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Nitella [24]
2 years ago
14

You are a technical consultant for many businesses in your community. One of your clients, a small law firm, has a single Active

Directory domain. They have two servers running Windows Server 2012 R2. Both servers are configured as domain controllers while also serving as file and printer servers.This client is calling you on a regular basis because users are deleting or damaging their files. You must visit the client's site and restore the files from backup. Your client has asked you to create an alternate solution.
Computers and Technology
1 answer:
stellarik [79]2 years ago
8 0

COMPLETE QUESTION:

You are a technical consultant for many businesses in your community. One of your clients, a small law firm, has a single Active Directory domain. They have two servers running Windows Server 2012 R2. Both servers are configured as domain controllers while also serving as file and printer servers.This client is calling you on a regular basis because users are deleting or damaging their files. You must visit the client's site and restore the files from backup. Your client has asked you to create an alternate solution. What should you do?

Answer:

Use the Windows VSS to create shadow copies on important data

Explanation:

Volume Snapshot Service (VSS) is a service available to Windows Operating Systems which allows safe backup of open files as well as locked ones. It does this by taking a snapshot the state of the drive, this snapshot information will be provided to the backup application whenever needed. It is important to mention that this service is only available on files which are in the NTFS format

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When should a technician record a system's baseline data?
bija089 [108]

Baseline data is a record of all the system performance specifications when the system is operating as designed. A technician should record a system's baseline data On a new system after it is installed.


3 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
2 years ago
How did josh norman and mike keller provide coverage of katrina?
Digiron [165]
<span>They wrote live updates to a blog, so D. The blog was called "Eye of the Storm." It was written as a series of personal musings and photographs of places where the storm hit- first hand accounts of the disaster. They received a journalism award for it.</span>
6 0
2 years ago
Read 2 more answers
A computer has 4 GB of RAM of which the operating system occupies 512 MB. The processes are all 256 MB (for simplicity) and have
Afina-wow [57]

Answer:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

Explanation:

Previous concepts

Input/output operations per second (IOPS, pronounced eye-ops) "is an input/output performance measurement used to characterize computer storage devices like hard disk drives (HDD)"

Solution to the problem

For this case since we have 4GB, but 512 MB are destinated to the operating system, we can begin finding the available RAM like this:

Available = 4096 MB - 512 MB = 3584 MB

Now we can find the maximum simultaneous process than can use with this:

\frac{3584 MB}{256 MB/proc}= 14 processes

And then we can find the maximum wait I/O that can be tolerated with the following formula:

1- p^{14}= rate

The expeonent for p = 14 since we got 14 simultaneous processes, and the rate for this case would be 99% or 0.99, if we solve for p we got:

1-p^{14} = 0.99

p^{14}= 0.01

p =(0.01)^{1/14}= 0.720

So then the value of the maximum I/O wait that can be tolerated is 0.720 or 72 %

3 0
2 years ago
Assume that a program uses the named constant PI to represent the value 3.14. The program uses the named constant in several sta
mihalych1998 [28]

Answer:

The advantage for the above condition is as follows:-

Explanation:

  • If a user creates a defined constant variable and assigns a value on its and then uses that variable instead of the value, then it will a great advantage.
  • It is because when there is a needs to change the value of that variable, then it can be done when the user changes the value in one place. There is no needs to change the vale in multiple places.
  • But if there is a value in multiple places instead of a variable and there is no constant variable, then the user needs to change the value in multiple places.
7 0
2 years ago
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