(8pts) Consider a system with 128G virtual memory, 8G physical memory. The syste
ID: 664159 • Letter: #
Question
(8pts) Consider a system with 128G virtual memory, 8G physical memory. The system uses 2 level paging. The number of page table entries in level 1 page table is 4K. The page size is 16K.
Identify
Partitioning of the virtual address
L1=
L2=
Offset =
Partitioning of Physical Address
Number of Level 2 page tables
Minimum size for page table entry in bits (include P/M but no other control bits).
Minimum size for page table entry if it is required to be 1, 2, 4, 8, 16, … bytes
Total size of ALL level 1 page table
Total size of ALL level 2 page tables.
Number of level 1 page tables
L1=
L2=
Offset =
Explanation / Answer
Partitioning of the virtual address
L1=10bits
L2=10bits
Offset =12bits
Partitioning of Physical Address
512bytes
8192 bytes
Pages are 4K VPN is 20 bits (220 VPNs), offset is 12 bits Virtual address is 0x7468 Virtual page is 0x7, offset is 0x468 Page table entry 0x7 contains 0x2 Page frame base is 0x2 * 0x1000 (4K) = 0x2000 Seventh virtual page is at address 0x2000 (3rd physical page) Physical address = 0x2000 + 0x468 = 0x2468
L1=10bits
L2=10bits
Offset =12bits
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.