process all five layers. 26. a) Virus Requires some form of human interaction to spread. Classic example: E-mail viruses. b) Worms No user replication needed. Worm in infected host scans IP addresses and port numbers, looking for vulnerable processes to infect. 27. Creation of a botnet requires an attacker to find vulnerability in some application or system (e.g. exploiting the buffer overflow vulnerability that might exist in an application). After finding the vulnerability, the attacker needs to scan for hosts that are vulnerable. The target is basically to compromise a series of systems by exploiting that particular vulnerability. Any system that is part of the botnet can automatically scan its environment and propagate by exploiting the vulnerability. An important property of such botnets is that the originator of the botnet can remotely control and issue commands to all the nodes in the botnet. Hence, it becomes possible for the attacker to issue a command to all the nodes, that target a single node (for example, all nodes in the botnet might be commanded by the attacker to send a TCP SYN message to the target, which might result in a TCP SYN flood attack at the target). 28. Trudy can pretend to be Bob to Alice (and vice-versa) and partially or completely modify the message(s) being sent from Bob to Alice. For example, she can easily change the phrase “Alice, I owe you $1000” to “Alice, I owe you $10,000”. Furthermore, Trudy can even drop the packets that are being sent by Bob to Alice (and vise-versa), even if the packets from Bob to Alice are encrypted. Chapter 1 Problems Problem 1 There is no single right answer to this question. Many protocols would do the trick. Here's a simple answer below: Messages from ATM machine to Server Msg name purpose -------- ------- HELO Let server know that there is a card in the ATM machine ATM card transmits user ID to Server PASSWD User enters PIN, which is sent to server BALANCE User requests balance WITHDRAWL User asks to withdraw money BYE user all done Messages from Server to ATM machine (display) Msg name purpose -------- ------- PASSWD Ask user for PIN (password) OK last requested operation (PASSWD, WITHDRAWL) OK ERR last requested operation (PASSWD, WITHDRAWL) in ERROR AMOUNT sent in response to BALANCE request BYE user done, display welcome screen at ATM Correct operation: client server HELO (userid) --------------> (check if valid userid) <------------- PASSWD PASSWD --------------> (check password) <------------- AMOUNT WITHDRAWL --------------> check if enough $ to cover withdrawl (check if valid userid) <------------- PASSWD PASSWD --------------> (check password) <------------- AMOUNT WITHDRAWL --------------> check if enough $ to cover withdrawl <------------- BYE Problem 2 At time N*(L/R) the first packet has reached the destination, the second packet is stored in the last router, the third packet is stored in the next-to-last router, etc. At time N*(L/R) + L/R, the second packet has reached the destination, the third packet is stored in the last router, etc. Continuing with this logic, we see that at time N*(L/R) + (P-1)*(L/R) = (N+P-1)*(L/R) all packets have reached the destination. Problem 3 a) A circuit-switched network would be well suited to the application, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be reserved for each application session without significant waste. In addition, the overhead costs of setting up and tearing down connections are amortized over the lengthy duration of a typical application session. b) In the worst case, all the applications simultaneously transmit over one or more network links. However, since each link has sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur. Given such generous link capacities, the network does not need congestion control mechanisms. Problem 4 Between the switch in the upper left and the switch in the upper right we can have 4 connections. Similarly we can have four connections between each of the 3 other pairs of adjacent switches. Thus, this network can support up to 16 connections. We can 4 connections passing through the switch in the upper-right-hand corner and another 4 connections passing through the switch in the lower-left-hand corner, giving a total of 8 connections. Yes. For the connections between A and C, we route two connections through B and two connections through D. For the connections between B and D, we route two connections through A and two connections through C. In this manner, there are at most 4 connections passing through any link. Problem 5 Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth services a car at a rate of one car every 12 seconds. a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first tollbooth to service the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km) before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 47 minutes. The whole process repeats itself for traveling between the second and third tollbooths. It also takes 2 minutes for the third tollbooth to service the 10 cars. Thus the total delay is 96 minutes. b) Delay between tollbooths is 8*12 seconds plus 45 minutes, i.e., 46 minutes and 36 seconds. The total delay is twice this amount plus 8*12 seconds, i.e., 94 minutes and 48 seconds. Problem 6 a) seconds. b) seconds. c) seconds. d) The bit is just leaving Host A. e) The first bit is in the link and has not reached Host B. f) The first bit has reached Host