向各位高手请教:删除记录前怎么加提示“are you sure want to delete this record?”(带"YEs" or "No"按钮)

ylljony 2004-08-20 02:23:47
删除记录前怎么提示“are you sure want to delete this record?”(带"YEs" or "No"按钮)
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ylljony 2004-08-23
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怎么回事,我怎么写都是不能弹出来呢?都试过了,如下所示:
Public Sub btDel_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btDel.Click
'btDel.Attributes.Add("onclick", "javascript: return confirm('确定要删除此项吗?');")
btDel.Attributes.Add("onclick", "if(confirm('are you sure...?')){return true;}else{return false;}")
end Sub
yanga 2004-08-20
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test.vb
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
Button1.Attributes.Add("onclick", "if(confirm('are you sure...?')){return true;}else{return false;}")

End Sub
为什么,总是第一次没反应,要第二次单击才弹出对话框
yanga 2004-08-20
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我把
button.attributes.add("onclick","if(confirm('are you sure...?')){return true;}else{return false;}")
写在后置代码的单击事件里,为什么总是第一次点击没反应,要点第二次才会弹出对话框啊
BoroSoft 2004-08-20
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button.attributes.add("onclick","return confirm('')")
drason 2004-08-20
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button.attributes.add("onclick","if(confirm('are you sure...?')){return true;}else{return false;}")
mfkygazb 2004-08-20
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用c#怎么写啊,
要怎么用啊?
Reeezak 2004-08-20
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同意 smoothwood(吾谁与归) 的办法
jianli0108 2004-08-20
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private void DataGrid2_ItemDataBound(object sender, System.Web.UI.WebControls.DataGridItemEventArgs e)
{
e.Item.Cells[4].Attributes.Add("onclick","javascript: return confirm('确定要删除此项吗?');");
}
我这个是在DATEGRID中删除的时候的提示确认
jijl2001 2004-08-20
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这样子,能行吗?有没有试过,
ylljony 2004-08-20
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<SCRIPT LANGUAGE="JavaScript">
<!--
function confirm_del()
{
if(window.confirm('are you sure want to delete this record?'))
{return true;}
else
{return false;}
}
</SCRIPT>
javascript函数这样写没错吧

在vb里面:
btDel.Attributes.Add("onclick", "JavaScript: return confirm_del();")
ylljony 2004-08-20
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我试过很多,就是没有提示?烦啊
ylljony 2004-08-20
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用vb举例吧
flyingFisher 2004-08-20
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win还是web?
如果是win 就很简单。
如果是web就最好在html文件里写一个javascript函数来实现,很简单的,而且用起来也很方便,在CS文件里直接判断就可以了
brightheroes 2004-08-20
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button.attributes.add("onclick","return confirm(''are you sure...?')");
Hero4444 2004-08-20
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window.confirm('are you sure want to delete this record?')
smoothwood 2004-08-20
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button.attributes.add("onclick","if(confirm('are you sure...?')){return true;}else{return false;}")
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Fundamentals of the JavaMail API Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Table of Contents If you're viewing this document online, you can click any of the topics below to link directly to that section. 1. Tutorial tips 2 2. Introducing the JavaMail API 3 3. Reviewing related protocols 4 4. Installing JavaMail 6 5. Reviewing the core classes 8 6. Using the JavaMail API 13 7. Searching with SearchTerm 21 8. Exercises 22 9. Wrapup 32 Fundamentals of the JavaMail API Page 1 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 1. Tutorial tips Should I take this tutorial? Looking to incorporate mail facilities into your platform-independent Java solutions? Look no further than the JavaMail API, which offers a protocol-independent model for working with IMAP, POP, SMTP, MIME, and all those other Internet-related messaging protocols. With the help of the JavaBeans Activation Framework (JAF), your applications can now be mail-enabled through the JavaMail API. Concepts After completing this module you will understand the: * Basics of the Internet mail protocols SMTP, POP3, IMAP, and MIME * Architecture of the JavaMail framework * Connections between the JavaMail API and the JavaBeans Activation Framework Objectives By the end of this module you will be able to: * Send and read mail using the JavaMail API * Deal with sending and receiving attachments * Work with HTML messages * Use search terms to search for messages Prerequisites Instructions on how to download and install the JavaMail API are contained in the course. In addition, you will need a development environment such as the JDK 1.1.6+ or the Java 2 Platform, Standard Edition (J2SE) 1.2.x or 1.3.x. A general familiarity with object-oriented programming concepts and the Java programming language is necessary. The Java language essentials tutorial can help. copyright 1996-2000 Magelang Institute dba jGuru Contact jGuru has been dedicated to promoting the growth of the Java technology community through evangelism, education, and software since 1995. You can find out more about their activities, including their huge collection of FAQs at jGuru.com . To send feedback to jGuru about this course, send mail to producer@jguru.com . Course author: Formerly with jGuru.com , John Zukowski does strategic Java consulting for JZ Ventures, Inc. His latest book is titled Java Collections from Apress . Fundamentals of the JavaMail API Page 2 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 2. Introducing the JavaMail API What is the JavaMail API? The JavaMail API is an optional package (standard extension) for reading, composing, and sending electronic messages. You use the package to create Mail User Agent (MUA) type programs, similar to Eudora, pine, and Microsoft Outlook. The API's main purpose is not for transporting, delivering, and forwarding messages; this is the purview of applications such as sendmail and other Mail Transfer Agent (MTA) type programs. MUA-type programs let users read and write e-mail, whereas MUAs rely on MTAs to handle the actual delivery. The JavaMail API is designed to provide protocol-independent access for sending and receiving messages by dividing the API into two parts: * The first part of the API is the focus of this course --basically, how to send and receive messages independent of the provider/protocol. * The second part speaks the protocol-specific languages, like SMTP, POP, IMAP, and NNTP. With the JavaMail API, in order to communicate with a server, you need a provider for a protocol. The creation of protocol-specific providers is not covered in this course because Sun provides a sufficient set for free. Fundamentals of the JavaMail API Page 3 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 3. Reviewing related protocols Introduction Before looking into the JavaMail API specifics, let's step back and take a look at the protocols used with the API. There are basically four that you'll come to know and love: * SMTP * POP * IMAP * MIME You will also run across NNTP and some others. Understanding the basics of all the protocols will help you understand how to use the JavaMail API. While the API is designed to be protocol agnostic, you can't overcome the limitations of the underlying protocols. If a capability isn't supported by a chosen protocol, the JavaMail API doesn't magically add the capability on top of it. (As you'll soon see, this can be a problem when working with POP.) SMTP The Simple Mail Transfer Protocol (SMTP) is defined by RFC 821 . It defines the mechanism for delivery of e-mail. In the context of the JavaMail API, your JavaMail-based program will communicate with your company or Internet Service Provider's (ISP's) SMTP server. That SMTP server will relay the message on to the SMTP server of the recipient(s) to eventually be acquired by the user(s) through POP or IMAP. This does not require your SMTP server to be an open relay, as authentication is supported, but it is your responsibility to ensure the SMTP server is configured properly. There is nothing in the JavaMail API for tasks like configuring a server to relay messages or to add and remove e-mail accounts. POP POP stands for Post Office Protocol. Currently in version 3, also known as POP3, RFC 1939 defines this protocol. POP is the mechanism most people on the Internet use to get their mail. It defines support for a single mailbox for each user. That is all it does, and that is also the source of a lot of confusion. Much of what people are familiar with when using POP, like the ability to see how many new mail messages they have, are not supported by POP at all. These capabilities are built into programs like Eudora or Microsoft Outlook, which remember things like the last mail received and calculate how many are new for you. So, when using the JavaMail API, if you want this type of information, you have to calculate it yourself. IMAP IMAP is a more advanced protocol for receiving messages. Defined in RFC 2060 , IMAP stands for Internet Message Access Protocol, and is currently in version 4, also known as IMAP4. When using IMAP, your