Multiprocessing Memory Management

1. Multiprocessing Memory Management • Partitioning • Relocation • Fragmentation • Relocation • Virtual Memory • Swapping

2. Partitioning • Every process including OS needs its own part of memory • Memory partitioning: One partition per process • static (fixed) vs. dynamic partitioning • static disadvantage: waste, large programs can’t run • Relocation: After assigning a partition to a process you have to adjust all addresses in the program • Security: Every process has only access to its own memory • Bounds register

3. Fixed Partition Example

4. Swapping • Example: multiprocessing environment and a round robin scheduling • 2 big processes to execute that need partition 3 • either only allow one of them in the ready queue • or whenever you start the next process you have to liberate partition 3 to make room for the next one • “Swapping”: moving of memory content from and to disk as processes switch • Can be very expensive

5. Dynamic Partitioning • No fixed size partition • Before a process starts running it allocates as much memory as it needs • You waste less memory, instead of waiting for partition 3, the second big process can have a combination of 1 and 2. • Same problem with swapping as before if the total size of all processes is bigger than total memory • Additional Problem: Fragmentation

6. Round Robin and Swapping

7. Free Process C Process B Process A Fragmentation A Finishes D starts B Finishes E starts C Finishes F starts What about G? Free Process F Free Process E Free G Process D

8. B C D Fragmentation continued • Happens in all forms of memory that you can alter (RAM, disk …) • Hard drive solution: instead of looking for a block that is big enough you start saving and if you run into some used part you leave a link to the next B A_1 C A_2 D A_rest

9. B C D Fragmentation • Problem of piecewise storage: slow access • More general solution: Rearrange memory content from time to time B C D

10. Virtual Memory • Motivation • More flexibility with memory partitioning • Independence of Physical memory size and address space • physical memory (RAM) can be smaller than the total memory requirement of a process • address space is limited (bus size and ISA) than physical memory • Address is not a real physical address but has to be translated

11. VM Example: Less physical memory than needed • RAM: 32K with address size is 15 bit • Job needs 64 K (“virtual” address size is 16 bit) • You could store half in RAM and second half on hard drive • You need data from 0010000000000100 • Determine: Is this location in RAM or on hard drive? • If RAM: find corresponding physical address and read • If hard drive: Load other half into RAM (done by OS), find corresponding physical address and read

12. VM: Technical Solution • Instead of taking half’s you make smaller slices called pages • Page table keeps track which page is in physical memory and where • OS loads and stores pages from disk into physical memory (RAM) if a page miss happens • Address translation in Memory Management Unit (MMU) • each virtual address gets translated into a physical one

13. Memory access

14. Page Table • Task: • Virtual address is: 0010000000000100 • How do you find correct entry in page table? • How do you find the correct corresponding address?

15. Simple Address translation Some address size computation: • Virtual address size: 16 • Physical address size: 15 • Table size: 16 entries • address size for table = 4 • Page size: 4K • How many bit do you need to find something within the page: 12 • Number of pages in RAM: 8 • You need 3 bit to find page

16. Example • Virtual Address: 0010000000000100 • Split address: • first 4 bit: 0010 = address in page table • last 12 bit: 000000000100 = page offset • Look up physical page address in page table at address 0010 (entry 2): 110 • Recombine physical page address and page offset to physical address: • 110000000000100

17. More Examples 00 0 00 0 01 1 00 1 00 0 00 0 10 1 00 0 00 0 11 1 00 0 00 0 00 0 00 0 00 0 00 0 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 • Convert the following virtual addresses given the page table: 01101111 11000100 10010010 Questions: • How big is physical memory? • How big is a page? • How many pages are in physical memory?

18. Solutions • 16 entries in page table = 4 bit of virtual address to locate page will be replaces by 2 bit (content of page table) • Physical address has 6 bit, size= 26 • virtual address was 8 bit of which 4 were for page table: 4 bit remaining for page offset • Page size = 16 • 4 pages in physical memory 01101111=111111 11000100=000100 10010010=100010

19. Page replacement strategy • If process requests an address that is not in physical memory the present bit in page table is 0 • OS interrupt: read page that contains that address into physical memory and save one old page to disk • Very similar to cache • Objective: • Keep pages that are likely to be used again • Minimize computational overhead • Strategies: • Least recently used • Oldest

20. What has to be done in case of page replacement? • Write old page from physical memory to disk • Load new page into physical memory • Update page table: • Set present bit of old page to 0 • Insert physical page address in table at position given by first 4 bits of virtual address

21. Example of Table Update 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 00 0 00 0 01 1 00 1 00 0 00 0 10 1 00 0 00 0 11 1 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 01 0 00 1 00 0 00 0 10 1 00 0 00 0 11 1 00 0 00 0 01 0 00 0 00 0 00 0 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 You request address: 00110111 This results a page miss Assume page 13 (01) is the least used

22. Virtual Memory Summary • Allow programs to run with some of their pages missing • Hardware generates an interrupt if page table entry is missing (add column(s)to page table for missing entry) • External page table used to locate page on disk • Page paged in, and instruction re-executed • Can now run programs larger than physical memory • Programmers (usually) do not need to know size of real memory.