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Page size = 8 KB. So offset = 13 bits. So for frames we need 32-13 = 19 bits.

So each PTE (page table entry) is of 19 frame bits + 1 valid bit = 20 bits

Now,maximum possible number of pages = maximum number PTEs = size of page table / size of a PTE = $20 * 8 * 2^{20}$ / $20$ = $2^{23}$

So total size of virtual address space = $2^{23} * 2^{13}$ = $2^{36}$.

So length of the virtual address space supported by the system should be 36 bits.
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It should be 36 .....

Size of the PT = Number of Pages *Frame bits

20MB =  (2^x) * 20/8   x=23 bits  Total Bits in VA  23 + 13 =36
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Given that,

Page Size = 8KB = 213B.

Physical Address Space = 32bits.

Number of frames in main memory = $\frac{2^{32}}{2^{13}}$ = 219.

So, number of bits to address each frame = 19.

Page table entry contains 1 valid bit and 19 bits for address translation, ie 20 bits or 2.5B.

Now given, Page Table Size = 20MB = 20 * 220B.

Page Table Size = Page table entry size * Number of pages

So, Number of pages = Page table size/Page table entry size = $\frac{20 * 2^{20}B}{2.5B}$ = 223.

Logical Address Space = Number of Pages * Page Size = 223 * 213 = 236B

So, 36 bits are required to address a word of Logical Address Space.