B. g) Want km. Problem 7 Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires sec=7msec. The time required to transmit the packet is sec= sec. Propagation delay = 10 msec. The delay until decoding is 7msec + sec + 10msec = 17.224msec A similar analysis shows that all bits experience a delay of 17.224 msec. Problem 8 a) 20 users can be supported. b) . c) . d) . We use the central limit theorem to approximate this probability. Let be independent random variables such that . “21 or more users” when is a standard normal r.v. Thus “21 or more users” . Problem 9 10,000 Problem 10 The first end system requires L/R1 to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of dproc; after receiving the entire packet, the packet switch connecting the first and the second link requires L/R2 to transmit the packet onto the second link; the packet propagates over the second link in d2/s2. Similarly, we can find the delay caused by the second switch and the third link: L/R3, dproc, and d3/s3. Adding these five delays gives dend-end = L/R1 + L/R2 + L/R3 + d1/s1 + d2/s2 + d3/s3+ dproc+ dproc To answer the second question, we simply plug the values into the equation to get 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec. Problem 11 Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus, dend-end = L/R + d1/s1 + d2/s2+ d3/s3 For the values in Problem 10, we get 6 + 20 + 16 + 4 = 46 msec. Problem 12 The arriving packet must first wait for the link to transmit 4.5 *1,500 bytes = 6,750 bytes or 54,000 bits. Since these bits are transmitted at 2 Mbps, the queuing delay is 27 msec. Generally, the queuing delay is (nL + (L - x))/R. Problem 13 The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets is: (L/R + 2L/R + ....... + (N-1)L/R)/N = L/(RN) * (1 + 2 + ..... + (N-1)) = L/(RN) * N(N-1)/2 = LN(N-1)/(2RN) = (N-1)L/(2R) Note that here we used the well-known fact: 1 + 2 + ....... + N = N(N+1)/2 It takes seconds to transmit the packets. Thus, the buffer is empty when a each batch of packets arrive. Thus, the average delay of a packet across all batches is the average delay within one batch, i.e., (N-1)L/2R. Problem 14 The transmission delay is . The total delay is Let . Total delay = For x=0, the total delay =0; as we increase x, total delay increases, approaching infinity as x approaches 1/a. Problem 15 Total delay . Problem 16 The total number of packets in the system includes those in the buffer and the packet that is being transmitted. So, N=10+1. Because , so (10+1)=a*(queuing delay + transmission delay). That is, 11=a*(0.01+1/100)=a*(0.01+0.01). Thus, a=550 packets/sec. Problem 17 There are nodes (the source host and the routers). Let denote the processing delay at the th node. Let be the transmission rate of the th link and let . Let be the propagation delay across the th link. Then . Let denote the average queuing delay at node . Then . Problem 18 On linux you can use the command traceroute www.targethost.com and in the Windows command prompt you can use tracert www.targethost.com In either case, you will get three delay measurements. For those three measurements you can calculate the mean and standard deviation. Repeat the experiment at different times of the day and comment on any changes. Here is an example solution: Traceroutes between San Diego Super Computer Center and www.poly.edu The average (mean) of the round-trip delays at each of the three hours is 71.18 ms, 71.38 ms and 71.55 ms, respectively. The standard deviations are 0.075 ms, 0.21 ms, 0.05 ms, respectively. In this example, the traceroutes have 12 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed through four ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Traceroutes from www.stella-net.net (France) to www.poly.edu (USA). The average round-trip delays at each of the three hours are 87.09 ms, 86.35 ms and 86.48 ms, respectively. The standard deviations are 0.53 ms, 0.18 ms, 0.23 ms, respectively. In this example, there are 11 routers in the path at each of the three hours. No, the paths didn’t change during any of the hours. Traceroute packets passed three ISP networks from source to destination. Yes, in this experiment the largest delays occurred at peering interfaces between adjacent ISPs. Problem 19 An example solution: Traceroutes from two different cities in France to New York City in United States In these traceroutes from two different cities in France to the same destination host in United States, seven links are in common including the transatlantic link. In this example of traceroutes from one city in France and from another city in Germany to the same host in United States, three links are in common including the transatlantic link. Traceroutes to two different cities in China from same host in United States Five links are common in the two traceroutes. The two traceroutes diverge before reaching China Problem 20 Throughput = min{Rs, Rc, R/M} Problem 21 If only use one path, the max throughput is given by: . If use all paths, the max throughput is given by . Problem 22 Probability of successfully receiving a packet is: ps= (1-p)N. The number of transmissions needed to be performed until the packet is successfully received by the client is a geometric random variable with success probability ps. Thus, the average number of transmissions needed is given by: 1/ps . Then, the average number of re-transmissions needed is given by: 1/ps -1. Problem 23 Let’s call the first packet