Consider a logical address space of 1K pages with a 4KB page size, mapped onto a physical memory of 1. How many bits are required in the logical address? Logical bits- 2. How many bits are required in the physical address? Physical bis=

Given a virtual memory of size 4 GiB, physical memory of size 1 GiB, and page size equal to 256 KiB.  How many bits are used to specify a virtual page number?

Assume a logical address space consists of 4 pages of 4096 words each and is mapped to a physical memory space of 256 frames. What is the size of the physical address space? 12 bits 14 bits 20 bits 8 bits

16-bit virtual addresses

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 0x2D map if the computer uses direct mapping? Blocks are numbered starting at 0

I don't need a plagiarised response. Send it original.

Suppose a computer system uses 16 but addresses for both virtual and physical address. In addition, assume each page (and frame) has a size of 256 bytes. -How many bits are used for page number? -How many bits for offset? -what is the max number of pages a process can have?