Tag: Memory Management

Segmentation and Paging | Segmented Paging

Operating System

Segmentation and Paging-

 

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

 

We have discussed-

  • Paging and Segmentation are the non-contiguous memory allocation techniques.
  • Paging divides the process into equal size partitions called as pages.
  • Segmentation divides the process into unequal size partitions called as segments.

 

Segmented Paging-

 

Segmented paging is a scheme that implements the combination of segmentation and paging.

 

Working-

 

In segmented paging,

  • Process is first divided into segments and then each segment is divided into pages.
  • These pages are then stored in the frames of main memory.
  • A page table exists for each segment that keeps track of the frames storing the pages of that segment.
  • Each page table occupies one frame in the main memory.
  • Number of entries in the page table of a segment = Number of pages that segment is divided.
  • A segment table exists that keeps track of the frames storing the page tables of segments.
  • Number of entries in the segment table of a process = Number of segments that process is divided.
  • The base address of the segment table is stored in the segment table base register.

 

Translating Logical Address into Physical Address-

 

  • CPU always generates a logical address.
  • A physical address is needed to access the main memory.

 

Following steps are followed to translate logical address into physical address-

 

Step-01:

 

CPU generates a logical address consisting of three parts-

  1. Segment Number
  2. Page Number
  3. Page Offset

 

 

  • Segment Number specifies the specific segment from which CPU wants to reads the data.
  • Page Number specifies the specific page of that segment from which CPU wants to read the data.
  • Page Offset specifies the specific word on that page that CPU wants to read.

 

Step-02:

 

  • For the generated segment number, corresponding entry is located in the segment table.
  • Segment table provides the frame number of the frame storing the page table of the referred segment.
  • The frame containing the page table is located.

 

Step-03:

 

  • For the generated page number, corresponding entry is located in the page table.
  • Page table provides the frame number of the frame storing the required page of the referred segment.
  • The frame containing the required page is located.

 

Step-04:

 

  • The frame number combined with the page offset forms the required physical address.
  • For the generated page offset, corresponding word is located in the page and read.

 

 

Diagram-

 

The following diagram illustrates the above steps of translating logical address into physical address-

 

 

Advantages-

 

The advantages of segmented paging are-

  • Segment table contains only one entry corresponding to each segment.
  • It reduces memory usage.
  • The size of Page Table is limited by the segment size.
  • It solves the problem of external fragmentation.

 

Disadvantages-

 

The disadvantages of segmented paging are-

  • Segmented paging suffers from internal fragmentation.
  • The complexity level is much higher as compared to paging.

 

Next Article- Practice Problems On Segmented Paging

 

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Segmentation in OS | Practice Problems

Operating System

Segmentation in OS-

 

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

 

We have discussed-

  • Like Paging, Segmentation is another non-contiguous memory allocation technique.
  • Segmentation divides the process into meaningful segments.
  • Segment table stores the information about each segment of the process.

 

In this article, we will discuss a practice problem based on segmentation.

 

PRACTICE PROBLEM BASED ON SEGMENTATION-

 

Problem-

 

Consider the following segment table-

 

Segment No. Base Length
0 1219 700
1 2300 14
2 90 100
3 1327 580
4 1952 96

 

Which of the following logical address will produce trap addressing error?

  1. 0, 430
  2. 1, 11
  3. 2, 100
  4. 3, 425
  5. 4, 95

 

Calculate the physical address if no trap is produced.

 

Solution-

 

In a segmentation scheme, the generated logical address consists of two parts-

  1. Segment Number
  2. Segment Offset

 

We know-

  • Segment Offset must always lie in the range [0, limit-1].
  • If segment offset becomes greater than or equal to the limit of segment, then trap addressing error is produced.

 

Option-A: 0, 430-

 

Here,

  • Segment Number = 0
  • Segment Offset = 430

 

We have,

  • In the segment table, limit of segment-0 is 700.
  • Thus, segment offset must always lie in the range = [0, 700-1] = [0, 699]

 

Now,

  • Since generated segment offset lies in the above range, so request generated is valid.
  • Therefore, no trap will be produced.
  • Physical Address = 1219 + 430 = 1649

 

Option-B: 1, 11-

 

Here,

  • Segment Number = 1
  • Segment Offset = 11

 

We have,

  • In the segment table, limit of segment-1 is 14.
  • Thus, segment offset must always lie in the range = [0, 14-1] = [0, 13]

 

Now,

  • Since generated segment offset lies in the above range, so request generated is valid.
  • Therefore, no trap will be produced.
  • Physical Address = 2300 + 11 = 2311

 

Option-C: 2, 100-

 

Here,

  • Segment Number = 2
  • Segment Offset = 100

 

We have,

  • In the segment table, limit of segment-2 is 100.
  • Thus, segment offset must always lie in the range = [0, 100-1] = [0, 99]

 

Now,

  • Since generated segment offset does not lie in the above range, so request generated is invalid.
  • Therefore, trap will be produced.

