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Letamp;#39;s look at the solution to our flat page table size quiz. We are asked what the total size of the page tables in the system is. There are two processes in the system, so there will be two page tables. We need to add up their sizes. The page size is 4 kilobytes. The applications have a 32-bit virtual address space. So we have 2 processes times 2 to the 32nd bytes of address space divided by 2 to the 12 bytes in a page. This is how many pages we have times the size of an entry is 8 bytes, which gives us two times eight is 16 x 2 to the 20th. 2 to the 20th bytes is a megabyte, So we have 16 megabytes. Note that it didnamp;#39;t matter how much memory the processes are actually using. The flat page table needs to have an entry for every possible page in the virtual address space, not only for the pages we are actually using. Also, it doesnamp;#39;t matter how much physical memory we actually have. What matters is what the size of the entry is. So for example, this entry was de
