Outline of Lecture 10

1
Outline of Lecture 10

• Cache Replacement in Multithreaded Architectures
• Page Replacement
• Stack Replacement Algorithms and Their Properties

2
Context Selection in Multithreaded Architectures
Context Selection
Functional Units
Memory
Interconnection Network
I/O
3
Context Selection, States
Software Hardware
Hardware Controls
Registers
Status word
Program Counter
Running
Waiting
Ready
Context Selection
4
Threads
Processor
Memory
Program Counter and Status Register
Ready Queue
Suspended Queue
...
...
Unloaded Ready Thread
Loaded Thread, it could be ready or suspended
CP
Context Pointer
Unloaded Suspended Thread
Register frames
5
Context Switching
One-thread execution
Two-thread execution
Three-thread execution
Time
Context-switching
6
Processor Efficiency
1.0
0.9
Linear Region
Saturation region
0
5
11
15
10
Number of Contexts
7
Efficiency Saturation Point
Factors influencing saturation point - context
switching time, ts - cache loading time, tc -
cache miss probability, pm Saturation point
happens at 1tc/(ts1/pm) number of
contexts. For small ts this is about 1tc
pm Large cache line yields small pm Large cache
line or slow memory result in large tc.
8
Context Switching Analysis
Each gray block lasts 1/pm instructions and
accesses 1/pm-1 data items. There is also s
register sets, so s threads are ready to execute
with single instruction context switch.
One-thread execution
1/pm
Each gap lasts tc cycles
Two-thread execution, s2
tc -ts-1/pm
ts1/pm
ts1/pm
In general with ts1 tc -(s-1)(1/pm1)

• We assume that ts1, costs one instruction, so in
  each of s threads lasts 1/pm instructions
• and accesses 1/pm-1 (1-pm)/pm data items. Two
  cases needs to be considered
• s(1/pm1)lt tc 1/pm (gap is not filled) then in
  time tc 1/pm the system accesses
• s(1-pm)/pm data items so the ADDT is (tc
  1/pm)/(s(1-pm)/pm) (1 tcpm)/(s-spm)
• otherwise, the gap is filled so in time 11/pm
  the system executes 1/pm-1 data
• accesses and ADDT is (1/pm1/(1/pm-1)
  (1pm)/(1-pm)
• To combine this two formulas, we notice that
  condition s(1/pm1)lt tc 1/pm means that
• s(1pm) lt1 tcpm so (1 tcpm)/sgt1 pm and (1
  tcpm)/(s-spm)gt(1 pm)/(1- pm). But then
• the following expression gives us the correct
  value
• max((1 tcpm)/(s-spm),(1 pm)/(1- pm))
  max((1 tcpm)/s,1 pm)/(1- pm).

9
Performance in Virtual Memory

• The performance of a virtual memory management
  system depends on the total number of page
  faults, which depend on
• The paging policies, including frame allocation
• Static allocation - the number of frames
  allocated to a process is fixed
• Dynamic allocation - the number of frames
  allocated to a process changes
• The frame allocation policies

10
Page Replacement

• When there is a page fault, the referenced page
  must be loaded
• If there is no available frame in memory one page
  is selected for replacement
• If the selected page has been modified, it must
  be copied back to disk (swapped out)
• Same pages may be referenced several times, so
  for good performance a good replacement algorithm
  will strive to cause minimum number of page
  faults.

11
Paging Policies

• Fetch policy -- decides when a page should be
  loaded into memory -gt demand paging
• Replacement policy -- decides which page in
  memory should be replaced -gt difficult
• Placement policy -- decides where in memory
  should a page be loaded -gt easy for paging

12
Page Faults and Performance Issues

• A page fault requires the operating system to
  carry out the page fault service. The total time
  it takes to service a page fault includes several
  time components
• The time interval to service the page fault
  interrupt - system
• The time interval to store back (swap out) the
  replaced page to the secondary storage device
  process/cleaning
• The time interval to load (swap in) the
  referenced page from the secondary storage device
  (disk unit) - process
• Delay in queuing for the secondary storage
  device - process
• Delay in scheduling the process with the
  referenced page - process

13
Demand Paging

• In demand paging, a page fault occurs when a
  reference is made to a page not in memory. The
  page fault may occur while
• fetching an instruction, or
• fetching an operand of an instruction.

