MadeEasy Test Series: Operating System - Memory Management

Question

Suppose that you wish to design a virtual memory system with the following characteristics:
• Size of page table entry is 4 bytes.
• Each page table must fit into a single page frame.
• System must be able to support virtual address space of 4 GB.
You decided to use a multi-level paging scheme with no more than 2 levels of page tables. What is the minimum page size that the system must have?

Answer 1

32 bits --generated address , k bits for the page frame 

size of page be 2^k

so for the 32 bits (4GB) of generated address , to identify the page will be 32-k bits

again 2 level of page tables so it will be like this

x | 32-k-x | k ---this sum will be equal to 32 bits

now the condition is each page table is fitted into a single frame so 

2^x * 4 = 2^k and 2 ^(32-k-x) * 4 = 2^k

solving this you will get k =12 bits --- so the page frame size will be 2¹² B

Comments

Would you please explain this line 2^x * 4 = 2^k and 2 ^(32-k-x) * 4 = 2^k    .

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here $2^x$ entries are there in the first level so each is 4 Bytes as mentioned in the ques so it must be equal to a page size which is $2^k$ similarly for the other one too

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Thanks,
I got it

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pls explain this answer