Inverted Page Table

Question

Consider a machine with 64 MB physical memory and 34 bit virtual address space.If the page size is 4 KB ,the approximate size of conventional and inverted page table will be??

Answer 1

for Conventional Page table ->                                      

no. of pages = virtual Memory size / page size               

no. of pages = 2$^{34}$  B /   2$^{12}$  B                                   ( assuming Memory is Byte Addressable)

                    =  2$^{22}$  

Page table size  = no. of pages * PTE size 

                         =  2$^{22}$  *  2 B => 8 MB                    ( becoz frame no. bits = 14 bits ~ 2 Bytes)

for Inverted  Page table ->   

no. of frames = Physical memory size / page size 

                      =   2$^{26}$  B /   2$^{12}$  B   

                      =  2$^{14}$ 

Page table size = no. of  frames * PTE size

                            = 2$^{14}$  *  3 B                                   ( becoz page no bits = 22 bits ~ 3 Bytes)

                           = 48 KB

Answer 2

for conventional table 
no of page table enteries =2³²  / 4KB
                                         =2²⁰ 
                      frame bits =14 bits taking approximate as 16 bits means 2 bytes so 2²⁰ *2= 2MB

for inverted page table 
we have enteries as no of frame so no frame = PAS/ page size 
                                                                   =2²⁶ /4KB
                                                                    =2¹⁴ *(PTE)

^{but for calculating inverted page table size there should be size page table entry size(PTE) as in  conventinal page table we know that each entry contains the corresponding frame no !!} 
 

                                                 

Comments

are there any standard PTE...if such questions come in gate?

 

 

And Correct it is not 32 it is 34....

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Virtual address space is 34 bit right

So no of entries=2^34/2^12=2^22