mail server must support the protocol. You can't just change your program to use IMAP instead of POP and expect everything in IMAP to be supported. Assuming your mail server supports IMAP, your JavaMail-based program can take Fundamentals of the JavaMail API Page 4 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks advantage of users having multiple folders on the server and these folders can be shared by multiple users. Due to the more advanced capabilities, you might think IMAP would be used by everyone. It isn't. It places a much heavier burden on the mail server, requiring the server to receive the new messages, deliver them to users when requested, and maintain them in multiple folders for each user. While this does centralize backups, as users' long-term mail folders get larger and larger, everyone suffers when disk space is exhausted. With POP, saved messages get offloaded from the mail server. MIME MIME stands for Multipurpose Internet Mail Extensions. It is not a mail transfer protocol. Instead, it defines the content of what is transferred: the format of the messages, attachments, and so on. There are many different documents that take effect here: RFC 822 , RFC 2045 , RFC 2046 , and RFC 2047 . As a user of the JavaMail API, you usually don't need to worry about these formats. However, these formats do exist and are used by your programs. NNTP and others Because of the split of the JavaMail API between provider and everything else, you can easily add support for additional protocols. Sun maintains a list of third-party providers that take advantage of protocols for which Sun does not provide out-of-the-box support. You'll find support for NNTP (Network News Transport Protocol) [newsgroups], S/MIME (Secure Multipurpose Internet Mail Extensions), and more. Fundamentals of the JavaMail API Page 5 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 4. Installing JavaMail Introduction There are two versions of the JavaMail API commonly used today: 1.2 and 1.1.3. All the examples in this course will work with both. While 1.2 is the latest, 1.1.3 is the version included with the 1.2.1 version of the Java 2 Platform, Enterprise Edition (J2EE), so it is still commonly used. The version of the JavaMail API you want to use affects what you download and install. All will work with JDK 1.1.6+, Java 2 Platform, Standard Edition (J2SE) version 1.2.x, and J2SE version 1.3.x. Note: After installing Sun's JavaMail implementation, you can find many example programs in the demo directory. Installing JavaMail 1.2 To use the JavaMail 1.2 API, download the JavaMail 1.2 implementation, unbundle the javamail-1_2.zip file, and add the mail.jar file to your CLASSPATH. The 1.2 implementation comes with an SMTP, IMAP4, and POP3 provider besides the core classes. After installing JavaMail 1.2, install the JavaBeans Activation Framework. Installing JavaMail 1.1.3 To use the JavaMail 1.1.3 API, download the JavaMail 1.1.3 implementation, unbundle the javamail1_1_3.zip file, and add the mail.jar file to your CLASSPATH. The 1.1.3 implementation comes with an SMTP and IMAP4 provider, besides the core classes. If you want to access a POP server with JavaMail 1.1.3, download and install a POP3 provider. Sun has one available separate from the JavaMail implementation. After downloading and unbundling pop31_1_1.zip, add pop3.jar to your CLASSPATH, too. After installing JavaMail 1.1.3, install the JavaBeans Activation Framework. Installing the JavaBeans Activation Framework All versions of the JavaMail API require the JavaBeans Activation Framework. The framework adds support for typing arbitrary blocks of data and handling it accordingly. This doesn't sound like much, but it is your basic MIME-type support found in many browsers and mail tools today. After downloading the framework, unbundle the jaf1_0_1.zip file, and add the activation.jar file to your CLASSPATH. For JavaMail 1.2 users, you should now have added mail.jar and activation.jar to your CLASSPATH. For JavaMail 1.1.3 users, you should now have added mail.jar, pop3.jar, and activation.jar to your CLASSPATH. If you have no plans of using POP3, you don't Fundamentals of the JavaMail API Page 6 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks need to add pop3.jar to your CLASSPATH. If you don't want to change the CLASSPATH environment variable, copy the jar files to your lib/ext directory under the Java Runtime Environment (JRE) directory. For instance, for the J2SE 1.3 release, the default directory would be C:\jdk1.3\jre\lib\ext on a Windows platform. Using JavaMail with the Java 2 Enterprise Edition If you use J2EE, there is nothing special you have to do to use the basic JavaMail API; it comes with the J2EE classes. Just make sure the j2ee.jar file is in your CLASSPATH and you're all set. For J2EE 1.2.1, the POP3 provider comes separately, so download and follow the steps to include the POP3 provider as shown in the previous section "Installing JavaMail 1.1.3." J2EE 1.3 users get the POP3 provider with J2EE so do not require the separate installation. Neither installation requires you to install the JavaBeans Activation Framework. Exercise Exercise 1. How to set up a JavaMail environment on page 22 Fundamentals of the JavaMail API Page 7 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 5. Reviewing the core classes Introduction Before taking a how-to approach at looking at the JavaMail classes in depth, this section walks you through the core classes that make up the API: Session, Message, Address, Authenticator, Transport, Store, and Folder. All these classes are found in the top-level package for the JavaMail API, javax.mail, though you'll frequently find yourself using subclasses found in the javax.mail.internet package. Session The Session class defines a basic mail session. It is through this session that everything else works. The Session object takes advantage of a java.util.Properties object to get information like mail server, username, password, and other information that can be shared across your entire application. The constructors for the class are private. You can get a single default session that can be shared with the getDefaultInstance() method: Properties props = new Properties(); // fill props with any information Session session = Session.getDefaultInstance(props, null); Or, you can create a unique session with getInstance(): Properties props = new Properties(); // fill props with any information Session session = Session.getDefaultInstance(props, null); In both cases, the null argument is an Authenticator object that is not being used at this time. In most cases, it is sufficient to use the shared session, even if working with mail sessions for multiple user mailboxes. You can add the username and password combination in at a later step in the communication process, keeping everything separate. Message Once you have your Session object, it is time to move on to creating the message to send. This is done with a type of Message . Because Message is an abstract class, you must work with a subclass, in most cases javax.mail.internet.MimeMessage .A MimeMessage is an e-mail message that understands MIME types and headers, as defined in the different RFCs. Message headers are restricted to US-ASCII characters only, though non-ASCII characters can be encoded in certain header fields. To create a Message, pass along the Session object to the MimeMessage constructor: MimeMessage message = new MimeMessage(session); Fundamentals of the JavaMail API Page 8 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Note: There are other constructors, like for creating messages from RFC822-formatted input streams. Once you have your message, you can set its parts, as Message implements the Part interface (with MimeMessage implementing MimePart ). The basic mechanism to set the content is the setContent() method, with arguments for the content and the mime type: message.setContent("Hello", "text/plain"); If, however, you know you are working with a MimeMessage and your message is plain text, you can use its setText() method, which only requires the actual content, defaulting to the MIME type of text/plain: message.setText("Hello"); For plain text messages, the latter form is the preferred mechanism to set the content. For sending other kinds of messages, like HTML messages, use the former. For setting the subject, use the setSubject() method: message.setSubject("First"); Address Once you've created the Session and the Message, as well as filled the message with content, it is time to address your letter with an Address . Like Message, Address is an abstract class. You use the javax.mail.internet.InternetAddress class. To create an address with just the e-mail address, pass the e-mail address to the constructor: Address address = new InternetAddress("president@whitehouse.gov"); If you want a name to appear next to the e-mail address, you can pass that along to the constructor, too: Address address = new InternetAddress("president@whitehouse.gov", "George Bush"); You will need to create address objects for the message's from field as well as the to field. Unless your mail server prevents you, there is nothing