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TiliK225 [7]
3 years ago
12

Archie wants to optimize an XFS filesystem and minimize the chance of future corruption. Which of the following commands will co

mpact or otherwise improve the layout of the contiguous blocks of file data for an XFS filesystem?A. defrag
B. xfs_fsr
C. fsck
D. xfs_db
Computers and Technology
2 answers:
svlad2 [7]3 years ago
6 0

Answer:

The best answer is "B"

xfs_fsr

Explanation:

The is a typical format of the command to use:

xfs_metadump -o /dev/sdb1 /reviewxfs

Roman55 [17]3 years ago
4 0

Answer:

B. xfs_fsr

Explanation:

<em>xfs_fsr</em> is used for  Defragmentaion of xfs files. It works by improving the organization of the disk extents of every file to ensure that contiguous space is being used.  This tool performs online defragmentation meaning that it can be run on a file system that is currently mounted and in use, just be aware of the increased I/O usage that will come as a result , it may be best to run it during a period of low disk activity if possible on a production system.

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"air piracy"

the unauthorized use or reproduction of another's work.

Explanation:

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To create a multi-column and multi-row box for users to type into in an html form, use the ____________ tag pair
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8.7 Code Practice: Question 1 edhesive
Ronch [10]

Answer:

here got from a friend

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7 0
2 years ago
Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

4 0
3 years ago
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