L-5.9: What is Paging | Memory management | Operating System

So in Paging we split a process into equally sized pages and insert it into the frames of main memory Let's assume we have a process P1 whose process size is 4 bytes and page size is 2 bytes So total size of the process, as we have assumed, is 4 bytes and size of a single page is 2 bytes And number of pages per process will be 4/2 bytes whose answer is 2 bytes Now I have shown it here that the process P1 is split into two pages 0 and 1 in which the numbers 0, 1, 2 and 3 We have seen before So there are 4 bytes in the process If we start numbering the bits as per binary system because binary numbering starts from 0 I have represented these bytes (here) as 2 bytes per page thus dividing the process So in Main Memory too We have 16 byte Main Memory And we have discussed earlier On why page size and frame size should be same So that if we divide a process into pages then one of the pages can be taken and fitted in the frame in compact Means the page should fit in the frame as it is So that's why we have taken the page size and frame size same always The number of frames can be calculated by dividing the memory size (16 Bytes) by the frame size (2 Bytes) That I have already represented here But always remember that number of frames (here) is 8 and whenever we start binary numbering or even in arrays We have numbered the pages 0-7 Now what is it representing internally? We have 16 Bytes of Main Memory which means I can store data upto 16 bytes Now I have just named these bytes We know that Main Memory is Byte Addressable Which means we represent the memory through bytes only That is whenever CPU demands Now we easily take the bytes out and give to the CPU through which we get to be proved that This is just a representation of dividing process into pages and Memory into frames Here remember an important point that CPU has started executing P1 When CPU is executing P1 there arises a problem that CPU doesn't know that you have created process pages and fitted them into memory frames Whenever CPU executes a process it just calls the data or instructions of the process Means, among the bytes 0,1,2 and 3 of P1 if CPU says This byte - But Where is it actually? It is in the Main Memory in pages form as we have seen that So here let's say What I am doing is let's say these The blocks or frames 1 and 2 are already filled So the Page 0 of P1 has come here and Page 1 of P1 has come here Let's say this one is also filled Let us fill these frames too so that you will get cleared on the concept of inserting data into frames So we have assumed that Page 0 and Page 1 of P1 has been inserted into Frame 2 and Frame 4 of the Main Memory respectively Which implies the bytes 0 and 1 of P1 have occupied the places of bytes 4 and 5 of Frame 2 and bytes 2 and 3 of P1 have occupied the places of bytes 8 and 9 of Frame 4 And that the bytes in P1 have been stored in their respective places in Main Memory Now irrespective of the knowledge about our paging strategy CPU is simply asking for byte 3 of P1 So if we want to give byte 3 to CPU from Main Memory it may be allocated anywhere in random which means it can be stored anywhere in the Main Memory Now here we need Mapping The address generated by the CPU That is not the Absolute address CPU needs data from here but it has generated the address 3. So I need such a technique which converts the address of 3 to address to 9 and give me that data This is what we call as Mapping This Mapping is getting done by the Memory Management Unit (MMU) The MMU converts the address generated by the CPU into Absolute address For this conversion the MMU uses Page Tables You can check it out how Every If every process has pages then I need more than one page tables We use a simple concept here We are creating page table of P1 and how many entries will it have? How many entries will a page table have? It is the number of pages in the process In the sense how many pages are in P1? Then how many entries will be here? This entry is Page 0 Now where is Page 0, look at it carefully, where is it? In frame 2 of Main Memory So what does it contain? The frame number So whenever I want to access a page of a process I will get to know of the frame number from which I will fetch its data Likewise I will access Page 1 of P1 from Frame 4 Now if the CPU asks for any byte, for example byte 3 here, we don't know where it is located First of all we have to locate the page number of byte 3 So that I can go to that page, let's say it is located either at Page 0 or at Page 1 If it is at Page 1 then I should check Frame 4 and fetch the data Now how does it work? CPU generates the address called Logical Address, which is very important point Made of two factors How much is the page size? How many bits do we need to represent two bytes? If we have two numbers, what may be the numberNs? How many bits do I require to represent 0 and 1? Only 1 One is 0 and other is 1 Similarly let's say we have total 8 numbers 8 numbers means the numbers that I have are 0 to 7 If I have 8 numbers the numeric range I have now is from 0 to 7 So zero is represented as 000 and seven is represented as 111 which shows that I need at most 3 bits to represent numbers of this range But if I want to represent numbers beyond 8 I need 4 bits The numbers in 0 to 7 are 8 And to represent this numeric range A very important point on numbers on which we are discussing, say, we have 16 numbers My numeric range is between 