stopping you from sending a message that appears to be from anyone. Once you've created the addresses, you connect them to a message in one of two ways. For identifying the sender, you use the setFrom() and setReplyTo() methods. message.setFrom(address) If your message needs to show multiple from addresses, use the addFrom() method: Fundamentals of the JavaMail API Page 9 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Address address[] = ...; message.addFrom(address); For identifying the message recipients, you use the addRecipient() method. This method requires a Message.RecipientType besides the address. message.addRecipient(type, address) The three predefined types of address are: * Message.RecipientType.TO * Message.RecipientType.CC * Message.RecipientType.BCC So, if the message was to go to the vice president, sending a carbon copy to the first lady, the following would be appropriate: Address toAddress = new InternetAddress("vice.president@whitehouse.gov"); Address ccAddress = new InternetAddress("first.lady@whitehouse.gov"); message.addRecipient(Message.RecipientType.TO, toAddress); message.addRecipient(Message.RecipientType.CC, ccAddress); The JavaMail API provides no mechanism to check for the validity of an e-mail address. While you can program in support to scan for valid characters (as defined by RFC 822) or verify the MX (mail exchange) record yourself, these are all beyond the scope of the JavaMail API. Authenticator Like the java.net classes, the JavaMail API can take advantage of an Authenticator to access protected resources via a username and password. For the JavaMail API, that resource is the mail server. The JavaMail Authenticator is found in the javax.mail package and is different from the java.net class of the same name. The two don't share the same Authenticator as the JavaMail API works with Java 1.1, which didn't have the java.net variety. To use the Authenticator, you subclass the abstract class and return a PasswordAuthentication instance from the getPasswordAuthentication() method. You must register the Authenticator with the session when created. Then, your Authenticator will be notified when authentication is necessary. You could pop up a window or read the username and password from a configuration file (though if not encrypted is not secure), returning them to the caller as a PasswordAuthentication object. Properties props = new Properties(); // fill props with any information Authenticator auth = new MyAuthenticator(); Session session = Session.getDefaultInstance(props, auth); Transport The final part of sending a message is to use the Transport class. This class speaks the Fundamentals of the JavaMail API Page 10 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks protocol-specific language for sending the message (usually SMTP). It's an abstract class and works something like Session. You can use the default version of the class by just calling the static send() method: Transport.send(message); Or, you can get a specific instance from the session for your protocol, pass along the username and password (blank if unnecessary), send the message, and close the connection: message.saveChanges(); // implicit with send() Transport transport = session.getTransport("smtp"); transport.connect(host, username, password); transport.sendMessage(message, message.getAllRecipients()); transport.close(); This latter way is best when you need to send multiple messages, as it will keep the connection with the mail server active between messages. The basic send() mechanism makes a separate connection to the server for each method call. Note: To watch the mail commands go by to the mail server, set the debug flag with session.setDebug(true). Store and folder Getting messages starts similarly to sending messages with a Session. However, after getting the session, you connect to a Store , quite possibly with a username and password or Authenticator. Like Transport, you tell the Store what protocol to use: // Store store = session.getStore("imap"); Store store = session.getStore("pop3"); store.connect(host, username, password); After connecting to the Store, you can then get a Folder , which must be opened before you can read messages from it: Folder folder = store.getFolder("INBOX"); folder.open(Folder.READ_ONLY); Message message[] = folder.getMessages(); For POP3, the only folder available is the INBOX. If you are using IMAP, you can have other folders available. Note: Sun's providers are meant to be smart. While Message message[] = folder.getMessages(); might look like a slow operation reading every message from the server, only when you actually need to get a part of the message is the message content retrieved. Once you have a Message to read, you can get its content with getContent() or write its content to a stream with writeTo(). The getContent() method only gets the message content, while writeTo() output includes headers. Fundamentals of the JavaMail API Page 11 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks System.out.println(((MimeMessage)message).getContent()); Once you're done reading mail, close the connection to the folder and store. folder.close(aBoolean); store.close(); The boolean passed to the close() method of folder states whether or not to update the folder by removing deleted messages. Moving on Essentially, understanding how to use these seven classes is all you need for nearly everything with the JavaMail API. Most of the other capabilities of the JavaMail API build off these seven classes to do something a little different or in a particular way, like if the content is an attachment. Certain tasks, like searching, are isolated and are discussed later. Fundamentals of the JavaMail API Page 12 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 6. Using the JavaMail API Introduction You've seen how to work with the core parts of the JavaMail API. In the following sections you'll find a how-to approach for connecting the pieces to do specific tasks. Sending messages Sending an e-mail message involves getting a session, creating and filling a message, and sending it. You can specify your SMTP server by setting the mail.smtp.host property for the Properties object passed when getting the Session: String host = ...; String from = ...; String to = ...; // Get system properties Properties props = System.getProperties(); // Setup mail server props.put("mail.smtp.host", host); // Get session Session session = Session.getDefaultInstance(props, null); // Define message MimeMessage message = new MimeMessage(session); message.setFrom(new InternetAddress(from)); message.addRecipient(Message.RecipientType.TO, new InternetAddress(to)); message.setSubject("Hello JavaMail"); message.setText("Welcome to JavaMail"); // Send message Transport.send(message); You should place the code in a try-catch block, as setting up the message and sending it can throw exceptions. Exercise: Exercise 2. How to send your first message on page 23 Fetching messages For reading mail, you get a session, get and connect to an appropriate store for your mailbox, open the appropriate folder, and get your messages. Also, don't forget to close the connection when done. String host = ...; String username = ...; String password = ...; // Create empty properties Properties props = new Properties(); // Get session Session session = Session.getDefaultInstance(props, null); Fundamentals of the JavaMail API Page 13 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks // Get the store Store store = session.getStore("pop3"); store.connect(host, username, password); // Get folder Folder folder = store.getFolder("INBOX"); folder.open(Folder.READ_ONLY); // Get directory Message message[] = folder.getMessages(); for (int i=0, n=message.length; iSubject()); } // Close connection folder.close(false); store.close(); What you do with each message is up to you. The above code block just displays whom the message is from and the subject. Technically speaking, the list of from addresses could be empty and the getFrom()[0] call could throw an exception. To display the whole message, you can prompt the user after seeing the from and subject fields, and then call the message's writeTo() method if the user wants to see it. BufferedReader reader = new BufferedReader ( new InputStreamReader(System.in)); // Get directory Message message[] = folder.getMessages(); for (int i=0, n=message.length; iSubject()); System.out.println("Do you want to read message? " + "[YES to read/QUIT to end]"); String line = reader.readLine(); if ("YES".equals(line)) { message[i].writeTo(System.out); } else if ("QUIT".equals(line)) { break; } } Exercise: Exercise 3. How to check for mail on page 25 Deleting messages and flags Deleting messages involves working with the Flags associated with the messages. There are different flags for different states, some system-defined and some user-defined. The predefined flags are defined in the inner class Flags.Flag and are listed below: * Flags.Flag.ANSWERED * Flags.Flag.DELETED * Flags.Flag.DRAFT * Flags.Flag.FLAGGED * Flags.Flag.RECENT * Flags.Flag.SEEN * Flags.Flag.USER Fundamentals of the JavaMail API Page 14 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Just