Operating System

Question

Assume a demand paged memory management system with the following characteristics:

page size = 2^12 bytes

physical address space = 2^24 bytes

logical address space = 2^32 bytes

TLB size = 2^6 bytes

How many page table entries can fit in the TLB if each entry only contains the information needed for logical to physical translation?

A) 16

B)32

C) 22

D)24

Comments

is 16 answer ?

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Option A) 16 .. ?

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16 is the correct answer

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How.
Please explain sir

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@SHUBHAM SHASTRI

@Manas Mishra

page table entry contain only the number of bits require to identify a frame & some other bits

like catching disabled,modified,referenced etc.....in this question it should be 12 only..

why 12+20?

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@himgta here asked abt the information needed to convert from logical to physical adress , so page tabel entry will contain both page no as well as frame no ....... so that by seeing frame no we  will knw to which frame this page has been assigned to and by seeing page no we will fetch that page

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@Manas Mishra

Any reference to read it?

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https://stackoverflow.com/questions/39032348/translating-from-logical-to-physical-addresses

Answer 1

page size = 2^12

LAS = 2^32

PAS = 2^24

no of pages =LAS/page size = 2^ 20 , so 20 bits required to adreess a page

frame = PAS/page size = 2^24/2^12 = 2^12 , so 12 bits needed to adreess frame

page table entry will contain page and frame  so total bits in PTE = 12+20 = 32 bits

TLB size = 2^6 . so total no of PTE that can fit in it = 2^6/(32/8)    {as 32 is in bits so divide by 8 to convert to bytes }

= 16

Comments

First calculate the number of pages in the system. From that we will get PTE.Round this off to next byte.

Second divide  TLB size./ PTE

if i am wrong .please correct me.

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https://stackoverflow.com/questions/39032348/translating-from-logical-to-physical-addresses

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Here TLB is given and not Page table so TLb page table entry has page no + frame no.

Answer 2

yeah i too got confused but after studying about TLB i got cleared.

TLB stores the recently or frequently accessed frame number(adress-base+number).

TLB check whether it is hit or miss we need tag bits

TLB entry =tag bits + frame no bits

here tag bits is page no bits

therefore TLB entry = page no bits + frame no bits

pagenumber =32(PSE)-12(offset)=20

frame number=24-12=12

tlb entry contains 32bits=4bytes

there fore no of enteries in TLB is 2^6/4=16

Answer 3

LA = 2³²

PA =2²⁴  Page Size = 2¹²

Number of pages = 2³² / 2¹² = 20 bits willbe used to address pages

Number of frame =  2²⁴ / 2¹² =12 bits will be used to address number of frames

Since TLB is used so its Page Table Entry = page no + Frame no

                                                                         = 20+12 = 32 bits = 4 Bytes

Page Table size = No of entries * Page Table Entry size

2⁶ Bytes = No of entries * 4 B

No of Entries = 16