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Consider a system implements 4 KB pages and 24 bits physical address space. Each page table entry contains a valid bits, a dirty bits, 2 permission bits and translation. If the maximum page table size of process is 30 KB. Then the size of virtual memory supported by the system is ________ (MB).
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64MB?
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7.5MB
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Can you please tell the approach you followed.
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how?
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Please someone solve it
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@yankur9 Is the ans 7.5MB?

+3

Answer is 60 MB.

Page size = $2^{12} B$ 

$24$ bits PAS, Hence frames = $\frac{2^{24}}{2^{12}} = 2^{12}$

Now each Page table entry = $12 + 1 + 1 + 2$ bits $= \  16$ bits $= \ 2$ B

Now PTsize = Pages * PTE

$30KB = x * 2B$

Hence $15K$ pages in Virtual memory.

Hence size of virtual memory = $15K*4KB = 60 MB$

1 Answer

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There are total 4K frames which require 12 bits to represent.
For page table entry,
each entry size  = 12 bits (no of bits require to represent a frame ) + 4 bits (extra required bits are given) = 16 bits = 2 bytes.
total no of entries in the page table  = page table size / each entry  size = 30 KB / 2 B = 15 K

so There 15 k pages which are 4KB in size each. Total virtual memory can be support  = 15 K * 4 KB = 60MB

Is there any misconception?
Let me know.
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