because a flag exists doesn't mean the flag is supported by all mail servers or providers. For instance, except for deleting messages, the POP protocol supports none of them. Checking for new mail is not a POP task but a task built into mail clients. To find out what flags are supported, ask the folder with getPermanentFlags(). To delete messages, you set the message's DELETED flag: message.setFlag(Flags.Flag.DELETED, true); Open up the folder in READ_WRITE mode first though: folder.open(Folder.READ_WRITE); Then, when you are done processing all messages, close the folder, passing in a true value to expunge the deleted messages. folder.close(true); There is an expunge() method of Folder that can be used to delete the messages. However, it doesn't work for Sun's POP3 provider. Other providers may or may not implement the capabilities. It will more than likely be implemented for IMAP providers. Because POP only supports single access to the mailbox, you have to close the folder to delete the messages with Sun's provider. To unset a flag, just pass false to the setFlag() method. To see if a flag is set, check it with isSet(). Authenticating yourself You learned that you can use an Authenticator to prompt for username and password when needed, instead of passing them in as strings. Here you'll actually see how to more fully use authentication. Instead of connecting to the Store with the host, username, and password, you configure the Properties to have the host, and tell the Session about your custom Authenticator instance, as shown here: // Setup properties Properties props = System.getProperties(); props.put("mail.pop3.host", host); // Setup authentication, get session Authenticator auth = new PopupAuthenticator(); Session session = Session.getDefaultInstance(props, auth); // Get the store Store store = session.getStore("pop3"); store.connect(); You then subclass Authenticator and return a PasswordAuthentication object from the getPasswordAuthentication() method. The following is one such implementation, with a single field for both. (This isn't a Project Swing tutorial; just enter the two parts in the one field, separated by a comma.) Fundamentals of the JavaMail API Page 15 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks import javax.mail.*; import javax.swing.*; import java.util.*; public class PopupAuthenticator extends Authenticator { public PasswordAuthentication getPasswordAuthentication() { String username, password; String result = JOptionPane.showInputDialog( "Enter 'username,password'"); StringTokenizer st = new StringTokenizer(result, ","); username = st.nextToken(); password = st.nextToken(); return new PasswordAuthentication(username, password); } } Because the PopupAuthenticator relies on Swing, it will start up the event-handling thread for AWT. This basically requires you to add a call to System.exit() in your code to stop the program. Replying to messages The Message class includes a reply() method to configure a new Message with the proper recipient and subject, adding "Re: " if not already there. This does not add any content to the message, only copying the from or reply-to header to the new recipient. The method takes a boolean parameter indicating whether to reply to only the sender (false) or reply to all (true). MimeMessage reply = (MimeMessage)message.reply(false); reply.setFrom(new InternetAddress("president@whitehouse.gov")); reply.setText("Thanks"); Transport.send(reply); To configure the reply-to address when sending a message, use the setReplyTo() method. Exercise: Exercise 4. How to reply to mail on page 27 Forwarding messages Forwarding messages is a little more involved. There is no single method to call, and you build up the message to forward by working with the parts that make up a message. A mail message can be made up of multiple parts. Each part is a BodyPart , or more specifically, a MimeBodyPart when working with MIME messages. The different body parts get combined into a container called Multipart or, again, more specifically a MimeMultipart . To forward a message, you create one part for the text of your message and a second part with the message to forward, and combine the two into a multipart. Then you add the multipart to a properly addressed message and send it. That's essentially it. To copy the content from one message to another, just copy over its Fundamentals of the JavaMail API Page 16 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks DataHandler , a class from the JavaBeans Activation Framework. // Create the message to forward Message forward = new MimeMessage(session); // Fill in header forward.setSubject("Fwd: " + message.getSubject()); forward.setFrom(new InternetAddress(from)); forward.addRecipient(Message.RecipientType.TO, new InternetAddress(to)); // Create your new message part BodyPart messageBodyPart = new MimeBodyPart(); messageBodyPart.setText( "Here you go with the original message:\n\n"); // Create a multi-part to combine the parts Multipart multipart = new MimeMultipart(); multipart.addBodyPart(messageBodyPart); // Create and fill part for the forwarded content messageBodyPart = new MimeBodyPart(); messageBodyPart.setDataHandler(message.getDataHandler()); // Add part to multi part multipart.addBodyPart(messageBodyPart); // Associate multi-part with message forward.setContent(multipart); // Send message Transport.send(forward); Working with attachments Attachments are resources associated with a mail message, usually kept outside of the message like a text file, spreadsheet, or image. As with common mail programs like Eudora and pine, you can attach resources to your mail message with the JavaMail API and get those attachments when you receive the message. Sending attachments: Sending attachments is quite like forwarding messages. You build up the parts to make the complete message. After the first part, your message text, you add other parts where the DataHandler for each is your attachment, instead of the shared handler in the case of a forwarded message. If you are reading the attachment from a file, your attachment data source is a FileDataSource . Reading from a URL, it is a URLDataSource . Once you have your DataSource, just pass it on to the DataHandler constructor, before finally attaching it to the BodyPart with setDataHandler(). Assuming you want to retain the original filename for the attachment, the last thing to do is to set the filename associated with the attachment with the setFileName() method of BodyPart. All this is shown here: // Define message Message message = new MimeMessage(session); message.setFrom(new InternetAddress(from)); message.addRecipient(Message.RecipientType.TO, new InternetAddress(to)); message.setSubject("Hello JavaMail Attachment"); // Create the message part BodyPart messageBodyPart = new MimeBodyPart(); // Fill the message messageBodyPart.setText("Pardon Ideas"); Fundamentals of the JavaMail API Page 17 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Multipart multipart = new MimeMultipart(); multipart.addBodyPart(messageBodyPart); // Part two is attachment messageBodyPart = new MimeBodyPart(); DataSource source = new FileDataSource(filename); messageBodyPart.setDataHandler(new DataHandler(source)); messageBodyPart.setFileName(filename); multipart.addBodyPart(messageBodyPart); // Put parts in message message.setContent(multipart); // Send the message Transport.send(message); When including attachments with your messages, if your program is a servlet, your users must upload the attachment besides telling you where to send the message. Uploading each file can be handled with a form encoding type of multipart/form-data.
Note: Message size is limited by your SMTP server, not the JavaMail API. If you run into problems, consider increasing the Java heap size by setting the ms and mx parameters. Exercise: Exercise 5. How to send attachments on page 28 Getting attachments: Getting attachments out of your messages is a little more involved then sending them because MIME has no simple notion of attachments. The content of your message is a Multipart object when it has attachments. You then need to process each Part, to get the main content and the attachment(s). Parts marked with a disposition of Part.ATTACHMENT from part.getDisposition() are clearly attachments. However, attachments can also come across with no disposition (and a non-text MIME type) or a disposition of Part.INLINE. When the disposition is either Part.ATTACHMENT or Part.INLINE, you can save off the content for that message part. Just get the original filename with getFileName() and the input stream with getInputStream(). Multipart mp = (Multipart)message.getContent(); for (int i=0, n=multipart.getCount(); i"; message.setContent(htmlText, "text/html")); On the receiving end, if you fetch the message with the JavaMail API, there is nothing built into the API to display the message as HTML. The JavaMail API only sees it as a stream of bytes. To display the message as HTML, you must either use the Swing JEditorPane or some third-party HTML viewer component. if (message.getContentType().equals("text/html")) { String content = (String)message.getContent(); JFrame frame = new