A and call the second packet B. If the bottleneck link is the first link, then packet B is queued at the first link waiting for the transmission of packet A. So the packet inter-arrival time at the destination is simply L/Rs. If the second link is the bottleneck link and both packets are sent back to back, it must be true that the second packet arrives at the input queue of the second link before the second link finishes the transmission of the first packet. That is, L/Rs + L/Rs + dprop = L/Rs + dprop + L/Rc Thus, the minimum value of T is L/Rc  L/Rs . Problem 24 40 terabytes = 40 * 1012 * 8 bits. So, if using the dedicated link, it will take 40 * 1012 * 8 / (100 *106 ) =3200000 seconds = 37 days. But with FedEx overnight delivery, you can guarantee the data arrives in one day, and it should cost less than $100. Problem 25 160,000 bits 160,000 bits The bandwidth-delay product of a link is the maximum number of bits that can be in the link. the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field s/R Problem 26 s/R=20000km, then R=s/20000km= 2.5*108/(2*107)= 12.5 bps Problem 27 80,000,000 bits 800,000 bits, this is because that the maximum number of bits that will be in the link at any given time = min(bandwidth delay product, packet size) = 800,000 bits. .25 meters Problem 28 ttrans + tprop = 400 msec + 80 msec = 480 msec. 20 * (ttrans + 2 tprop) = 20*(20 msec + 80 msec) = 2 sec. Breaking up a file takes longer to transmit because each data packet and its corresponding acknowledgement packet add their own propagation delays. Problem 29 Recall geostationary satellite is 36,000 kilometers away from earth surface. 150 msec 1,500,000 bits 600,000,000 bits Problem 30 Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the top of the protocol stack. When the passenger checks in, his/her bags are checked, and a tag is attached to the bags and ticket. This is additional information added in the Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or separating the passengers and baggage on the sending side, and then reuniting them (hopefully!) on the destination side. When a passenger then passes through security and additional stamp is often added to his/her ticket, indicating that the passenger has passed through a security check. This information is used to ensure (e.g., by later checks for the security information) secure transfer of people. Problem 31 Time to send message from source host to first packet switch = With store-and-forward switching, the total time to move message from source host to destination host = Time to send 1st packet from source host to first packet switch = . . Time at which 2nd packet is received at the first switch = time at which 1st packet is received at the second switch = Time at which 1st packet is received at the destination host = . After this, every 5msec one packet will be received; thus time at which last (800th) packet is received = . It can be seen that delay in using message segmentation is significantly less (almost 1/3rd). Without message segmentation, if bit errors are not tolerated, if there is a single bit error, the whole message has to be retransmitted (rather than a single packet). Without message segmentation, huge packets (containing HD videos, for example) are sent into the network. Routers have to accommodate these huge packets. Smaller packets have to queue behind enormous packets and suffer unfair delays. Packets have to be put in sequence at the destination. Message segmentation results in many smaller packets. Since header size is usually the same for all packets regardless of their size, with message segmentation the total amount of header bytes is more. Problem 32 Yes, the delays in the applet correspond to the delays in the Problem 31.The propagation delays affect the overall end-to-end delays both for packet switching and message switching equally. Problem 33 There are F/S packets. Each packet is S=80 bits. Time at which the last packet is received at the first router is sec. At this time, the first F/S-2 packets are at the destination, and the F/S-1 packet is at the second router. The last packet must then be transmitted by the first router and the second router, with each transmission taking sec. Thus delay in sending the whole file is To calculate the value of S which leads to the minimum delay, Problem 34 The circuit-switched telephone networks and the Internet are connected together at "gateways". When a Skype user (connected to the Internet) calls an ordinary telephone, a circuit is established between a gateway and the telephone user over the circuit switched network. The skype user's voice is sent in packets over the Internet to the gateway. At the gateway, the voice signal is reconstructed and then sent over the circuit. In the other direction, the voice signal is sent over the circuit switched network to the gateway. The gateway packetizes the voice signal and sends the voice packets to the Skype user. Chapter 2 Review Questions The Web: HTTP; file transfer: FTP; remote login: Telnet; e-mail: SMTP; BitTorrent file sharing: BitTorrent protocol Network architecture refers to the organization of the communication process into layers (e.g., the five-layer Internet architecture). Application architecture, on the other hand, is designed by an application developer and dictates the broad structure of the application (e.g., client-server or P2P). The process which initiates the communication