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Consider a computer with 8 Mbytes of main memory and a 128 K cache. The cache block size is 4 K. It uses a direct mapping scheme for cache management. How many different main memory blocks can map onto a given physical cache block?
- a)2048
- b)256
- c)64
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Consider a computer with 8 Mbytes of main memory and a 128 K cache. Th...
Number of block in cash 
Number of block in main memory
Since the system uses direct mapping hence different memory blocks that can map into the cache block.
= Number of blocks in MM% Number of blocks in cache
Number of block in main memory
Since the system uses direct mapping hence different memory blocks that can map into the cache block.
= Number of blocks in MM% Number of blocks in cache
Most Upvoted Answer
Consider a computer with 8 Mbytes of main memory and a 128 K cache. Th...
Solution:
Given,
Main memory size = 8 Mbytes
Cache size = 128 K
Cache block size = 4 K
Direct mapping scheme is used
Number of blocks in the cache = (Cache size)/(Cache block size)
= 128 K/4 K
= 32
Hence, there are 32 physical cache blocks.
Number of blocks in main memory = (Main memory size)/(Cache block size)
= 8 Mbytes/4 K
= 2048
Hence, there are 2048 main memory blocks.
Since direct mapping scheme is used, each main memory block can map to only one physical cache block.
Therefore, the number of different main memory blocks that can map onto a given physical cache block is:
(Number of blocks in main memory)/(Number of blocks in the cache)
= 2048/32
= 64
Hence, the correct option is (c) 64.
Given,
Main memory size = 8 Mbytes
Cache size = 128 K
Cache block size = 4 K
Direct mapping scheme is used
Number of blocks in the cache = (Cache size)/(Cache block size)
= 128 K/4 K
= 32
Hence, there are 32 physical cache blocks.
Number of blocks in main memory = (Main memory size)/(Cache block size)
= 8 Mbytes/4 K
= 2048
Hence, there are 2048 main memory blocks.
Since direct mapping scheme is used, each main memory block can map to only one physical cache block.
Therefore, the number of different main memory blocks that can map onto a given physical cache block is:
(Number of blocks in main memory)/(Number of blocks in the cache)
= 2048/32
= 64
Hence, the correct option is (c) 64.
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