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3 years ago

Describe how data center storage applications drive the development of SAN technology.

Computers and Technology

1 answer:

Ede4ka [16]3 years ago

5 0

Answer:

The storage area network basically apply in the networking model for storage various application in the data center. The storage area network (SAN) provide the common pathway between the server and the storage device.

The storage area network technology basically implemented in the fiber channel configuration over the ethernet. SAN provide access to the data that shared by the personal resources. It is basically simple and dedicated network which provide for the data storage.

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How should students approach the decision to take the SAT or the ACT?

Vikentia [17]

Answer:

I believe that the only way to really make an informed decision as to which test is a better fit for you is to take a full-length diagnostic exam for both the ACT and the SAT before doing any prep. I also would look at the pros and cons for each test.

I personally took the ACT because there is <em>NO PENALTY </em>for guessing on the test. An educated guess won't hurt your score on the ACT.

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3 years ago

Does the security burden fall primarily on the user? On

liubo4ka [24]

Answer:

yes and no because of the security

Explanation:

yes and no because of the security

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3 years ago

Select the correct answer.

Dvinal [7]

Answer:

Explanation:

because when you make a presentation so you are already use technology

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3 years ago

Which of the following is NOT a best practice to protect data on your mobile computing device?

OleMash [197]

<u>Lock your device screen when not in use and require a password to reactivate</u> is not a best practice to protect data on your mobile computing device.

<h3>What is a mobile computing device?</h3>

Any device that was built using mobile parts, such as mobile hardware and software, is referred to as a mobile computing device. Portable devices that can function like a typical computing device in terms of operation, execution, and provision of services and applications are known as mobile computing devices.

Portable and handheld computing devices are other names for mobile computing devices.

Modern handheld devices that have the hardware and software needed to run common desktop and Web applications are generally referred to as mobile computing devices. Similar hardware and software elements found in personal computers, such as processors, random memory and storage, Wi-Fi, and an operating system, are also found in mobile computing devices. They are made specifically for mobile architecture and portability, which sets them apart from PCS.

Learn more about mobile computing devices

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4 0

1 year ago

Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,

nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
P5 will not be allocated to any of the available space because none can contain it.
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0

3 years ago