Assume the address format for a fully-associative cache is as follows: 6 bits 2 bits Tag Offset 8bits Given the cache directory is as shown in the diagram below, indicate whether the memory reference Ox5E results in a cache hits or a miss. Tag valid Block 000 110110 001 000001 010 000010 1 011 000101 100 001000 101 100010 110 010111 111 110110
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- 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 entries does the cache have? Tag Index Offset Block offset | Byte offset 31–12 11-6 5-2 1-0In 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
- Assume the address format for a fully-associative cache is as follows: Tag Offset b Given the cache directory is as shown in the diagram below, indicate whether the memory reference OxDA results in a cache hits or a miss. Tag valid Block 000 110110 1 000001 1 001 010 ? 000010 011 000101 100 001000 101 100010 110 010111 110110 O Miss Hit D 111 1 1 1 0 0 1For 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?The table below shows a 2-way set associative cache. What is the maximum size of the memory in Kbytes? First Way byte offset (binary) 01 Second Way byte offset (binary) Tag (binary) | (8 bits) Index Tag |(8 bits) | Valid Valid 00 10 11 00 01 10 11 000 AF 1 ВС 86 42 19 6D 1 7F B3 74 83 001 25 3B 69 FD 62 3B 1 D2 6C 68 DC 010 DO 1 44 53 23 62 C3 45 38 87 DA 011 15 1 83 13 48 AB 9B 1 41 8E 90 7A 100 DD 1 36 F4 ЕС ЕВ C5 1 7B ЕВ 41 69 101 4A 72 5B EA 8A 1A 34 3B 54 8E 110 8C 1 11 B1 C8 2E 4B 56 52 5B 27 11 55 A6 61 33 4B FO 48 СА А4 2E О а. 16 O b. 64 О с. 32 O d. 10 O e. 4 O f. 1 O g. 2 O h. 8 Clear my choice
- 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. We assume that each word has 4 bytes. How many entries does the cache have?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
- A cache memory system with capacity of N words and block size of B words is to be designed. If it is designed as a direct mapped cache, the length of the TAG field is 14 bits. If it is designed as a 4-way set associative cache, the length of the TAG field will be ………… bits.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. 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?