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performing the join To choose the best overall algorithm, we must consider even nonoptimal algorithms for individual operations Thus, in addition to considering alternative expressions for a query, we must also consider alternative algorithms for each operation in an expression We can use rules much like the equivalence rules to de ne what algorithms can be used for each operation, and whether its result can be pipelined or must be materialized We can use these rules to generate all the query-evaluation plans for a given expression Given an evaluation plan, we can estimate its cost using statistics estimated by the techniques in Section 142 coupled with cost estimates for various algorithms and evaluation methods described in 13 That still leaves the problem of choosing the best evaluation plan for a query There are two broad approaches: The rst searches all the plans, and chooses the best plan in a cost-based fashion The second uses heuristics to choose a plan We discuss these approaches next Practical query optimizers incorporate elements of both approaches

A cost-based optimizer generates a range of query-evaluation plans from the given query by using the equivalence rules, and chooses the one with the least cost For a complex query, the number of different query plans that are equivalent to a given plan can be large As an illustration, consider the expression r1

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No organization is an island Every organization has critical suppliers without which it could not carry out its critical functions The ability to recover from a disaster also frequently requires the support of one or more service providers or suppliers The IT auditor should examine contracts for all critical suppliers and consider the following guidelines: Does the contract support the organization s requirements for delivery of services and supplies, even in the event of a local or regional disaster Determine whether the service provider has its own disaster recovery capabilities that will ensure its ability to deliver critical services during a disaster Determine the recourse available should the supplier be unable to provide goods or services during a disaster

where the joins are expressed without any ordering With n = 3, there are 12 different join orderings: r1 r2 r3

1 (r2 1 r3 ) 1 (r1 1 r3 ) 1 (r1 1 r2 )

1 (r3 1 r2 ) 1 (r3 1 r1 ) 1 (r2 1 r1 )

The IT auditor should examine the organization s insurance policies related to the loss of property and assets supporting critical business processes Insurance coverage should cover the actual cost of recovery, or a lesser amount if the organization s executive management has accepted a lower amount The IT auditor should obtain documentation that includes cost estimates for various disaster recovery scenarios, including equipment replacement, business interruption, and the cost of performing business functions and operating IT systems in alternate sites These cost estimates should be compared with the value of insurance policies

In general, with n relations, there are (2(n 1))!/(n 1)! different join orders (We leave the computation of this expression for you to do in Exercise 1410) For joins involving small numbers of relations, this number is acceptable; for example, with n = 5, the number is 1680 However, as n increases, this number rises quickly With n = 7, the number is 665280; with n = 10, the number is greater than 176 billion! Luckily, it is not necessary to generate all the expressions equivalent to a given expression For example, suppose we want to nd the best join order of the form (r1

which represents all join orders where r1 , r2 , and r3 are joined rst (in some order), and the result is joined (in some order) with r4 and r5 There are 12 different join orders for computing r1 1 r2 1 r3 , and 12 orders for computing the join of this result with r4 and r5 Thus, there appear to be 144 join orders to examine However, once we have found the best join order for the subset of relations {r1 , r2 , r3 }, we can use that order for further joins with r4 and r5 , and can ignore all costlier join orders of r1 1 r2 1 r3 Thus, instead of 144 choices to examine, we need to examine only 12 + 12 choices

Summary

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The McGraw Hill Companies, 2001

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