Consider a computer system with a cache and a byte addressable main memory. The cache is a 2-way set associative cache with block size of 4 bytes, tag size of 1 byte and 8 indexes. There is a valid bit (denoted as V) associated with each block, and a LRU bit (denoted as B) associated with each index. The contents of the cache are shown as follows, with all numbers given in hex notation.
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Problem 6. Suppose we have a computer with 32 megabytes of main memory, 256 bytes of cache, and a block size of 16 bytes. For each configuration below, determine the memory address format, indicating the number of bits needed for each appropriate field (i.e. tag, block, set, offset). Show any relevant calculations. Direct cache mapping and memory is byte-addressable a) Direct cache mapping and memory is word-addressable with a word size of 16 bits b) c) 2-way set associative cache...
Q2. Consider a four-way set associative cache with a data size of 64 KB. The CPU generates a 32-bit byte addressable memory address. Each memory word contains 4 bytes. The block size is 16 bytes. Show the logical partitioning of the memory address into byte offset, cache index, and tag components.
please answer $5 UXIF map in the computer uses direct mapping Question 18 5 pts Suppose we have a byte-addressable computer using 2-way set associative mapping with 16-bit main memory addresses and 32 blocks of cache. Suppose also that each block contains 8 bytes. The size of the block offset field is bits, the bits. size of the set field is bits, and the size of the tag field is 5 pts Question 19 Suppose we have a byte-addressable computer...
1. A cache holds 64 words where each word is 4 bytes. Assume a 32 bit address. There are four different caches a. A direct-mapped cache with block size = 16 words b. 2-way set-associative cache with block size = 8 words c. 4-way set-associative cache with block size=4 words d. A fully associative cache with block size = 16 words. Complete the table for each cache. Cache a Cache be Cache Cache de 16 Number of bits needed for...
Consider a processor that has a 20-bit address and a 1K Byte Cache. The cache and main memory are divided into blocks where each block is 256 Bytes. If direct mapping is used, what is the tag size of each block in cache and how many tag comparisons are made for a one-cache access? Repeat part (1) for fully associative mapping. Repeat part (1) for 2 way set-associative cache. For the direct map find out which of the following accesses...
Suppose a computer using a fully associative cache has 232 bytes of byte-addressable main memory and a cache of 1024 blocks, were each cache block contains 32 bytes. Consider a memory address as seen by the cache. How many bits are in the tag field?
Assume the cache can hold 64 kB. Data are transferred between main memory and the cache in blocks of 4 bytes each. This means that the cache is organized as 16K=2^14 lines of 4 bytes each. The main memory consists of 16 MB, with each byte directly addressable by a 24-bit address (2^24 =16M). Thus, for mapping purposes, we can consider main memory to consist of 4M blocks of 4 bytes each. Please show illustrations too for all work. Part...
Assume the following about a computer with a cache: .. The memory is byte addressable. • Memory accesses are to 1-byte words (not to 4-byte words). .. Addresses are 8 bits wide. .. The cache is 2-way associative cache (E=2), with a 2-byte block size (B=2) and 4 sets (5=4). • The cache contents are as shown below (V="Valid"): Set #Way #0 Way #1 V=1;Tag=0x12; Data = v=1;Tag=0x10; Data = Ox39 0x00 0x26 Ox63 V=1;Tag=0x09; Data = v=1;Tag=0x11; Data =...
Text: Explain how a 32-bit byte memory address should be divided into Tag/Index/Offset fields for each of the cache configurations below. Note: 1KB = 210 bytes. You must explain how many bits to assign to each field and the ordering of the three fields. You get at most 50% of the credit if you give the length of each field without an explanation. 1) A fully associative cache with cache block size = 2 words and cache size = 512KB....
A processor outputs the hexadecimal address: AA0893BF, with the following partitioning: AA0 (Tag) 890 (Index) BF (Offset) Assume that the memory space is Byte addressable. Note that the parts below are not related to each other, i.e treat them separately. a. If Direct Mapped Cache is used, how many bytes are in the cache? b. If Two-Way Set Associative Cache is used, how many sets are in the entire cache? c. If Two-Way Set Associative Cache is used, how many...