Demand Paging




Demand Paging

Consider how an executable program might be loaded from disk into memory. One option is to load the entire program in physical memory at program execution time. However, a problem with this approach, is that we may not initially need the entire program in memory. Consider a program that starts with a list of available options from which the user is to select. Loading the entire program into memory results in loading the executable code for all options, regardless of whether an option is ultimately selected by the user or not. An alternative strategy is to initially load pages only as they are needed. This technique is known as demand paging and is commonly used in virtual memory systems.

Demand Paging

With demand-paged virtual memory, pages are only loaded when they are demanded during program execution; pages that are never accessed are thus never loaded into physical memory. A demand-paging system is similar to a paging system with swapping (Figure 9.4) where processes reside in secondary memory (usually a disk). When we want to execute a process, we swap it into memory. Rather than swapping the entire process into memory, however, we use a lazy swapper. A lazy swapper never swaps a page into memory unless that page will be needed. Since we are now viewing a process as a sequence of pages, rather than as one large contiguous address space, use of the term swapper is technically incorrect. A swapper manipulates entire processes, whereas a pager is concerned with the individual pages of a process. We thus use pager, rather than swapper, in connection with demand paging.

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Basic Concepts

When a process is to be swapped in, the pager guesses which pages will be used before the process is swapped out again. Instead of swapping in a whole process, the pager brings only those necessary pages into memory. Thus, it avoids reading into memory pages that will not be used anyway, decreasing the swap rime and the amount of physical memory needed. With this scheme, we need some form of hardware support to distinguish between the pages that are in memory and the pages that are on the disk. The valid-invalid bit scheme described in Section 8.5 can be used for this purpose. This time, however, when this bit is set to "valid/" the associated page is both legal and in memory.

Demand Paging

 If the bit is set to "invalid," the page either is not valid (that is, not in the logical address space of the process) or is valid but is currently on the disk. The page-table entry for a page that is brovight into memory is set as usual, but the page-table entry for a page that is not currently in memory is either simply marked, invalid or contains the address of the page on disk. This situation is depicted in Figure 9.5.

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Demand Paging

 Notice that marking a page invalid will have no effect if the process never attempts to access that page. Hence, if we guess right and page in all and only those pages that are actually needed, the process will run exactly as though we had brought in all pages. While the process executes and accesses pages that are memory resident, execution proceeds normally.

But what happens if the process tries to access a page that was not brought into memory? Access to a page marked invalid causes a page-fault trap. The paging hardware, in translating the address through the page table, will notice that the invalid bit is set, causing a trap to the operating system. This trap is the result of the operating system's failure to bring the desired page into memory. The procedure for handling this page fault is straightforward (Figure 9.6):

1. We check an internal table (usually kept with the process control block) for this process to determine whether the reference was a valid or an invalid memory access.

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 2. If the reference was invalid, we terminate the process. If it was valid, but we have not yet brought in that page, we now page it in.

3. We find a free frame (by taking one from the free-frame list, for example).

4. We schedule a disk operation to read the desired page into the newly allocated frame.

5. When the disk read is complete, we modify the internal table kept with the process and the page table to indicate that the page is now in memory.

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6. We restart the instruction that was interrupted by the trap. The process can now access the page as though it had always been in memory. In the extreme case, we can start executing a process with no pages in memory. When the operating system sets the instruction pointer to the first instruction of the process, which is on a non-memory-resident page, the process immediately faults for the page. After this page is brought into memory, the process continues to execute, faulting as necessary until every page that it needs is in memory. At that point, it can execute with no more faults. This scheme is pure demand paging: Never bring a page into memory until it is required. Theoretically, some programs could access several new pages of memory with each instruction execution (one page for the instruction and many for data), possibly causing multiple page faults per instruction.

This situation would result in unacceptable system performance. Fortunately, analysis of running processes shows that this behavior is exceedingly unlikely. Programs tend to have locality of reference, described in Section 9.6.1, which results in reasonable performance from demand paging. The hardware to support demand paging is the same as the hardware for paging and swapping:

• Page table. This table has the ability to mark an entry invalid through a valid-invalid bit or special value of protection bits.

 • Secondary memory. This memory holds those pages that are not present in main memory. The secondary memory is usually a high-speed disk. It is known as the swap device, and the section of disk used for this purpose is known as swap space.

