Question

Page Table is divided into 8K pages and each page contain 2k entries. Page table entry size is 4 bytes and page size is 1 K words

Q find out

a)Logical address

b)no of pages in LAS

c)1st level page table size

d)2nd level page table size

Comments

Didn't notice page size is given in "word", so  here is updated answer
Since it's given page table is divided into 8K pages each page contain 2k entries, size of each entries will be 4 bytes, so page size will be 2k * 4 bytes = 8 KB. So we can conclude here 1 word = 8 Bytes
a) 13 (1st level page table) + 11 (2nd level page table offset) + 13 (page size) = 37 bits
b) No of pages in LAS =$\frac{2^{37}}{2^{10}}$ = 2²⁷
c) 8 k * 4 bytes = 32 KB
d) 8 k * 2k * 4 bytes = 64 MB

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@stblue

page size = 2¹⁰ bytes

inner page table size = 2¹¹*4 bytes = 2¹³ bytes, shouldn't be inner page table fit in one page!! it is occupying more than one page.

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@joshi_nitish page size should be 2¹³ Bytes, even i didn't notice its given in words, when i first read the question. Yes logically we need one more page table, for proper addressing. Right now inner and outer page table occupies  more than one page . 

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@stblue

i think it should be like 

2  (1st)

11 (2nd)

11 (3rd)

13 (offset)

1st page table need 2 index bits and second page table needs 11 bits, because it is nowhere mentioned that only 2 page tables are present, moreover taking only 2 page tables with above configuration is invalid.

Answer 1

page size is given in words , it first has to be converted to bytes right?