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3 years ago

____________ is a widely accepted international best practices framework for implementing information systems security.

Computers and Technology

1 answer:

balandron [24]3 years ago

8 0

Answer:

Control Objectives for Information and related Technology (COBIT)

Explanation:

COBIT is a framework developed by the ISACA for managing and supervising Information Technology processes. Such kinds of frameworks allow companies to position themselves with the expectations that they and their clients expect. COBIT's aim is to provide management and business process owners with an IT control and governance model that helps deliver value from IT and identify and handle the IT-related risks. It helps bridge the gap between business needs, control requirements and technical issues. It is a paradigm of control for addressing the IT governance requirements and maintaining the security of knowledge and information systems. COBIT emphasizes on what allows processes to work well.

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yes a computer can evaluate expressions depends on the computer for the maybe expression

Explanation:

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2 years ago

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Direct Mapped Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a direct-mapped cache.

xz_007 [3.2K]

The question given is incomplete and by finding it on internet i found the complete question as follows:

Direct Mapped Cache.

Memory is byte addressable

Fill in the missing fields based upon the properties of a direct-mapped cache. Click on "Select" to access the list of possible answers Main Memory Size Cache Size Block Size Number of Tag Bits 3 1) 16 KiB 128 KiB 256 B 20 2) 32 GiB 32 KiB 1 KiB 3) 64 MiB 512 KiB 1 KiB Select] 4 KiB 4) 16 GiB 10 Select ] Select ] 5) 10 64 MiB [ Select ] 6) Select] 512 KiB 7

For convenience, the table form of the question is attached in the image below.

Answers of blanks:

1. 3 bits

2. 20 bits

3. 64 MB

4. 16 MB

5. 64 KB

6. 64 MB

Explanation:

Following is the solution for question step-by-step:

Part 1:

No. of Tag bits = No. of bits to represent

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 128 KB = $2^7 * 2^{10} = 2^{17}$

Cache Memory = 16 KB = $2^4 * 2^{10}= 2^{14}$

Putting values in A:

Tag bits = 17 - 14 = 3 bits

Part 2:

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 32 GB = $2^5 * 2^{30} = 2^{35}$

Cache Memory = 16 KB = $2^5 * 2^{10}= 2^{15}$

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

Main Memory = $2^6 * 2^{20} = 64 MB$

Part 4:

Given that:

Main Memory Size = $2^4 * 2^{30} = 2^{34}$

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size = $2^4 * 2^{20} = 16 MB$

Part 5:

Given that:

Main Memory Size = 64 MB = $2^6 * 2^{20}$

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size = $2^{16} = 2^6 * 2^{10} = 64 KB$

Part 6:

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size = $2^{26} = 2^6 * 2^20 = 64 MB$

i hope it will help you!

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2 years ago

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2 years ago

How does an operating system manage peripheral devices?

azamat

Answer:

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Explanation:

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2 years ago