 

Option-D: 3, 425-

 

Here,

  • Segment Number = 3
  • Segment Offset = 425

 

We have,

  • In the segment table, limit of segment-3 is 580.
  • Thus, segment offset must always lie in the range = [0, 580-1] = [0, 579]

 

Now,

  • Since generated segment offset lies in the above range, so request generated is valid.
  • Therefore, no trap will be produced.
  • Physical Address = 1327 + 425 = 1752

 

Option-E: 4, 95-

 

Here,

  • Segment Number = 4
  • Segment Offset = 95

 

We have,

  • In the segment table, limit of segment-4 is 96.
  • Thus, segment offset must always lie in the range = [0, 96-1] = [0, 95]

 

Now,

  • Since generated segment offset lies in the above range, so request generated is valid.
  • Therefore, no trap will be produced.
  • Physical Address = 1952 + 95 = 2047

 

Thus, Option-(C) is correct.

 

To watch video solution and practice other problems,

Watch this Video Lecture

 

Next Article- Segmented Paging in OS

 

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Segmentation in OS | Segmentation and Paging

Operating System

Non-Contiguous Memory Allocation-

 

Before you go through this article, make sure that you have gone through the previous article on Non-Contiguous Memory Allocation.

 

We have discussed-

  • Non-contiguous memory allocation is a memory allocation technique.
  • It allows to store parts of a single process in a non-contiguous fashion.

 

Techniques-

 

There are two popular techniques used for non-contiguous memory allocation-

 

 

In this article, we will discuss about Segmentation.

 

Segmentation-

 

  • Like Paging, Segmentation is another non-contiguous memory allocation technique.
  • In segmentation, process is not divided blindly into fixed size pages.
  • Rather, the process is divided into modules for better visualization.

 

Characteristics-

 

  • Segmentation is a variable size partitioning scheme.
  • In segmentation, secondary memory and main memory are divided into partitions of unequal size.
  • The size of partitions depend on the length of modules.
  • The partitions of secondary memory are called as segments.

 

Example-

 

Consider a program is divided into 5 segments as-

 

 

Segment Table-

 

  • Segment table is a table that stores the information about each segment of the process.
  • It has two columns.
  • First column stores the size or length of the segment.
  • Second column stores the base address or starting address of the segment in the main memory.
  • Segment table is stored as a separate segment in the main memory.
  • Segment table base register (STBR) stores the base address of the segment table.

 

For the above illustration, consider the segment table is-

 

 

Here,

  • Limit indicates the length or size of the segment.
  • Base indicates the base address or starting address of the segment in the main memory.

 

In accordance to the above segment table, the segments are stored in the main memory as-

 

 

Translating Logical Address into Physical Address-

 

  • CPU always generates a logical address.
  • A physical address is needed to access the main memory.

 

Following steps are followed to translate logical address into physical address-

 

Step-01:

 

CPU generates a logical address consisting of two parts-

  1. Segment Number
  2. Segment Offset

 

 

  • Segment Number specifies the specific segment of the process from which CPU wants to read the data.
  • Segment Offset specifies the specific word in the segment that CPU wants to read.

 

Step-02:

 

  • For the generated segment number, corresponding entry is located in the segment table.
  • Then, segment offset is compared with the limit (size) of the segment.

 

Now, two cases are possible-

 

Case-01: Segment Offset >= Limit

 

  • If segment offset is found to be greater than or equal to the limit, a trap is generated.

 

Case-02: Segment Offset < Limit

 

  • If segment offset is found to be smaller than the limit, then request is treated as a valid request.
  • The segment offset must always lie in the range [0, limit-1],
  • Then, segment offset is added with the base address of the segment.
  • The result obtained after addition is the address of the memory location storing the required word.

