GATE CSE 2015 | Set 2 | Question: 25

Question

A computer system implements a $40\;\text{-bit}$ virtual address, page size of $8\;\text{kilobytes}$, and a $128\text{-entry}$ translation look-aside buffer $\text{(TLB)}$ organized into $32$ sets each having $4$ ways. Assume that the $\text{TLB}$ tag does not store any process id. The minimum length of the $\text{TLB}$ tag in bits is ______.

Comments

Number of virtual pages: Page number bits = 27 → total pages = 2²⁷

Finding the set index bits:

• There are 32 sets → needs 5 bits (since 2⁵ = 32).
• Each page maps to a unique set, based on these 5 bits.

How to calculate the tag bits:

So, the tag bits come from the remaining page number bits after removing the set index bits.

Hence: Tag bits=Page number bits−Set index bits=27−5=22

• In set-associative mapping, each page of virtual memory maps to one specific set.

• Inside that set, there are multiple blocks (or lines) where it could be placed.

• To identify which page currently occupies a line inside the set, we use the tag.

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TLB stores some of the enteries of page table which is it acts like cache for the page table.
what does page table store? for every virtual page number it stores corresponding frame number .
how many total pages do we have here? 2^40 / 2^13 = 2^27 which means 2^27 enteries are there in page table now they need to be cached in TLB.

Here TLB se set associative with a total of 32 sets .
But what does tag bits tell ? if 2  different entries map to the same set then there set index are same but they are diffrenetiated with the help of tag bits. So if I have x number of tag bits then it means 2^x different entries map to the same set of TLB . 

Since there are 32 sets for 2^27 enteries therefore 2^27/2^5 = 2^22 which means 2^22 enteries map to the same set of TLB so we need 22 bits to differentiate those page table enteries hence 
ANS= 22 bits 

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Virtual address is divided into VPN and offset. TLB is 128 entries, organized into 32 sets each having 4 ways (32*4=128). So it's a set-associative TLB. The TLB tag is used to identify which virtual page maps to a TLB entry.

Virtual address = 40 bits

page size 8KB => offset 13 bits

VPN = 27 bits.

The TLB is indexed by some bits from the VPN.

Since it has 32 sets, TLB has 32 sets, so index = 5 bits.

So the VPN is divided into: tag (remaining bits) and set index (5 bits).
tag = 27-5 = 22 bits.

VPN (27 bits) = [ Tag: 22 bits | Set Index: 5 bits ]

So answer 22.

Answer 1

page size is 8kb

d=13 bit

p=27 bit

tlb set not 128/4=2^5

 

set no =5 bit

 

tag bit =40-5-13=22 bit

Comments

easiest explanation

Answer 2

8-9 this type of question asked in gate exam

Answer 3

Comments

easy explaination