CSC264 Comp Org/Arch - Virtual Memory

Virtual Memory

Example 1

virtual address space 2⁸ words (this is the amount of memory used by the app)
virtual addresses range from 00 to FF (0 to 2⁸-1)
4 page frames allocated to our app
each page/page frame is 32 words

Therefore we have the following:

There are 2² page frames so 2 bits are required to choose the frame.
Our app uses 2⁸ words, each page is 2⁵ words so that is 2³ pages.
Each page/frame is 2⁵ words so 5 bits are required to choose the address of a word within the page.

This means that our virtual addresses are 8 bits (3 bits to choose the page, 5 bits to locate the word within the page) and our physical addresses are 7 bits (2 bits to choose the frame, 5 bits to locate the word within the page).

Suppose the processor generates the address 72₁₆ = 0111 0010₂. The first 3 bits give the page number (3₁₆) and the last 5 bits give the offset within the page (12₁₆). To get the physical address, use the page number as the subscript in the page table. Page table entry 3 specifies that the page is in RAM and that it is in frame 0. Therefore the physical address is 001 0010 (frame 0, offset 10010).

Suppose the processor generates the address BA₁₆ = 1011 1010₂. The first 3 bits give the page number (5₁₆) and the last 5 bits give the offset within the page (1A₁₆). To get the physical address, use the page number as the subscript in the page table. Page table entry 5 specifies that the page is not in RAM. It needs to be loaded into a page frame. Suppose it is loaded into frame 2, which means that page 0, which is currently in frame 2 is swapped out to disk. Then the physical address is 101 1010 (frame 2, offset 11010). Also the page table must be updated so that entry 0 has a valid bit of 0 (page 0 is now on disk) and entry 5 specifies that the page is in frame 2 and the valid bit is set to 1.

Example 2

virtual address space 8KB = 2¹³ bytes (this is the amount of memory used by the app)
physical memory is 4KB
each page/page frame is 1KB

Therefore we have the following:

Virtual address is 13 bits.
There are 2² page frames so 2 bits are required to choose the frame.
Our app uses 8KB = 2¹³ bytes, and each page is 1KB = 2¹⁰ bytes so that is 2³ pages.
Each page/frame is 2¹⁰ bytes so 10 bits are required to choose the address of a byte within the page.

This means that our virtual addresses are 13 bits (3 bits to choose the page, 10 bits to locate the word within the page) and our physical addresses are 12 bits (2 bits to choose the frame, 10 bits to locate the word within the page).

Suppose the processor generates virtual address 1553₁₆ = 101 01 0101 0011₂. The first 3 bits give the page number (5₁₆) and the last 10 bits give the offset within the page (153₁₆). To get the physical address, use the page number as the subscript in the page table. Page table entry 5 specifies that the page is in RAM and that it is in frame 1. Therefore the physical address is 01 01 0101 0011 (frame 1, offset 01 0101 0011).

Suppose the processor generates virtual address 802₁₆ = 010 00 0000 0010₂. The first 3 bits give the page number (2₁₆) and the last 10 bits give the offset within the page (2₁₆). To get the physical address, use the page number as the subscript in the page table. Page table entry 2 specifies that the page is in RAM and that it is in frame 0. Therefore the physical address is 00 00 0000 0010 (frame 0, offset 00 0000 0010).

Suppose the processor generates virtual address 1004₁₆ = 100 00 0000 0100₂. The first 3 bits give the page number (4₁₆) and the last 10 bits give the offset within the page (4₁₆). To get the physical address, use the page number as the subscript in the page table. Page table entry 4 specifies that the page is not in RAM, since the valid bit is 0. This results in a page fault.