Final answer:
In a computer with 8 Mbytes of main memory and a 128 K cache with a 4 K cache block size using direct mapping, 2,048 different main memory blocks can map onto a given physical cache block. The correct answer is a) 2048.
Step-by-step explanation:
To find out how many different main memory blocks can map onto a given physical cache block in a direct mapping cache management scheme, we need to understand the full capacity of the main memory, the size of the cache, the block size of the cache, and how these relate to each other.
The main memory is 8 Mbytes and the cache block size is 4 Kbytes. Since each main memory block has to be mapped to a specific cache line/block, we divide the main memory size by the block size to determine the number of different blocks in main memory.
The multiplicative prefixes such as kilo (K) and mega (M) should be clear, as they are frequently used in computer memory. A kilobyte (KB) is 1,024 bytes, and a megabyte (MB) is 1,024 KB. Therefore, 8 MB is the same as 8 x 1,024 KB, which equals 8,192 KB.
When we divide 8,192 KB (the size of the main memory) by 4 KB (the cache block size), we get 2,048. Thus, 2,048 different main memory blocks can map to a single cache block in this scenario.
The correct answer is a) 2048.
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