is the client; the process that waits to be contacted is the server. No. In a P2P file-sharing application, the peer that is receiving a file is typically the client and the peer that is sending the file is typically the server. The IP address of the destination host and the port number of the socket in the destination process. You would use UDP. With UDP, the transaction can be completed in one roundtrip time (RTT) - the client sends the transaction request into a UDP socket, and the server sends the reply back to the client's UDP socket. With TCP, a minimum of two RTTs are needed - one to set-up the TCP connection, and another for the client to send the request, and for the server to send back the reply. One such example is remote word processing, for example, with Google docs. However, because Google docs runs over the Internet (using TCP), timing guarantees are not provided. a) Reliable data transfer TCP provides a reliable byte-stream between client and server but UDP does not. b) A guarantee that a certain value for throughput will be maintained Neither c) A guarantee that data will be delivered within a specified amount of time Neither d) Confidentiality (via encryption) Neither SSL operates at the application layer. The SSL socket takes unencrypted data from the application layer, encrypts it and then passes it to the TCP socket. If the application developer wants TCP to be enhanced with SSL, she has to include the SSL code in the application. A 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. 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. When the user first visits the site, the server creates a unique identification number, creates an entry in its back-end database, and returns this identification number as a cookie number. This cookie number is stored on the user’s host and is managed by the browser. During each subsequent visit (and purchase), the browser sends the cookie number back to the site. Thus the site knows when this user (more precisely, this browser) is visiting the site. Web caching can bring the desired content “closer” to the user, possibly to the same LAN to which the user’s host is connected. Web caching can reduce the delay for all objects, even objects that are not cached, since caching reduces the traffic on links. Telnet is not available in Windows 7 by default. to make it available, go to Control Panel, Programs and Features, Turn Windows Features On or Off, Check Telnet client. To start Telnet, in Windows command prompt, issue the following command > telnet webserverver 80 where "webserver" is some webserver. After issuing the command, you have established a TCP connection between your client telnet program and the web server. Then type in an HTTP GET message. An example is given below: Since the index.html page in this web server was not modified since Fri, 18 May 2007 09:23:34 GMT, and the above commands were issued on Sat, 19 May 2007, the server returned "304 Not Modified". Note that the first 4 lines are the GET message and header lines inputed by the user, and the next 4 lines (starting from HTTP/1.1 304 Not Modified) is the response from the web server. FTP uses two parallel TCP connections, one connection for sending control information (such as a request to transfer a file) 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 information out of band. The message is first sent from Alice’s host to her mail server over HTTP. Alice’s mail server then sends the message to Bob’s mail server over SMTP. Bob then transfers the message from his mail server to his host over POP3. 17. Received: from 65.54.246.203 (EHLO bay0-omc3-s3.bay0.hotmail.com) (65.54.246.203) by mta419.mail.mud.yahoo.com with SMTP; Sat, 19 May 2007 16:53:51 -0700 Received: from hotmail.com ([65.55.135.106]) by bay0-omc3-s3.bay0.hotmail.com with Microsoft SMTPSVC(6.0.3790.2668); Sat, 19 May 2007 16:52:42 -0700 Received: from mail pickup service by hotmail.com with Microsoft SMTPSVC; Sat, 19 May 2007 16:52:41 -0700 Message-ID: Received: from 65.55.135.123 by by130fd.bay130.hotmail.msn.com with HTTP; Sat, 19 May 2007 23:52:36 GMT From: "prithula dhungel" To: prithula@yahoo.com Bcc: Subject: Test mail Date: Sat, 19 May 2007 23:52:36 +0000 Mime-Version: 1.0 Content-Type: Text/html; format=flowed Return-Path: prithuladhungel@hotmail.com Figure: A sample mail message header Received: This header field indicates the sequence in which the SMTP servers send and receive the mail message including the respective timestamps. In this example there are 4 “Received:” header lines. This means the mail message passed through 5 different SMTP servers before being delivered to the receiver’s mail box. The last (forth) “Received:” header indicates the mail message flow from the SMTP server of the sender to the second SMTP server in the chain of servers. The sender’s SMTP server is at address 65.55.135.123 and the second SMTP server in the chain is by130fd.bay130.hotmail.msn.com. The third “Received:” header indicates the mail message flow from the second SMTP server in the chain to the third server, and so on. Finally, the first “Received:” header indicates the flow of the mail messages from the forth SMTP server to the last SMTP server (i.e. the receiver’s mail server) in the chain. Message-id: The message has been given this number BAY130-F26D9E35BF59E0D18A819AFB9310@phx.gbl (by bay0-omc3-s3.bay0.hotmail.com. Message-id is a unique string assigned by the