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.
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Consider a 2-way set associative cache consisting of 8 blocks total of byte-addressable memory with 4...
A direct-mapped cache consists of 8 blocks. Byte-addressable main memory contains 4K blocks of 8 bytes each. Access time for the cache is 22ns, and the time required to fill a cache slot from main memory is 300ns. (This time allows us to determine the block is missing and bring it into cache.) Assume a request is always started in parallel to both cache and to main memory(so if it is not found in cache, we do not have to...
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.
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?
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?
) 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...
Suppose a computer using direct mapped cache has 232 byte of byte-addressable main memory, and a cache of 1024 blocks, where each cache block contains 32 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, block, and offset fields? c.To which cache block will the memory address 0x000063FA map?
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 we have a byte-addressable computer with a cache that holds 8 blocks of 4 bytes each. Assuming that each memory address has 8 bits and cache is originally empty, for the cache mapping technique, two-way set associative, trace how cache is used when a program accesses the following series of addresses in order: 0x01, 0x04, 0x09, 0x05, 0x14, 0x21, and 0x01.
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...
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...