0 and 15 If 15 is the largest number of the range then it is represented with 4 bits (ie) 1111 We are not talking about values but the numbers are 16 in the range If 15 is the largest number of the range then it is represented with 4 bits If my page size is 2 bytes then I need only 1 bit to represent it Here I have Page number and given number of Pages per process is also 2. So I only need 1 bit to represent it (ie) 0 and 1 If I have 2 bits 0 and 1 I need only 1 bit space to represent it and we are given space here Now the point is the CPU generated address is 3 So to represent 3 in Binary I have written the page number here as 1 Here the main point is I have to access the Page 1 of the page table in P1 We have 2 bytes in each page I have to select zeroth byte or first byte I have reached the Page number but I have to choose among the zeroth byte or the first byte to fetch That is represented in this byte where I have written 1 which means I have to select the second byte This might be a little confusing but one point to remove it is we should start numbering from 0 For example if I have 2 rows of students where there are 2 students at row 0: student 0 and student 1 Start numbering from 0 now here for byte 3 we have to go to page 1 and access the second byte To represent second byte we use 1 in Binary numbering In the sense if we go to Page 1 we select second byte in it (ie) byte 1 and yes we have fetched this byte To fetch this byte we have to convert logical address to absolute address Means I have to convert the logical address 3 to absolute address 9 To convert to 9 we have to use As we discussed earlier that logical address generated by CPU should be converted into physical address Now that I know about logical address which consists of page number and page offset Let's see about physical address Physical address represents the location of the byte to be searched The number of bits to represent the physical address depends on the size of the Main Memory Memory address is byte addressable so nothing to be done as byte is minimum Let's take 16. To represent 16 in binary we need 4 bits as discussed before So total bits of physical address is 4 bits In this the first one represents frame number and the next one represents frame offset Frame offset means frame size To represent 2 bytes we need 1 bit in the sense And if you have an iota of doubt you can check 8 numbers in 0-7 where for 0 it is 000 and for 7 it is 111 so in total we need 3 bits which is used here Total bits to represent physical address are 4 in number in which 3 of them are for frame number and 1 is for frame offset or frame size So pay attention here that the address 3 generated by CPU has to be converted into physical address 9 To convert we know we have to go to frame 4 in page 1 of P1 It is given page 1 and in there we have frame 4 which we accessed too Here we accessed the page 1 successfully How do we represent 4 in binary? Frame 4 is theoretical but CPU takes only binary values as per which 4 is represented as 100 And frame offset which means frame size Moreover the bits of page offset and frame offset should be same always Because if we want the second page of page 1 of P1 then it will also be located in the second position in the respective frame This is a simple concept In this page this is first roll number and this is second roll number So if we shift places of these two to right here then their places will be retained as it is here too Here without changing their sequence we shifted their places to somewhere which is unknown distant So in normal terms I need the second byte and in binary terms I need the byte 1 from the frame 4 This represents frame 1 and not frame 0 Therefore I will fetch data from byte 9 and give it to CPU So it is summarised that CPU generates the logical address CPU just tells the byte number it needs which we should convert into logical address to access the page number which again should be converted into physical address to access the frame number Now let's review this concept even faster with a simplest example CPU says I need byte 1 It needs byte 1 Now that the CPU has generated the address of byte 1 how can we represent it? If we have two bit space for representation we represent it as 01 Which means We need byte 1 in page 0 (ie) this one which we wanted After we went to the page table to check the page 0, it says that it is located at frame 2 in Main memory to which we bounced again In that, I need to pick the byte 5 We got to know that it is located at frame 2 in which 2 can be represented as 010 in binary 2 can be represented as 010 when represented in 3 bits And this bit will remain as it is of page offset as we are not changing the sequence of the bits So can you guess what value comes in decimal? Now I will get the data from the same location as that of byte 5 So we need Page tables to map from logical address to physical address of a byte Now after CPU generates a byte address, we find the corresponding page numbers in which we see the corresponding frame number to which we jump and fetch the required byte and give it to the CPU back And the mapping from 1 to 5 is done by the Memory Management Unit and how? Just remember this always Which byte in it There are two factors in physical address These two bits will not change because Bytes are placed as it is without any change

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Id: 6c-mOFZwP_8

Length: 25min 53sec (1553 seconds)

Published: Mon Apr 02 2018