mail system when the message is first created. From: This indicates the email address of the sender of the mail. In the given example, the sender is “prithuladhungel@hotmail.com” To: This field indicates the email address of the receiver of the mail. In the example, the receiver is “prithula@yahoo.com” Subject: This gives the subject of the mail (if any specified by the sender). In the example, the subject specified by the sender is “Test mail” Date: The date and time when the mail was sent by the sender. In the example, the sender sent the mail on 19th May 2007, at time 23:52:36 GMT. Mime-version: MIME version used for the mail. In the example, it is 1.0. Content-type: The type of content in the body of the mail message. In the example, it is “text/html”. Return-Path: This specifies the email address to which the mail will be sent if the receiver of this mail wants to reply to the sender. This is also used by the sender’s mail server for bouncing back undeliverable mail messages of mailer-daemon error messages. In the example, the return path is “prithuladhungel@hotmail.com”. With download 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 (office PC, home PC, etc.). In the download and keep configuration, messages are not deleted after the user retrieves the messages. 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 (including very old messages). Yes an organization’s mail server and Web server can have the same alias for a host name. The MX record is used to map the mail server’s host name to its IP address. You should be able to see the sender's IP address for a user with an .edu email address. But you will not be able to see the sender's IP address if the user uses a gmail account. It is not necessary that Bob will also provide chunks to Alice. Alice has to be in the top 4 neighbors of Bob for Bob to send out chunks to her; this might not occur even if Alice provides chunks to Bob throughout a 30-second interval. Recall that in BitTorrent, a peer picks a random peer and optimistically unchokes the peer for a short period of time. Therefore, Alice will eventually be optimistically unchoked by one of her neighbors, during which time she will receive chunks from that neighbor. The overlay network in a P2P file sharing system consists of the nodes participating in the file sharing system and the logical links between the nodes. There is a logical link (an “edge” in graph theory terms) from node A to node B if there is a semi-permanent TCP connection between A and B. An overlay network does not include routers. Mesh DHT: The advantage is in order to a route a message to the peer (with ID) that is closest to the key, only one hop is required; the disadvantage is that each peer must track all other peers in the DHT. Circular DHT: the advantage is that each peer needs to track only a few other peers; the disadvantage is that O(N) hops are needed to route a message to the peer that is closest to the key. 25. File Distribution Instant Messaging Video Streaming Distributed Computing With the UDP server, there is no welcoming socket, and all data from different clients enters the server through this one socket. With the TCP server, there is a welcoming socket, and each time a client initiates a connection to the server, a new socket is created. Thus, to support n simultaneous connections, the server would need n+1 sockets. For the TCP application, as soon as the client is executed, it attempts to initiate a TCP connection with the server. If the TCP server is not running, then the client will fail to make a connection. For the UDP application, the client does not initiate connections (or attempt to communicate with the UDP server) immediately upon execution Chapter 2 Problems Problem 1 a) F b) T c) F d) F e) F Problem 2 Access control commands: USER, PASS, ACT, CWD, CDUP, SMNT, REIN, QUIT. Transfer parameter commands: PORT, PASV, TYPE STRU, MODE. Service commands: RETR, STOR, STOU, APPE, ALLO, REST, RNFR, RNTO, ABOR, DELE, RMD, MRD, PWD, LIST, NLST, SITE, SYST, STAT, HELP, NOOP. Problem 3 Application layer protocols: DNS and HTTP Transport layer protocols: UDP for DNS; TCP for HTTP Problem 4 The document request was http://gaia.cs.umass.edu/cs453/index.html. The Host : field indicates the server's name and /cs453/index.html indicates the file name. The browser is running HTTP version 1.1, as indicated just before the first pair. The browser is requesting a persistent connection, as indicated by the Connection: keep-alive. This is a trick question. This information is not contained in an HTTP message anywhere. So there is no way to tell this from looking at the exchange of HTTP messages alone. One would need information from the IP datagrams (that carried the TCP segment that carried the HTTP GET request) to answer this question. Mozilla/5.0. The browser type information is needed by the server to send different versions of the same object to different types of browsers. Problem 5 The status code of 200 and the phrase OK indicate that the server was able to locate the document successfully. The reply was provided on Tuesday, 07 Mar 2008 12:39:45 Greenwich Mean Time. The document index.html was last modified on Saturday 10 Dec 2005 18:27:46 GMT. There are 3874 bytes in the document being returned. The first five bytes of the returned document are : Errors will occur. Problem 32 In the original program, UDPClient does not specify a port number when it