# Andrew S. Tanenbaum (OS) Edition 4 Exercise 3 Question 19 (Page No. 256)

1 vote
81 views
A computer with a $32-bit$ address uses a two-level page table. Virtual addresses are split into a $9-bit$ top-level page table field, an $11-bit$ second-level page table field, and an offset. How large are the pages and how many are there in the address space?

$\underline{\mathbf {Answer:}}$

Number of bits used for virtual page numbers $=20$

Number of bits used for offset $=12$

$\therefore$ Size of Page $=\;\mathbf{4K} \;$.

$20$ bits for Virtual Page means $2^{20}$ pages.

edited by
0
can you plz explain the meaning of “how many are there in the address space”

## Related questions

1 vote
1
161 views
A machine has $48-bit$ virtual addresses and $32-bit$ physical addresses. Pages are $8\: KB.$ How many entries are needed for a single-level linear page table?
A computer has $32-bit$ virtual addresses and $4-KB$ pages. The program and data together fit in the lowest page $(0–4095)$ The stack fits in the highest page. How many entries are needed in the page table if traditional (one-level) paging is used? How many page table entries are needed for two-level paging, with $10$ bits in each part?
Section $3.3.4$ states that the Pentium Pro extended each entry in the page table hierarchy to $64$ bits but still could only address only $4\: GB$ of memory. Explain how this statement can be true when page table entries have $64$ bits.
Suppose that a machine has $438-bit$ virtual addresses and $32-bit$ physical addresses. What is the main advantage of a multilevel page table over a single-level one? With a two-level page table, $16-KB$ pages, and $4-byte$ entries, how many bits should be allocated for the top-level page table field and how many for the next level page table field? Explain.