Suppose a computer using direct mapped cache has 2^24 bytes of byte- addressable main memory and a cache size of 64K bytes, and each cache block contains 32 bytes. (Note: 64K = 26 * 210) a) How many blocks of main memory are there? b) What is the format of a memory address as seen by cache, i.e., what are the sizes of the tag, block, and offset fields?

In the table below, to have 8 memory addresses where each memory address holds one byte of data (byte-addressable), we need 3 bits to index every address.  (check image)  If you have 64 GiB of RAM memory, then how many bits are needed to have memory  addresses for every byte-addressable memory location?

Given a memory address, say 12A1H, find the Block number(i) the address   belongs. Assume, the size of the Block is 8Bytes.   Also, find the starting address of the block to which the address belongs.   Find the capacity of RAM.   Find the number of addressable locations in RAM.   Find the total number of blocks RAM has.   Assume RAM is byte-addressable.

Problem 0. The following table gives some of the parameters for a number of different hardware caches. Fill in the table with the values of the missing parameters. Recall that m is the number of physical address bits, C is the cache size in bytes, B is the block size in bytes, E is the associativity, i.e., lines per set, S is the number of sets, t is the number of tag bits, s is the number of set index bits, and b is the number of block offset bits. Cache m 48 32030 (b) 48 32 24 24 C B E 16384 32 16 32768 64 65536 128 512 32 1024 8 8 4 14 1 4 S b S t

Direct Mapping Example: CPU is searching an Instruction stored at RAM address 1110011010 in cache and doesn't find it. What happens then? Tag 0000000000 ?? 0000000001 ?? Block:0, (j=0) 0000000010 Instruction-1 Line1 = 1 Tag Tag 0000000011 Instruction-2 0000000100 Instruction-3 Instruction-4 Block:1, G=1) Solution: 0000000101 Line 2 (= 2) Tag 0000000110 Instruction-5 0000000111 Instruction-6 0000001100 Instruction-7 0000001101 Instruction-8 Block:2 0000001110 Data-1 (j=2) 0000001111 Data-2 Line.3 (3) 0000011100 Data-3 0000011101 Data-4 Tag 0000011110 Data-5 0000011111 Data-6 Cache Size = 64B Size of a line = 4B Total number of Lines in Cache: m 16 RAM Size = 1KB %3D 1111111100 %3D Size of a Block= 4B Total number of Blocks, M = 256 j = 0, 1, 2, ..255 (M-1) Block 255. G =255) 1111111101 1111111110 i = 0, 1, 2, ... 15 (m-1) 1111111111

HELP WITH PART A,B, AND C

Consider the following table that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): ADDRESS OxFFFF OXOCOE OXOCOD Ox0C0C OXOCOB OXOCOA 0x0C09 0x0000 CONTENTS 1111 1111 1111 1111 1111 1110 1101 1100 0001 1011 1100 0101 0110 0101 1000 0111 1100 0000 0100 0000 0011 0001 0101 0010 0000 1100 0000 1101 0000 0000 0000 0000 The table above shows the addresses in hex (base 16) and the contents at the corresponding address in binary (base 2). A.) What are the contents in hex of the memory location at following address in binary: 0000 1100 0000 1110? (Enter hex like the following example: Ox2A3F)

Part 1: Preliminary Summarize the difference between binary machine language, assembly language, and high-level source files like C. Contrast the difference between a compiler, assembler, and linker What is the difference between the data segment and the text segment of in an assembly language program? What register is used as the stack pointer? What is a ebreak instruction in the sample program? How would you define a constant named 'BLUE' and assign it to the value OX00F in assembly language? Describe how to implement the psuedo-instruction li t1, e using a native instruction.

help with part A B AND C.

A memory system has 4 KB byte-addressable main memory and a direct-mapped cache that consists of 8 blocks with 16 bytes per block. The following shows the main memory address format that allows us to map addresses from main memory to cache. Note: 12 bit address, 4 bit offset, 3 bit $block, and 5 bit tag Assume the cache directory shown below: