Paging Addressing: Logical vs Physical Bits and Page Table Size

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Sep 12, 2026

In a paging memory-management scheme, a logical (virtual) address is split into a keywordpage number and a keywordpage offset. The keywordframe number in the physical address is used to access the corresponding keywordframe containing that offset.

Given:

Physical memory size: $2^{32}$ bytes
Page size: $2^{10}$ bytes
Logical address space: $2^{16}$ pages

We compute: (i) logical address bits, (ii) frame size, (iii) physical address frame bits, and (iv) number of page table entries.

Key idea:

Offset bits are determined solely by page/frame size: $2^{10}$ bytes $\Rightarrow$ offset is $10$ bits.
Logical address bits = bits for page number + bits for offset.
Physical address bits = bits for frame number + bits for offset.

Virtual Memory Paging: Address Translation Basics

Computing the components

We will use the standard relation:

Number of pages (logical) = $2^{16}$
Page offset bits = $\log_2(2^{10}) = 10$

Then:

Virtual address page-number bits = $\log_2(2^{16}) = 16$
Virtual address total bits = $16 + 10 = 26$
Physical frame bits depend on number of frames:
- Total frames = physical memory / frame size = $2^{32}/2^{10} = 2^{22}$
- Frame-number bits = $\log_2(2^{22}) = 22$

Also:

Page table entries = number of logical pages = $2^{16}$ .

Step-by-step address bit calculation

1
Step 1
Page size is $2^{10}$ bytes, so the offset within a page requires $\log_2(2^{10}) = 10$ bits.
2
Step 2
Logical address space has $2^{16}$ pages, so the page number requires $\log_2(2^{16}) = 16$ bits.
3
Step 3
Logical address bits = page-number bits + offset bits = $16 + 10 = 26$ .
4
Step 4
In paging, each frame holds exactly one page, so frame size equals page size = $2^{10}$ bytes.
5
Step 5
Number of frames = $2^{32} / 2^{10} = 2^{22}$ , so frame-number bits = $\log_2(2^{22}) = 22$ .
6
Step 6
A page table has one entry per logical page, so entries = $2^{16}$ .

Bit breakdown (virtual vs physical)

Offset is fixed by page size; page/frame number bits depend on page/frame counts.

Answers (i)–(iv) in one place

Pro Tip: Offset bits never change

Once the page size is fixed, the page offset always uses exactly $\log_2(\text{page size})$ bits. That’s why both virtual and physical addresses share the same offset-width.

Watch out for confusing address-space size vs number of pages

Here the prompt gives logical address space as $2^{16}$ pages (not $2^{16}$ bytes). Always convert page counts into bits using $\log_2(\text{#pages})$ , then add the offset bits.

Address translation lifecycle in paging

Split virtual address

Use page number and page offset (offset width = 10 bits)."

Page table lookup

Index the page table with the page number to retrieve the frame number."

Form physical address

Combine frame number with the same offset to create the physical address."

Paging essentials: quick checks

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Question · Term

If page size is $2^{10}$ bytes, how many offset bits?

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Answer · Definition

Offset bits = $\log_2(2^{10}) = 10$ .

Knowledge Check

Question 1 of 4

Q1Single choice

Given page size $2^{10}$ bytes, how many bits are in the page offset?

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An inverted page table (IPT) is a global paging structure that keeps one entry per physical frame, recording the virtual page, process ID, and status bits, thus reducing page‑table memory.

Fields: frame index, VPN, PID/ASID, control bits, hash/link.
Entries = Physical memory ÷ page size (e.g., $4\text{GB}/4\text{KB}=2^{20}$ ).
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Benefit: memory usage depends on frames, not on all virtual pages.
Cost: slower translation than direct indexing; relies on hashing/TLB.

Integrated Diagram: TLB and Cache Operations for a Logical/Virtual Address

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