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Elden [556K]
3 years ago
7

MORE FREEE POINTS AYEEE

Computers and Technology
2 answers:
Lerok [7]3 years ago
6 0

Answer:

Hehehehehe

Explanation:

Ludmilka [50]3 years ago
4 0

Answer:

thank u

Explanation:

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Identify and explain the three tiers of web-based applications.
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This was on google i take no credit for this work

Btw I don´t know if this is what u want

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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.
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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:

<u>Part 1:</u>

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

<u>Part 2:</u>

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

<u>Part 3:</u>

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

OR

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

<u>Part 4:</u>

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

<u>Part 5:</u>

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

<u>Part 6:</u>

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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