 

Diagram-

 

The following diagram illustrates the above steps of translating logical address into physical address-

 

 

Advantages-

 

The advantages of segmentation are-

  • It allows to divide the program into modules which provides better visualization.
  • Segment table consumes less space as compared to Page Table in paging.
  • It solves the problem of internal fragmentation.

 

Disadvantages-

 

The disadvantages of segmentation are-

  • There is an overhead of maintaining a segment table for each process.
  • The time taken to fetch the instruction increases since now two memory accesses are required.
  • Segments of unequal size are not suited for swapping.
  • It suffers from external fragmentation as the free space gets broken down into smaller pieces with the processes being loaded and removed from the main memory.

 

To gain better understanding about Segmentation,

Watch this Video Lecture

 

Next Article- Practice Problems On Segmentation

 

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Page Replacement Algorithms | Important Results

Operating System

Page Replacement Algorithms-

 

Before you go through this article, make sure that you have gone through the previous article on Page Replacement Algorithms.

 

We have discussed-

  • When a page fault occurs, the required page has to be brought from the secondary memory.
  • If all the frames of main memory are already occupied, then a page has to be replaced.
  • Page replacement algorithms help to decide which page should be replaced.

 

In this article, we will discuss some important results.

 

Important Results-

 

Result-01: For a Reference String Consisting Of Repeated Sequence Of Page Numbers-

 

For such a reference string,

  • Optimal page replacement algorithm replaces the most recently used page to minimize the page faults.
  • This is because most recently page will be required after the longest time.
  • Thus, Optimal page replacement algorithm acts as Most Recently Used (MRU) page replacement algorithm.

 

In other words,

  • MRU page replacement algorithm will behave as Optimal page replacement algorithm.
  • MRU page replacement algorithm gives the optimal performance.

 

Example-

 

Consider the following reference string consisting of a repeated sequence of page numbers-

1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6

 

Here,

  • Both Optimal and MRU page replacement algorithm replaces the most recently used page.
  • Most recently used page will be required after the longest time.
  • Hence, both these algorithms give the optimal performance.

 

 

Total number of page faults occurred = 12

 

NOTES-

 

Note-01:

 

  • These are the minimum number of page faults that are possible.
  • If any other page replacement algorithm is used, it will surely give rise to more number of page faults.
  • FIFO and LRU page replacement algorithm give rise to 24 number of page faults.

 

Note-02:

 

  • CPU may generate such a reference string when executing loops.
  • This is because while executing loops, same set of instructions have to be executed again and again.

 

Note-03:

 

  • Principle of locality states that the next reference would be most probably from the most recently used pages.
  • Here, the behavior of Optimal page replacement algorithm contradicts with the principle of locality.
  • Here, Optimal algorithm replaces the most recently used page to give the optimal performance.
  • So, from here we can infer that principle of locality does not hold true in certain scenarios.

 

Result-02: For a Reference String Where First Half String Is a Mirror Image Of Other Half-

 

For such a reference string,

  • Optimal page replacement algorithm replaces the least recently used page or firstly arrived page to minimize page faults.
  • This is because such a page will be required after the longest time.
  • Thus, Optimal page replacement algorithm acts as LRU and FIFO page replacement algorithm.

 

In other words,

  • LRU and FIFO page replacement algorithm will behave as Optimal page replacement algorithm.
  • LRU and FIFO page replacement algorithm gives the optimal performance.

 

Example-

 

Consider the following reference string where first half string is a mirror image of other half string-

1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 8, 7, 6, 5, 4, 3, 2, 1

 

Here,

  • Optimal, LRU and FIFO page replacement algorithms replaces the least recently used or firstly arrived page.
  • Least recently used or firstly arrived page will be required after the longest time.
  • Hence, all these algorithms give the optimal performance.

 

 

Total number of page faults occurred = 14

 

  • These are the minimum number of page faults that are possible.
  • If any other page replacement algorithm is used, it will surely give rise to more number of page faults.

 

Remember-

 

  • Optimal page replacement algorithm always gives the best performance in every scenario.
  • This is because it foresees the future and takes the form of an algorithm which gives the best performance in that scenario.

 

Next Article- Segmentation in OS

 

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Belady’s Anomaly | Page Fault | Paging

Operating System

Page Fault in OS-

 

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

 

We have discussed-

  • When a page fault occurs, the required page has to be brought from the secondary memory.
  • If all the frames of main memory are already occupied, then a page has to be replaced.
  • Page Replacement Algorithms help to decide which page should be replaced.

