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Now consider the following form of our example query: customer -name (( branch -city = Brooklyn (branch) 1 account) 1 depositor) When we compute the subexpression ( branch -city = Brooklyn (branch) 1 account) we obtain a relation whose schema is (branch-name, branch-city, assets, account-number, balance) We can eliminate several attributes from the schema, by pushing projections based on equivalence rules 8a and 8b The only attributes that we must retain are those that either appear in the result of the query or are needed to process subsequent operations By eliminating unneeded attributes, we reduce the number of columns of the intermediate result Thus, we reduce the size of the intermediate result In our example, the only attribute we need from the join of branch and account is accountnumber Therefore, we can modify the expression to customer -name ( ( account -number (( branch -city = Brooklyn (branch)) 1 account)) 1 depositor) The projection account -number reduces the size of the intermediate join results

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Notice that in the preceding example this unit is the secondary, and the other unit, the primary, is performing the active role

A good ordering of join operations is important for reducing the size of temporary results; hence, most query optimizers pay a lot of attention to the join order As mentioned in 3 and in equivalence rule 6a, the natural-join operation is associative Thus, for all relations r1 , r2 , and r3 , (r1

ciscoasa(config)# ciscoasa(config)# ciscoasa(config)# ciscoasa(config)# failover link logical_if_name failover replication http [no] monitor-interface logical_if_name failover polltime [unit | interface] [msec] time [holdtime time]

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Although these expressions are equivalent, the costs of computing them may differ Consider again the expression We could choose to compute account 1 depositor rst, and then to join the result with However, account 1 depositor is likely to be a large relation, since it contains one tuple for every account In contrast, branch -city = Brooklyn (branch) 1 account is probably a small relation To see that it is, we note that, since the bank has a large number of widely distributed branches, it is likely that only a small fraction of the bank s customers have accounts in branches located in Brooklyn Thus, the preceding expression results in one tuple for each account held by a resident of Brooklyn Therefore, the temporary relation that we must store is smaller than it would have been had we computed account 1 depositor rst branch -city = Brooklyn (branch) customer -name (( branch -city = Brooklyn (branch)) 1 account 1 depositor)

23:

The McGraw Hill Companies, 2001

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There are other options to consider for evaluating our query We do not care about the order in which attributes appear in a join, since it is easy to change the order before displaying the result Thus, for all relations r1 and r2 , r1

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That is, natural join is commutative (equivalence rule 5) Using the associativity and commutativity of the natural join (rules 5 and 6), we can consider rewriting our relational-algebra expression as customer -name ((( branch -city = Brooklyn (branch)) 1 depositor) 1 account) That is, we could compute ( branch -city = Brooklyn (branch)) 1 depositor rst, and, after that, join the result with account Note, however, that there are no attributes in common between Branch-schema and Depositor-schema, so the join is just a Cartesian product If there are b branches in Brooklyn and d tuples in the depositor relation, this Cartesian product generates b d tuples, one for every possible pair of depositor tuple and branches (without regard for whether the account in depositor is maintained at the branch) Thus, it appears that this Cartesian product will produce a large temporary relation As a result, we would reject this strategy However, if the user had entered the preceding expression, we could use the associativity and commutativity of the natural join to transform this expression to the more ef cient expression that we used earlier

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