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Virtual Memory 2

Consider 27 b...

Question

Consider 27 bit logical address and page frame size is 1 KB, if each PTE size is 4 Byte then the size of page table (in KB) of virtual program is

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Solution

$Virtual program size = 2^{27} B y t e s$
$Number of pages in virtual program = \frac{2^{27} B}{2^{10} B} = 2^{17} p a g e s$
$Size of page table = 2^{17} * 4 B y t e = 2^{19} B y t e s .$

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