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Here is how to run the test using two analyzers: 1Configure the analyzers to only capture packets from the client and server computers You can use Microsoft's Network Monitor or any third-party protocol analyzer 2Synchronize your analyzer with the person running the other analyzer so that both analyzers start capturing packets at the same time 3Use the client to request information from the server 4Stop capturing when the conversation is complete 5Repeat these steps for each kind of transaction (requesting a record, depositing a record, changing a record, and so on) When you have finished, you will have two sets of analyzer trace files that hold the packets for each type of transaction It is essential that you keep track of which files belong to what transaction and which analyzer You will examine each trace file to determine the number of packets traveling in each direction, the number of bytes traveling in each direction, and delays at each computer and on the network The traces from both analyzers should be almost identical, except for the times when packets were seen coming in To make sense of it all, you need a simple spreadsheet Table 10-4 illustrates a spreadsheet with five conversations Listed are the start and stop times, the application name, its protocol, the host and destination addresses, and packet and byte counts for traffic in both directions Table 10-4: A Sample Spreadsheet Showing Five Conversations

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The selection predicate may include comparisons between two attributes To illustrate, consider the relation loan-of cer that consists of three attributes: customer-name, banker-name, and loan-number, which speci es that a particular banker is the loan of cer for a loan that belongs to some customer To nd all customers who have the same name as their loan of cer, we can write customer -name = banker -name (loan-o cer )

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Suppose we want to list all loan numbers and the amount of the loans, but do not care about the branch name The project operation allows us to produce this relation The project operation is a unary operation that returns its argument relation, with certain attributes left out Since a relation is a set, any duplicate rows are eliminated Projection is denoted by the uppercase Greek letter pi ( ) We list those attributes that we wish to appear in the result as a subscript to The argument relation follows in parentheses Thus, we write the query to list all loan numbers and the amount of the loan as loan-number , amount (loan) Figure 311 shows the relation that results from this query

"11/20/1999 13:30:54" "11/20/1999 13:31:24" "11/20/1999

The fact that the result of a relational operation is itself a relation is important Consider the more complicated query Find those customers who live in Harrison We write: customer -name ( customer -city = Harrison (customer )) Notice that, instead of giving the name of a relation as the argument of the projection operation, we give an expression that evaluates to a relation In general, since the result of a relational-algebra operation is of the same type (relation) as its inputs, relational-algebra operations can be composed together into loan-number L-11 L-14 L-15 L-16 L-17 L-23 L-93 Figure 311 amount 900 1500 1500 1300 1000 2000 500

The McGraw Hill Companies, 2001

Result of Restrict Operation on the Sample Offering Table (TABLE 34)

"11/20/1999 14:00:53" "11/20/1999 14:00:33" "11/20/1999

a relational-algebra expression Composing relational-algebra operations into relational-algebra expressions is just like composing arithmetic operations (such as +, , , and ) into arithmetic expressions We study the formal de nition of relationalalgebra expressions in Section 322

Consider a query to nd the names of all bank customers who have either an account or a loan or both Note that the customer relation does not contain the information, since a customer does not need to have either an account or a loan at the bank To answer this query, we need the information in the depositor relation (Figure 35) and in the borrower relation (Figure 37) We know how to nd the names of all customers with a loan in the bank: customer -name (borrower ) We also know how to nd the names of all customers with an account in the bank: customer -name (depositor ) To answer the query, we need the union of these two sets; that is, we need all customer names that appear in either or both of the two relations We nd these data by the binary operation union, denoted, as in set theory, by So the expression needed is customer -name (borrower ) customer -name (depositor ) The result relation for this query appears in Figure 312 Notice that there are 10 tuples in the result, even though there are seven distinct borrowers and six depositors This apparent discrepancy occurs because Smith, Jones, and Hayes are borrowers as well as depositors Since relations are sets, duplicate values are eliminated customer-name Adams Curry Hayes Jackson Jones Smith Williams Lindsay Johnson Turner Figure 312 Names of all customers who have either a loan or an account

CourseNo IS320 IS480 OffTerm SPRING SPRING OffYear 2006 2006 OffLocation BLM214 BLM302 OffTime 8:30 AM 10:30 AM FacSSN 098-76-5432 987-65-4321 OffDays MW MW

The McGraw Hill Companies, 2001

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