JFrame(); JEditorPane text = new JEditorPane("text/html", content); text.setEditable(false); JScrollPane pane = new JScrollPane(text); frame.getContentPane().add(pane); frame.setSize(300, 300); frame.setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE); frame.show(); } Including images with your messages: On the other hand, if you want your HTML content message to be complete, with embedded images included as part of the message, you must treat the image as an attachment and reference the image with a special cid URL, where the cid is a reference to the Content-ID header of the image attachment. The process of embedding an image is quite similar to attaching a file to a message, the only Fundamentals of the JavaMail API Page 19 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks difference is you have to tell the MimeMultipart that the parts are related by setting its subtype in the constructor (or with setSubType()) and set the Content-ID header for the image to a random string which is used as the src for the image in the img tag. The following demonstrates this completely. String file = ...; // Create the message Message message = new MimeMessage(session); // Fill its headers message.setSubject("Embedded Image"); message.setFrom(new InternetAddress(from)); message.addRecipient(Message.RecipientType.TO, new InternetAddress(to)); // Create your new message part BodyPart messageBodyPart = new MimeBodyPart(); String htmlText = "

Hello

" + ""; messageBodyPart.setContent(htmlText, "text/html"); // Create a related multi-part to combine the parts MimeMultipart multipart = new MimeMultipart("related"); multipart.addBodyPart(messageBodyPart); // Create part for the image messageBodyPart = new MimeBodyPart(); // Fetch the image and associate to part DataSource fds = new FileDataSource(file); messageBodyPart.setDataHandler(new DataHandler(fds)); messageBodyPart.setHeader("Content-ID","memememe"); // Add part to multi-part multipart.addBodyPart(messageBodyPart); // Associate multi-part with message message.setContent(multipart); Exercise: Exercise 6. How to send HTML messages with images on page 29 Fundamentals of the JavaMail API Page 20 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 7. Searching with SearchTerm Introduction The JavaMail API includes a filtering mechanism found in the javax.mail.search package to build up a SearchTerm . Once built, you then ask a Folder what messages match, retrieving an array of Message objects: SearchTerm st = ...; Message[] msgs = folder.search(st); There are 22 different classes available to help you build a search term. * AND terms (class AndTerm) * OR terms (class OrTerm) * NOT terms (class NotTerm) * SENT DATE terms (class SentDateTerm) * CONTENT terms (class BodyTerm) * HEADER terms (FromTerm / FromStringTerm, RecipientTerm / RecipientStringTerm, SubjectTerm, etc..) Essentially, you build up a logical expression for matching messages, then search. For instance the following term searches for messages with a (partial) subject string of ADV or a from field of friend@public.com. You might consider periodically running this query and automatically deleting any messages returned. SearchTerm st = new OrTerm( new SubjectTerm("ADV:"), new FromStringTerm("friend@public.com")); Message[] msgs = folder.search(st); Fundamentals of the JavaMail API Page 21 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 8. Exercises About the exercises These exercises are designed to provide help according to your needs. For example, you might simply complete the exercise given the information and the task list in the exercise body; you might want a few hints; or you may want a step-by-step guide to successfully complete a particular exercise. You can use as much or as little help as you need per exercise. Moreover, because complete solutions are also provided, you can skip a few exercises and still be able to complete future exercises requiring the skipped ones. Each exercise has a list of any prerequisite exercises, a list of skeleton code for you to start with, links to necessary API pages, and a text description of the exercise goal. In addition, there is help for each task and a solutions page with links to files that comprise a solution to the exercise. Exercise 1. How to set up a JavaMail environment In this exercise you will install Sun's JavaMail reference implementation. After installing, you will be introduced to the demonstration programs that come with the reference implementation. Task 1: Download the latest version of the JavaMail API implementation from Sun. Task 2: Download the latest version of the JavaBeans Activation Framework from Sun. Task 3: Unzip the downloaded packages. You get a ZIP file for all platforms for both packages. Help for task 3: You can use the jar tool to unzip the packages. Task 4: Add the mail.jar file from the JavaMail 1.2 download and the activation.jar file from the JavaBeans Activation Framework download to your CLASSPATH. Help for task 4: Copy the files to your extension library directory. For Microsoft Windows, using the default installation copy, the command might look like the following: cd \javamail-1.2 copy mail.jar \jdk1.3\jre\lib\ext cd \jaf-1.0.1 copy activation.jar \jdk1.3\jre\lib\ext If you don't like copying the files to the extension library directory, detailed instructions are available from Sun for setting your CLASSPATH on Windows NT. Task 5: Go into the demo directory that comes with the JavaMail API implementation and compile the msgsend program to send a test message. Help for task 5: javac msgsend.java Fundamentals of the JavaMail API Page 22 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Task 6: Execute the program passing in a from address with the -o option, your SMTP server with the -M option, and the to address (with no option). You'll then enter the subject, the text of your message, and the end-of-file character (CTRL-Z) to signal the end of the message input. Help for task 6: Be sure to replace the from address, SMTP server, and to address. java msgsend -o from@address -M SMTP.Server to@address If you are not sure of your SMTP server, contact your system administrator or check with your Internet Service Provider. Task 7: Check to make sure you received the message with your normal mail reader (Eudora, Outlook Express, pine, ...). Exercise 1. How to set up a JavaMail environment: Solution Upon successful completion, the JavaMail reference implementation will be in your CLASSPATH. Exercise 2. How to send your first message In the last exercise you sent a mail message using the demonstration program provided with the JavaMail implementation. In this exercise, you'll create the program yourself. For more help with exercises, see About the exercises on page 22 . Prerequisites: * Exercise 1. How to set up a JavaMail environment on page 22 Skeleton code: * MailExample.java Task 1: Starting with the skeleton code , get the system Properties. Help for task 1: Properties props = System.getProperties(); Task 2: Add the name of your SMTP server to the properties for the mail.smtp.host key. Fundamentals of the JavaMail API Page 23 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Help for task 2: props.put("mail.smtp.host", host); Task 3: Get a Session object based on the Properties. Help for task 3: Session session = Session.getDefaultInstance(props, null); Task 4: Create a MimeMessage from the session. Help for task 4: MimeMessage message = new MimeMessage(session); Task 5: Set the from field of the message. Help for task 5: message.setFrom(new InternetAddress(from)); Task 6: Set the to field of the message. Help for task 6: message.addRecipient(Message.RecipientType.TO, new InternetAddress(to)); Task 7: Set the subject of the message. Help for task 7: message.setSubject("Hello JavaMail"); Task 8: Set the content of the message. Help for task 8: message.setText("Welcome to JavaMail"); Task 9: Use a Transport to send the message. Help for task 9: Transport.send(message); Task 10: Compile and run the program, passing your SMTP server, from address, and to address on the command line. Fundamentals of the JavaMail API Page 24 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Help for task 10: java MailExample SMTP.Server from@address to@address Task 11: Check to make sure you received the message with your normal mail reader (Eudora, Outlook Express, pine, ...). Exercise 2. How to send your first message: Solution The following Java source file represents a solution to this exercise: * Solution/MailExample.java Exercise 3. How to check for mail In this exercise, create a program that displays the from address and subject for each message and prompts to display the message content. For more help with exercises, see About the exercises on page 22 . Prerequisites: * Exercise 1. How to set up a JavaMail environment on page 22 Skeleton Code * GetMessageExample.java Task 1: Starting with the skeleton code , get or create a Properties object. Help for task 1: Properties props = new Properties(); Task 2: Get a Session object based on the Properties. Help for task 2: Session session = Session.getDefaultInstance(props, null); Task 3: Get a Store for your e-mail protocol, either pop3 