1.)
block offset field size = 5 bits
set field size = 4 bits
tag field = 15 bits
2.)
pls both 31 Question 31 10 pts Suppose we have a byte-addressable computer using 4-way set...
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...
Question 28 7 pts Consider a byte-addressable computer with 24-bit addresses, a cache capable of storing a total of 64K bytes of data, and blocks of 32 bytes. If the computer uses direct mapping, the format of the memory address is as follows: bits for the tag field, bits for the cache block number, and bits for the block offset.
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...
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?
29 pls clesr ans
the average access time for the processor to access an item? 29ns Question 29 7 pts Consider a byte-addressable computer with 24-bit addresses, a cache capable of storing a total of 64K bytes of data, and blocks of 32 bytes. If the computer uses direct mapping, the format of the memory address is as follows: bits for the tag field, bits for the cache block number, and bits for the block offset. Question 30 7 pts...
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.
Question 29 7 pts 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, to which cache block would the hexadecimal address Ox1F map if the computer uses direct mapping?
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
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?
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.