Test by Bikram | Mock GATE | Test 2 | Question: 34

Question

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 ________ $bytes$.

Comments

I got answer 2^15 how 2^20?

Page size =frame size .

Here page size  = 2^32/2^12 = 2^20

No of Frames = 2^30/2^20 = 2^10

Total no of bits to identify the entry = 20+12 = 32 that means 5 bits.

So size of inverted table  = 2^10 * 2^5 = 2^15

what is the mistake here??

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How is the answer 2^20?According to me,

Page offset= 20 bits.

No of pages= 4KB=2^12, therefore 12 bits are required for the page number.

So,number of bits for each entry =12+12(overhead)=24bits=3 bytes.

Number of frames= 2^10

Hence ,size of page table= no entries*size of each entry= 2^10 * 3 bytes =3072 bytes.

Correct me please,where am i going wrong?

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look at my answer,i provided the explanantion

Answer 1

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

Comments

Doesn't 4KB pages means number of pages = 2^12 here?

cause this is what I took and did the further calculations based on this

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4Kb means,page size here

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If the queston is written in this way i.e. 4KB pages,then do we always consider this as page size?

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It would have been much simpler, if the wordings would have been page size is 4KB.