Segmented Paging | Practice Problems

Segmented Paging-

 

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

 

We have discussed-

• Segmented Paging is a scheme that implements the combination of Segmentation and Paging.
• First, segmentation divides the process into segments.
• Then, paging divides each segment into pages.

 

In this article, we will discuss practice problems based on segmented paging.

 

PRACTICE PROBLEMS BASED ON SEGMENTED PAGING-

 

Problem-01:

 

A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 2¹⁶ bytes each. The virtual address space is divided into 8 non-overlapping equal size segments. The memory management unit (MMU) has a hardware segment table, each entry of which contains the physical address of the page table for the segment. Page tables are stored in the main memory and consists of 2 byte page table entries. What is the minimum page size in bytes so that the page table for a segment requires at most one page to store it?

 

Solution-

 

Given-

• Virtual Address Space = Process size = 2¹⁶ bytes
• Physical Address Space = Main Memory size = 2¹⁶ bytes
• Process is divided into 8 equal size segments
• Page table entry size = 2 bytes

 

Let page size = n bytes.

Now, since page table has to be stored into a single page, so we must have-

Size of page table <= Page size

 

Size of Each Segment-

 

Size of each segment

= Process size / Number of segments

= 2¹⁶ bytes / 8

= 2¹⁶ bytes / 2³

= 2¹³ bytes

= 8 KB

 

Number of Pages Of Each Segment-

 

Number of pages each segment is divided

= Size of segment / Page size

= 8 KB / n bytes

= (8K / n) pages

 

Size of Each Page Table-

 

Size of each page table

= Number of entries in page table x Page table entry size

= Number of pages the segment is divided x 2 bytes

= (8K / n) x 2 bytes

= (16K / n) bytes

 

Page Size-

 

Substituting values in the above condition, we get-

(16K / n) bytes <= n bytes

(16K / n) <= n

n² >= 16K

n² >= 2¹⁴

n >= 2⁷

 

Thus, minimum page size possible = 2⁷ bytes = 128 bytes.

 

Problem-02:

 

Considering problem-01, give the division of virtual address.

 

Solution-

 

Number of Bits Required For Segment Number-

 

Number of segments the process is divided

= 8

= 2³

Thus, Number of bits required to identify a particular segment in segment table = 3 bits.

 

Number of Bits Required For Page Number-

 

Number of pages a segment is divided

= Segment size / Page size

= 8KB / 128 bytes

= 2¹³ bytes / 2⁷ bytes

= 2⁶ pages

Thus, Number of bits required to identify a particular page in table = 6 bits.

 

Number of Bits Required For Page Offset-

 

Page size

= 128 bytes

= 2⁷ bytes

Thus, Number of bits required for page offset = 7 bits.

 

Thus, virtual address is divided as-

 

 

Problem-03:

 

A certain computer system has the segmented paging architecture for virtual memory. The memory is byte addressable. Both virtual and physical address spaces contain 2¹⁶ bytes each. The virtual address space is divided into 8 non-overlapping equal size segments. The memory management unit (MMU) has a hardware segment table, each entry of which contains the physical address of the page table for the segment. Page tables are stored in the main memory and consists of 2 byte page table entries.. Assume that each page table entry contains (besides other information) 1 valid bit, 3 bits for page protection and 1 dirty bit. How many bits are available in page table entry for storing the aging information for the page? Assume that page size is 512 bytes.

 

Solution-

 

Given-

• Virtual Address Space = Process size = 2¹⁶ bytes
• Physical Address Space = Main Memory size = 2¹⁶ bytes
• Process is divided into 8 equal size segments
• Page table entry size = 2 bytes = 16 bits
• Page table entry besides other information contains 1 valid bit, 3 protection bits, 1 dirty bit
• Page size = 512 bytes

 

Number of Frames in Main Memory-

 

Number of frames in main memory

= Size of main memory / Page size

= 2¹⁶ bytes / 512 bytes

= 2¹⁶ bytes / 2⁹ bytes

= 2⁷ frames

Thus, Number of bits required for frame identification in page table entry = 7 bits

 

Number Of Bits Available For Storing Aging Information-

 

Number of bits available for storing aging information

= Number of bits in page table entry – ( Number of bits required for frame identification + 1 valid bit + 3 protection bits + 1 dirty bit)

= 16 bits – (7 + 1 + 3 + 1) bits

= 16 bits – 12 bits

= 4 bits

 

Next Article- Disk Scheduling Algorithms

 

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