creates the socket. In this case, the code lets the underlying operating system choose a port number. With the additional line, when UDPClient is executed, a UDP socket is created with port number 5432 . UDPServer needs to know the client port number so that it can send packets back to the correct client socket. Glancing at UDPServer, we see that the client port number is not “hard-wired” into the server code; instead, UDPServer determines the client port number by unraveling the datagram it receives from the client. Thus UDP server will work with any client port number, including 5432. UDPServer therefore does not need to be modified. Before: Client socket = x (chosen by OS) Server socket = 9876 After: Client socket = 5432 Problem 33 Yes, you can configure many browsers to open multiple simultaneous connections to a Web site. The advantage is that you will you potentially download the file faster. The disadvantage is that you may be hogging the bandwidth, thereby significantly slowing down the downloads of other users who are sharing the same physical links. Problem 34 For an application such as remote login (telnet and ssh), a byte-stream oriented protocol is very natural since there is no notion of message boundaries in the application. When a user types a character, we simply drop the character into the TCP connection. In other applications, we may be sending a series of messages that have inherent boundaries between them. For example, when one SMTP mail server sends another SMTP mail server several email messages back to back. Since TCP does not have a mechanism to indicate the boundaries, the application must add the indications itself, so that receiving side of the application can distinguish one message from the next. If each message were instead put into a distinct UDP segment, the receiving end would be able to distinguish the various messages without any indications added by the sending side of the application. Problem 35 To create a web server, we need to run web server software on a host. Many vendors sell web server software. However, the most popular web server software today is Apache, which is open source and free. Over the years it has been highly optimized by the open-source community. Problem 36 The key is the infohash, the value is an IP address that currently has the file designated by the infohash. Chapter 3 Review Questions Call this protocol Simple Transport Protocol (STP). At the sender side, STP accepts from the sending process a chunk of data not exceeding 1196 bytes, a destination host address, and a destination port number. STP adds a four-byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number. The segment now has two header fields: a source port field and destination port field. At the sender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address, a source port number, and a destination port number. STP creates a segment which contains the application data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number. No, the transport layer does not have to do anything in the core; the transport layer “lives” in the end systems. For sending a letter, the family member is required to give the delegate the letter itself, the address of the destination house, and the name of the recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an envelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate then gives the letter to the family member with this name. No, the mail service does not have to open the envelope; it only examines the address on the envelope. Source port number y and destination port number x. An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP. Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls. Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however. Yes, both segments will be directed to the same socket. For each received segment, at the socket interface, the operating system will provide the process with the IP addresses to determine the origins of the individual segments. For each persistent connection, the Web server creates a separate “connection socket”. Each connection socket is identified with a four-tuple: (source IP address, source port number, destination IP address, destination port number). When host C receives and IP datagram, it examines these four fields in the datagram/segment to determine to which socket it should pass the payload of the TCP segment. Thus, the requests from A and B pass through different sockets. The identifier for both of these sockets has 80 for the destination port; however, the identifiers for these sockets have different values for source IP addresses. Unlike UDP, when the transport layer passes a TCP segment’s payload to the application process, it does not specify the source IP address, as this is implicitly specified by the socket identifier. Sequence numbers are required for a receiver to find out whether an arriving packet contains new data or is a retransmission. To handle losses in the channel. If the ACK for a transmitted packet is not received within the duration of the timer for the packet, the packet (or its ACK or NACK) is assumed to have been lost. Hence, the packet is retransmitted. A timer would still be necessary in the protocol rdt 3.0. If the round trip time is known then the only advantage will be that, the sender knows for sure that either the packet or the ACK (or NACK) for the packet has been lost, as compared to the real scenario, where the ACK (or NACK) might still be on the way to the sender, after the timer expires. However, to detect the loss, for each packet, a timer of constant duration will still be necessary at the sender. The packet loss caused a time out after which all the five packets were retransmitted. Loss of an ACK didn’t trigger any retransmission as Go-Back-N uses cumulative acknowledgements. The sender was unable to send sixth packet as the send window size is fixed to 5. When the packet was lost, the received four packets were buffered the receiver. After the timeout, sender retransmitted the lost packet and receiver delivered the buffered packets to application in correct order. Duplicate ACK was sent by the receiver for the lost ACK. The sender was unable to send sixth packet as the send win

