Problem 2

Subproblem 2A

physical address bits = log2(8M) = 23

Subproblem 2B

page size is 4kB and log2(4K)
there are 32 address bits, so max number of address is 2^32 or 4G
max number of virutal pages is 4G/4k or 1M

Subproblem 2C

Number of physical pages is 8MB/4kB or 2k

Subproblem 2D

bits of virtual page numbers = 32 - log(4k) = 20
bits of physical page numbers = 23 - log(4k) = 11

Subproblem 2E

Number of virtual pages to each physical page = 1M/2k = 512
Direct map is 512 virtual for each physical, that's a lot

Subproblem 2F

Would need PTE for every page, or 1M PTE's

Subproblem 2G

PTE needs 11 bits for physical page number and 2 for V and D, 13 per PTE
2 Bytes