Computer System Architecture - Pipeline and Vector Processing
A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple (c, h, s), where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as (0, 0, 0), the 1st sector as (0, 0, 1), and so on.
41. The address (400, 16, 29) corresponds to sector number :
- Option : C
- Explanation : Sector no = 400 * 20 * 63 + 16 * 63 + 29 = 505037.
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A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple (c, h, s), where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as (0, 0, 0), the 1st sector as (0, 0, 1), and so on.
42. The address of 1039th sector is
- Option : C
- Explanation : For option
(a) sector no. = 0 * 20 * 63 + 15 * 63 + 31 = 976 sector.
(b) Sector no = 0 * 20 * 63 + 16 * 63 + 30 = 1038
(c) Sector no = 0 * 20 * 63 + 16 * 63 + 31 = 1039 sector
Hence answer is (c).
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- Option : B
- Explanation : 6000 rotations ________ 60 sec
1 rotation _________ 10 ms
Rotational latency = 5 ms
Time for one disk access = 15 ms
Time to load all libraries = 15 * 100
= 1500 ms = 1.5 sec
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- Option : D
- Explanation : Since starting on disk starts from <1200, 9, 40> so no of sectors left on 9th surface is 24
so, on 9th surface total storage of “12288” B is possible Now, a part from 9th surface, on cylinder no. 1200 only 6 surface is left.
To storage possible on these 6 surface are
= 6 * 26 * 29 ￫ storage on each sector.
ᛎ
= 196608 B
So total on cylinder no. 1200, storage possible
= 196608 + 12288 = 208896 B
So Since file size is 42797 kB and out of which 208896 B are stored on cylinder no. 1200. So we are left only with only 43615232 B.
Since in 1 cylinder, Storage Possible is
= 24 * 26 * 29 B = 524288 B
So we need about
= 43615232B / 524288B
= 83.189 more cylinders.
Hence we’ll need the 1284th Cylinder to completely store the file. C02 after 1283rd cylinder we will left with data which will need 189
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- Option : D
- Explanation :
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