median cache misses per entry using 64-byte cache blocks and no prefetching.
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- Is there a way to find out which parts of a log entry a certain log processing function can read? The following code determines the typical number of cache misses per entry while using 64-byte cache blocks and no prefetching.This function will be able to determine the fields in a log entry it needs to look at in order to do its job properly. The following line of code computes the average number of cache misses for each item in the cache when 64-byte cache blocks are used without any prefetching being performed by the program.For a direct-mapped cache design with 64-bit addresses, the following bits of the address are used to access the cache: Tag Index Offset 63-13 12-4 3-0 a. What is the cache block size (in bytes)?b. What is the cache size (in bytes)?c. What is the total number of bits (including valid bit, tag bits and data array bits) to implement this cache?d. For the same block and cache sizes, you want to implement a 4-way set-associative cache, what is the number of index bit and the number of tag bits?
- 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?Suppose a computer using fully associative cache has 4 GB of byte-addressable main memory and a cache of 256 blocks, where each block contains 256 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 0X1A1B1C1D map?Here is the question: A direct-mapped cache consists of 8 blocks. A byte-addressable main memory contains 4K blocks of eight bytes each. Access time for the cache is 20 ns and the time required to fill a cache slot from main memory is 300 ns. Assume a request is always started in sequential to cache and then to main memory. If a block is missing from cache, the entire block is brought into the cache and the access is restarted. Initially, the cache is empty. b) Compute the hit ratio for a program that loops 3 times from address 0 to 75 (base 10) in memory. For b, another example has been provided in regards to a previous problem: A direct-mapped cache consists of eight blocks. Main memory contains 4K blocks of eight words each. Access time for the cache is 22 ns and the time required to fill a cache slot from main memory is 300ns (this time will allow 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…
- instruction is in the first picture cacheSim.h #include<stdlib.h>#include<stdio.h>#define DRAM_SIZE 1048576typedef struct cb_struct {unsigned char data[16]; // One cache block is 16 bytes.u_int32_t tag;u_int32_t timeStamp; /// This is used to determine what to evict. You can update the timestamp using cycles.}cacheBlock;typedef struct access {int readWrite; // 0 for read, 1 for writeu_int32_t address;u_int32_t data; // If this is a read access, value here is 0}cacheAccess;// This is our dummy DRAM. You can initialize this in anyway you want to test.unsigned char * DRAM;cacheBlock L1_cache[2][2]; // Our 2-way, 64 byte cachecacheBlock L2_cache[4][4]; // Our 4-way, 256 byte cache// Trace points to a series of cache accesses.FILE *trace;long cycles;void init_DRAM();// This function print the content of the cache in the following format for an N-way cache with M Sets// Set 0 : CB1 | CB2 | CB 3 | ... | CB N// Set 1 : CB1 | CB2 | CB 3 | ... | CB N// ...// Set M-1 : CB1 | CB2 | CB…You play the role of the instructor of CPSC440. Your CPSC440 students need more practice on the cache memory concept. You, as a professor of the CPSC440 course, decide to solve one more cache memory access problem (in addition to two problems in our Assignment). The new problem is: You are given a direct-mapped cache of 4 blocks with four-word per block (a total of 16 words in the cache). The main memory size is 64 words. We have the following memory access sequence: Word1, Word 8, Word0, Word 17, Word 14, Word 62, Word 55, Word 25, Word 16, and Word 15. You need to write an essay that explains how you solve this problem. You also need to show your students the final cache content using the given table shown below.You also need to explain to your students the result of hit or miss for each word access.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?
- A cache is set up with a block size of 32 words. There are 64 blocks in cache and set up to be 4-way set associative. You have byte address 0x8923. Show the word address, block address, tag, and index Show each access being filled in with a note of hit or miss. You are given word address and the access are: 0xff, 0x08, 0x22, 0x00, 0x39, 0xF3, 0x07, 0xc0.The padding technique is effective to remove false sharing, but it requires consuming more memory in order to fill in the first cache line to enforce y to move to another cache line. If the memory consumption is a concern, what would you suggest as an alternative possible solution the programmer can do to avoid false sharing?In the event that a request cannot be fulfilled by the cache, the processor will submit a request to main memory while the write buffer delivers the relevant data block. What steps need to be taken here?