Data Mgmt Lab_Part_43

# Data Mgmt Lab_Part_43 - External Sorting 127 3 Since each...

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Unformatted text preview: External Sorting 127 3. Since each page is read and written once per pass, the total number of page I/Os for sorting the file is 2 ∗ N ∗ (# passes ): (a) 2*10000*13 = 260000. (b) 2*20000*7 = 280000. (c) 2*2000000*6 = 24000000. 4. In Pass 0, d N/B e runs are produced. In Pass 1, we must be able to merge this many runs; i.e., B − 1 ≥ d N/B e . This implies that B must at least be large enough to satisfy B ∗ ( B − 1) ≥ N ; this can be used to guess at B , and the guess must be validated by checking the first inequality. Thus: (a) With 10000 pages in the file, B = 101 satisfies both inequalities, B = 100 does not, so we need 101 buffer pages. (b) With 20000 pages in the file, B = 142 satisfies both inequalities, B = 141 does not, so we need 142 buffer pages. (c) With 2000000 pages in the file, B = 1415 satisfies both inequalities, B = 1414 does not, so we need 1415 buffer pages. Exercise 13.2 Answer Exercise 13.1 assuming that a two-way external sort is used. Answer 13.2 Answer omitted. Exercise 13.3 Suppose that you just finished inserting several records into a heap file and now want to sort those records. Assume that the DBMS uses external sort and makes eﬃcient use of the available buffer space when it sorts a file. Here is some potentially useful information about the newly loaded file and the DBMS software available to operate on it: The number of records in the file is 4500. The sort key for the file is 4 bytes long. You can assume that rids are 8 bytes long and page ids are 4 bytes long. Each record is a total of 48 bytes long. The page size is 512 bytes. Each page has 12 bytes of control information on it. Four buffer pages are available....
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Data Mgmt Lab_Part_43 - External Sorting 127 3 Since each...

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