14
Problems to be Solved within Demand Paging

• Two major problems must be solved to implement
  demand paging
• Each process needs a minimum number of frames.
  This minimum number is based on the machine
  architecture.
• 1. Frame allocation - decide how many frames to
  allocate to each process, usually needed only for
  loading the process to the memory initially.
• 2. Page replacement - select which pages are to
  be replaced when a page fault occurs.

15
Page Replacement Algorithms

• Paging System may be characterized by 3 items
• The Reference String
• The Page Replacement Algorithm
• The number of page frames available in memory, m
• A page replacement algorithm is said to satisfy
  the inclusion property or is called a stack
  algorithm if the set of pages in a k-frame memory
  is always a subset of the pages in a (k
  1)-frame memory.

16
Page Reference

• A page reference string is a sequence of page
  numbers in order of reference
• An example of a sequence of page references is
• lt 3,6,2,1,4,7,3,5,8,9,2,8,10,7 gt
• The sequence of page references represents the
  behavior of a process during execution
• Every Process generates a sequence of memory
  references as it runs
• Each memory reference corresponds to a specific
  virtual page
• A process memory access may be characterized by
  an ordered list of page numbers, referred to as
  the reference string

17
Reference String

• w r1 r2 ...rT-1 rT sequence of virtual page
  references
• M0 - initial memory state
• M0, M1, ... MT real memory state
• Mt under request for page rt is Mt-1 Xt - Yt
• where Xt pages brought in,
• and Yt pages moved out in t-th step

For demand driven fetching
18
Cost of Fetching

• f(k) - cost of fetching k pages,
• f(1) 1 tseek ttransfer
• f(0) 0
• f(k1) gt f(k). The cost C(m,w) is

for demand replacement with it
simplifies to C(m,w) p x f(1)p,
where p denotes number of page faults in 1...T.
19
Demand Policy Optimality

• For a disk

for electronic auxiliary memory (electronic
disk) f(k) kf(1)k (ignoring costs
of page fault interrupt). When f(k)k, there is a
demand replacement algorithm with a cost function
that does at least as well for all memory sizes
and reference strings as any other algorithm,
nice result but not very useful when disks are
used! Conclusion pre-fetching helps by bringing
more than one page per page fault but it is
difficult to predict which pages to pre-fetch!
20
Replacement Policies

• OPT (Optimal) - remove the page with next
  reference most distant into the future,
• FIFO (First In First Out)- circulating pointer,
  often inefficient,
• LIFO (Last In First Out) - special situations
  (sequential access),
• LFU (Least Frequently Used) good performance
• MRU (Most Recently Used), not the same as Last In
  First Out, e.g., consider string 12314 with m3,
  then after 1231, MRU will replace 1 while LIFO 3
• Example w (123)r (456)s, m3.
• LFU - 3 x (1min(r,s)), MRU - 3 x (1s) page
  faults.

21
Stack Algorithm Definition

• M(m,w) is a state of real memory after
  referencing string w in m frames, where M(m,0)
  0.
• Inclusion property characterizing stack
  algorithms

Given w, there is a permutation of the virtual
pages labeled 1,2,...,n, called stack, S(w)
s1 (w), ..., sn (w) such that M(m,w) s1
(w), ..., sm (w) For a sequence of references,
there is a sequence of stacks. dp(w) is the
distance of page p from the top of the stack
after processing string w.
22
Stack Updating via Priority

• Consider stack-updating procedure through
  priority list with the following properties
• Priority is independent of the number of frames
  m.
• The currently referenced page has highest
  priority.
• 3. The resident page with the lowest priority is
  moved up in stack (from real memory) only when
  necessary and only to its priority level.
• With these three properties, class of priority
  algorithms is the same as class of stack
  algorithms, since only removal can brake stack
  property and if M is the stack of m-frames and y
  is the page in m1 frame, then

Selecting a victim with m1 frames Selecting
a victim with m frames Although minminM,y
may be y, the stack is the same with m and m1
frames, because in such a case m1st frame will
contain minM.