Assume a 32-bit memory address, and a 128KB direct-mapped cache with 64-byte blocks. Show how the memory address is divided into tag, index and offset. Indicate clearly how many bits are in each.
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) Assume a 32-bit memory address, and a 128KB direct-mapped cache with 64-byte blocks.
Show how the memory address is divided into tag, index and offset. Indicate clearly how many bits are in each.
1b) Consider the memory address 0x2c0868. For the cache in part a, what are its tag and index? Show in binary.
1c) Suppose an access to 0x2c0868 is a cache miss. For the cache in part a, what are the addresses of the (aligned) words that are brought into the cache?
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- Computer Science Consider a direct-mapped cache with 8 lines, each holding 16 bytes of data. The cache is byte-addressable and the main memory consists of 64 KB, which is also byte-addressable. Assume that a program reads 16KB of memory sequentially. Answer the following questions:a) How many bits are required for the tag, index, and offset fields of a cache address?b) What is the cache size in bytes?c) What is the block size in bytes?d) What is the total number of blocks in main memory?e) How many cache hits and misses will occur for the program, assuming that the cache is initially empty?f) What is the hit ratio?g) Give an example virtual address (in BINARY) that will be placed in cache line 5.Assume 32 bit memory addresses , which are byte addresses. You have direct mapped cache with block-size of 8 bytes and size of 128 blocks. How many bits will the index , block offset, and tag be ? How many total bits are required for the cache , including the data ?a) Assume that a direct mapped cache memory stores 1024 blocks and 64 bytes per block.i) What is storage capacity of the cache memory in bytes?ii) What is length of tag bits if 20 bits addressing is used in CPU?b) Draw a table indicating the initial state (Valid, Tag bits and Data) of a direct mapped cache with 8-blocks and 1 word per block. (Assume that words have 5-bits address length). Then, fill / modify the table (valid, tag and data columns) by accessing word-addresses below in given order. (e.g. old value → new value → ...)5, 10, 26, 5, 10, 23, 20, 5, 23, 10(Show memory accesses by writing Mem[Word Address] in “Data” column)c) Which accesses are “hit” and which accesses are “miss” in question (b)?5, 10, 26, 5, 10, 23, 20, 5, 23, 10
- Assume that we have a computer with a cache memory of 512 blocks with a total size of 128K bits. Knowing that the computer uses a word addressable mode and the format of the memory address as seen by the Fully associative cache scheme is as shown below, answer the below questions: Fully Associative Cache Format 19 5 1- How many words do we have in each cache block? 2- What is the size of each word? 3- What is the size of the main memory? 4- How many blocks are there in the main memory? 5- Draw the format of the memory address as seen by the Direct Mapped Cache scheme, showing the fields as well as their sizes.Determine which bits in a 32-bit address are used for selecting the byte (B), selecting the word (W), indexing the cache (I), and the cache tag (T), for each of the following caches: A. 64-line, direct-mapped, write-through, 8 byte lineA computer is using a fully associative cache and has 216 bytes of main memory (byte addressable) and a cache of 64 blocks, where each block contains 32 bytes. a. How many blocks of main memory are there? b. What will be the sizes of the tag, index, and byte offset fields? c. To which cache set will the memory address 0xF8C9(hexadecimal) map?
- Please help with detailed explanation for problem C, don't copy solutions from other sources. Consider a byte addressing architecture with 64-bit memory addresses. a)Which bits of the address would be used in the tag, index and offset in a direct-mapped cache with 512 1-word blocks. b)Which bits of the address would be used in the tag, index and offset in a direct-mapped cache with 64 8-word blocks. c)What is the ratio of bits used for storing data to total bits stored in the cache in each of the above cases a and b?Suppose a computer using fully associative cache has 4G bytes of byte-addressable main memory and a cache of 512 blocks, where each cache block contains 128 bytes. a) How many blocks of main memory are there? b) What is the format of a memory address as seen by the cache, i.e., what are the sizes of the tag and offset fields? c) To which cache block will the memory address 0x018072 map?Memory Hierarchy and Cache Suppose that we have a computer that uses a memory address of 12-bits. This computer has a 64-byte cache with 16 bytes per frame. The computer accesses a number of memory locations throughout the course of a running program. Suppose this computer uses direct-mapped cache. The system accesses the following memory addresses (given in hex) in this exact order: F2E, A17, 2E0, 44E, 34F, 341, B50, B58 a. What is the hit ratio for the memory reference sequence given above? b. Show the content of each cache frame following each memory reference (frame content to be shown as tag +frame index) c.If we keep the same cache size and the same frame size but switch to a 2-way set associative cache mapping scheme. Given the memory address reference 555, indicate where we would look in the cache to find this data. Indicate which fields will be used to find the exact location?
- Assume that we have a computer with a cache memory of 512blocks with a total size of 128K bits. Knowing that the computer uses a word addressable mode and the format of the memory address as seen by the Fully associative cache scheme is as shown below, answer the below questions: Fully Associative Cache Format 1- How many words do we have in each cache block? 2- What is the size of each word? 3- What is the size of the main memory? 4- How many blocks are there in the main memory? 5- Draw the format of the memory address as seen by the Direct Mapped Cache scheme, showing the fields as well as their sizes.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 88Assume we have a computer with 512 blocks of cache memory with a total capacity of 128K bits. Answer the following questions knowing that the computer operates in a word addressable mode and that the format of the memory address as perceived by the Fully associative cache scheme is as shown below: Cache Format with Full Associativity 19 5 1- How many words are in each cache block? 2- How big are the letters in each word? 3- How large is the primary memory? 4- What is the total number of blocks in main memory? 5- Draw the memory address format as observed by the Direct Mapped Cache technique, including the fields and their sizes.