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An 8 kbyte direct mapped write-back cached is organized as multiple blocks, each of size 32 byte. The processor generates 32-bit addresses. The cache controller maintains the tag information for each cache block comprising of the following 1 Valid bit 1 Modified bit As many bits as the minimum needed to identify the memory block mapped in the cache. What is the total size of memory needed at the cache controller to store metadata (tags) for the cache ?
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- 4864 bit
- 6144 bit
- 6656 bit
- 5376 bit
- 4864 bit
Correct Option: D
Direct mapped cache => number of sets = 8KB/32B = 2565 bits for offset within a block & 8 bits for identifying set. So remaining 32 – 8 – 5 = 19 bits for tag.1 tag/32B block it includes 2 more bits. So 21 bits/ 32 B block Number of such blocks = 256 So total tag mem = 256 × 21 = 5376
