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.

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 =

Cache Memory  = 16 KB =

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 =

Cache Memory  = 16 KB =

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB =

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

OR

Main Memory =

Part 4:

Given that:

Main Memory Size =

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size =

Part 5:

Given that:

Main Memory Size  = 64 MB =

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size =

Part 6:

Cache Memory = 512 KB =

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size =

i hope it will help you!

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