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ciscoasa(config)# object-group icmp-type icmp_traffic ciscoasa(config-icmp-type)# icmp-object echo-reply ciscoasa(config-icmp-type)# icmp-object source-quench ciscoasa(config-icmp-type)# icmp-object unreachable ciscoasa(config-icmp-type)# icmp-object time-exceeded ciscoasa(config-icmp-type)# exit ciscoasa(config)# object-group network ALL_servers ciscoasa(config-network)# group-object web_servers ciscoasa(config-network)# group-object ftp_servers ciscoasa(config-network)# exit ciscoasa(config)# access-list PERMIT_IN permit tcp any object-group web_servers eq 80

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require to hold on Lending-schema, and show that each one can be tested in at least one relation in the decomposition We can test the functional dependency: branch-name branch-city assets using Branch-schema = (branch-name, branch-city, assets) We can test the functional dependency: loan-number amount branch-name using Loan-schema = (branch-name, loan-number, amount) If each member of F can be tested on one of the relations of the decomposition, then the decomposition is dependency preserving However, there are cases where, even though the decomposition is dependency preserving, there is a dependency in F that cannot be tested in any one relation in the decomposition The alternative test can therefore be used as a suf cient condition that is easy to check; if it fails we cannot conclude that the decomposition is not dependency preserving, instead we will have to apply the general test We now give a more ef cient test for dependency preservation, which avoids computing F + The idea is to test each functional dependency in F by using a modi ed form of attribute closure to see if it is preserved by the decomposition We apply the following procedure to each in F result = while (changes to result) do for each Ri in the decomposition t = (result Ri )+ Ri result = result t The attribute closure is with respect to the functional dependencies in F If result contains all attributes in , then the functional dependency is preserved The decomposition is dependency preserving if and only if all the dependencies in F are preserved Note that instead of precomputing the restriction of F on Ri and using it for computing the attribute closure of result, we use attribute closure on (result Ri ) with respect to F , and then intersect it with Ri , to get an equivalent result This procedure takes polynomial time, instead of the exponential time required to compute F +.

ciscoasa(config)# access-list PERMIT_IN permit tcp any object-group ftp_servers eq 21 ciscoasa(config)# access-list PERMIT_IN permit icmp any any object-group icmp_traffic ciscoasa(config)# access-list PERMIT_IN permit icmp any object-group ALL_servers echo ciscoasa(config)# access-list PERMIT_IN deny ip any any ciscoasa(config)# access-group PERMIT_IN in interface outside

Today s compensation administrator can choose from among many automation alternatives including: desktop applications, custom solutions, dedicated applications, and program suite optionsThe selection of the right automation choice depends on information processing needs and costs Of course, as processing needs arise, so do costs both direct (purchase price) and indirect (staff and field time)

221 For each of the SQL aggregate functions sum, count, min and max, show how to compute the aggregate value on a multiset S1 S2 , given the aggregate values on multisets S1 and S2 Based on the above, give expressions to compute aggregate values with grouping on a subset S of the attributes of a relation r(A, B, C, D, E), given aggregate values for grouping on attributes T S, for the following aggregate functions: a sum, count, min and max b avg c standard deviation 222 Show how to express group by cube(a, b, c, d) using rollup; your answer should have only one group by clause 223 Give an example of a pair of groupings that cannot be expressed by using a single group by clause with cube and rollup 224 Given a relation S(student, subject, marks), write a query to nd the top n students by total marks, by using ranking 225 Given relation r(a, b, d, d), Show how to use the extended SQL features to generate a histogram of d versus a, dividing a into 20 equal-sized partitions (that is, where each partition contains 5 percent of the tuples in r, sorted by a)