edited by
826 views
2 votes
2 votes
Suppose that a certain computer with paged virtual memory has $4 KB$ pages, a $32-bit$ byte addressable virtual address space, and a $30-bit$ byte-addressable physical address space.

The system manages an inverted page table, where each entry includes the page number plus $12$ overhead bits (such as flags and identifiers). Then the size of the basic inverted page table, including page numbers and overhead bits is ________ $bytes$.
edited by

1 Answer

Best answer
3 votes
3 votes

Concept of Inverted Page table is the opposite opposite of normal page table. as it stores one entry for each adress of physical memory(in case of Normal paging it is doen for Virtual Memory)

Lets say what happens for a Normal PT

# of entries=VAS/Page Size

but in case of inverted instead of virtual adress, Physical adress is taken in consideration so here

# of PT entry=PAS/PS=230/212=218

and in normal PT, Page table entry size is=(bit in PAS- Bit in page size)

But in Inverted,just the opposite,so PTE=(bit in VAS-Bit in Page size=(32-12)=20 bits

total pte=20 bit+12 overhead bit=32 bit=4byte

so PT size =# of entry*PTE

=218*4 byte

=220 byte

selected by
Answer:

Related questions

4 votes
4 votes
2 answers
2
Bikram asked Jan 24, 2017
469 views
The value of the expression $1 - 6 + 2 - 7 + 3 - 8 +$……… to $100$ terms is __________.
5 votes
5 votes
2 answers
4
Bikram asked Jan 24, 2017
803 views
$HCF$ of two numbers is $23$ and other two factors of their $LCM$ are $13 $&$ 14$. The largest of two numbers is _________