If a physical address is 32 bits and 10 bits are needed for

If a physical address is 32 bits and 10 bits are needed for the page offset, how many bits are in the physical page number? 10 22 32 If a virtual address is 32 bits and 10 bits are needed for the page offset, how many entries are there in the page table? 32 2^32 2^22 22 Even if I did confuse this on the video at least once, what is the true name of the cache that holds page table entries? table-lookahead buffer translation lookup buffer translation-lookaside buffer table-lookaside buffer

Solution

1.

physical page Number bits = Physical Address - Offset .

32-10 = 22

option 2 is correct choice.

2.

entries in page table = 2³²/2¹⁰ = 2^22.

option 3 is correct choice.

3.

table look ahead bigger :

It is more useful translations from virtual to physical addresses done very fastly via caching.

it is efficiently stores page table address.

Option 1 is correct choice.