数据库复习题2(答案)

1、资料收集于网络如有侵权请联系网站删除谢谢复习题（ 2）1、 试分别判断下列图中 G1和G2是否互模拟 (bisimulation) ，并说明理由 :aaaG1=G2=bcbcG1G2ababcccddd答案：(1) 在图中标出各点的状态，我们构造关系，可知 G2 可以模拟G1，下面我们讨论精品文档资料收集于网络如有侵权请联系网站删除谢谢是否可模拟，在G2 中有一个 a 变换可对应到G1 中 2 个变换，即，。但有两个变换b，c，而在 G1 中仅存在只有b 或只有 c 的状态点，可知 G1 和 G2 不能互模拟。(2) 如图，标出各状态点，构造有关系可知其中 G1 中的点均可由G2 中的点模拟，

2、下面我们考虑可知同样其中G2 中的点均可由G1 中的点模拟 . 所以 G1 和 G2 为互模拟的。2、 给定如下数据图 (Data Graph)：r1personcompanypersoncompanypersonmanagesworks-forc1employeep1ceop2c2ceop3nameposition works-fornameworks-fornamephonenameaddresspositionaddressnames0s2s3s4s5s6urls7s8s9s1“Widget”“Trenton ”“Jones ”“Gadget”“Paris ”“Dupont”“Sales

3、 ”“Smith ”“5552121 ”“Manager ”s10“www.gp.fr”试给出其 Strong DataGuide 图精品文档资料收集于网络如有侵权请联系网站删除谢谢答案：r1personworks-formanagesemployeec1,c2p1,p2,p3p2ceonamephonep1,p3urlpositionnameaddresss0,s4,s8s1,s9s5s2,s6s3,s7s10Strong DataGuide 图3、 Consider the relation,r , shown in Figure 5.27 . Give the result of the

4、 following query :Figure 5.27Query 1:select building , room number, time_slo_ id , count(*)from rgroup by rollup (building , room number , time_slo_ id )Query 1:select building , room number, time_slo_ id , count(*)from rgroup by cube (building , room number , time_slo_ id )答案：Query 1返回结果集：为以下四种分组统计

5、结果集的并集且未去掉重复数据。buildingroom numbertime_slo_ idcount(*)产生的分组种数：4 种；精品文档资料收集于网络如有侵权请联系网站删除 谢谢第一种： group by A,B,CGarfield359A1Garfield359B1Saucon651A1Saucon550C1Painter705D1Painter403D1第二种： group by A,BGarfield359A2Garfield359B2Saucon651A1Saucon550C1Painter705D1Painter403D1第三种： group by AGarfield359A2G

6、arfield359B2Saucon651A2Saucon550C2Painter705D2Painter403D2第四种： group by NULL。 本没有 group by NULL的写法，在这里指是为了方便说明，而采用之。含义是：没有分组，也就是所有数据做一个统计。例如聚合函数是SUM 的话，那就是对所有满足条件的数据进行求和。Garfield359A6Garfield359B6Saucon651A6Saucon550C6Painter705D6Painter403D6Query 2:group by 后带 rollup 子句与 group by 后带 cube 子句的 唯一区别 就

7、是：带 cube 子句的 group by 会产生更多的分组统计数据。cube 后的列有多少种组合（注意组合是与顺序无关的）就会有多少种分组。返回结果集：为以下八种分组统计结果集的并集且未去掉重复数据。buildingroom numbertime_slo_ idcount(*)产生的分组种数：8 种第一种： group by A,B,CGarfield359A1Garfield359B1Saucon651A1精品文档资料收集于网络如有侵权请联系网站删除谢谢Saucon550C1Painter705D1Painter403D1第二种： group by A,BGarfield359A2Garf

8、ield359B2Saucon651A1Saucon550C1Painter705D1Painter403D1第三种： group by A,CGarfield359A1Garfield359B1Saucon651A1Saucon550C1Painter705D2Painter403D2第四种： group by B,CGarfield359A2Garfield359B2Saucon651A1Saucon550C1Painter705D1Painter403D1第五种： group by AGarfield359A2Garfield359B2Saucon651A2Saucon550C2Pain