Clever Internet Suite version 9.1 These Internet components give you everything you need to jumpstart your Internet development without any external dependencies. The suite of Internet Components contain over fifty components which are constantly refined and improved. The Clever Internet Suite components allow you to compose and parse messages in MIME format with multiple file attachments, build and send Web Form POST requests and many other features. The SMTP, POP3 and IMAP clients allow you to send and retrieve email messages over the Internet. All server components represent the fully functional multithreaded servers with the ability to customize the connection settings, support the most common protocols extensions. Using the Clever Internet Suite you can add instant SSL / TLS / SSH security to your Internet applications and implement many useful Internet-related features: Downloading, uploading and submitting of the Internet-resources. Sending and receiving e-mail messages with DKIM signatures. HTTP, FTP, SMTP, POP3, IMAP and NNTP client / server solutions. secure channel with X509 certificates support. OAUTH 2.0 authorization. SOAP Security and many more. In the new version 9.1 we have updated the MailMessage, SoapMessage and SFtp components, fixed issues in the TLS and cryptography engines. Starting from the version 8.0 the library was splitted on design-time and runtime packages. Starting from now, the components can be used by RAD Studio form designer in 64-bit projects. The basic socket components were replaced and improved. The class inheritance was changed. Please see the Help documentation for more details. The Help documentation is included to the Clever Internet Suite installation and also available at our website: Clever Internet Suite downloads What's new in version 9.1 The full RAD Studio 10.2 and 10.2.1 Tokyo support. MailMessage - RFC 5987 support added, file names in UTF-8 format implemented. New CryptEncoder component - provides methods for data encoding and decoding in PEM and SSH2 cryptographic formats. SFTP client - the public key authorization was implemented, the following new algorithms were added: diffie-hellman-group-exchange-sha256 and diffie-hellman-group14-sha1 for key exchange functionality; rsa-sha2-256 - server host key algorithm; aes256-ctr, aes192-ctr, aes128-ctr, aes256-cbc, aes192-cbc ,aes128-cbc - data encryption algorithms; hmac-sha2-256 - for message authentication algorithms; sha2-256 - hash calculation algorithms. DKIM component - the Config property was added. This property allows you to configure the used encryption and signature algorithms. The signature verification is disabled when retrieving the message header. HTTP client - the TLS1.2 option was enabled by default. EPP server - the UTF-8 encoding was added. All TCP-based server components (SMTP, POP3, IMAP4, FTP, etc.) - the new CharSet property allows you to specify the desired character encoding for received commands and server responses. All UDP-based client components (DNS query, etc.) - the new CharSet property allows you to specify the desired character encoding transmitted datagrams. HTTP request component - was redesigned, the Items indexed property now represents the HTTP request items collection object. The source code structure was changed. starting from now, the sources are placed in the following folders: common, design, packages, resources and ssh. To compile the source code project, navigate to the packages folder, choose the corresponding IDE version subfolder, e.g., D102 for Delphi 10.2 Tokyo, open and compile both the clinetsuite_102.dproj and dcl_clinetsuite_102.dproj projects. Finally, install dcl_clinetsuite_102.dproj to your RAD Studio IDE. The SSH and SFTP component files were moved to the main clinetsuite package. The clinetsuitessh package was removed from the installation. Implemented Features HTTP RIO component with SOAP Security support. Simple HTTP Server component. Complete Client / Server solutions with full SSL / TLS support for FTP, SMTP, POP3 and IMAP4 protocols. TLS / SSL support for all protocol components. SSH support for the SFTP protocol. Royalty free licensing. Free Unlimited Email support. Fully-indexed help documentation. Delphi and C++Builder demos code. Send / receive HTTP requests in JSON format. International domain names for HTTP components. The ability to verify the client and server credentials and certificates. HTTP Client - HTTPS (SSL / TLS) protocol, NTLM and Negotiate authentication. FTP, SMTP, POP3, IMAP, NNTP clients - three new TLS modes - implicit, explicit and automatic. STARTTLS command support. NTLM authentication