程序代写代做代考 Version 1:

Version 1:
Disk Calculations – 6 marks
Suppose we have a disk drive with:
– 8192 tracks on each side of 4 double-sided platters – each track has 512 sectors on it
– each page has 4 sectors
– each sector has 1024 bytes
– it takes 0.1 ms to read one page
– average seek time is 10 ms
a) Compute the time it takes to read one track, assuming you’re already on the correct cylinder. b) Compute the RPM of the disk drive.
c) Compute the number of pages on a track.
ANSWERS:
a) [2 marks] 512 sectors/track / 4 sectors/page = 128 pages/track
Then, 128 pages/track * 0.1 ms/page = 12.8 ms/track = 12.8 ms to read one track
b) [can give 2 if the calculation works using their answer to (a)] Answer from (a) is 12.8 ms/rev. So, that’s 1 rev / 12.8 ms.
Then, 1 rev / 12.8 ms * 1000 ms/sec * 60 sec/min = about 4688 RPM
c) [2 marks] From their calculation in (a), there are 128 pages/track.

Version 2:
Suppose we have a disk drive with:
– 8192 tracks on each side of 4 double-sided platters – each track has 512 sectors on it
– each page has 4 sectors
– each sector has 1024 bytes
– it takes 0.1 ms to read one page
– average seek time is 10 ms
a) Compute the time it takes to read one cylinder, assuming you’re already on the correct cylinder.
b) Compute the RPM of the disk drive.
c) Compute the number of pages on a cylinder.
ANSWERS:
a) [3 marks, can give 2 marks for saying 12.8 ms to read one track]
512 sectors/track / 4 sectors/page = 128 pages/track
Then, 128 pages/track * 0.1 ms/page = 12.8 ms/track = 12.8 ms to read one track. Then, 8 tracks/cyl, * 12.8 ms/track = 102.4 ms/cyl.
b) [2 marks, can give 2 if the calculation works using their wrong, but inferred, answer to (a)] Answer from (a) is 12.8 ms/rev. So, that’s 1 rev / 12.8 ms.
Then, 1 rev / 12.8 ms * 1000 ms/sec * 60 sec/min = about 4688 RPM
c) [1 mark] There are 128 pages/track and 8 tracks/cyl = 1024 pages/cyl.
If they lost 1 mark for the cylinder calculation in (a), they would lose one mark for (c); so, the total loss would be 2 marks for the cylinder calculation, which would be consistent with the other students’ randomized questions. For example, they get 0 if they thought there were 4 tracks/cyl.

Version 3:
Suppose we have a disk drive with:
– 8192 tracks on each side of 8 double-sided platters – each track has 1024 sectors on it
– each page has 4 sectors
– each sector has 1024 bytes
– it takes 0.03 ms to read one page
– average seek time is 8 ms
a) Compute the time it takes to read one track, assuming you’re already on the correct cylinder. b) Compute the RPM of the disk drive.
c) Compute the number of pages on two tracks.
ANSWERS:
a) [2 marks]
1024 sectors/track / 4 sectors/page = 256 pages/track
Then, 256 pages/track * 0.03 ms/page = 7.68 ms/track = 7.68 ms to read one track
b) [2 marks, can give 2 if the calculation works using their answer to (a)] Answer from (a) is 7.68 ms/rev. So, that’s 1 rev / 7.68 ms.
Then, 1 rev / 7.68 ms * 1000 ms/sec * 60 sec/min = about 7813 RPM
c) [2 marks, can give 2 if the calculation works using their answer to (a)] From their calculation in (a), there are 256 pages/track * 2 tracks = 512 pages for 2 tracks.

Version 4:
Suppose we have a disk drive with:
– 8192 tracks on each side of 8 double-sided platters – each track has 1024 sectors on it
– each page has 4 sectors
– each sector has 1024 bytes
– it takes 0.04 ms to read one page
– average seek time is 8 ms
a) Compute the time it takes to read one cylinder, assuming you’re already on the correct cylinder.
b) Compute the RPM of the disk drive.
c) Compute the number of pages on one cylinder.
ANSWERS:
a) [3 marks, can give 2 marks for saying 10.24 ms to read one track; 0 otherwise]
1024 sectors/track / 4 sectors/page = 256 pages/track
Then, 256 pages/track * 0.04 ms/page = 10.24 ms/track = 10.24 ms to read one track. Then, 16 tracks/cyl, * 10.24 ms/track = about 163.84 ms/cyl.
b) [2 marks, can give 2 if the calculation works using their wrong, but inferred, answer to (a)] Answer from (a) is 10.24 ms/rev. So, that’s 1 rev / 10.24 ms.
Then, 1 rev / 10.24 ms * 1000 ms/sec * 60 sec/min = about 5859 RPM.
c) [1 mark] There are 256 pages/track and 16 tracks/cyl = 4096 pages/cyl.
If they lost 1 mark for the cylinder calculation in (a), they would lose one mark for (c); so, the total loss would be 2 marks for the cylinder calculation, which would be consistent with the other students’ randomized questions. For example, they get 0 if they thought there were 8 tracks/cyl.