Line 1 Set index 2-way set associative cache Line 0 Tag Valid Byte 0 Byte 1...
) Consider an 8-way associative 64 Kilo Byte cache with 32 byte cache lines. Assume memory addresses are 32 bits long. a). Show how a 32-bit address is used to access the cache (show how many bits for Tag, Index and Byte offset). b). Calculate the total number of bits needed for this cache including tag bits, valid bits and data c). Translate the following addresses (in hex) to cache set number, byte number and tag (i) B2FE3053hex (ii) FFFFA04Ehex...
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.
Consider a 2-way set associative cache consisting of 8 blocks total of byte-addressable memory with 4 bytes per block. Assume that the cache is initially empty. Given the following address sequence, fill in the table below. Time Access Tag Set Offset 3 10010001 11001001 10110110 10101011 10110010
Set-Associative Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a set-associative cache. Click on "Select" to access the list of possible answers. Set Block Size Number of Tag Bits Select] Select] Main Memory Size Cache Size 256 B 1) 128 KiB 16 KiB 2) 32 GiB 32 KiB 1 KiB 3) [Select ] 512 KiB 1 KiB [Select ] 10 16 GiB 4 KiB Select ] I Select ] 5) 10 64 MiB...
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 =...
For a 2-way set associative cache with an 8-bit byte addressable address and 8- byte blocks, what is the maximum number of sets in the cache?
You have a 2-way set associative L1 cache that is 8KB, with 4-word cache lines. You get the following sequence of writes to the cache --each is a 32-bit address in hexadecimal 0x32E4 0x8000 0x1F50 0x8004 0x72EC OxDOOC 0x800C 0x72E8 0x4008 OxD000 0x82E0 a) [7 Pts] How many cache misses occur with an LFU (Least Frequently Used) policy? Give a detailed answer and fill in the table below for each address reference Set Index (in hex) Memory address(in hex) 0x32E4...
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 2-way set associative cache consists of four sets 0, 1, 2, 3. The main memory is word addressable (i.e. treat the memory as an array of words indexed by the address). It contains 2048 blocks 0 through 2047, and each block has eight words. (a) How many bits are needed to address the main memory? (b) Show how a main memory address will be translated into a tag, a set number, and an offset within a block. Illustrate this...
Find a wrong description about associative cache. 1-way set associative cache is identical to the direct mapped cache Each cache block contains one valid bit and one tag regardless of the number of data blocks Fully associative cache requires all entries to be searched at once Associative cache can decrease miss rate compared to direct mapped cache