數(shù)據(jù)庫復(fù)習(xí)題2(答案)-展示頁
2025-07-05 06:07本頁面
【正文】 the cost using the following three ponents:Seek time: This access seeks the initial position of the file (whose cost can be approximated using the average seek time) and then seeks between adjacent tracks 496 times (whose cost is the tracktotrack seek time). So the seek time is + 496* = seconds.Rotational delay: The transfer time of one track of data is 1/ (7200/60) = seconds.For this question, we use (other numbers between 0 and are also fine). So the rotational delay for 497 seeks is * 497 = .Transfer time: It takes *(500000/63) = seconds to transfer data in 500,000 sectors.Therefore, total access time is + + = seconds.(d) number of pages = 6250time cost per page: (seek) + (rotational delay) + *8/63 (transfer) = secondstotal cost = 6250 * = seconds [Disk Page Layout] The figure below shows a page containing variable length records. The page size is 1KB (1024 bytes). It contains 3 records, some free space, and a slot directory in that order. Each record has its record id, in the form of Rid=(page id, slot number), as well as its start and end addresses in the page, as shown in the figure.Now a new record of size 200 bytes needs to be inserted into this page. Apply the record insertion operation with page paction, if necessary. Show the content of the slot directory after the new record is inserted. Assume that you have only the page, not any other temporary space, to work with.Answer: Content of the slot directory, from left to right, is:[(650, 200), (0, 200), (500, 150), (200, 300)], 4,
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