What is the size of the cache in bytes for: data: 32KB 1 block = 8 words 1 block = 1 cache line word addressing word size = address size = 32 bits draw a picture; give the address parsing; brief explanation
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What is the size of the cache in bytes for:
data: 32KB
1 block = 8 words
1 block = 1 cache line
word addressing
word size = address size = 32 bits
draw a picture; give the address parsing; brief explanation
Step by step
Solved in 2 steps with 1 images
- Mapping the main memory bit address to the cache address is called address mapping. What are the common cache mapping methods?For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. a. What is the cache block size in words? b. How many entries does the cache have? Tag 31-13 Index 12-6 Offset 5-0For a cache memory of size 32 KB, how many cache lines (blocks) does the cache hold for block sizes of 32 or 64 bytes?
- A cache is set up with a block size of 32 words. There are 64 blocks in cache and set up to be 4-way set associative. You have byte address 0x8923. Show the word address, block address, tag, and index Show each access being filled in with a note of hit or miss. You are given word address and the access are: 0xff, 0x08, 0x22, 0x00, 0x39, 0xF3, 0x07, 0xc0.What Is the Definition of Cache Performance?What is the cache block size (in words)?
- A cache system is to be designed to store data from a 256 MB memory space. If each block of main memory contains 16 words, determine the number of blocks that are needed and draw the logical organization of the full address identifying the block ID portion and the word (offset) portionFor a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache: Tag Index Offset 31-10 9-6 5-0 1. What is the cache block size (in words)? 2. How many entries does the cache have? 3. What is the ratio between total bits required for such a cache implementation over the data storage bits?A direct-matched cache has a structure where the first (left) 12 bits are the label, the next 15 are the line, and the last 5 are the word. In which line the address 1878112 should be stored? a) 32167 b) 572 c) 14988 d) 25923
- Identify the line number, tag, and word position for each of the 30-bit addresses shown below if they are stored in a cache using the direct mapping method. a.) Address: 23D94EA616 Lines in cache: 4K Block size: 2b.) Address: 1A54387F6 Lines in cache: 8K Block size: 4For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. Tag 31-10 Index 9-5 a. What is the cache block size (in words)? b. How many entries does the cache have? Offset 4-0 c. What is the ratio between total bits required for such a cache implementation over the data storage bits?In a Direct Mapped Cache Memory Physical Address format the Cache line offset field size and word offset field size are same (with word size of one Byte). The number of tag bits in the Physical Address format is equal to the number of blocks in Cache Memory. If the Tag field Size is Mega words. 16 bits, the size of the physical Memory is