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Swap-space allocation is discussed in Chapter 12. A crucial requirement for demand paging is the need to be able to restart any instruction after a page fault. Because we save the state (registers, condition code, instruction counter) of the interrupted process when the page fault occurs, we must be able to restart the process in exactly the same place and state, except that the desired page is now in memory and is accessible. In most cases, this requirement is easy to meet. A page fault may occur at any memory reference. If the page fault occurs on the instruction fetch, we can restart byfetching the instruction again. If a page fault occurs while we are fetching an operand, we must fetch and decode the instruction then adding again. However, there is not much repeated work (less than one complete instruction), and the repetition is necessary only when a page fault occurs. The major difficulty arises when one instruction may modify several different locations. For example, consider the IBM. System 360/370 MVC (move character) instruction., which can move up to 256 bytes from one location to another (possibly overlapping) location. If either block (source or destination) straddles a page boundary, a page fault might occur after the move is partially done. In addition, if the source and destination blocks overlap, the source block may have been modified, in which case we cannot simply restart the instruction. This problem can be solved in two different ways. In one solution, the microcode computes and attempts to access both ends of both blocks. If a page fault is going to occur, it will happen at this step, before anything is modified. The move can then take place; w r e know that no page fault can occur, since all the relevant pages are in memory. The other solution uses temporary registers to hold the values of overwritten locations. If there is a page fault, all the old values are written back into memory before the trap occurs.

This action restores memory to its state before the instruction was started, so that the instruction can be repeated. This is by no means the only architectural problem resulting from adding paging to an existing architecture to allow demand paging, but it illustrates some of the difficulties involved. Paging is added between the CPU and the memory in a computer system. It should be entirely transparent to the user process. Thus, people often assume that paging can be added to any system. Although this assumption is true for a non-demand-paging environment, where a page fault represents a fatal error, it is not true where a page fault means only that an additional page must be brought into memory and the process restarted

Performance of Demand Paging

 Demand paging can significantly affect the performance of a computer system. To see why, let's compute the effective access time for a demand-paged memory. For most computer systems, the memory-access time, denoted ma, ranges from 10 to 200 nanoseconds. As long as we have no page faults, the effective access time is equal to the memory access time. If, however, a page fault occurs, we must first read the relevant page from disk and then access the desired word. Let p be the probability of a page fault (0 s p 5 1). We would expect p to be close to zero—that is, we would expect to have only a few page faults. The effective access time is then effective access time = (1 - p) x ma + p x page fault time. To compute the effective access time, we must know how much time is needed to service a page fault. A page fault causes the following sequence to occur:

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1. Trap to the operating system

2. Save the user registers and process state.

3. Determine that the interrupt was a page fault.

' 4. Check that the page reference was legal and determine the location of the page on the disk.

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5. Issue a read from the disk to a free frame: a. Wait in a queue for this device until the read request is serviced. b. Wait for the device seek and /or latency time. c. Begin the transfer of the page to a free frame.

6. While waiting, allocate the CPU to some other user (CPU scheduling, optional).

7. Receive an interrupt from the disk I/O subsystem (I/O completed).

8. Save the registers and process state for the other user (if step 6 is executed).

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9. Determine that the interrupt was from the disk. 10. Correct the page table and other tables to show that the desired page is now in memory.

11. Wait for the CPU to be allocated to this process again.

12. Restore the user registers, process state, and new page table, and then resume the interrupted instruction. Not all of these steps are necessary in every case. For example, we are assuming that, in step 6, the CPU is allocated to another process while the I/O occurs. This arrangement allows multiprogramming to maintain CPU utilization but requires additional time to resume the page-fault service routine when the I/O transfer is complete. In any case, we are faced with three major components of the page-fault service time:

 1. Service the page-fault interrupt.

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2. Read in the page.

 3. Restart the process.

The first and third tasks can be reduced, with careful coding, to several hundred instructions. These tasks may take from 1 to 100 microseconds each. The page-switch time, however, will probably be close to 8 milliseconds. A typical hard disk has an average latency of 3 milliseconds, a seek of 5 milliseconds, and a transfer time of 0.05 milliseconds. Thus, the total paging time is about 8 milliseconds, including hardware and software time. Remember also that we are looking at only the device-service time. If a queue of processes is waiting for the device (other processes that have caused page faults), we have to add device-queueing time as we wait for the paging device to be free to service our request, increasing even more the time to swap.

