CGS 3763 Operating Systems Concepts Spring 2013

CGS 3763 Operating Systems Concepts Spring 2013 Dan C. Marinescu Office: HEC 304 Office hours: M-Wd 11:30 - 12:30 AM

Last time: Answers to student questions Page replacement algorithms Today Page replacement algorithms Implementation of paging Hit and miss ratios Performance penalties Next time Virtual memory Reading assignments Chapters 8 and 9 of the textbook Lecture 37 – Monday, April 15, 2013 Lecture 37

FIFO Page replacement algorithm PS: Primary storage Lecture 37

OPTIMAL page replacement algorithm Lecture 37

LRU page replacement algorithm Lecture 37

LRU, OPTIMAL, MRU LRU looks only at history OPTIMAL “knows” not only the history but also the future. In some particular cases Most Recently Used Algorithm performs better than LRU. Example: primary device with 4 cells. Reference string 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 LRU F FFFFFFFFFFFFFF MRU F FFF - - - - F - - - F - - Lecture 37

The OPTIMAL replacement policy keeps in the 3-blocks primary memory the same pages as it does in case of the 4-block primary memory. Lecture 37

The FIFO replacement policy does not keep in the 3-blocks primary memory the same pages as it does in case of the 4-block primary memory. Lecture 37

The LRU replacement policy keeps in the 3-blocks primary memory the same pages as it does in case of the 4-block primary memory. Lecture 37

The FIFO replacement policy does not keep in the 3-blocks primary memory the same pages as it does in case of the 4-block primary memory Lecture 37

Implementation of Page Table • Page table is kept in main memory • Page-table base register (PTBR) points to the page table • Page-table length register (PRLR) indicates size of the page table • In this scheme every data/instruction access requires two memory accesses. One for the page table and one for the data/instruction. • To reduce the access time cache the most recently used frame addresses for pages. Use of a special fast-lookup hardware cache the translation look-aside buffers (TLBs) implemented as an associative memory. • Before a process A gets control of the CPU • the PTBR register is loaded with the address of the page table of process A • the PRLR is loaded with the size of the page table of process A. Lecture 37

Paging hardware Lecture 37

Content addressable or associative memory Page # Frame # • Allows parallel search. • Instead an address you search in parallel for a cell with a given contents. Address translation (p, d) • If p is in associative register, get frame # out • Otherwise get frame # from page table in memory Lecture 37

TLB hits and misses Lecture 37

Dynamic address translation Lecture 37

Hit ratio and miss ratios The primary memory is several orders of magnitude faster than the secondary memory. If a virtual address is within a page P previously brought in the primary memory we have a “hit,” otherwise we have a “miss” and before we access the data we have to bring page P from the secondary storage in, a time consuming operation. Hit ratio  the fraction of the total number of memory references when we had a hit. Miss ratio  the fraction of the total number of memory references when we had a miss. Lecture 37

Two-level memory system • To understand the effect of the hit/miss ratios we consider only a two level memory system • P  a faster and smaller primary memory • S  A slower and larger secondary memory. • The two levels could be: P  L1 cache and S  main memory P  main memory and S disk Lecture 37

The performance of a two level memory • The latency Lp << LS • LP latency of the primary device e.g., 10 nsec for RAM • LS latency of the secondary device, e.g., 10 msec for disk • Hit ratio h the probability that a reference will be satisfied by the primary device. • Average Latency (AS)  AS = h x LP + (1-h) LS. • Example: • LP = 10 nsec (primary device is main memory) • LS = 10 msec (secondary device is the disk) • Hit ratio h= 0.90  AS= 0.9 x 10 + 0.1 x 10,000,000 = 1,000,000.009 nsec~ 1000 microseconds = 1 msec • Hit ratio h= 0.99  AS= 0.99 x 10 + 0.01 x 10,000,000 = 100,000.0099 nsec~ 100 microseconds = 0.1 msec • Hit ratio h= 0.999  AS= 0.999 x 10 + 0.001 x 10,000,000 = 10,000.0099 nsec~ 10 microseconds = 0.01 msec • Hit ratio h= 0.9999  AS= 0.999 0x 10 + 0.001 x 10,000,000 = 1,009.99 nsec~ 1 microsecond This considerable slowdown is due to the very large discrepancy (six orders of magnitude) between the primary and the secondary device. Lecture 37

CGS 3763 Operating Systems Concepts Spring 2013