Question

4. Assume that we have a machine with the following memory specifications: Virtual addresses are 32 bits wide Physical addresses are 26 bits wide . Page size is 16 Kbytes 4/A) How many pages are in the virtual memory space? 4/B) How many page frames are in the physical memory space? 4/C) If each page table entry consists of a physical frame number, 1 present/absent bit Answers Pages .Page Frames and 1 dirty/clean bit (which shows if the page has changed since loaded in physical memory), what is the size of the page table in Kbytes? Kbytes Write your calculations here: In the machine specified in question 4, the first few rows of the page table looks like this

Answer 1

Solution :

4/A :

Number of pages = size of virtual memory / size of a page
= 2^32 / 16 KBytes
= 2^32 / 2^14
= 2^18

4/B :
Number of page frames = size of physical memory / size of a page
= 2^26 / 16 KBytes
= 2^26 / 2^14
= 2^12

4/C :
Number of bits used for each page entry = number of bits used for page frame number + number of extra bits
= 12 + 2 (one is dirty bit and another is present/absent bit)
= 14 bits

Size of page table = number of pages in virtual memory * number of bits used for each entry
= 2^18 * 14 bits
$\small \cong$ 2^18 * 16 bits
= 2^18 * 2^4
= 2^22 bits
= 2^19 bytes
= 512 KBytes.

Answer : ( around )512 KB

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