. If the size of the MAR is 20 bits, it can address __________________ unique memory...
. If the size of the MAR is 20 bits, it can address __________________ unique memory locations
Homework Answers
If the size of the MAR is 20 bits, it can address 
unique memory locations
=
1048576
1048576
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. If the size of the MAR is 20 bits, it can address
__________________ unique memory...
1. For a 512 k × 32 memory system, how many unique address locations are there?...
1. For a 512 k × 32 memory system, how many unique address locations are there? Give the exact number. 2. For a 512 k × 32 memory system, what is the capacity in bits? 3. For a 512 k × 32 memory system, how wide does the incoming address bus need to be in order to access every unique address location?
The physical address of memory on a machine is 32 bits. The machine has a direct...
The physical address of memory on a machine is 32 bits. The machine has a direct mapped cache of size 512 KB with a block size of 8 bytes. What is the size of the tag field in bits? Can you please explain this step by step and what the importance of direct mapped is?
2. A computer uses a memory with addresses of 8 bits. (What's the size of the...
2. A computer uses a memory with addresses of 8 bits. (What's the size of the MM?) This computer has a 16-byte cache with 4 bytes per block. (How many blocks in the cache?) The computer accesses a number of memory locations throughout the course of running a program. Suppose this computer uses direct-mapped cache. a. What's the format of a memory address as seen by the cache ? Tag ? bits Block ? bits Offset ? bits b. The...
Memory Sizing NOTE: K (kilo) means 1024, not 1000. A byte (B) is 8 bits. A kilobyte...
Memory Sizing NOTE: K (kilo) means 1024, not 1000. A byte (B) is 8 bits. A kilobyte (KB) is therefore 8 x 1024 = 8192 bits. a) A 32 KB (kilobytes) memory has a 16 bit wordsize. How many words total can be stored in this memory? _________words b) A 256 KB memory has a 32 bit wordsize. How many bits are required to address this memory? _________ bits c) A computer memory has a 128 bit wordsize. It is made up...
How many address lines are needed for memory with 512 locations? How many bits are present...
How many address lines are needed for memory with 512 locations? How many bits are present in 4k x 8 SARM IC?
11. In a paged virtual memory system, can the computer 's physical memory address space be...
11. In a paged virtual memory system, can the computer 's physical memory address space be larger than a process's virtual memory address space? Explain your answer. (Note: A computer 's "physical memory address space" is the number of bits for the frame number plus the number of bits foir the offset. A process's "virtual memory address space" is the number of bits for a page number plus the number of bits for the offset.)
A primary memory system consists of 128 address bits. Each block within the cache is 256...
A primary memory system consists of 128 address bits. Each block within the cache is 256 bytes. The total cache size is 131,072 bytes. What are the sizes of the tag and index in the cache? First, find the number of blocks. 128,000 bytes # Blocks = 256 bytes per block bort = 512 blocks Where does the 128,000 come from?? Now find the number of bits required to distinguish all the blocks. This is the index aka set size....
A digital computer has a memory unit with 24 bits per word.
A digital computer has a memory unit with 24 bits per word. the instructions set consists of 150 different operations. all instructions have an operation codepart(opcode) and an address part (allowing for only one address) Each instruction is stored in one word of memory.a. how many bits are needed for the opcodeb. how many bits are left for the address part of the instruction.c. What is the maximum allowable size of memory.d. what is the largest unsigned binary number that...
A digital computer has a memory unit with 24 bits per word. The instruction set consists...
A digital computer has a memory unit with 24 bits per word. The instruction set consists of 199 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. [] How many bits are needed for the opcode? How many bits are left for the address part of the instruction? What is the maximum allowable size for memory? What is the largest signed...
A digital computer has a memory unit with 24 bits per word. The instruction set consists...
A digital computer has a memory unit with 24 bits per word. The instruction set consists of 110 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. (10 points) How many bits are needed for the opcode? How many bits are left for the address part of the instruction? What is the maximum allowable size for memory? What is the largest...
2. A computer uses a memory with addresses of 8 bits. (What's the size of the MM?) This computer has a 16-byte cache with 4 bytes per block. (How many blocks in the cache?) The computer accesses a number of memory locations throughout the course of running a program. Suppose this computer uses direct-mapped cache. a. What's the format of a memory address as seen by the cache ? Tag ? bits Block ? bits Offset ? bits b. The...
11. In a paged virtual memory system, can the computer 's physical memory address space be larger than a process's virtual memory address space? Explain your answer. (Note: A computer 's "physical memory address space" is the number of bits for the frame number plus the number of bits foir the offset. A process's "virtual memory address space" is the number of bits for a page number plus the number of bits for the offset.)
A primary memory system consists of 128 address bits. Each block within the cache is 256 bytes. The total cache size is 131,072 bytes. What are the sizes of the tag and index in the cache? First, find the number of blocks. 128,000 bytes # Blocks = 256 bytes per block bort = 512 blocks Where does the 128,000 come from?? Now find the number of bits required to distinguish all the blocks. This is the index aka set size....
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