数据库系统实现(第二版 英文版)部分答案ppt课件.ppt

数据库系统实现(第二版_英文版)部分答案ppt课件Solutions for Database System Implementation
Gan Lin
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完整版PPT课件
Chapter 13
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完整版PPT课件
The capacity of the disk?
The disk has 8 × 100,000 = 800,000 tracks. The average track has 2000 × 1024 = 2048,000 bytes. Thus, the capacity is 214 × 108 bytes
 
The maximum seek time?
The maximum seek time occurs when the heads have to move across all the tracks. Thus, substitute 100,000 tracts (really 99999) for n in the formula 1+.0003n to get 31()ms.
 
The maximum rotational latency?
The maximum rotational latency is one full revolution. Since the disk rotates at 6,000 rpm, it takes 1/6000 of a minute, or to rotate.
Exercise ()
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完整版PPT课件
Transfer time of a block?
Since a track has 2000 sectors and 2000 gaps, and 10% of a track is used for gaps, there are 36o for gaps and 324o for sectors of a track circle. So there are 36o/2000 for each gap and 324o/2000 for each sector. This block is 65546 bytes (. 64 sectors), the heads must therefore pass over 63 gaps and 64 gaps, . 63×36o/2000 + 64×324o/2000. As the maximum rotational latency is , we can have the transfer time of the block is (63×36o/2000 + 64×324o/2000) × o= or .
Exercise ()
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The average seek time?
The average n is tracks/3 of a surface, . 100,000/3. So the average seek time is
1+.0003×n =1+.0003×100,000/3=11ms.
The average rotational latency?
The average rotational latency is =
The average density of bits in the sectors of a outer track？
A track has 2000 sectors, each sector holds 1024 bytes, and every byte takes 8 bits, so a track is 2000×1024×8 bits. For the inch disk, the density of the outer track is
2000×1024×8/ ( × π × 90%) = bit/inch.
Exercise ()
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完整版PPT课件
The average tracks the heads have to move:
[(1+2+……+4095)+(1+2+……(65536-4096)]/65536≈28928
Seek time: 1+28928/4000=
Tran