 

Effect of Increasing Number of Frames-

 

  • The number of page faults should either decrease or remain constant on increasing the number of frames in main memory.
  • But sometimes the unusual behavior is observed.
  • Sometimes, on increasing the number of frames in main memory, the number of page faults also increase.

 

Belady’s Anomaly-

 

Belady’s Anomaly is the phenomenon of increasing the number of page faults on increasing the number of frames in main memory.

 

Following page replacement algorithms suffer from Belady’s Anomaly-

  • FIFO Page Replacement Algorithm
  • Random Page Replacement Algorithm
  • Second Chance Algorithm

 

NOTE-

 

In the above discussion,

  • “Algorithms suffer from Belady’s Anomaly” does not mean that always the number of page faults will increase on increasing the number of frames in main memory.
  • This unusual behavior is observed only sometimes.

 

Reason Behind Belady’s Anomaly-

 

  • An algorithm suffers from Belady’s Anomaly if and only if it does not follow stack property.
  • Algorithms that follow stack property are called as stack based algorithms.
  • Stack based algorithms do not suffer from Belady’s Anomaly.
  • This is because these algorithms assign priority to a page for replacement that is independent of the number of frames in the main memory.

 

Examples-

 

Following page replacement algorithms are stack based algorithms-

  1. LRU Page Replacement Algorithm
  2. Optimal Page Replacement Algorithm

Hence, they do not suffer from Belady’s Anomaly.

 

Stack Property-

 

Consider-

  • Initially, we had ‘m’ number of frames in the main memory.
  • Now, the number of frames in the main memory is increased to ‘m+1’.

 

According to stack property-

  • At each stage, the set of pages that were present in the main memory when number of frames is ‘m’ will be compulsorily present in the corresponding stages in main memory when the number of frames is increased to ‘m+1’.

 

The following illustrations explain it clearly.

 

Illustration-01: For Optimal Page Replacement Algorithm-

 

Consider the reference string is-

0, 1, 2, 3, 0, 1, 4, 0, 1, 2, 3, 4

 

Now, observe the following two cases-

 

Case-01: When frame size = 3

 

 

Number of page faults = 7

 

Case-02: When frame size = 4

 

 

Number of page faults = 6

 

From here, we can observe-

  • At all the stages in case-02, main memory compulsorily contains the set of pages that are present in the corresponding stages in case-01.
  • Thus, optimal page replacement algorithm follows the stack property.
  • Hence, it does not suffer from Belady’s Anomaly.
  • As a proof, number of page faults decrease when the number of frames is increased from 3 to 4.

 

Illustration-02: For LRU Page Replacement Algorithm-

 

Consider the reference string is-

0, 1, 2, 3, 0, 1, 4, 0, 1, 2, 3, 4

 

Now, observe the following two cases-

 

Case-01: When frame size = 3

 

 

Number of page faults = 10

 

Case-02: When frame size = 4

 

 

Number of page faults = 8

 

From here, we can observe-

  • At all the stages in case-02, main memory compulsorily contains the set of pages that are present in the corresponding stages in case-01.
  • Thus, LRU page replacement algorithm follows the stack property.
  • Hence, it does not suffer from Belady’s Anomaly.
  • As a proof, number of page faults decrease when the number of frames is increased from 3 to 4.

 

Illustration-03: For FIFO Page Replacement Algorithm-

 

Consider the reference string is-

0, 1, 2, 3, 0, 1, 4, 0, 1, 2, 3, 4

 

Now, observe the following two cases-

 

Case-01: When frame size = 3

 

 

Number of page faults = 9

 

Case-02: When frame size = 4

 

 

Number of page faults = 10

 

From here, we can observe-

  • At stage-07 and stage-08 in case-02, main memory does not contain the set of pages that are present in the corresponding stages in case-01.
  • Thus, FIFO page replacement algorithm does not follow the stack property.
  • Hence, it suffers from Belady’s Anomaly.
  • As a proof, number of page faults increase when the number of frames is increased from 3 to 4.

 

NOTE-

 

In the above illustration,

  • FIFO Page Replacement Algorithm suffers from Belady’s Anomaly.
  • It would be wrong to say that “FIFO Page Replacement Algorithm always suffer from Belady’s Anomaly”.

 

To gain better understanding about Belady’s Anomaly,

Watch this Video Lecture

 

Next Article- Page Replacement Algorithms Important Results

 

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