Consider a system with paging-based memory management, whose architecture allows for a 4GB virtual address space for processes. The size of logical pages and physical frames is 4KB. The system has 8GB of physical RAM. The system allows a maximum of 1K processes to run concurrently. Assuming the OS uses multi-level paging, calculate the maximum memory space required to store the page tables of all processes in the system? Assume that each page table entry requires an additional 10 bits (beyond the frame number) to store various flags. Assume page table entries are rounded up to the nearest byte. Consider the memory required for both outer and inner page tables in your calculations.
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We have a paging system with 4 gigabyte virtual address space per process, 4 kilobyte pages, 8 gigabyte physical RAM, and up to 1024 concurrent processes. First, let's calculate how many pages each process has. With 4 gigabyte virtual space and 4 kilobyte pages, each process has 2 to the power of 20 pages, which is about 1 million pages.
Next, we calculate the size of each page table entry. With 8 gigabyte physical RAM and 4 kilobyte frames, we have 2 to the power of 21 frames, requiring 21 bits for the frame number. Adding 10 flag bits gives us 31 total bits per entry. Rounding up to the nearest byte, each page table entry is 4 bytes.
Now we determine the page table structure. With 4 kilobyte pages and 4-byte entries, each page can hold 1024 page table entries, requiring 10 bits for indexing. Since we have 20 bits for page numbers, we can split this into two 10-bit indices, creating a two-level page table structure. Level 1 has 1024 entries pointing to Level 2 tables, and each Level 2 table has 1024 entries pointing to physical frames.
Now let's calculate the maximum memory needed for a single process's page tables. The Level 1 table has 1024 entries, taking 4096 bytes or 1 page. To address all 2 to the power of 20 pages, we need 1024 Level 2 tables, each containing 1024 entries and taking 1 page. So the total is 1 page for Level 1 plus 1024 pages for Level 2, giving us 1025 pages per process, which is about 4.1 megabytes.
Finally, we calculate the total memory for all processes. With 1024 concurrent processes, each requiring 1025 pages, we need a total of 1,049,600 pages. Converting to bytes and then to gigabytes, we get 4,291,506,176 bytes, which equals exactly 4 gigabytes. Therefore, the maximum memory space required to store all page tables in the system is 4 gigabytes.