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In a virtual memory system, size of virtual address is 32-bit, size of physical address is 30-bit, page size is 4 Kbyte and size of each page table entry is 32-bit. The main memory is byte addressable.

i. What is the size of main memory?

ii. Find out, number of frames in main memory.

iii. Find out, number of bits used for storing other information in frame.
in Operating System by (5 points) | 38 views
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$1GB,2^{18}=256K,14-bits$
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can you elaborate?

1 Answer

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a)

size of physical address is 30-bit

Size of MM=$2^{30}=1GB$

b)

Page Size/frame Size=4KB

# of frames$=\frac{2^{30}B}{2^{12}B}=2^{18}=256K$

c)

PTE=32 bit

From b) we calculated # of bits for frame identification and that is 18-bits.

So remaining bits will be used for storing other information. So $32-18=14 -bits$
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