If we take an average page-fault service time of 8 milliseconds and a memory-access time of 200 nanoseconds, then the effective access time in nanoseconds is effective access time = (1 - p) x (200) + p (8 milliseconds) = (1 - p) x 200 + p x 8.00(1000 = 200 + 7,999,800 x p. We see, then, that the effective access time is directly proportional to the page-fault rate. If one access out of 1,000 causes a page fault, the effective access time is 8.2 microseconds. The computer will be slowed down by a factor of 40 because of demand paging! If we want performance degradation to be less than 10 percent, we need 220 > 200 + 7,999,800 x p, 20 > 7,999,800 x p, p < 0.0000025. That is, to keep the slowdown due to paging at a reasonable level, we can allow fewer than one memory access out of 399,990 to page-fault. In sum, it is important to keep the page-fault rate low in a demand-paging system. Otherwise, the effective access time increases, slowing process execution dramatically. An additional aspect of demand paging is the handling and overall use of swap space. Disk I/O to swap space is generally faster than that to the file system. It is faster because swap space is allocated in much larger blocks, and file lookups and indirect allocation methods are not used (Chapter 12).

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The system can therefore gain better paging throughput by copying an entire file image into the swap space at process startup and then performing demand paging from the swap space. Another option is to demand pages from the file system initially but to write the pages to swap space as they are replaced. This approach will ensure that only needed pages are read from the file system but that all subsequent paging is done from swap space. Some systems attempt to limit the amount of swap space used through demand paging of binary files. Demand pages for such files are brought directly from the file system. However, when page replacement is called for, these frames can simply be overwritten (because they are never modified), and the pages can be read in from the file system, again if needed. Using this approach, the file system itself serves as the backing store. However, swap space must still be used for pages not associated with a file; these pages include the stack and heap for a process. This method appears to be a good compromise and is used in several systems, including Solaris and BSD UNIX.

 



