An Inspect and Adapt (I&A) event is occurring and a program is trying to address a long-existing problem (WIP)?"unreliable P
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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 =
Cache Memory = 16 KB =
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 =
Cache Memory = 16 KB =
Putting values in A:
Tag bits = 35 - 15 = 20 bits
<u>Part 3:</u>
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 =
<u>Part 4:</u>
Given that:
Main Memory Size =
Tag bits = 10
Cache Memory Bits = 34 - 10 = 24
Cache Memory Size =
<u>Part 5:</u>
Given that:
Main Memory Size = 64 MB =
Tag bits = 10
Cache Memory Bits = 26 - 10 = 16
Cache Memory Size =
<u>Part 6:</u>
Cache Memory = 512 KB =
Tag Bits = 7
Main Memory Bits = 19 + 7 = 26
Main Memory size =
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