PW Testseries

Question

Answer 1

TLB size: 32KB

Page table entry size: 4 bytes

 

Number of entries in the TLB: 32KB / 4 bytes =8K = 8192 entries

Set associativity: 8-way so Number of sets: 8192 entries / 8 ways = 1024 sets

Bits required for set index= 10 bit

Page size: 2KB = 2^11 bytes so Bits required for page offset: 11 bits

Minimum tag length: 17 bits 

Virtual address size = 17 bits + 10 bits + 11 bits = 38 bits

 

Therefore, the virtual address space is 256 GB.

Comments

Yes it's correct.

Answer 2

Page size = 2 KB = 2^11 B

TLB = 32 KB = 2^15 B

TLB = 8-way set associative

Each page table entry = 4 B = 2^2 B

TLB tag = 17 bits 
 

Page size = 2 KB = 2^11 B , 

 ----->  So  Page Offset = 11 bits

 

 2^2B -------------------->  1 entry

 2^15B---------------------- No. of  entries =  2^15 / 2^2  = 2^13 entries  

 

2^3 entries(8 - way set Asso.) --------------------> 1 set

2^13 entries -----------------------------------> No. of sets = 2^13 / 2^3 = 2^10 sets

--------> so , Set index = 10 bits

--------> TLB tag = 17 bits 

 

Tag = 17 bits	Set index = 10 bits	page offset = 11 bits

 

Virtual address size = 17 bits + 10 bits + 11 bits = 38 bits

                                    =2^38 / 2^30 = 2^8GB = 256 GB

 

 

Answer 3

0.5 GB

We will have

tag = 17 bits

page offfset = log[2kB] = 11 bits

Set offset = log[cache size / page size * set associativity] = log[32KB/ 2KB * 8] = 1 bit

total = 17 + 1 + 11 = 29 bits

VAS = 2^29 ~ 0.5 * 2^30 = 0.5 GB