Given a 16 - bit logical address space, with a page size of 1KByte (2^10), and a physical memory size of 32KByte (2^15).
What are the number of entries in the page table? 16-10 = 6, 2^6 entries
What size (in Bytes) is a page table entry? I think the answer is 15 bits? but it's asking for bytes so maybe this is wrong?
NoPages = locations/pagesize = (2^16)/(2^10) = 2^6 so there are 2^6 entries in the page table
NoFrames = physicalMemory/pagesize = (2^15)/(2^10) = 2^5 so there are 2^5 frames each entry in the page table is 5 bits long therefore the size of an entry is 2^5 = 32 bytes