羅左欣 BE STRONG TO BE USEFUL

20170608 [學習筆記] Linux 系統程式 (14)


一、作業系統

(一) Background

  • Code needs to be in memory to execute, but entire program rarely used
    • Error code, unusual routines, large data structures
  • Entire program code not needed at same time
  • Consider ability to execute partially-loaded program
    • Program no longer constrained by limits of physical memory
    • Program and programs could be larger than physical memory
  • Virtual memory – separation of user logical memory from physical memory
    • Only part of the program needs to be in memory for execution
    • Logical address space can therefore be much larger than physical address space
    • Allows address spaces to be shared by several processes
    • Allows for more efficient process creation
    • More programs running concurrently
    • Less I/O needed to load or swap processes
  • Virtual memory can be implemented via
    • Demand paging
    • Demand segmentation

(二) Virtual Address Space

  • Enables sparse address spaces with holes left for growth, dynamically linked libraries, etc
  • System libraries shared via mapping into virtual address space
  • Shared memory by mapping pages read-write into virtual address space
  • Pages can be shared during fork(), speeding process creation

(三) Demand Paging

  • Could bring entire process into memory at load time
  • Or bring a page into memory only when it is needed
    • Less I/O needed, no unnecessary I/O
    • Less memory needed
    • Faster response
    • More users
  • Page is needed » reference to it
    • invalid reference » abort
    • not-in-memory » bring to memory
  • Lazy swapper – never swaps a page into memory unless page will be needed
    • Swapper that deals with pages is a pager

1. Valid-Invalid Bit

  • With each page table entry a valid–invalid bit is associated (v » in-memory – memory resident, i » not-in-memory)
  • Initially valid–invalid bit is set to i on all entries

2. Example

  • Memory access time = 200 nanoseconds
  • Average page-fault service time = 8 milliseconds

    EAT = (1 – p) x 200 + p (8 milliseconds) = (1 – p) x 200 + p x 8,000,000 = 200 + p x 7,999,800

  • If one access out of 1,000 causes a page fault, then EAT = 8.2 microseconds. (This is a slowdown by a factor of 40)
  • If want performance degradation < 10 percent
    • 220 > 200 + 7,999,800 x p
    • 20 > 7,999,800 x p
    • p < .0000025 (< one page fault in every 400,000 memory accesses)

3. Optimizations

  • Copy entire process image to swap space at process load time
    • Then page in and out of swap space
    • Used in older BSD Unix
  • Demand page in from program binary on disk, but discard rather than paging out when freeing frame
    • Used in Solaris and current BSD

(四) Page Fault

  • If there is a reference to a page, first reference to that page will trap to operating system
    • page fault
  • Operating system looks at another table to decide
    • Invalid reference » abort
    • Just not in memory
  • Get empty frame
  • Swap page into frame via scheduled disk operation
  • Reset tables to indicate page now in memory Set validation bit = v
  • Restart the instruction that caused the page fault

1. Aspects of Demand Paging

  • Extreme case – start process with no pages in memory
    • OS sets instruction pointer to first instruction of process, non-memory-resident -> page fault
    • And for every other process pages on first access
    • Pure demand paging
  • Actually, a given instruction could access multiple pages -> multiple page faults
    • Pain decreased because of locality of reference
  • Hardware support needed for demand paging
    • Page table with valid / invalid bit
    • Secondary memory (swap device with swap space)
    • Instruction restart

2. Instruction Restart

  • Consider an instruction that could access several different locations
    • block move
    • auto increment/decrement location
    • Restart the whole operation?
      • What if source and destination overlap?
Performance of Demand Paging
  • Stages in Demand Paging
    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
    5. Issue a read from the disk to a free frame: (1) Wait in a queue for this device until the read request is serviced (2) Wait for the device seek and/or latency time (3) Begin the transfer of the page to a free frame
    6. While waiting, allocate the CPU to some other user
    7. Receive an interrupt from the disk I/O subsystem (I/O completed)
    8. Save the registers and process state for the other user
    9. Determine that the interrupt was from the disk
    10. Correct the page table and other tables to show 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

