Question 2 Single-Level Page Table Consider a computer with 4 GB of physical mem
ID: 3894635 • Letter: Q
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Question 2 Single-Level Page Table Consider a computer with 4 GB of physical memory, where the page For simplicity assume than 1GB-1000MB, 1 MB-1000KB, 1 KB-1000 bytes size is 4 KB. (a) How many pages must be tracked by a single-level page if the computer has 4GB of physical memory and the page table size is 4 KB? (b) How many bits are required for the virtual page number (VPN) to address any page within this 4GB memory space? (c) Assuming that the smallest addressable unit of memory within a page is a byte (8-bits) How many bits are required for the offset to refer to any byte in the 4 KB page? (d) Assuming each page table entry (PTE) requires 4 bytes of memory, how much memory is required to store the page table for one process (in MB)? (e) Using this memory requirement, how many processes would fill the memory with page table data on a 4GB computer?Explanation / Answer
(a) Number of Pages = Size of physical Memory / Size of Page
Here,
Physical memory = 4 GB = 232
Page size = 4 KB = 212
since Number of Pages = 232 / 212 = 220.
Number of Pages = 220.
(b) No. of bits required for VPN to address any page within this 4 GB is = ( log2 PA ) bits.
where PA = size of physical memory.
Physical memory = 4 GB = 232
No. of bits = ( log2 232) = 32 bits.
No. of bits required for VPN to address any page within this 4 GB is = 32 bits.
(c) bits required for offset to refer any byte = ( log2 P ) bits.
where P = size of page.
Page size = 4 KB = 212
offset = ( log2 212) = 12.
Therefore no.of bits required for offset to refer any byte is = 12 bits.
(d) Page table size = No.of pages * Page table entry size
Number of Pages = 220
Page table entry size = 4 bytes.
Page table size = 220 * 22 = 222 = 4 MB.
Page table size = 4 MB.
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