or imap. Help for task 3: Store store = session.getStore("pop3"); Task 4: Connect to your mail host's store with the appropriate username and password. Fundamentals of the JavaMail API Page 25 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Help for task 4: store.connect(host, username, password); Task 5: Get the folder you want to read. More than likely, this will be the INBOX. Help for task 5: Folder folder = store.getFolder("INBOX"); Task 6: Open the folder read-only. Help for task 6: folder.open(Folder.READ_ONLY); Task 7: Get a directory of the messages in the folder. Save the message list in an array variable named message. Help for task 7: Message message[] = folder.getMessages(); Task 8: For each message, display the from field and the subject. Help for task 8: System.out.println(i + ": " + message[i].getFrom()[0] + "\t" + message[i].getSubject()); Task 9: Display the message content when prompted. Help for task 9: System.out.println(message[i].getContent()); Task 10: Close the connection to the folder and store. Help for task 10: folder.close(false); store.close(); Task 11: Compile and run the program, passing your mail server, username, and password on the command line. Answer YES to the messages you want to read. Just hit ENTER if you don't. If you want to stop reading your mail before making your way through all the messages, enter QUIT. Help for task 11: java GetMessageExample POP.Server username password Fundamentals of the JavaMail API Page 26 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Exercise 3. How to check for mail: Solution The following Java source file represents a solution to this exercise. * Solution/GetMessageExample.java Exercise 4. How to reply to mail In this exercise, create a program that creates a canned reply message and attaches the original message if it's plain text. For more help with exercises, see About the exercises on page 22 . Prerequisites: * Exercise 3. How to check for mail on page 25 Skeleton Code: * ReplyExample.java Task 1: The skeleton code already includes the code to get the list of messages from the folder and prompt you to create a reply. Task 2: When answered affirmatively, create a new MimeMessage from the original message. Help for task 2: MimeMessage reply = (MimeMessage)message[i].reply(false); Task 3: Set the from field to your e-mail address. Task 4: Create the text for the reply. Include a canned message to start. When the original message is plain text, add each line of the original message, prefix each line with the "> " characters. Help for task 4: To check for plain text messages, check the messages MIME type with mimeMessage.isMimeType("text/plain"). Task 5: Set the message's content, once the message content is fully determined. Task 6: Send the message. Task 7: Compile and run the program, passing your mail server, SMTP server, username, password, and from address on the command line. Answer YES to the messages you want to send replies. Just hit ENTER if you don't. If you want to stop going through your mail before Fundamentals of the JavaMail API Page 27 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks making your way through all the messages, enter QUIT. Help for task 7: java ReplyExample POP.Server SMTP.Server username password from@address Task 8: Check to make sure you received the message with your normal mail reader (Eudora, Outlook Express, pine, ...). Exercise 4. How to reply to mail: Solution The following Java source file represents a solution to this exercise. * Solution/ReplyExample.java Exercise 5. How to send attachments In this exercise, create a program that sends a message with an attachment. For more help with exercises, see About the exercises on page 22 . Prerequisites: * Exercise 2. How to send your first message on page 23 Skeleton Code: * AttachExample.java Task 1: The skeleton code already includes the code to get the initial mail session. Task 2: From the session, get a Message and set its header fields: to, from, and subject. Task 3: Create a BodyPart for the main message content and fill its content with the text of the message. Help for task 3: BodyPart messageBodyPart = new MimeBodyPart(); messageBodyPart.setText("Here's the file"); Task 4: Create a Multipart to combine the main content with the attachment. Add the main content to the multipart. Help for task 4: Multipart multipart = new MimeMultipart(); multipart.addBodyPart(messageBodyPart); Fundamentals of the JavaMail API Page 28 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Task 5: Create a second BodyPart for the attachment. Task 6: Get the attachment as a DataSource. Help for task 6: DataSource source = new FileDataSource(filename); Task 7: Set the DataHandler for the message part to the data source. Carry the original filename along. Help for task 7: messageBodyPart.setDataHandler(new DataHandler(source)); messageBodyPart.setFileName(filename); Task 8: Add the second part of the message to the multipart. Task 9: Set the content of the message to the multipart. Help for task 9: message.setContent(multipart); Task 10: Send the message. Task 11: Compile and run the program, passing your SMTP server, from address, to address, and filename on the command line. This will send the file as an attachment. Help for task 11: java AttachExample SMTP.Server from@address to@address filename Task 12: Check to make sure you received the message with your normal mail reader (Eudora, Outlook Express, pine, ...). Exercise 5. How to send attachments: Solution The following Java source file represents a solution to this exercise. * Solution/AttachExample.java Exercise 6. How to send HTML messages with images In this exercise, create a program that sends an HTML message with an image attachment where the image is displayed within the HTML message. Fundamentals of the JavaMail API Page 29 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks For more help with exercises, see About the exercises on page 22 . Prerequisites: * Exercise 5. How to send attachments on page 28 Skeleton code: * logo.gif * HtmlImageExample.java Task 1: The skeleton code already includes the code to get the initial mail session, create the main message, and fill its headers (to, from, subject). Task 2: Create a BodyPart for the HTML message content. Task 3: Create a text string of the HTML content. Include a reference in the HTML to an image () that is local to the mail message. Help for task 3: Use a cid URL. The content-id will need to be specified for the image later. String htmlText = "

Hello

" + ""; Task 4: Set the content of the message part. Be sure to specify the MIME type is text/html. Help for task 4: messageBodyPart.setContent(htmlText, "text/html"); Task 5: Create a Multipart to combine the main content with the attachment. Be sure to specify that the parts are related. Add the main content to the multipart. Help for task 5: MimeMultipart multipart = new MimeMultipart("related"); multipart.addBodyPart(messageBodyPart); Task 6: Create a second BodyPart for the attachment. Task 7: Get the attachment as a DataSource, and set the DataHandler for the message part to the data source. Task 8: Set the Content-ID header for the part to match the image reference specified in the HTML. Help for task 8: messageBodyPart.setHeader("Content-ID","memememe"); Task 9: Add the second part of the message to the multipart, and set the content of the Fundamentals of the JavaMail API Page 30 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks message to the multipart. Task 10: Send the message. Task 11: Compile and run the program, passing your SMTP server, from address, to address, and filename on the command line. This will send the images as an inline image within the HTML text. Help for task 11: java HtmlImageExample SMTP.Server from@address to@address filename Task 12: Check if your mail reader recognizes the message as HTML and displays the image within the message, instead of as a link to an external attachment file. Help for task 12: If your mail reader can't display HTML messages, consider sending the message to a friend. Exercise 6. How to send HTML messages with images: Solution The following Java source files represent a solution to this exercise. * Solution/logo.gif * Solution/HtmlImageExample.java Fundamentals of the JavaMail API Page 31 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Section 9. Wrapup In summary The JavaMail API is a Java package used for reading, composing, and sending e-mail messages and their attachments. It lets you build standards-based e-mail clients that employ various Internet mail protocols, including SMTP, POP, IMAP, and MIME, as well as related protocols such as NNTP, S/MIME, and others. The API divides naturally into two parts. The first focuses on sending, receiving, and managing messages independent of the protocol used, whereas the second focuses on specific use of the protocols. The purpose of this tutorial was to show how to use the first part of the API, without attempting to deal with protocol providers. The core JavaMail API consists of seven classes --Session, Message, Address, Authenticator, Transport, Store, and Folder --all of which are found in javax.mail, the top-level