Given the following logical addresses (in decimal), convert them to a format? As

Question

Given the following logical addresses (in decimal), convert them to a format? Assume the page size to be 4Kbytes (4096 bytes). a. 234 b. 23456 c. 234567 d. 2345678 Consider the following segment table: What are the physical addresses for the following logical addresses? a. 0, 430 b. 1, 2056 c. 2, 5024 d. 3, 7024 e. 4, 512 f. 5, 12384 An experimental operating system has a 30-bit virtual address, yet on certain embedded devices, it has only an 16-bit physical address. It also has a 4-KB page size. How many entries are there in each of the following? a. A conventional, single-level page table b. An inverted page table Consider a logical address space of 512 pages with a 4-KB page size, mapped onto a physical memory of 256 frames. a. How many bits are required in the logical address? b. How many bits are required in the physical address?

Explanation / Answer

Solution:

The page size is 4 KB which means that 12 bits are required to show the logical address.

a)

<0, 234>

b)

<23, 456>

c)

<234, 567>

d)

<2345, 678>

Question: 2

Check logical address, it must be less than length.

Formula:

Physical Address = Logical Address + Base Address

a)

For Segment 1

430<512

Physical address= 430+128= 558

b)

For Segment 2

2056 !< 2048

This means that no physical address.

c)

For Segment 3

5024 !< 4096

This means that no physical address.

d)

For segment 3

7024<8192

Physical address= 7024+16384= 23408

e)

For segment 4

512<1024

Physical address= 512+32768= 33280

f)

12384<16384

Physical address= 12384+65536= 77920

I hope this helps.

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