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

Shouldn't $24GB$ be answer?

$2^{32}*48$bit = $24GB$

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They assume page table entry size same evrey where, which is 16 bits .

so 2³² * 16 bits = 8 GB

Not sure.

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i think 24GB should be rt..

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yes, no information to determine the PTE size of page table 1.

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Answer given $ = 32GB$

Explaination : $2^{32}*64$bit $= 32GB$

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ACE test series I thought they will even use 100 bits

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There was one more question $\Rightarrow$

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(B) is sure shot answer, but (D) is also invalid as base add for $S_{2}$ is $40950$

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@Manish, how did you approach demand paging question ?

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page fault rate $= \frac{200}{1000} = 0.2$

$T_{avg} = 0.8*0.02 + 0.2*2 = 0.416$ (for $1$ instruction, so for $1000$ instuctions multiplied by $1000$.

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They meant "effective average memory access time"

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And i calculated average total time for $1000$ instructions. my bad! i should have thought that "average access time for system" will remain between $\color{olive}{0.02 - 2}$ and a rough idea about which side can be guessed by Page hit ratio. $0.8$

Thanks get it.

What about $3^{rd}$ question?

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Not really- unless they put the word "average" the question is ambiguous.

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Manish, i think you should go for virtual gate and made easy.

Ace questions are very faulty .

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@Kapilp giving those too.

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@Kapilp just gave a virtual gate test. See this $\Rightarrow$

my ans : set = 10, tag = 15, offset = 6

Clearly, solving equation we get

$ 1 =  10999p\Rightarrow p= 0.00009091735$

Don't get how they did this ?

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First 2 questions you are right. But for first 256 K cache means size is 256 KB?

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Questions are right, but iam taking about answer.

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I meant your answer only

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@Arjun sir, 2nd question 0.01 is right .

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yes, correct to 2 decimal place if we round, we get 0.01%.

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@Arjun sir, in $2^{nd}$ question nothing is mentioned about rounding, and on solving equation we get page fault ratio to be $0.00009091735$

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yes, but it says 2 decimal places and is in %. So, 0.00-0.01 should get marks.

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Thanks sir, what about $3^{rd}$ question.

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No idea.

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@mcjoshi 11 is rt..

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@Manish, u can use this 

http://cs.stackexchange.com/questions/19911/exponential-averaging-for-sjf-cpu-scheduling

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T2=0.5*10 + 0.5*8 = 9

T3=0.5*9 + 0.5 *7 = 8

T4=0.5*8+0.5*4=6

T5=0.5*6+0.5*16=11

by this way getting 11. correct me!

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@Kapilp the link does this :

Suppose even if it is considered that at a particular time instant, we know bust time of all processes till now. Why $16$ is not considered here (nothing is mentioned is question that we are required to predict time for last process)

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@Arjun sir, @Gabbar @Kapilp @habibkhan

One more question(virtual gate). Nothing is mentioned about semaphores(binary or counting), not initialized

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Manish since they are make many quetion , there may be chance that some quetion they think correct but not clear to all. yes this is one of them . ignore that .

BY looking at option u may guess its binary semaphore with s = 1 and u = 0.

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@Anirudh, i too tried a lot. but if $s = 1$, $C$ should be printed only once as we are not performing UP on $s$.

As, $C$ is printed more than once, so they are using counting semaphore and In this case we need to know their values too.

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yes .....

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@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 .

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

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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.

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@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.

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@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