For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. We assume that each word has 4 bytes. How many entries does the cache have? Tag Index Offset Block offset Byte offset 31–12 11-6 5-2 1-0
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- For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. We assume that each word has 4 bytes. How many entries does the cache have?For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. We assume that each word has 4 bytes. What is the ratio between total bits required for such a cache implementation over the data storage bits?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.
- For a direct-mapped cache design with a 32-bit address, the following bits of the address are used to access the cache. We assume that each word has 4 bytes. How many words of data are included in one cache line?For 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?For a direct-mapped cache design with 64-bit addresses, the following bits of the address are used to access the cache: Tag Index Offset 63-13 12-4 3-0 a. What is the cache block size (in bytes)?b. What is the cache size (in bytes)?c. What is the total number of bits (including valid bit, tag bits and data array bits) to implement this cache?d. For the same block and cache sizes, you want to implement a 4-way set-associative cache, what is the number of index bit and the number of tag bits?
- For a direct-mapped cache design with a 32-bit address, the following bitsof the address are used to access the cache. Use the table below. a. What is the cache block size (in words)?b. How many entries does the cache have?c. What is the ration between total bits required for such a cache implementation overthe data storage bit?For 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?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 5-0
- For a direct-mapped cache design with a 64-bit address, the following bits of the address are used to access the cache. Tag Index Offset 63-9 8-5 4-0 Beginning from power on, the following byte-addressed cache references are recorded. Нех 00 04 10 84 E8 AO 400 1E 8C Cic B4 884 (A) For cach reference, list (i) its teg, index, end offset, (ii) whether it is a hit or a miss, and (iii) which bytes were replaced (if any). (B) what is the hit ratio? (C) List the final state of the cache, with each valid entry represented as a record of . For example,3. The table below represents five lines from a cache that uses fully associative mapping with a block size of 8. Identify the address of the shaded data, 0xE6, first in binary and then in hexadecimal. The tag numbers and word id bits are in binary, but the content of the cache (the data) is in hexadecimal. Word id bits Tag 000 001 010 011 100 101 110 111 ------------------------------------------ 1011010 10 65 BA 0F C4 19 6E C3 1100101 21 76 CB 80 D5 2A 7F B5 0011011 32 87 DC 91 E6 3B F0 A6 1100000 43 98 ED A2 F7 4C E1 97 1111100 54 9A FE B3 08 5D D2 88A direct-mapped cache consists of 16 blocks. A byte-addressable main memory contains 4K blocks of 16 bytes each. Access time for the cache is 30 ns and the time required to fill a cache slot from main memory is 250 ns. Assume a request is always started in sequential to cache and then to main memory. If a block is missing from cache, the entire block is brought into the cache and the access is restarted. Initially, the cache is empty. a) Show the main memory address format that allows us to map addresses from main memory to cache. Be sure to include the fields as well as their sizes. b) Compute the hit ratio for a program that loops 4 times from locations 0 to 42 (base 10) in memory. c) Compute the effective access time for this program.