9、ter705D2Painter403D2第六种： group by BGarfield359A2Garfield359B2Saucon651A1Saucon550C1Painter705D1Painter403D1第七种： group by CGarfield359A2Garfield359B1Saucon651A2Saucon550C1精品文档资料收集于网络如有侵权请联系网站删除谢谢Painter705D2Painter403D2第八种： group by NULLGarfield359A6Garfield359B6Saucon651A6Saucon550C6Painter705D6Pain

10、ter403D64、 Disks and Access TimeConsider a disk with a sector扇区 size of 512 bytes,63 sectors per track磁道 , 16,383 tracks per surface盘面 , 8 double-sided platters柱面 (i.e., 16 surfaces). The disk platters rotate at 7,200 rpm (revolutions perminute). The average seek time is 9 msec, whereas the track-to

11、-track seek time is 1 msec.Suppose that a page size of 4096 bytes is chosen. Suppose that a file containing 1,000,000 records of 256 bytes each is to be stored on such a disk. No record is allowed to span two pages (use these numbers in appropriate places in your calculation).(a) What is the capacit

12、y of the disk?(b) If the file is arranged sequentially on the disk, how many cylinders are needed?(c) How much time is required to read this file sequentially?(d) How much time is needed to read 10% of the pages in the file randomly?Answer:(a) Capacity = sector size * num. of sectors per track * num

13、. of tracks per surface * num of surfaces = 512 * 63 * 16383 * 16 = 8 455 200 768(b) File: 1,000,000 records of 256 bytes eachNum of records per page: 4096/256 = 161,000,000/ 16 = 62,500 pages or 62,500 * 8 = 500,000 sectors Each cylinder has 63 * 16 = 1,008 sectorsSo we need 496.031746 cylinders.(c

14、) We analyze the cost using the following three components: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 adjacenttracks 496 times (whose cost is the track-to-track seek time). So the seek time is 0.00

15、9 + 496*0.001 = 0.505 seconds.Rotational delay:The transfer time of one track of data is 1/ (7200/60) = 0.0083 seconds.For this question, we use 0.0083/2 as an estimate of the rotational delay (other numbers between 0 and 0.00415 are also fine). So the rotational delay for 497 seeks is 0.00415 * 497

16、 = 2.06255.Transfer time: It takes 0.0083*(500000/63) = 65.8730159 seconds to transfer data in 500,000 sectors.精品文档资料收集于网络如有侵权请联系网站删除谢谢Therefore, total access time is 0.505 + 2.06255 + 65.8730159 = 68.4405659 seconds. (d) number of pages = 6250time cost per page: 0.009 (seek) + 0.0083/2 (rotational

17、delay) + 0.0083*8/63 (transfer) = 0.0142 secondstotal cost = 6250 * 0.0142 = 88.77 seconds5、 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 h

18、as 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 compaction, if necessary. Show the content of

19、 the slot directory after the new record is inserted. Assume that you have only the page, not any other temporary space, to work with.精品文档资料收集于网络如有侵权请联系网站删除谢谢Answer:Content of the slot directory, from left to right, is:(650, 200), (0, 200), (500, 150), (200, 300), 4, 8506、 Buffer Management for File

20、 and Index Accesses Consider the following two relations:student(snum:integer, sname:char(30), major:char(25), standing:char(2), age:integer)enrolled(snum:integer, cname:char(40)The following index is available:A B+ Tree index on the <snum> attribute of the student relation.Assume that the buf

21、fer size is large enough to store multiple paths of each B+Tree but not an entire tree.(a) Consider Query 1 and Query 2 that retrieve the snum s of students who havetaken DatabaseSystems and Operating Systems , respectively, from theenrolled table. We know that Query 1 will be executed before Query

22、2, and bothqueries are executed using a file scan of the enrolled table.Which replacement policy would you recommend for the buffer manager to use tosupport this workload?(b) Now assume that wehave retrieved the snum s of students who have taken DatabaseSystems from the enrolled table. In the exact

23、order of the retrieved snum s (not necessarily in sorted order),we then retrieve the names of those students via repeated lookups in the B+ Tree on <snum>.For these repeated accesses to the index on student.snum, which replacement policy would you recommend for efficient buffer management?精品文档资料收集于网络如有侵权请联系网站删除谢谢Query 1:selectsnumfrom student s, enrolled ewhere s.snum=e.snum and cname like DatabaseSystems ;Query 2:selectsnumfrom student s, enrolled ewhere s.snum=e