package for the JavaMail API. We used these classes to work through a number of common e-mail-related tasks, including sending messages, retrieving messages, deleting messages, authenticating, replying to messages, forwarding messages, managing attachments, processing HTML-based messages, and searching or filtering mail lists. Finally, we provided a number of step-by-step exercises to help illustrate the concepts presented. Hopefully, this will help you add e-mail functionality to your platform-independent Java applications. Resources You can do much more with the JavaMail API than what's found here. The lessons and exercises found here can be supplemented by the following resources: * Download the JavaMail 1.2 API from the JavaMail API home page . * The JavaBeans Activation Framework is required for versions 1.2 and 1.1.3 of the JavaMail API. * The JavaMail-interest mailing list is a Sun-hosted discussion forum for developers. * Sun's JavaMail FAQ addresses the use of JavaMail in applets and servlets, as well as prototol-specific questions. * Tutorial author John Zukowski maintains jGuru's JavaMail FAQ . * Want to see how others are using JavaMail? Check out Sun's list of third-party products. * If you want more detail about JavaMail, read Rick Grehan's "How JavaMail keeps it simple" (Lotus Developer Network, June 2000). * Benoit Marchal shows how to use Java and XML to produce plain text and HTML newsletters in this two-part series, "Managing e-zines with JavaMail and XSLT" Part 1 (developerWorks, March 2001) and Part 2 (developerWorks, April 2001). * "Linking Applications with E-mail" (Lotus Developer Network, May 2000) discusses how groupware can facilitate communication, collaboration, and coordination among applications. Fundamentals of the JavaMail API Page 32 Presented by developerWorks, your source for great tutorials ibm.com/developerWorks Feedback Please let us know whether this tutorial was helpful to you and how we could make it better. We'd also like to hear about other tutorial topics you'd like to see covered. Thanks! For questions about the content of this tutorial, contact the author John Zukowski ( jaz@zukowski.net ) Colophon This tutorial was written entirely in XML, using the developerWorks Toot-O-Matic tutorial generator. The Toot-O-Matic tool is a short Java program that uses XSLT stylesheets to convert the XML source into a number of HTML pages, a zip file, JPEG heading graphics, and PDF files. Our ability to generate multiple text and binary formats from a single source file illustrates the power and flexibility of XML. Fundamentals of the JavaMail API Page 33
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 : Yes. (From RFC 2616) “A client might have started to send a new request at the same time that the server has decided to close the "idle" connection. From the server's point of view, the connection is being closed while it was idle, but from the client's point of view, a request is in progress.” Problem 7 The total amount of time to get the IP address is . Once the IP address is known, elapses to set up the TCP connection and another elapses to request and receive the small object. The total response time is Problem 8 . . Problem 9 The time to transmit an object of size L over a link or rate R is L/R. The average time is the average size of the object divided by R:  = (850,000 bits)/(15,000,000 bits/sec) = .0567 sec The traffic intensity on the link is given by =(16 requests/sec)(.0567 sec/request) = 0.907. Thus, the average access delay is (.0567 sec)/(1 - .907)  .6 seconds. The total average response time is therefore .6 sec + 3 sec = 3.6 sec. The traffic intensity on the access link is reduced by 60% since the 60% of the requests are satisfied within the institutional network. Thus the average access delay is (.0567 sec)/[1 – (.4)(.907)] = .089 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .6); the average response time is .089 sec + 3 sec = 3.089 sec for cache misses (which happens 40% of the time). So the average response time is (.6)(0 sec) + (.4)(3.089 sec) = 1.24 seconds. Thus the average response time is reduced from 3.6 sec to 1.24 sec. Problem 10 Note that each downloaded object can be completely put into one data packet. Let Tp denote the one-way propagation delay between the client and the server. First consider parallel downloads using non-persistent connections. Parallel downloads would allow 10 connections to share the 150 bits/sec bandwidth, giving each just 15 bits/sec. Thus, the total time needed to receive all objects is given by: (200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + (200/(150/10)+Tp + 200/(150/10) +Tp + 200/(150/10)+Tp + 100,000/(150/10)+ Tp ) = 7377 + 8*Tp (seconds) Now consider a persistent HTTP connection. The total time needed is given by: (200/150+Tp + 200/150 +Tp + 200/150+Tp + 100,000/150+ Tp ) + 10*(200/150+Tp + 100,000/150+ Tp ) =7351 + 24*Tp (seconds) Assuming the speed of light is 300*106 m/sec, then Tp=10/(300*106)=0.03 microsec. Tp is therefore negligible compared with transmission delay. Thus, we see that persistent HTTP is not significantly faster (less than 1 percent) than the non-persistent case with parallel download. Problem 11 Yes, because Bob has more connections, he can get a larger share of the link bandwidth. Yes, Bob still needs to perform parallel downloads; otherwise he will get less bandwidth than the other four users. Problem 12 Server.py from socket import * serverPort=12000 serverSocket=socket(AF_INET,SOCK_STREAM) serverSocket.bind(('',serverPort)) serverSocket.listen(1) connectionSocket, addr = serverSocket.accept() while 1: sentence = connectionSocket.recv(1024) print 'From Server:', sentence, '\n' serverSocket.close() Problem 13 The MAIL FROM: in SMTP is a message from the SMTP client that identifies the sender of the mail message to the SMTP server. The From: on the mail message itself is NOT an SMTP message, but rather is just a line in the body of the mail message. Problem 14 SMTP uses a line containing only a period to mark the end of a message body. HTTP uses “Content-Length header field” to indicate the length of a message body. No, HTTP cannot use the method used by SMTP, because HTTP message could be binary data, whereas in SMTP, the message body must be in 7-bit ASCII format. Problem 15 MTA stands for Mail Transfer Agent. A host sends the message to an MTA. The message then follows a sequence of MTAs to reach the receiver’s mail reader. We see that this spam message follows a chain of MTAs. An honest MTA should report where it receives the message. Notice that in this message, “asusus-4b96 ([58.88.21.177])” does not report from where it received the email. Since we assume only the originator is dishonest, so “asusus-4b96 ([58.88.21.177])” must be the originator. Problem 16 UIDL abbreviates “unique-ID listing”. When a POP3 client issues the UIDL command, the server responds with the unique message ID for all of the messages present in the user's mailbox. This command is useful for “download and keep”. By maintaining a file that lists the messages retrieved during earlier sessions, the client can use the UIDL command to determine which messages on the server have already been seen. Problem 17 a) C: dele 1 C: retr 2 S: (blah blah … S: ………..blah) S: . C: dele 2 C: quit S: +OK POP3 server signing off b) C: retr 2 S: blah blah … S: ………..blah S: . C: quit S: +OK POP3 server signing off C: list S: 1 498 S: 2 912 S: . C: retr 1 S: blah ….. S: ….blah S: . C: retr 2 S: blah blah … S: ………..blah S: . C: quit S: +OK POP3 server signing off Problem 18 For a given input of domain name (such as ccn.com), IP address or network administrator name, the whois database can be used to locate the corresponding registrar, whois server, DNS server, and so on. NS4.YAHOO.COM from www.register.com; NS1.MSFT.NET from ww.register.com Local Domain: www.mindspring.com Web servers : www.mindspring.com 207.69.189.21, 207.69.189.22, 207.69.189.23, 207.69.189.24, 207.69.189.25, 207.69.189.26, 207.69.189.27, 207.69.189.28 Mail Servers : mx1.mindspring.com (207.69.189.217) mx2.mindspring.com (207.69.189.218) mx3.mindspring.com (207.69.189.219) mx4.mindspring.com (207.69.189.220) Name Servers: itchy.earthlink.net (207.69.188.196) scratchy.earthlink.net (207.69.188.197) www.yahoo.com Web Servers: www.yahoo.com (216.109.112.135, 66.94.234.13) Mail Servers: a.mx.mail.yahoo.com (209.191.118.103) b.mx.mail.yahoo.com (66.196.97.250) c.mx.mail.yahoo.com (68.142.237.182, 216.39.53.3) d.mx.mail.yahoo.com (216.39.53.2) e.mx.mail.yahoo.com (216.39.53.1) f.mx.mail.yahoo.com (209.191.88.247, 68.142.202.247) g.mx.mail.yahoo.com (209.191.88.239, 206.190.53.191) Name Servers: ns1.yahoo.com (66.218.71.63) ns2.yahoo.com (68.142.255.16) ns3.yahoo.com (217.12.4.104) ns4.yahoo.com (68.142.196.63) ns5.yahoo.com (216.109.116.17) ns8.yahoo.com (202.165.104.22) ns9.yahoo.com (202.160.176.146) www.hotmail.com Web Servers: www.hotmail.com (64.4.33.7, 64.4.32.7) Mail Servers: mx1.hotmail.com (65.54.245.8, 65.54.244.8, 65.54.244.136) mx2.hotmail.com (65.54.244.40, 65.54.244.168, 65.54.245.40) mx3.hotmail.com (65.54.244.72, 65.54.244.200, 65.54.245.72) mx4.hotmail.com (65.54.244.232, 65.54.245.104, 65.54.244.104) Name