Calculate the total number of lines of "direct mapping" cache, If a main memory is 1G words divided Into 128 Blocks, and the number of words in a memory block is 32 words.
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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.A 2-way set associative cache consists of four sets. Main memory contains 2K blocks of eight words each. 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. Compute the hit ratio for a program that loops 6 times from locations 8 to 51 in main memory. You may leave the hit ratio in terms of a fraction. Please show details how you obtain the result.Q1 Calculate the total number of lines of "direct mapping" cache, If a main memory is 1G words divided into 128 Blocks, and the number of words in a memory block is 32 words.
- 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 is3. 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 88Given that a 4-way set associative cache memory has 64 KB data and each block contains 32 bytes. The main memory capacity is 4 GB. a. Find the number of bits for the main memory address. ANSWER: bits b. How many blocks are there in a set? ANSWER: blocks c. How many sets the cache has? ANSWER: d. The main memory address format is => | Tag: e. Which set will be mapped by the main memory address 458195h. ANSWER: sets bits | Set: bits | Word: bits | (in decimal)
- 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?Given that a 4-way set associative cache memory has 64 KB data and each block contains 32 bytes. The main memory capacity is 4 GB. a. Find the number of bits for the main memory address. ANSWER: b. How many blocks are there in a set? ANSWER: c. How many sets the cache has? ANSWER: d. The main memory address format is => | Tag: bits | blocks sets bits | Set: e. Which set will be mapped by the main memory address 458195h. ANSWER: decimal) bits bits | Word: (inFor 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 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 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 1-0Q2: a cache memory consists of 512 blocks, and if the last word in the block is 1111111. I the last Tag is 1111. Then the total number of words in main memory is Tags, and The main memory consists of blocks. The total number of bits in address fields i . where number of bits for tag field is .... and the remaining number of bits for word field i ..