ACE test series(OS):

Question

Consider a $64$bit processor. A process has a $64$ bit logical address space with two-level paging. The outer table is indexed using 32-bits and the inner table using $16$bits. What is the size of outer page table size ?

a). $4GB$                              b). $8GB$

c). $32MB$                            d). $32GB$

Comments

@Manish, you can try that question like this ( Actually i solve these type of questions like this only ) => 

T₁ (Expected) = T₁ (Given) + T₀ (Expected) = 0.5 * 8 + 0.5 * 10 = 9

T₂ (Expected) = T₂ (Given) + T₁ (Expected) = 0.5 * 7 + 0.5 * 9 = 8

T₃ (Expected) = T₃ (Given) + T₂ (Expected) = 0.5 * 4 + 0.5 * 8 = 6

T₄ (Expected) = T₄ (Given) + T₃ (Expected) = 0.5 * 16 + 0.5 * 6 = 11 

PS : It would be better if u post questions in Q/A section , not in the comments .

---

Thanks @Kapilp, i too got answer as $11$(same way)

---

I think answer should be 4GB

Answer 1

32 bits for indexing outer page table.
So no of outer page entry=2^32.

Now among 64 bit virtual address 32 bit for outer page table index,16 bit for inner page table index..and rest 16 bit offset.

So page size=frame size=2^16 .

So among 64 bit physical address (64-16)=48 bit for frame no/indexing frame.

Now outer page table will contain address of inner page table i.e particular frame address.

So each outer page table entry size=48 bit.

So,Outer page table size=2^32*48=24GB.(No Option Matching)

Comments

Yes..that i assume as nothing specified about it.So we can consider that physical and logical both are of 64 bit.

---

@Kishalay. The bits for indexing only tell you about the offset within the page table at that level. Here, we are not able to infer anything about physical address.

---

@arjun sir 32bit processers have RAM limit of 4GB, by that logic 64bit processor should have 64bit physical address if nothing given ?

Answer 2

given local address of 64 bit (32 bit outer page table+16 bit inner page table+16 bit offset)

in an  outer page each entry takes 16 bit(takes 2 byte)

so size 2^32 *2 byte=8GB