Consider a three-level page table to translate a 39-bit virtual address to a physical address as shown below. The page size is 4KB (1KB = $2^{10}$ bytes) and page table entry size at every level is 8 bytes. A process $P$ is currently using 2GB (1GB = $2^{30}$ bytes) virtual memory which is mapped to 2GB of physical memory. The minimum amount of memory required for the page table of $P$ across all levels is ________ KB.

To find the minimum memory required for the page tables across all levels, we assume that the process's 2GB virtual memory is allocated contiguously.

1. Basic Parameters:

   • Page Size $= 4\text{ KB} = 2^{12}\text{ bytes}$

   • Page Table Entry (PTE) size $= 8\text{ bytes} = 2^3\text{ bytes}$

   • Entries per page table page $= \frac{4\text{ KB}}{8\text{ B}} = 2^9 = 512\text{ entries}$

   • Each page table page at any level takes $4\text{ KB}$ of memory.

2. Number of Pages Required by Process $P$:

   • Process virtual memory size $= 2\text{ GB} = 2^{31}\text{ bytes}$

   • Total number of pages $= \frac{2^{31}\text{ bytes}}{2^{12}\text{ bytes}} = 2^{19}\text{ pages}$

3. Page Table Hierarchy Calculations:

   • Level 3 Page Tables: Each Level 3 table can map $2^9 = 512$ pages.

     $$\text{Number of Level 3 tables} = \left\lceil \frac{2^{19}}{2^9} \right\rceil = 2^{10} = 1024\text{ tables}$$

   • Level 2 Page Tables: Each Level 2 table can point to $2^9 = 512$ Level 3 tables.

     $$\text{Number of Level 2 tables} = \left\lceil \frac{1024}{512} \right\rceil = \left\lceil \frac{2^{10}}{2^9} \right\rceil = 2\text{ tables}$$

   • Level 1 Page Table (Root): There is always $1$ root page table at Level 1, which points to the Level 2 tables.

     $$\text{Number of Level 1 tables} = 1\text{ table}$$

4. Total Memory Required:

   • Total page table pages $= 1024 + 2 + 1 = 1027\text{ pages}$

   • Total memory $= 1027 \times 4\text{ KB} = 4108\text{ KB}$

Thus, the minimum amount of memory required is 4108 KB.