OS Memory Management

Question

Consider a paging system with page size of 8 KB. If a process of size 34 KB is in logical address space, find the internal fragmentation in Kbytes.

a - 4

b- 5

c -6

d -7

 

Internal fragmentation is always half the page size I guess So the answer should be 4 KB but it is 6 KB

How????

Comments

why half of page size is wasted?, I have read it somewhere

Answer 1

Page size is 8KB  and process size is 34KB

so  require  5 pages =   4 pages having no internal fragmentation  and 1 page for remaining 2KB of the process

so 6 KB is  wasted in the last page

Comments

Yup Missed it :)

Answer 2

Internal Fragmentation  arises when the memory allocated to a program is not fully utilized by it.

Now here process size is 34kb and page size is 8kb

hence process requires 5 pages (4 page for 32kb and 1 more page for 2kb more i.e. 4+1=5 pages )for process out of which 6kb will not be utilized by the process hence internal fragmentation will be 6kb

Comments

Yup Missed it :)

Answer 3

Page size=8kB

Process size=34KB

number of the pages=34KB/8KB=5

but 4 pages will take proper space but one last process will not fit in 2kB.

So required space which is waste is =4KB+2KB=6KB.

Answer 4

You can do this without pen/paper see we have 8kb of the page size which can be equally distributed towards 32kb but here the size is 34 kb so every time it will have +2 more internal fragmentation so 1/2(8)+2.