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- instruction is in the first picture please give me only implementation of int L1lookup(u_int32_t address) and int L2lookup(u_int32_t address) 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…Code a descriptor that describes a data memory segment that grows upward and begins at location 03000000H and ends at location 05FFFFFFH and can be written. The memory has not been accessed, and can be accessed with the lowest privilege level. Assume that the segment is available (i.e., AV=1) and that the instructions are 32 bits (i.e., D=1).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. c) Compute the effective access time for this program. Show me how to solve using this equation: EAT = H x AccessC + (1 – H) x AccessMM where H is the cache hit rate and AccessC and AccessMM are the access times for cache and main memory, respectively. thanks
- Computer Science Consider a direct-mapped cache with 8 lines, each holding 16 bytes of data. The cache is byte-addressable and the main memory consists of 64 KB, which is also byte-addressable. Assume that a program reads 16KB of memory sequentially. Answer the following questions:a) How many bits are required for the tag, index, and offset fields of a cache address?b) What is the cache size in bytes?c) What is the block size in bytes?d) What is the total number of blocks in main memory?e) How many cache hits and misses will occur for the program, assuming that the cache is initially empty?f) What is the hit ratio?g) Give an example virtual address (in BINARY) that will be placed in cache line 5.consider someone is using direct-mapped cache memory, each external memory address maps to what? a specific cache memory location one of many different cache memory locations a unique cache memory location, not shared with other external addresses a randomly-selected cache memory location(MAC) In the MAC address table, MAC addresses are stored in a dynamic RAM. Assume the table was 8000 bytes in size and the memory access time was 100 nanoseconds. What would be the worst-case look-up time if a binary search was employed to locate an address?
- Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. Could you hand draw the page table, if possible a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page tableCreate a program in C++ which simulates a direct cache. The memory array that contains the data to becached is byte addressable and can contain 256 single byte entries or lines. The cache has only 8 entriesor lines. The Data field in each line of the cache is 8 bits. Since the data stored in each line of the cache isonly 8 bits, there is no need for a line field. Only a tag field is needed which is log2(256) = 8 bits.The memory array can be filled with any values of your choice. The program should work by taking userinput of a memory address (index). This input represents the memory data that should be cached.Check the cache to see if the item is already cached. If it is not, your program should counta cache miss, and then replace the item currently in the cache with the data from the inputted address.Allow the user to input addresses (in a loop), until they so choose to end the program. The program should output the number of cache misses upon ending.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?
- Code a descriptor that describes a memory segment that begins at location 0005CF00h and ends at location 00060EFFh. The memory segment is a data segment that grows upward in the memory system and can be written. The segment has a user level privilege (lowest) and has not been accessed. The descriptor is for an 80386 microprocessor.A direct-mapped cache consists of 16 blocks. A byte-addressable main memory contains 4K blocks of 16 bytes each. Access time for the cache is 30 ns and the time required to fill a cache slot from main memory is 250 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. a) Show the main memory address format that allows us to map addresses from main memory to cache. Be sure to include the fields as well as their sizes. b) Compute the hit ratio for a program that loops 4 times from locations 0 to 42 (base 10) in memory. c) Compute the effective access time for this program.Memory Hierarchy and Cache Suppose that we have a computer that uses a memory address of 12-bits. This computer has a 64-byte cache with 16 bytes per frame. The computer accesses a number of memory locations throughout the course of a running program. Suppose this computer uses direct-mapped cache. The system accesses the following memory addresses (given in hex) in this exact order: F2E, A17, 2E0, 44E, 34F, 341, B50, B58 a. What is the hit ratio for the memory reference sequence given above? b. Show the content of each cache frame following each memory reference (frame content to be shown as tag +frame index) c.If we keep the same cache size and the same frame size but switch to a 2-way set associative cache mapping scheme. Given the memory address reference 555, indicate where we would look in the cache to find this data. Indicate which fields will be used to find the exact location?