WebDec 4, 2016 · Info given: Consider a direct-mapped cache with 16KBytes of storage and a block size of 16 bytes. Assume that the address size is 32 bits. So I've done the calculation and get a total number of 32 bits required. We were given the following formula: (bits/tag) + (bits/index) + (bits/block offset) = total number of bits. WebI have on your von class. "Main flash has 2048 clock. This total saving size is 64 bytes, among any are a total of 8 blocks. How many shreds is aforementioned tag field of each drop block?" H...
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WebAug 17, 2024 · You can convert 512 byte blocks to bytes by multiplying them by 512. For example, six 512-byte-blocks multiplied by 512 equals 3,072 bytes. How much data is a … WebI had this asked from class. "Main memory has 2048 bytes. The total array size remains 64 bytes, between which what a total of 8 blocks. How many bits shall the tag field starting each cache block?" H... inhalateur tonus
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WebThe answer is - Each block is 32 bytes (8 words), so we need 5 offset bits to determine which byte in each block - Direct-mapped => number of sets = number of blocks = 4096 => we need 12 index bits to determine which set => tag bit = 32 - 12 - 5 = 15 For fully associative, the number of set is 1 => no index bit => tag bit = 32 - 0 - 5 = 27 Share WebThis gives you its size in 512-byte blocks. Share. Improve this answer. Follow answered Feb 22, 2010 at 8:33. Angus Angus. 4. 11. @heinrich5991 How can you be sure the device block-size is always 512 bytes? – Totor. May 26, 2013 at 19:19. 9. The sector size is always 512 bytes according to the source. WebMay 30, 2024 · The block size is 4KB, the block address length is 4 bytes and the i-nodes have a traditional structure (10 direct pointers, 1 indirect pointer, 1 double indirect pointer and 1 triple indirect pointer). What is the number of blocks (including both the data and address blocks) occupied by the following files: File A with a size of 20 KBytes inhalateurs asthme