Operating Systems

Start Lecture #21

3.4.A Belady's Anomaly

Consider a system that has no pages loaded and that uses the FIFO PRU.
Consider the following reference string (sequences of pages referenced).

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

If we have 3 frames this generates 9 page faults (do it).

If we have 4 frames this generates 10 page faults (do it).

Theory has been developed and certain PRA (so called stack algorithms) cannot suffer this anomaly for any reference string. FIFO is clearly not a stack algorithm. LRU is.

Repeat the above calculations for LRU.

3.5 Design Issues for (Demand) Paging Systems

3.5.1 Local vs Global Allocation Policies

A local PRA is one is which a victim page is chosen among the pages of the same process that requires a new frame. That is the number of frames for each process is fixed. So LRU for a local policy means the page least recently used by this process. A global policy is one in which the choice of victim is made among all pages of all processes.

• Of course we can't have a purely local policy, why?
  Answer: A new process has no pages and, even if we didn't restrict the frame needed to a local one for the first page loaded, the process would remain with only one page.
• Perhaps wait until a process has been running a while before restricting it to existing frames or give the process an initial allocation of frames based on the size of the executable.

In general a global policy seems to work better. For example, consider LRU. With a local policy, the local LRU page might have been more recently used than many resident pages of other processes. A global policy needs to be coupled with a good method to decide how many frames to give to each process. By the working set principle, each process should be given |w(k,t)| frames at time t, but this value is hard to calculate exactly.

If a process is given too few frames (i.e., well below |w(k,t)|), its faulting rate will rise dramatically. If this occurs for many or all the processes, the resulting situation in which the system is doing very little useful work due to the high I/O requirements for all the page faults is called thrashing.

Page Fault Frequency (PFF)

An approximation to the working set policy that is useful for determining how many frames a process needs (but not which pages) is the Page Fault Frequency algorithm.

• For each process keep track of the page fault frequency, which is the number of faults divided by the number of references.
• Actually, must use a window or a weighted calculation since you are interested in the recent page fault frequency.
• Actually, it is too expensive to calculate the number of references so, as above, we approximate this by the amount of (virtual) time.
• If the PFF is exceptionally low, free some of this processes frames (e.g., limit victim selection to this process for a while).
• If the PFF is too high, allocate more frames to this process. Either
  1. Raise its number of frames if using a local policy; or
  2. Bar its frames from eviction (for a while) and use a global policy.
• What if there are not enough frames in the entire system? That is, what if the PFF is too high for all processes?
  Answer: Reduce the MPL as we now discuss.

3.5.2 Load Control

To reduce the overall memory pressure, we must reduce the multiprogramming level (or install more memory while the system is running, which is not possible with current technology). That is, we have a connection between memory management and process management. These are the suspend/resume arcs we saw way back when and are shown again in the diagram on the right.

When the PFF (or another indicator) is too high, we choose a process and suspend it, thereby swapping it to disk and releasing all its frames. When the frequency gets low, we can resume one or more suspended processes. We also need a policy to decide when a suspended process should be resumed even at the cost of suspending another.

This is called medium-term scheduling. Since suspending or resuming a process can take seconds, we clearly do not perform this scheduling decision every few milliseconds as we do for short-term scheduling. A time scale of minutes would be more appropriate.

3.5.3 Page Size

Page size must be a multiple of the disk block size. Why?
Answer: When copying out a page if you have a partial disk block, you must do a read/modify/write (i.e., 2 I/Os).

Characteristics of a large page size.

• Good for demand paging I/O:

  We will learn later this term that the total time for performing 8 I/O operations each of size 1KB is much larger that the time for a single 8KB I/O. Hence it is better to swap in/out one big page than several small pages.

  But if the page is too big you will be swapping in data that are not local and hence might well not be used.

• Large internal fragmentation (1/2 page size).
  
  
• Small page table (process size / page size * size of PTE).
  
  
• These last two can be analyzed together by setting the derivative of the sum equal to 0. The minimum overhead occurs at a page size of

          sqrt(2 * process size * size of PTE)
        

  Since the term inside the sqrt is typically megabytes, we see that modern practice of having the page size a few kilobytes is near the minimum point.
  
  
• A very large page size leads to very few pages. A process will have many faults if it references more regions than the number of (large) frames that the process has been allocated.

A small page size has the opposite characteristics.

