##### Question

In: Computer Science

# Given a system with 4M bytes of address space per process, 64M bytes of physical memory,...

Given a system with 4M bytes of address space per process, 64M bytes of physical memory, pages of size 1K bytes, and page table entries of 8 bytes. How many entries can fit in one page?

Select one:

a. 128

b. 1K

c. 8

d. 64K

Given a system with 16G bytes of address space per process, 8G bytes of physical memory, and pages of size 16K bytes. How many entries are there in a linear page table?

Select one:

a. 1M

b. 16K

c. 8G

d. 512K

## Solutions

##### Expert Solution

Solution:

(1)

Given,

=>Logical address space size = 4 MB

=>Physical address space size = 64 MB

=>Page size = 1 KB

=>Page table entry size = 8 B

The answer will be an option,

(a) 128

Explanation:

Calculating number of entries in one page:

=>Number of entries in one page = page size/entry size

=>Number of entries in one page = 1 KB/8 B

=>Number of entries in one page = 1*1024 B/8 B as 1 KB = 1024 B

=>Number of entries in one page = 128

=>Hence number of entries can fit in one page = 128

=>Hence option (a) is correct and other options are incorrect.

(2)

Given,

=>Logical address space size = 16 GB

=>Physical address space size = 8 GB

=>Page size = 16 KB

The answer will be an option,

(a) 1 M

Explanation:

 Page number Page offset

20 bits                                                                              14 bits

Calculating number of bits for page offset:

=>Number of bits for page offset = log2(page size in bytes)

=>Number of bits for page offset = log2(16 KB)

=>Number of bits for page offset = log2(16*2^10 B) as 1 KB = 2^10 B

=>Number of bits for page offset = log2(2^14)

=>Number of bits for page offset = 14 bits

Calculating number of bits for page number:

=>Number of bits for page number = log2(logical address space size in bytes) - number of bits for page offset

=>Number of bits for page number = log2(16 GB) - 14

=>Number of bits for page number = log2(16*2^30 B) - 14 as 1 GB = 2^30 B

=>Number of bits for page number = 34 - 14

=>Number of bits for page number = 20 bits

Calculating number of entries in linear page table:

=>Number of entries in linear page table = page table size/entry size

=>Number of entries in linear page table = number of pages*frame number bits/frame number bits

=>Number of entries in linear page table = number of pages

=>Number of entries in linear page table = 2^page number bits

=>Number of entries in linear page table = 2^20

=>Number of entries in linear page table = 1 M as 1 M = 2^20

=>Hence option (a) is correct and other options are incorrect.

I have explained each and every part with the help of statements attached to it.

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