(五) Copy-on-Write

  • Copy-on-Write (COW) allows both parent and child processes to initially share the same pages in memory
    • If either process modifies a shared page, only then is the page copied
  • COW allows more efficient process creation as only modified pages are copied
  • In general, free pages are allocated from a pool of zero-fill-on-demand pages
    • Why zero-out a page before allocating it?
  • vfork() variation on fork() system call has parent suspend and child using copy-on-write address space of parent
    • Designed to have child call exec()
    • Very efficient

(六) Page Replacement

  • Prevent over-allocation of memory by modifying page-fault service routine to include page replacement
  • Use modify (dirty) bit to reduce overhead of page transfers – only modified pages are written to disk
  • Page replacement completes separation between logical memory and physical memory – large virtual memory can be provided on a smaller physical memory

1. What Happens if There is no Free Frame?

  • Used up by process pages
  • Also in demand from the kernel, I/O buffers, etc
  • How much to allocate to each?
  • Page replacement – find some page in memory, but not really in use, page it out
    • Algorithm – terminate? swap out? replace the page?
    • Performance – want an algorithm which will result in minimum number of page faults
  • Same page may be brought into memory several times

2. Basic Page Replacement

  1. Find the location of the desired page on disk
  2. Find a free frame:

    a. If there is a free frame, use it b. If there is no free frame, use a page replacement algorithm to select a victim frame c. Write victim frame to disk if dirty

  3. Bring the desired page into the (newly) free frame; update the page and frame tables
  4. Continue the process by restarting the instruction that caused the trap
    • Note now potentially 2 page transfers for page fault – increasing EAT

3. Page and Frame Replacement Algorithms

  • Frame-allocation algorithm determines
    • How many frames to give each process
    • Which frames to replace
  • Page-replacement algorithm
    • Want lowest page-fault rate on both first access and re-access
  • Evaluate algorithm by running it on a particular string of memory references (reference string) and computing the number of page faults on that string
    • String is just page numbers, not full addresses
    • Repeated access to the same page does not cause a page fault
(1) First-In-First-Out (FIFO) Algorithm

(2) Optimal Algorithm
  • Replace page that will not be used for longest period of time
    • 9 is optimal for the example on the next slide
  • How do you know this?
    • Can’t read the future
  • Used for measuring how well your algorithm performs

(3) Least Recently Used (LRU) Algorithm
  • Use past knowledge rather than future
  • Replace page that has not been used in the most amount of time
  • Associate time of last use with each page

  • Counter implementation
    • Every page entry has a counter; every time page is referenced through this entry, copy the clock into the counter
    • When a page needs to be changed, look at the counters to find smallest value
      • Search through table needed
  • Stack implementation
    • Keep a stack of page numbers in a double link form:
    • Page referenced:
      • move it to the top
      • requires 6 pointers to be changed
    • But each update more expensive
    • No search for replacement
  • LRU and OPT are cases of stack algorithms that don’t have Belady’s Anomaly

(4) LRU Approximation Algorithms
  • LRU needs special hardware and still slow
  • Reference bit
    • With each page associate a bit, initially = 0
    • When page is referenced bit set to 1
    • Replace any with reference bit = 0 (if one exists)
      • We do not know the order, however
  • Second-chance algorithm
    • Generally FIFO, plus hardware-provided reference bit
    • Clock replacement
    • If page to be replaced has
      • Reference bit = 0 -> replace it
      • reference bit = 1 then:
        • set reference bit 0, leave page in memory
        • replace next page, subject to same rules

(5) Counting Algorithms
  • Keep a counter of the number of references that have been made to each page
    • Not common
  • LFU Algorithm: replaces page with smallest count
  • MFU Algorithm: based on the argument that the page with the smallest count was probably just brought in and has yet to be used
(6) Page-Buffering Algorithms
  • Keep a pool of free frames, always
    • Then frame available when needed, not found at fault time
    • Read page into free frame and select victim to evict and add to free pool
    • When convenient, evict victim
  • Possibly, keep list of modified pages
    • When backing store otherwise idle, write pages there and set to non-dirty
  • Possibly, keep free frame contents intact and note what is in them
    • If referenced again before reused, no need to load contents again from disk
    • Generally useful to reduce penalty if wrong victim frame selected

Similar Posts

Comments