Storage And File Structure

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CHAPTER10Storage and File StructurePractice Exercises10.1Answer: This arrangement has the problem that Pi and B4i 3 are on thesame disk. So if that disk fails, reconstruction of B4i 3 is not possible, sincedata and parity are both lost.10.2Answer:10.3a.It is stored as an array containing physical page numbers, indexed bylogical page numbers. This representation gives an overhead equalto the size of the page address for each page.b.It takes 32 bits for every page or every 4096 bytes of storage. Hence,it takes 64 megabytes for the 64 gigabyte of flash storage.c.If the mapping is such that, every p consecutive logical page numbers are mapped to p consecutive physical pages, we can store themapping of the first page for every p pages. This reduces the inmemory structure by a factor of p. Further, if p is an exponent of2, we can avoid some of the least significant digits of the addressesstored.Answer:a.To ensure atomicity, a block write operation is carried out as follows:i. Write the information onto the first physical block.ii. When the first write completes successfully, write the same information onto the second physical block.iii. The output is declared completed only after the second writecompletes successfully.During recovery, each pair of physical blocks is examined. If bothare identical and there is no detectable partial-write, then no furtheractions are necessary. If one block has been partially rewritten, thenwe replace its contents with the contents of the other block. If there1

2Chapter 10 Storage and File Structurehas been no partial-write, but they differ in content, then we replacethe contents of the first block with the contents of the second, or viceversa. This recovery procedure ensures that a write to stable storageeither succeeds completely (that is, updates both copies) or resultsin no change.The requirement of comparing every corresponding pair of blocksduring recovery is expensive to meet. We can reduce the cost greatlyby keeping track of block writes that are in progress, using a smallamount of nonvolatile RAM. On recovery, only blocks for whichwrites were in progress need to be compared.b.10.410.5The idea is similar here. For any block write, the information block iswritten first followed by the corresponding parity block. At the timeof recovery, each set consisting of the nth block of each of the disksis considered. If none of the blocks in the set have been partiallywritten, and the parity block contents are consistent with the contents of the information blocks, then no further action need be taken.If any block has been partially-written, it’s contents are reconstructedusing the other blocks. If no block has been partially-written, but theparity block contents do not agree with the information block contents, the parity block’s contents are reconstructed.Answer:a.Although moving record 6 to the space for 5, and moving record 7to the space for 6, is the most straightforward approach, it requiresmoving the most records, and involves the most accesses.b.Moving record 7 to the space for 5 moves fewer records, but destroysany ordering in the file.c.Marking the space for 5 as deleted preserves ordering and movesno records, but requires additional overhead to keep track of allof the free space in the file. This method may lead to too many“holes” in the file, which if not compacted from time to time, willaffect performance because of reduced availability of contiguousfree records.Answer: (We use “ i ” to denote a pointer to record “i”.)The original file of Figure 10.7.

Exercisesheaderrecord 0record 1record 2record 3record 4record 5record 6record 7record 8record 9record 10record 11a.SrinivasanComp. hysics 165000 44000095000 BrandtKimHistoryFinanceBiologyComp. Sci.Elec. Eng.6200080000720009200080000The file after insert (24556, Turnamian, Finance, 98000).headerrecord 0record 1record 2record 3record 4record 5record 6record 7record 8record 9record 10record zartEinsteinComp. Sci.FinanceMusicPhysics33456GoldPhysics 465000980004000095000 BrandtKimHistoryFinanceBiologyComp. Sci.Elec. Eng.6200080000720009200080000The file after delete record 2.3

4Chapter 10 Storage and File Structureheaderrecord 0record 1record 2record 3record 4record 5record 6record 7record 8record 9record 10record 111010124556SrinivasanTurnamianComp. Sci.Finance22222EinsteinPhysics33456GoldPhysics 26500098000 495000 BrandtKimHistoryFinanceBiologyComp. Sci.Elec. Eng.6200080000720009200080000The free record chain could have alternatively been from the headerto 4, from 4 to 2, and finally from 2 to 6.c.The file after insert (34556, Thompson, Music, 67000).headerrecord 0record 1record 2record 3record 4record 5record 6record 7record 8record 9record 10record sonEinsteinComp. Sci.FinanceMusicPhysics33456GoldPhysics 465000980006700095000 BrandtKimHistoryFinanceBiologyComp. Sci.Elec. Eng.6200080000720009200080000Answer:The relation section with three tuples is as follows.course idsec idsemesteryearbuildingroom numbertime slot PainterPackardTaylor5141013128AHCThe relation takes with five students for each section is as follows.

154321597627654376543785468972998988course 1CS-347BIO-301CS-101CS-347BIO-301BIO-301BIO-301sec 0201020105gradeAAACnullAFAAnullAAnullnullnullThe multitable clustering for the above two instances can be taken e a record is inserted/deleted, check if the usage of theblock has changed levels. In that case, update the corresponding

6Chapter 10 Storage and File Structurebits. Note that we don’t need to access the bitmaps at all unlessthe usage crosses a boundary, so in most of the cases there is nooverhead.b.When free space for a large record or a set of records is sought, thenmultiple free list entries may have to be scanned before finding aproper sized one, so overheads are much higher. With bitmaps, onepage of bitmap can store free info for many pages, so I/O spent forfinding free space is minimal. Similarly, when a whole block or alarge part of it is deleted, bitmap technique is more convenient forupdating free space information.10.8Answer: Hash table is the common option for large database buffers. Thehash function helps in locating the appropriate bucket, on which linearsearch is performed.10.9Answer:a.MRU is preferable to LRU where R1 1 R2 is computed by usinga nested-loop processing strategy where each tuple in R2 must becompared to each block in R1 . After the first tuple of R2 is processed,the next needed block is the first one in R1 . However, since it isthe least recently used, the LRU buffer management strategy wouldreplace that block if a new block was needed by the system.b.LRU is preferable to MRU where R1 1 R2 is computed by sortingthe relations by join values and then comparing the values by proceeding through the relations. Due to duplicate join values, it maybe necessary to “back-up” in one of the relations. This “backing-up”could cross a block boundary into the most recently used block,which would have been replaced by a system using MRU buffermanagement, if a new block was needed.Under MRU, some unused blocks may remain in memory forever. Inpractice, MRU can be used only in special situations like that of thenested-loop strategy discussed in Exercise 10.9a.

45678 CS-101 1 Fall 2009 F 54321 CS-101 1 Fall 2009 A-76543 CS-101 1 Fall 2009 A CS-347 1 Fall 2009 Taylor 3128 C 00128 CS-347 1 Fall 2009 A-12345 CS-347 1 Fall 2009 A 23856 CS-347 1 Fall 2009 A 54321 CS-347 1 Fall 2009 A 76543 CS-347 1 Fall 2009 A 10.7 Answer: a. Everytime a record is

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