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A very large database is an essential part of an enterprise and will be in continuous use by many people At the same time it will require many storage devices A very large database presents particular problems in its management, since it cannot be turned o without a ecting the operation and well-being of the enterprise into which it is integrated A formal de nition of a very large database is one where the time required for making a copy is greater than the permissible interval between update transactions performed to keep the database up to date An example of a very large database is found at a telephone company with 5 million subscribers [Karsner in Kerr75 ] Much larger yet are databases at the social security administration and other national systems

StdSSN 124-56-7890 234-56-7890

37:

To have a copy of the contents of a le frozen at a certain instant in time is important for many purposes, including periodic analysis of data, backup for reliability purposes, and auditing To avoid problems, it is best not to permit a le to be modi ed while it is being copied In a very large database these two considerations con ict This de nition imposes certain design constraints on the systems which we will be discussing

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intersection, and difference operators are shown in Tables 323 through 325, respectively Even though we can determine that two rows are identical from looking only at StdSSN, all columns are compared due to the way that the operators are designed Note that the result of Studentl DIFFERENCE Student2 would not be the same as Studentl DIFFERENCE Studentl The result of the latter (Studentl DIFFERENCE Studentl) is the second and third rows of Studentl (rows in Studentl but not in Studentl) Because of the union compatibility requirement, the union, intersection, and difference operators are not as widely used as other operators However, these operators have some im portant, specialized uses One use is to combine tables distributed over multiple locations For example, suppose there is a student table at Big State University (BSUStudent) and a stu dent table at University of Big State (UBSStudent) Because these tables have identical columns, the traditional set operators are applicable To find students attending either uni versity, you should use UBSStudent UNION BSUStudent To find students only attending Big State, you should use BSUStudent DIFFERENCE UBSStudent To find students attend ing both universities, you should use UBSStudent INTERSECT BSUStudent Note that the resulting table in each operation has the same number of columns as the two input tables The traditional operators are also useful if there are tables that are similar but not union compatible For example, the Student and Faculty tables have some compatible columns (StdSSN with FacSSN, StdLastName with FacLastName, and StdCity with FacCity), but other columns are different The union compatible operators can be used if the Student and Faculty tables are first made union compatible using the project operator presented in Section 341

template <class T> valarray<T> operator/(const valarray<T> ob, const T &v); template <class T> valarray<T> operator/(const T &v, const valarray<T> ob); template <class T> valarray<T> operator/(const valarray<T> ob1, const valarray<T> &ob2); template <class T> valarray<T> operator%(const valarray<T> ob, const T &v); template <class T> valarray<T> operator%(const T &v, const valarray<T> ob); template <class T> valarray<T> operator%(const valarray<T> ob1, const valarray<T> &ob2);

The bu ering associated with the bulk transfer rate also evens out any variations in response time, so that performance is not directly a ected by the periodic need to read a new block Insertion of a record into the main le requires Insert into a Sequential File insertion into the proper place according to the key The sequence would not be maintained if new records were to be added to the end For very small les, records beyond the point of insertion can be moved up to make space for putting the new record in place This e ort involves locating the insertion point by a fetch, and subsequently reading and rewriting the remainder of the le Each phase involves again half of the blocks of the le on the average, so T I = TF + 1 n R r (btt + TRW ) = n + n 2 Bfr t Bfr in place 3-20

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