Hello, I am doing some exercices to understand how the virtual memory and paging works, my question is as follows :

Suppose we use a paged memory with pages of 1024 bytes, the virtual address space is of 8 pages but the physical memory can only contain 4 frames of pages. Replacement policy is LRU.

1. What is the physical address in main memory that corresponds to virtual address 4096? and how do you get to that result?
2. Same thing as question 1 but with virtual address 1024
3. A page fault occurs when accessing a word in page 0, which page frame will be used to receive the virtual page 0?

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