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We are given a computer architecture with 32-bit virtual address and a page size of 4KB. Each page table entry is 8 Bytes. Let’s say we have 2 processes in the system, where Process #1 is using 1MB of its address space and Process #2 is using 1GB of its address space. What is the total number and size of the page tables in the system?
Expert Answer
we have a virtual address space of 2^32 and each page size is 2^12, we can store (2^x/2^12) =y pages. Since each entry into this page table has an address of size 8 bytes, then we have y*4 = 1MB.
y=1/4 MB
x-12=-2
x=14 pages
similiary for 1GB x-12=20
x=32 pages.
Each page entry occupies 8 bytes so size=(14+32)*8=46*8=368 bytes
