Explanation:

An inverted page table is a data structure used by the operating system to keep track of which virtual addresses are mapped to which physical addresses. In an inverted page table, instead of each process having its own page table, there is a single table that maps physical addresses to virtual addresses.

Given:

- 64 bit virtual addresses
- 4 KB page size
- 256 MB of RAM

To calculate the number of entries required for an inverted page table, we need to consider the following:

1. Virtual address space:

- 64 bit virtual addresses allow for a maximum of 2^64 possible addresses.
- Since the page size is 4 KB, each page can hold 2^12 bytes of memory.
- Therefore, the virtual address space can be divided into 2^64 / 2^12 = 2^52 pages.

2. Physical address space:

- There is 256 MB of RAM available.
- Since the page size is 4 KB, the physical address space can be divided into 256 MB / 4 KB = 2^18 pages.

3. Inverted page table:

- In an inverted page table, each entry corresponds to a physical page frame.
- Since there are 2^18 physical page frames available, the inverted page table requires 2^18 entries.

4. Hash table:

- Since the number of virtual pages (2^52) is much larger than the number of physical pages (2^18), a hash table is used to map virtual pages to physical page frames.
- The size of the hash table depends on the number of entries in the inverted page table.
- A commonly used formula for the size of the hash table is:

size of hash table = (number of entries in inverted page table) / 0.7

- Using this formula, the size of the hash table for our case is:

size of hash table = 2^18 / 0.7 = 245760.87

- Since the size of a hash table must be a power of 2, the next larger power of 2 is 2^18.

Therefore, the inverted page table requires 2^18 entries, or 262144 (216) entries.