for all mail clients was added: POP3, SMTP, IMAP. When the UseSasl option enabled, these components automatically determine and use the most secured authentication method. POP3, SMTP clients - working with GMAIL service with OAUTH 2.0 authorization. Fully functional HTTP / HTTPS client. FTP Client - SSL / TLS support, the ability to set FTP file attributes and permissions were added. FTP FXP (Site to Site) transfer mode. FTP Server - custom events for all operations with files and directories on the server, the possibility to customize all file-related operations including compressing and uncompressing functionality. Virtual File System. Events for implementing the mail sending / receiving progress with the ability to interrupt the process. Creating self-signed certificates, working with any type of certificate store including current user stores, local machine stores, file stores, registry stores and so on. Creating and parsing email messages in any format with file Attachments and Embedded pictures, the ability to determine the attachment size before saving it to the disk. POP3, SMTP, IMAP4 servers - SSL / TLS support, NTLM, APOP and CRAM-MD5 authentication methods, File System message storage. POP3, SMTP and IMAP4 clients - SSL / TLS support, NTLM and CRAM-MD5 authentication methods. Web DAV - manages remote files and folders on a web server. DNS Query - queries a DNS server for records. This component can retrieve the mail exchange domain (MX records), resolve the host IP, retrieve the information about name servers and many other information from the DNS server. DNS Server - implements fully functional DNS server and allows you to both manage handed DNS zones and cache non-authoritative DNS records. SMTP Relay - implements a SMTP Relay agent. With SMTP relay, a mail message may pass through a number of intermediate relay or gateway hosts on its path from sender to recipient. HTML Parser - parse HTML Tags: links, images, tables, meta tags and many more. GZip compressor component - provides compressing / uncompressing feature when transmitting data over the Internet, storing data in to a file, database and many more. RSS client - is used for creating, retrieving, and editing RSS feeds. Email validation components - allow you to validate recipients email addresses, mailbox availability and also handle bounced emails. Trace Logger - provides logging functionality for storing debug information and tracing your code. Also a set of additional classes and components are available: Mail Message, HTTP Request, GZip compressor, MIME Encoder, S/MIME message, HTML Parser. All these components make the application development process easy and clean. You can use these components separately from the main protocol components with any other library and even with your own socket implementation. Please visit our site Clever Components to learn more about our products. Feel free to join our Mail List Subscription at Mail List Subscription and stay tuned. Fixed bugs The HttpAuthorization is not thread-safe - fixed. SFTP file permissions parsing errors were fixed, IsDir file attribute works correctly. Installer runs with errors on Windows XP - fixed. TLS engine - INCOMPLETE_MESSAGE error occurred - fixed. RSA key for RSA-SHA-256 algorithm was imported incorrectly - fixed. SFTP large directory listing errors, the problems with receiving of incloplete SSH packets was fixed. SOAP Signing takes a lot of time - fixed. DKIM - the signature verification should be disabled when retrieving the message header. UDP server and all descendant components: there were issues with starting of listening socket (the server was hanged) - fixed. MailMessage - the message attachments and images lists were not automatically updated in the helper properties (Images, Attachments) when adding or deleting the body from the Bodies collection - fixed. MailMessage - the header fields with mixed encoding style were decoded incorrectly - fixed. DNS Server - some fixes in the server engine. IMAP client - the AppendMessage method did not work with GMail IMAP - fixed. FTP client - the file names with international symbols were obtained incorrectly - fixed. HTTP client - the resource URLs were not escaped - fixed. All HTTP clients (TclHttp, TclWebDav, TclDownloader, TclMultiDownloader, TclNewsChecker, TclWebUpdate) - the components incorrectly provided a list of supported compression algorithms - fixed. FTP server - the directory navigation worked incorrectly in case if the root folder is the root of the disk (e.g, c:\). The connected client cannot get back to the root folder after navigating to a subfolder - fixed. Known problems and restrictions FTP client and server - data connection uses IpV4 protocol only. Socks5 firewall - supports IpV4 only. The HTTP RIO component is available only starting from Delphi 2005 / C++Builder 2006. The SOAP encryption is available only on Win32 platform. The SOAP encryption requires an external library clcryptext.dll that is included to the Clever Internet Suite installation. This library represents managed code that is wrapped by Win32 functions. The source code for this library can be downloaded separately at www.clevercomponents.com website.

来源：http://stackoverflow.com/questions/23216981/server-giving-msxml3-dll-error-80072f7d-when-trying-to-access-secure-url For years I have used classic asp to connect to the s

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