Frequently Asked Questions

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Ans: Paging is a memory-management scheme that permits the physical address space of a process to be noncontiguous. Paging avoids the considerable problem of fitting memory chunks of varying sizes onto the backing store; most memory-management schemes used before the introduction of paging suffered from this problem. The problem arises because, when some code fragments or data residing in main memory need to be swapped out, space must be found on the backing store. view more..
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Ans: An important aspect of memory management that became unavoidable with paging is the separation of the user's view of memory and the actual physical memory. As we have already seen, the user's view of memory is not the same as the actual physical memory. The user's view is mapped onto physical memory. This mapping allows differentiation between logical memory and. physical memory. view more..
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Ans: Although semaphores provide a convenient and effective mechanism for process synchronization, using them incorrectly can result in timing errors that are difficult to detect, since these errors happen only if some particular execution sequences take place and these sequences do not always occur. We have seen an example of such errors in the use of counters in our solution to the producer-consumer problem view more..
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Ans: Demand Paging Consider how an executable program might be loaded from disk into memory. One option is to load the entire program in physical memory at program execution time. However, a problem with this approach, is that we may not initially need the entire program in memory. Consider a program that starts with a list of available options from which the user is to select. Loading the entire program into memory results in loading the executable code for all options, regardless of whether an option is ultimately selected by the user or not. An alternative strategy is to initially load pages only as they are needed. This technique is known as demand paging and is commonly used in virtual memory systems. view more..
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Ans: Thrashing If the number of frames allocated to a low-priority process falls below the minimum number required by the computer architecture, we must suspend, that process's execution. We should then page out its remaining pages, freeing all its allocated frames. This provision introduces a swap-in, swap-out level of intermediate CPU scheduling. In fact, look at any process that does not have ''enough" frames. If the process does not have the number of frames it needs to support pages in active use, it will quickly page-fault. At this point, it must replace some page. However, since all its pages are in active use, it must replace a page that will be needed again right away. Consequently, it quickly faults again, and again, and again, replacing pages that it must bring back in immediately. This high paging activity is called thrashing. A process is thrashing if it is spending more time paging than executing. view more..
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Ans: When a process running in user mode requests additional memory, pages are allocated from the list of free page frames maintained by the kernel. This list is typically populated using a page-replacement algorithm such as those discussed in Section 9.4 and most likely contains free pages scattered throughout physical memory, as explained earlier. Remember, too, that if a user process requests a single byte of memory, internal fragmentation will result, as the process will be granted, an entire page frame. Kernel memory, however, is often allocated from a free-memory pool different from the list used to satisfy ordinary user-mode processes. view more..
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Ans: We turn next to a description of the scheduling policies of the Solaris, Windows XP, and Linux operating systems. It is important to remember that we are describing the scheduling of kernel threads with Solaris and Linux. Recall that Linux does not distinguish between processes and threads; thus, we use the term task when discussing the Linux scheduler. view more..
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Ans: Overview of Mass-Storage Structure In this section we present a general overview of the physical structure of secondary and tertiary storage devices view more..
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Ans: Allocation of Frames We turn next to the issue of allocation. How do we allocate the fixed amount of free memory among the various processes? If we have 93 free frames and two processes, how many frames does each process get? The simplest case is the single-user system. Consider a single-user system with 128 KB of memory composed of pages 1 KB in size. This system has 128 frames. The operating system may take 35 KB, leaving 93 frames for the user process. Under pure demand paging, all 93 frames would initially be put on the free-frame list. When a user process started execution, it would generate a sequence of page faults. The first 93 page faults would all get free frames from the free-frame list. view more..
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Ans: The direct-access nature of disks allows us flexibility in the implementation of files, in almost every case, many files are stored on the same disk. The main problem is how to allocate space to these files so that disk space is utilized effectively and files can be accessed quickly. Three major methods of allocating disk space are in wide use: contiguous, linked, and indexed. Each method has advantages and disadvantages. Some systems (such as Data General's RDOS for its Nova line of computers) support all three. view more..
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Ans: Free-Space Management Since disk space is limited, we need to reuse the space from deleted files for new files, if possible. (Write-once optical disks only allow one write to any given sector, and thus such reuse is not physically possible.) To keep track of free disk space, the system maintains a free-space list. The free-space list records all free disk blocks—those not allocated to some file or directory. To create a file, we search the free-space list for the required amount of space and allocate that space to the new file. This space is then removed from the free-space list. view more..
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Ans: File Concept Computers can store information on various storage media, such as magnetic disks, magnetic tapes, and optical disks. So that the computer system will be convenient to use, the operating system provides a uniform logical view of information storage. The operating system abstracts from the physical properties of its storage devices to define a logical storage unit, the file. Files are mapped by the operating system onto physical devices. These storage devices are usually nonvolatile, so the contents are persistent through power failures and system reboots. view more..
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Ans: Modern disk drives are addressed as large one-dimensional arrays of logical blocks, where the logical block is the smallest unit of transfer. The size of a logical block is usually 512 bytes, although some disks can be low-level formatted to have a different logical block size, such as 1,024 bytes. view more..
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Ans: In the previous sections, we explored the motivation for file sharing and some of the difficulties involved in allowing users to share files. Such file sharing is very desirable for users who want to collaborate and to reduce the effort required to achieve a computing goal. Therefore, user-oriented operating systems must accommodate the need to share files in spite of the inherent difficulties. In this section, we examine more aspects of file sharing view more..
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Ans: When information is stored in a computer system, we want to keep it safe from physical damage (reliability) and improper access (protection). Reliability is generally provided by duplicate copies of files. Many computers have systems programs that automatically (or through computer-operator intervention) copy disk files to tape at regular intervals (once per day or week or month) to maintain a copy should a file system be accidentally destroyed. view more..
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Ans: Goal of systems analysis and design is to improve organizational systems. This process involves developing or acquiring application software and training employees. view more..
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Ans: Disk Scheduling One of the responsibilities of the operating system is to use the hardware efficiently. For the disk drives, meeting this responsibility entails having fast access time and large disk bandwidth. The access time has two major components. The seek time is the time for the disk arm to move the heads to the cylinder containing the desired sector. The rotational latency is the additional time for the disk to rotate the desired sector to the disk head. The disk bandwidth is the total number of bytes transferred, divided by the total time between the first request for service and the completion of the last transfer. We can improve both the access time and the bandwidth by scheduling the servicing of disk I/O requests in a good order. Whenever a process needs I/O to or from the disk, it issues a system call to the operating system view more..
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Ans: The operating system is responsible for several other aspects of disk management, too. Here we discuss disk initialization, booting from disk, and bad-block recovery. view more..




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