Homework: Consider a 32-bit address machine using paging with 8KB pages and 4 byte PTEs. How many bits are used for the offset and what is the size of the largest page table? Repeat the question for 128KB pages.

3.5.4 Separate Instruction and Data (I and D) Spaces

This was used when machine have very small virtual address spaces. Specifically the PDP-11, with 16-bit addresses, could address only 2₁₆ bytes or 64KB, a severe limitation. With separate I and D spaces there could be 64KB of instructions and 64KB of data.

Separate I and D are no longer needed with modern architectures having large address spaces.

3.5.5 Shared pages

Permit several processes to each have the same page loaded in the same frame. Of course this can only be done if the processes are using the same program and/or data.

3.5.6 Shared Libraries (Dynamic-Linking)

In addition to sharing individual pages, process can share entire library routines. The technique used is called dynamic linking and the objects produced are called shared libraries or dynamically-linked libraries (DLLs). (The traditional linking you did in lab1 is today often called static linking).

• With dynamic linking, frequently used routines are not linked into the program. Instead, just a stub is linked.
• When the routine is called (or when the process begins), the stub checks to see if the real routine has been loaded by another program).
  • If it has not been loaded, load it (really page it in as needed).
  • If it is already loaded, share it. The read-write data must be shared copy-on-write.
• Advantages of dynamic linking.
  • Saves RAM: Only one copy of a routing is in memory even when it is used concurrently by many processes. For example even a big server with hundreds of active processes will have only one copy of printf in memory. (In fact with demand paging only part of the routine will be in memory.)
  • Saves disk space: Files containing executable programs no longer contain copies of the shared libraries.
  • A bug fix to a dynamically linked library fixes all applications that use that library, without having to relink these applications.
• Disadvantages of dynamic linking.
  • New bugs in dynamically linked library infect all applications.
  • Applications change even when they haven't changed.
• A Technical Difficulty with dynamic linking. The shared library has different virtual addresses in each process so addresses relative to the beginning of the module cannot be used (they would need to be relocated to different addresses in the multiple copies of the module). Instead position-independent code must be used. For example, jumps within the module would use PC-relative addresses.

3.5.7 Mapped Files

The idea of memory-mapped files is to use the mechanisms in place for demand paging (and segmentation, if present) to implement I/O.

3.5.8 Cleaning Policy (Paging Daemons)

Done earlier

The only point to add is now that we know replacement algorithms one can suggest an implementation. If a clock-like algorithm is used for victim selection, one can have a two handed clock with one hand (the paging daemon) staying ahead of the other (the one invoked by the need for a free frame).

The front hand simply writes out any page it hits that is dirty and thus the trailing hand is likely to see clean pages and hence is more quickly able to find a suitable victim.

Unless specifically requested, you may ignore paging daemons when answering exam questions.

3.6 Implementation Issues

3.6.1 Operating System Involvement with Paging

When must the operating system be involved with paging?

• During process creation. The OS must guess at the size of the process and then allocate a page table and a region on disk to hold the pages that are not memory resident. A few pages of the process must be loaded.
  
  
• The Ready→Running transition. Real memory must be allocated for the page table if the table has been swapped out (which is permitted when the process is not running).
  Some hardware register(s) must be set to point to the page table. There can be many page tables resident, but the hardware must be told the location of the page table for the running process—the active page table.
  The MMU must be cleared (unless it contains a process id field).
  
  
• Processing a page fault. Lots of work is needed; see 3.6.2 just below.
  
  
• Process termination. Free the page table and the disk region for swapped out pages.

3.6.2 Page Fault Handling

What happens when a process, say process A, gets a page fault? Compare the following with the processing for a trap command and for an interrupt.

1. The hardware detects the fault and traps to the kernel (switches to supervisor mode and saves state).
   
   
2. Some assembly language code saves more state, establishes the C-language (or another programming language) environment, and calls the OS.
   
   
3. The OS determines that a page fault occurred and which page was referenced.
   
   
4. If the virtual address is invalid, process A is killed. If the virtual address is valid, the OS must find a free frame. If there is no free frames, the OS selects a victim frame. (Really, the paging daemon does this prior to the fault occurring, but it is easier to pretend that it is done here.) Call the process owning the victim frame, process B. (If the page replacement algorithm is local, then B=A.)
   
   
5. The PTE of the victim page is updated to show that the page is no longer resident.
   
   
6. If the victim page is dirty, the OS schedules an I/O write to copy the frame to disk and blocks A waiting for this I/O to occur.
   
