About Operating System, about page-table entries status bits

Question

In the movie The Social Network, when Mark Zuckberg was in class, the teacher asked this question:

Suppose we're given a computer, with a 16-bit virtual address, and a page size of 256-bytes,the system uses one-level page tables that start at address hex 400, may be you want DMA (Direct Memory Access) on your 16-bit system. Who knows? The first pages are reserved for hardware flags, etc. Assume page-table entries have eight status bits. The eight status bits would then be ...

Mark Zuckberg answered:

One valid bit, one modified bit, one reference bit and five permission bits.

How did he get this?

Answer 1

Intel nomenclature in parentheses. The 'valid' (present), 'modified' (dirty) and 'reference' (accessed) bits are the minimum set of bits you need for a demand paging manager and MMU. The 'valid' (present) bit is used by the MMU to know whether the page is mapped to a valid physical address.

The 'modified' (dirty) bit is used by the demand paging manager to determine if the page being evicted needs to be written to backing media. As accessing backing media can be considered an expensive operation, you really want to keep this to a minimum--especially when writing to it as that is generally slower than reading from it.

The 'reference' (accessed) bit is useful to the demand paging manager to figure out how to age the pages it controls. You don't want to evict the most frequently used pages as that would require saving and/or loading them repeatedly from backing store (which has already been stated as SLOW).

The remaining five bits are gravy. They are free to use as permission and/or option bits. For example, can the page be accessed by supervisor and/or user threads? Is the page available for write, or is it read-only? What is the caching strategy to be used on the page?

Hope this helps.

Sparky

Answer 2

http://chomaloma.blogspot.com.au/2011/02/social-network-inaccuracies-regarding.html

That does explain it a little