Consider a system which has 2 level paging.The page table is divided into 2K pages and each page is having 4K entries.Memory is word addressable and Physical address space is 64MW which is divided into 16K frames.Page table entry size is 2 words.
(A)Length of logical address
(B)Length of physical address
(C)$1^{st}$ level page table size
(D)one $2^{nd}$ level page table size
My Work
Page table of process is divided into 2K pages, means at second level we have 2K pages.So, this means we should have 2K entries at the first level page table and hence number of bits to index first level page table should be $p_1=log_2(2K)=11$
Each page has $4K$ entries and each such entry size is $2$ words.So, Page size must be $8KW$.Since, at second level, each page table is represented by a page, and each such page has $4K$ entries,so number of bits needed to index second level page table
$p_2=log_2(4K)=12$
Page Size=$8KW=2^{13}words \rightarrow d(page\,offset\,bits)=13$
Physical Address space=$64MW=2^{26}$
$f+d=26$
so $f=13$
That means maximum frames must be $2^{13}=8K$, but question mentions it's 16.
Have I done anything wrong or the question is poorly framed?