Cache Size is 64KB, Block size is 32B and the cache is Two-Way Set Associative. For a 32-bit physical address, give the division between Block Offset, Index and Tag.
Submitted by: Administrator64k/32 = 2000 blocks
2 way set assoc- 2000/2 = 1000 lines-> 10 bits for index
32B block-> 5 bits for block offset
32-10-5= 17 bits for tag
Submitted by: Administrator
2 way set assoc- 2000/2 = 1000 lines-> 10 bits for index
32B block-> 5 bits for block offset
32-10-5= 17 bits for tag
Submitted by: Administrator
Read Online Computer Architecture Job Interview Questions And Answers
Top Computer Architecture Questions
| ☺ | What is the difference between interrupt service routine and subroutine? |
| ☺ | For a pipeline with n stages, what is the ideal throughput? What prevents us from achieving this ideal throughput? |
| ☺ | Whats the difference between Write-Through and Write-Back Caches? Explain advantages and disadvantages of each? |
| ☺ | What is Cache Coherency? |
| ☺ | What are the components in a Microprocessor? |
Top PC Hardware Categories
| ☺ | Motherboard Interview Questions. |
| ☺ | Basic Computer Interview Questions. |
| ☺ | Embedded Systems Interview Questions. |
| ☺ | A + (Plus) Hardware Interview Questions. |
| ☺ | Computer Architecture Interview Questions. |