   
7. Assuming process A needed to be blocked (i.e., the victim page is dirty) the scheduler is invoked to perform a context switch.
   • Tanenbaum forgot some here.
   • The process selected by the scheduler (say process C) runs.
   • Perhaps C is preempted for D or perhaps C blocks and D runs and then perhaps D is blocked and E runs, etc.
   • When the I/O to write the victim frame completes, a disk interrupt occurs. Assume processes C is running at the time.
   • Hardware trap / assembly code / OS determines I/O done.
   • The scheduler marks A as ready.
   • The scheduler picks a process to run, maybe A, maybe B, maybe C, maybe another processes.
   • At some point the scheduler does pick process A to run. Recall that at this point A is still executing OS code.
   
   
8. Now the O/S has a free frame (this may be much later in wall clock time if a victim frame had to be written). The O/S schedules an I/O to read the desired page into this free frame. Process A is blocked (perhaps for the second time) and hence the process scheduler is invoked to perform a context switch.
   
   
9. Again, another process is selected by the scheduler as above and eventually a disk interrupt occurs when the I/O completes (trap / asm / OS determines I/O done). The PTE in process A is updated to indicate that the page is in memory.
   
   
10. The O/S may need to fix up process A (e.g., reset the program counter to re-execute the instruction that caused the page fault).
    
    
11. Process A is placed on the ready list and eventually is chosen by the scheduler to run. Recall that process A is executing O/S code.
    
    
12. The OS returns to the first assembly language routine.
    
    
13. The assembly language routine restores registers, etc. and returns to user mode.

The user's program running as process A is unaware that all this happened (except for the time delay).

3.6.3 Instruction Backup

A cute horror story. The hardware support for page faults in the original Motorola 68000 (the first microprocessor with a large address space) was so bad that an early demand paging system for the 68000, used two processors one running one instruction behind. If the first got a page fault, there wasn't always enough information to figure out what to do so (for example did a register pre-increment occur), the system switched to the second processor after bringing in the faulting page. The next generation machine, the 68010, provided extra information on the stack so the horrible 2-processor kludge was no longer necessary.

Don't worry about instruction backup; it is very machine dependent and modern implementations tend to get it right.

3.6.4 Locking (Pinning) Pages in Memory

We discussed pinning jobs already. The same (mostly I/O) considerations apply to pages.

3.6.5 Backing Store

The issue is where on disk do we put pages that are not in frames.

• For program text, which is presumably read only, a good choice is the file executable itself.
  
  
• What if we decide to keep the data and stack each contiguous on the backing store. Data and stack grow so we must be prepared to grow the space on disk, which leads to the same issues and problems as we saw with MVT.
  
  
• If those issues/problems are painful, we can scatter the pages on the disk.
  • That is we employ paging!
  • This is NOT demand paging.
  • Need a table to say where the backing space for each page is located.
    • This corresponds to the page table used to tell where in real memory a page is located.
    • The format of the memory page table is determined by the hardware since the hardware modifies/accesses it. It is machine dependent.
    • The format of the disk page table is decided by the OS designers and is machine independent.
    • If the format of the memory page table were flexible, then we might well keep the disk information in it as well. But normally the format is not flexible, and hence this is not done.
  
  
• What if we felt disk space was too expensive and wanted to put some of these disk pages on say tape?
  Ans: We use demand paging of the disk blocks! That way "unimportant" disk blocks will migrate out to tape and are brought back in if needed.
  Since a tape read requires seconds to complete (because the request is not likely to be for the sequentially next tape block), it is crucial that we get very few disk block faults.
  I don't know of any systems that did this.

Homework: Assume every memory reference takes 0.1 microseconds to execute providing the reference page is memory resident. Assume a page fault takes 10 milliseconds to service providing the necessary disk block is actually on the disk. Assume a disk block fault takes 10 seconds service. So the worst case time for a memory reference is 10.0100001 seconds. Finally assume the program requires that a billion memory references be executed.

1. If the program is always completely resident, how long does it take to execute?
2. If 0.1% of the memory references cause a page fault, but all the disk blocks are on the disk, how long does the program take to execute and what percentage of the time is the program waiting for a page fault to complete?
3. If 0.1% of the memory references cause a page fault and 0.1% of the page faults cause a disk block fault, how long does the program take to execute and what percentage of the time is the program waiting for a disk block fault to complete?