Suppose logical address space is **32 bit so total possible logical entries will be 2^32 and other hand suppose each page size is 4 byte then size of one page is *2^2*2^10=2^12...* now we know that no. of pages in page table is pages=total possible logical address entries/page size so pages=2^32/2^12 =2^20 Now suppose that each entry in page table takes 4 bytes then total size of page table in *physical memory will be=2^2*2^20=2^22=4mb***