Answer the following questions with the information provided: Suppose we have a

Question

Answer the following questions with the information provided:

Suppose we have a system with a page size of 1K bytes and a 4K virtual address space. The physical memory size is 16K.

A single load instruction is stored at virtual address 2048 (2K):

load 1024, r1

First, list the virtual addresses generated by the execution of this instruction.

How many bits are needed to encode each virtual address?

Show how to divide the virtual address into a virtual page number and offset.

How many bits are needed to encode each physical address?

5. Suppose we have a system with a page size of 1K bytes and a 4K virtual address space. The physical memory size is 16K A single load instrootion is stored at virtual address 2018 (2K load 1024, r1 First, list the virtual addresses generated by the execution ofthis instruction. How many bits are needed to encode each virtual addresa? Show how to divide the virtual address into a virtualpage number and offset How many bits are needed to encode cach physical address?

Explanation / Answer

Answer:

4K virtual address space means 12 bits for virtual address in which 10 bits for page offset. Right? & 14 bits to encode physical memory.

We know that Virtual memory bits = Page number +page offset bits

And Physical memory bits = frame number+page offset bits

Virtual space = 4kB = 2^2 * 2^10B = 2^12B

Thus 12 bits are needed to encode each virtual address.

Physical address space = 16KB = 2^4 *2^10B = 2^14

Thus 14 bits are needed to encode each physical address.

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