Servers: ns1.msft.net (207.68.160.190) ns2.msft.net (65.54.240.126) ns3.msft.net (213.199.161.77) ns4.msft.net (207.46.66.126) ns5.msft.net (65.55.238.126) d) The yahoo web server has multiple IP addresses www.yahoo.com (216.109.112.135, 66.94.234.13) e) The address range for Polytechnic University: 128.238.0.0 – 128.238.255.255 f) An attacker can use the whois database and nslookup tool to determine the IP address ranges, DNS server addresses, etc., for the target institution. By analyzing the source address of attack packets, the victim can use whois to obtain information about domain from which the attack is coming and possibly inform the administrators of the origin domain. Problem 19 The following delegation chain is used for gaia.cs.umass.edu a.root-servers.net E.GTLD-SERVERS.NET ns1.umass.edu(authoritative) First command: dig +norecurse @a.root-servers.net any gaia.cs.umass.edu ;; AUTHORITY SECTION: edu. 172800 IN NS E.GTLD-SERVERS.NET. edu. 172800 IN NS A.GTLD-SERVERS.NET. edu. 172800 IN NS G3.NSTLD.COM. edu. 172800 IN NS D.GTLD-SERVERS.NET. edu. 172800 IN NS H3.NSTLD.COM. edu. 172800 IN NS L3.NSTLD.COM. edu. 172800 IN NS M3.NSTLD.COM. edu. 172800 IN NS C.GTLD-SERVERS.NET. Among all returned edu DNS servers, we send a query to the first one. dig +norecurse @E.GTLD-SERVERS.NET any gaia.cs.umass.edu umass.edu. 172800 IN NS ns1.umass.edu. umass.edu. 172800 IN NS ns2.umass.edu. umass.edu. 172800 IN NS ns3.umass.edu. Among all three returned authoritative DNS servers, we send a query to the first one. dig +norecurse @ns1.umass.edu any gaia.cs.umass.edu gaia.cs.umass.edu. 21600 IN A 128.119.245.12 The answer for google.com could be: a.root-servers.net E.GTLD-SERVERS.NET ns1.google.com(authoritative) Problem 20 We can periodically take a snapshot of the DNS caches in the local DNS servers. The Web server that appears most frequently in the DNS caches is the most popular server. This is because if more users are interested in a Web server, then DNS requests for that server are more frequently sent by users. Thus, that Web server will appear in the DNS caches more frequently. For a complete measurement study, see: Craig E. Wills, Mikhail Mikhailov, Hao Shang “Inferring Relative Popularity of Internet Applications by Actively Querying DNS Caches”, in IMC'03, October 27­29, 2003, Miami Beach, Florida, USA Problem 21 Yes, we can use dig to query that Web site in the local DNS server. For example, “dig cnn.com” will return the query time for finding cnn.com. If cnn.com was just accessed a couple of seconds ago, an entry for cnn.com is cached in the local DNS cache, so the query time is 0 msec. Otherwise, the query time is large. Problem 22 For calculating the minimum distribution time for client-server distribution, we use the following formula: Dcs = max {NF/us, F/dmin} Similarly, for calculating the minimum distribution time for P2P distribution, we use the following formula: Where, F = 15 Gbits = 15 * 1024 Mbits us = 30 Mbps dmin = di = 2 Mbps Note, 300Kbps = 300/1024 Mbps. Client Server N 10 100 1000 u 300 Kbps 7680 51200 512000 700 Kbps 7680 51200 512000 2 Mbps 7680 51200 512000 Peer to Peer N 10 100 1000 u 300 Kbps 7680 25904 47559 700 Kbps 7680 15616 21525 2 Mbps 7680 7680 7680 Problem 23 Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of a rate of us/N. Note that this rate is less than each of the client’s download rate, since by assumption us/N ≤ dmin. Thus each client can also receive at rate us/N. Since each client receives at rate us/N, the time for each client to receive the entire file is F/( us/N) = NF/ us. Since all the clients receive the file in NF/ us, the overall distribution time is also NF/ us. Consider a distribution scheme in which the server sends the file to each client, in parallel, at a rate of dmin. Note that the aggregate rate, N dmin, is less than the server’s link rate us, since by assumption us/N ≥ dmin. Since each client receives at rate dmin, the time for each client to receive the entire file is F/ dmin. Since all the clients receive the file in this time, the overall distribution time is also F/ dmin. From Section 2.6 we know that DCS ≥ max {NF/us, F/dmin} (Equation 1) Suppose that us/N ≤ dmin. Then from Equation 1 we have DCS ≥ NF/us . But from (a) we have DCS ≤ NF/us . Combining these two gives: DCS = NF/us when us/N ≤ dmin. (Equation 2) We can similarly show that: DCS =F/dmin when us/N ≥ dmin (Equation 3). Combining Equation 2 and Equation 3 gives the desired result. Problem 24 Define u = u1 + u2 + ….. + uN. By assumption us <= (us + u)/N Equation 1 Divide the file into N parts, with the ith part having size (ui/u)F. The server transmits the ith part to peer i at rate ri = (ui/u)us. Note that r1 + r2 + ….. + rN = us, so that the aggregate server rate does not exceed the link rate of the server. Also have each peer i forward the bits it receives to each of the N-1 peers at rate ri. The aggregate forwarding rate by peer i is (N-1)ri. We have (N-1)ri = (N-1)(usui)/u = (us + u)/N Equation 2 Let ri = ui/(N-1) and rN+1 = (us – u/(N-1))/N In this distribution scheme, the file is broken into N+1 parts. The server sends bits from the ith part to the ith peer (i = 1, …., N) at rate ri. Each peer i forwards the bits arriving at rate ri to each of the other N-1 peers. Additionally, the server sends bits from the (N+1) st part at rate rN+1 to each of the N peers. The peers do not forward the bits from the (N+1)st part. The aggregate send rate of the server is r1+ …. + rN + N rN+1 = u/(N-1) + us – u/(N-1) = us Thus, the server’s send rate does not exceed its link rate. The aggregate send rate of peer i is (N-1)ri = ui Thus, each peer’s send rate does not exceed its link rate. In this distribution scheme, peer i receives bits at an aggregate rate of Thus each peer receives the file in NF/(us+u). (For simplicity, we neglected to specify the size of the file part for i = 1, …., N+1. We now provide that here. Let Δ = (us+u)/N be the distribution time. For i = 1, …, N, the ith file part is Fi = ri Δ bits. The (N+1)st file part is FN+1 = rN+1 Δ bits. It is straightforward to show that F1+ ….. + FN+1 = F.) The solution to this part is similar to that of 17 (c). We know from section 2.6 that Combining this with a) and b) gives the desired result. Problem 25 There are N nodes in the overlay network. There are N(N-1)/2 edges. Problem 26 Yes. His first claim is possible, as long as there are enough peers staying in the swarm for a long enough time. Bob can always receive data through optimistic unchoking by other peers. His second claim is also true. He can run a client on each host, let each client “free-ride,” and combine the collected chunks from the different hosts into a single file. He can even write a small scheduling program to make the different hosts ask for different chunks of the file. This is actually a kind of Sybil attack in P2P networks. Problem 27 Peer 3 learns that peer 5 has just left the system, so Peer 3 asks its first successor (Peer 4) for the identifier of its immediate successor (peer 8). Peer 3 will then make peer 8 its second successor. Problem 28 Peer 6 would first send peer 15 a message, saying “what will be peer 6’s predecessor and successor?” This message gets forwarded through the DHT until it reaches peer 5, who realizes that it will be 6’s predecessor and that its current successor, peer 8, will become 6’s successor. Next, peer 5 sends this predecessor and successor information back to 6. Peer 6 can now join the DHT by making peer 8 its successor and by notifying peer 5 that it should change its immediate successor to 6. Problem 29 For each key, we first calculate the distances (using d(k,p)) between itself and all peers, and then store the key in the peer that is closest to the key (that is, with smallest distance value). Problem 30 Yes, randomly assigning keys to peers does not consider the underlying network at all, so it very likely causes mismatches. Such mismatches may degrade the search performance. For example, consider a logical path p1 (consisting of only two logical links): ABC, where A and B are neighboring peers, and B and C are neighboring peers. Suppose that there is another logical path p2 from A to C (consisting of 3 logical links): ADEC. It might be the case that A and B are very far away physically (and separated by many routers), and B and C are very far away physically (and separated by many routers). But it may be the case that A, D, E, and C are all very close physically (and all separated by few routers). In other words, a shorter logical path may correspond to a much longer physical path. Problem 31 If you run TCPClient first, then the client will attempt to make a TCP connection with a non-existent server process. A TCP connection will not be made. UDPClient doesn't establish a TCP connection with the server. Thus, everything should work fine if you first run UDPClient, then run UDPServer, and then type some input into the keyboard. If you use different port numbers, then the client will attempt to establish a TCP connection with the wrong process or a non-existent process. 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

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