Question 3: Consider a 32-bit physical address memory system with block size 16 bytes and a...
Consider a direct-mapped cache with 32 blocks Cache is initially empty, Block size = 16 bytes The following memory addresses (in hexadecimal) are referenced: 0x2B4, 0x2B8, 0x2BC, 0x3E8, 0x3EC,0x4F0, 0x8F4, 0x8F8, 0x8FC. Map addresses to cache blocks and indicate whether hit or miss
For a 16K-byte, direct-mapped cache, suppose the block size is 32 bytes, draw a cache diagram. Indicate the block size, number of blocks, and address field decomposition (block offset, index, and tag bit width) assuming a 32-bit memory address.
For a direct-mapped cache with a 32-bit address and 32-bit words, the following address bits are used to access the cache. TAG INDEX OFFSET 31-15 14-8 7-0 a. What is the cache block size (in words)? [13 points] b. How many blocks does the cache have? [12 points]
use the following list of 32 bit memory address references, given as word addresses. Note that you will need to convert them to binary: 3, 180, 43, 2, 191, 88, 190, 14, 181, 44, 186, 253 4. Assume a direct-mapped cache with 16 one-word blocks. For each reference, list the binary address, the tag, the index, and if the reference is a hit or a miss, assuming the cache is initially empty.
32 bytes of memory. 16 bytes of 2-way setassociative cache, where blocks can go anywhere within the set. Block is 4 bytes, set in cache is two blocks. Populate memory starting with 0-9, then upper case letters. Hint- with full associativity in the set: each block has its own set of Tag bits in the cache. Memory is not organized by sets, though blocks get assigned to sets, and load in the cache per set. A) Complete: Bits in Address...
Question 33 10 pts For a direct mapped cache of 4 blocks with a cache block size of 1 byte, in which cache block will each memory location map to? The order of accesses if given by the operation number. Indicate if each access is a hit or a miss, and what the tag value is for each entry. Assume that the cache is initially empty, and the accesses are in order of appearance. REDRAW AND COMPLETE THE CACHE TABLE...
A primary memory system consists of 128 address bits. Each block within the cache is 256 bytes. The total cache size is 131,072 bytes. What are the sizes of the tag and index in the cache? First, find the number of blocks. 128,000 bytes # Blocks = 256 bytes per block bort = 512 blocks Where does the 128,000 come from?? Now find the number of bits required to distinguish all the blocks. This is the index aka set size....
Please help with this
Below is a list of 32-bit (1 word) memory address references a program makes, given as word addresses (not byte addresses): 2, 4, 5, 4, 6, 4, 12, 13, 2, 13, 4, 253 For each of these references, identify the tag and index, given a 16 word, direct-mapped cache which has 8 two-word blocks. Also, list if each reference is a hit or a miss, assuming the cache is initially empty. Your answer should be a...
The physical address of memory on a machine is 32 bits. The machine has a direct mapped cache of size 512 KB with a block size of 8 bytes. What is the size of the tag field in bits? Can you please explain this step by step and what the importance of direct mapped is?
A 256kiB (2^18 bytes) cache has a block size of 32 bytes and is 32-way set-associative. How many bits of a 32-bit address will be in the Tag, Index, and Bock Offset?