Database System Implementation第三章答案.pdf

Database System Implementation Solutions for Chapter 3 Solutions for Section 3.2 Solutions for Section 3.3 Solutions for Section 3.4 Solutions for Section 3.5 Solutions for Section 3.2 Exercise 3.2.1 First, note that SQL2 dates require 10 bytes, and SQL2 times require 8 bytes if there is no decimal point; this material is in Section 3.1.3. a) The bytes requirsed by each of the fields is 15 + 2 + 10 + 8 = 35. b) Round each of the four field lengths up to a multiple of 4, to get 16 + 4 + 12 + 8 = 40. c) Round up again, to a multiple of 8, to get 16 + 8 + 16 + 8 = 48. Exercise 3.2.3 The elements in the header also require rounding upward in (b) and (c). Thus, the lengths including header fields and adding the lengths for the bodies as calculated in Exercise 3.2.1 are: a) 4 + 4 + 1 + 35 = 44. b) 4 + 4 + 4 + 40 = 52. c) 8 + 8 + 8 + 48 = 72. Exercise 3.2.5 For parts (a) and (b), the header takes 40 bytes, while for (c) it takes 80 bytes. The calculations are thus: a) (4096-40)/44 = 92. b) (4096-40)/52 = 78. c) (4096-80)/72 = 55. Return to Top Solutions for Section 3.3 Exercise 3.3.1 The Megatron 747 has 8192 cylinders, so we require 2 bytes for the cylinder. There are 8 surfaces, or 8 tracks per cylinder, so 1 byte suffices for the track. The average number of sectors per track is 256, and we have chosen to use 8 sectors per block (which is not a property of the disk itself). If we used blocks consisting of a single sector, and the track with the greatest number of sectors had more than 256, then we would need 2 bytes for the sector. However, at 8 sectors/block, we could have wildly differing numbers of sectors per track, and still not get close to 256 blocks/track. Since 256 integers can be represented in 1 byte, we think that 1 byte is sufficient for the block within a track, and we think the correct answer is 4 bytes for the block address on a Megatron 747. Exercise 3.3.3(a) Assuming 4096-byte blocks,