# How do I find the physical address of a logical address?

1. If Logical Address = 31 bit, then Logical Address Space = 231 words = 2 G words (1 G = 230)
2. If Logical Address Space = 128 M words = 27 * 220 words, then Logical Address = log2 227 = 27 bits.
3. If Physical Address = 22 bit, then Physical Address Space = 222 words = 4 M words (1 M = 220)

## How do we generate physical address in segmentation?

To get total physical address, put the lower nibble 0H to segment address and add offset address. The figure shows the formation of 20-bit physical address. Suppose the Data Segment holds the Base Address as 1000h and the data you need is present in the 0020h memory location (Offset) of the Data Segment.

## What are the physical addresses for the following logical addresses A 0 430?

Segment 0, Logical Address is 430 – So physical address becomes – 219+430 = 649 b. Segment 1, Logical Addres is 10 – So physical address becomes – 2300+10 = 2310.

## How do you calculate total logical and physical address space?

we know that logical address spaces is = total no of bits required to represent total no of pages + bits required to map page offset . Hence total bits required = 3 (because total no of pages is 8 and to represent you need three bits) + 10 (page offset is 1024 so you need 10 bits) = 13 bits all total.

• First-fit.
• 212K is put in 500K partition.
• 417K is put in 600K partition.
• 112K is put in 288K partition(new partition 288K=500K-212K)
• 426K must wait.
• Best-fit.
• 212K is put in 300K partition.
• 417K is put in 500K partition.

## How do you calculate physical address from segment offset pairs?

In particular, given a [segment:offset] pair, a 20-bit external (or physical) address is produced by segment × 24 + offset, where segment × 24 is called the segment address, which has its 16 most significant bits from the 16-bit segment register, and its four LSbs are all zeros.

There are two ways to convert a virtual address to a physical address: by using the ! vtop extension, and by using the ! pte extension. For an overview of virtual address in Windows, see Virtual address spaces.

## How is a 20 bit address obtained if there are only 16-bit registers?

First, we shift the first address by four bits to the left: 7 2 3 A 0 which is called the base. Then, we add the the second address: 0 0 0 5 which is called the offset. The result of this operation is a new 20-bit address: 7 2 3 A 5 .

## What happens if the logical address is segment ID 2 and offset 1000 explain the answer?

Explanation: Segment-2 has a base address = 1527 and limit address = 498, So it can only access the memory location from 1527 till 1527 + 498, if the process tries to access the memory with offset 1000 then a segmentation fault trap will be generated.

## How much the number of bits calculation required for the logical address if you considered the segment size is 1 K bytes and there are 32 segments?

To specify a particular segment, 5 bits are required (since 2^5 = 32). Having selected a page, to select a particular byte one needs 10 bits (since 2^10= 1K byte). So, totally 5 + 10 =15 bits are needed.

## How do you calculate page numbers and offset of virtual address?

1. this is the distance from the beginning of the page.
2. e.g. address in the process, A = 10,000.
3. page size = 4k.
4. page offset = 10000 mod 4k = 10,000 mod 4096 = 1908.
5. this calculation is done quickly on the computer since the page size is power of 2, e.g., 4k = 2^12.

## What does a 10 bit physical page number imply?

How many memory pages does the Page Directory occupy? We are told that the Page Directory index is 10 bits, implying 210 = 1024 entries. Each entry occupies 4 bytes, so the total size of the of the Page Directory is 4*210 = 212 bytes, or exactly one page.

## How many bits are there in the physical address if a logical address space of 8 pages?

Since the logical address space consists of 8 = 23 pages, the logical addresses must be 10+3 = 13 bits.

## How many bits is the physical address?

Physical addresses are 44 bits and there are 4 protection bits per page.

## What is the difference between logical and physical address space?

Logical Address Space is the set of all logical addresses generated by CPU for a program whereas the set of all physical address mapped to corresponding logical addresses is called Physical Address Space.

## How do you find the physical page number?

In your example, page size is 16 KBytes, so log2(16*2^10) is 14; that is, page offset is 14 bits. Then, calculate Physical Page Number (PPN) size by subtracting page offset from total number of bits allocated for physical address. Since in your example, physical address is 32-bit, PPN = 32 – 14, or 18 bits.

## How can I get physical address in 8086?

in 8086 microprocessor a 20 bit address is divided in 16bit+4bit address in which 4 bit binary is the segment address.

## Can a specific physical address have more than two logical addresses?

In 8086 more than one logical addresses ( segment:offset ) can have the same physical address ( 001F:000F and 000F:010F are the same addresses, like 001F:000F = 01F0+000F = 01FF in the same way as 000F:010F = 01FF ). In physical memory, any programs see the logical memory address.

Whenever workloads access data in memory, the system needs to look up the physical memory address that matches the virtual address. This is what we refer to as memory translations or mappings. To map virtual memory addresses to physical memory addresses, page tables are used.

## Why we need mapping from logical address to physical address space?

The user program generates the logical address and thinks it is running in it, but the program needs physical memory for its execution. Therefore, the logical address must be mapped to the physical address by MMU before they are used.

The physical address is the 20-bit address that is actually put on the address pins of the 8086 microprocessor and decoded by the memory interfacing circuitry. The offset address is a location within a 64K-byte segment range. The logical address consists of a segment value and an offset address.

## How are the segment registers used to form 20-bit physical address?

5.1. A segment register changes the memory address accessed by 16 bits at a time, because its value is shifted left by 4 bits (or multiplied by 16) to cover the entire 20-bit address space. The segment register value is added to the addressing register’s 16-bit value to produce the actual 20-bit memory address.

## How does 8086 manage a 20-bit address in a 16-bit processor?

It is a 16-bit Microprocessor having 20 address lines and16 data lines that provides up to 1MB storage. It consists of powerful instruction set, which provides operations like multiplication and division easily. It supports two modes of operation, i.e. Maximum mode and Minimum mode.

## How would you show your understanding that the offset D of the logical address must be?

The offset d of the logical address must be between 0 and the segment limit. If it is not, we trap to the operating system (logical addressing attempt beyond end of segment). When an offset is legal, it is added to the segment base to produce the address in physical memory of the desired byte.