Optimal Page Size | Paging

Paging in OS-

 

Before you go through this article, make sure that you have gone through the previous article on Paging in OS.

 

We have discussed-

• Paging is a non-contiguous memory allocation technique.
• Page Table keeps track of the frames storing the pages of the process.

 

Overhead in Paging-

 

In paging scheme, there are mainly two overheads-

 

1. Overhead of Page Tables-

 

• Paging requires each process to maintain a page table.
• So, there is an overhead of maintaining a page table for each process.

 

2. Overhead of Wasting Pages-

 

• There is an overhead of wasting last page of each process if it is not completely filled.
• On an average, half page is wasted for each process.

 

Thus,

Total overhead for one process

= Size of its page table + (Page size / 2)

 

Optimal Page Size-

 

Optimal page size is the page size that minimizes the total overhead.

It is given as-

 

 

Also Read- Important Formulas Of Paging

 

Proof-

 

Total overhead due to one process

= Size of its page table + (Page size / 2)

= Number of entries x Page table entry size + (Page size / 2)

= Number of pages the process is divided x Page table entry size + (Page size / 2)

= (Process size / Page size) x Page table entry size + (Page size / 2)

 

Now,

For minimum overhead,

 

Keeping process size and page table entry size as constant, differentiating overhead with respect to page size, we get-

 

 

This page size minimizes the total overhead.

 

PRACTICE PROBLEMS BASED ON OPTIMAL PAGE SIZE-

 

Problem-01:

 

In a paging scheme, virtual address space is 4 KB and page table entry size is 8 bytes. What should be the optimal page size?

 

Solution-

 

Given-

• Virtual address space = Process size = 4 KB
• Page table entry size = 8 bytes

 

We know-

Optimal page size

= (2 x Process size x Page table entry size)^1/2

= (2 x 4 KB x 8 bytes)^1/2

= (2¹⁶ bytes x bytes)^1/2

= 2⁸ bytes

= 256 bytes

Thus, Optimal page size = 256 bytes.

 

Problem-02:

 

In a paging scheme, virtual address space is 16 MB and page table entry size is 2 bytes. What should be the optimal page size?

 

Solution-

 

Given-

• Virtual address space = Process size = 16 MB
• Page table entry size = 2 bytes

 

We know-

Optimal page size

= (2 x Process size x Page table entry size)^1/2

= (2 x 16 MB x 2 bytes)^1/2

= (2²⁶ bytes x bytes)^1/2

= 2¹³ bytes

= 8 KB

Thus, Optimal page size = 8 KB.

 

Problem-03:

 

In a paging scheme, virtual address space is 256 GB and page table entry size is 32 bytes. What should be the optimal page size?

 

Solution-

 

Given-

• Virtual address space = Process size = 256 GB
• Page table entry size = 32 bytes

 

We know-

Optimal page size

= (2 x Process size x Page table entry size)^1/2

= (2 x 256 GB x 32 bytes)^1/2

= (2⁴⁴ bytes x bytes)^1/2

= 2²² bytes

= 4 MB

Thus, Optimal page size = 4 MB.

 

Next Article- Practice Problems On Paging in OS

 

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