Transcript SSW01 - Admin

Overview

Assignment 5: hints
 Garbage

collection
Assignment 4: solution
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A5 Ex1 - Barriers
Explain the difference between a read and
a write barrier
 Show the instrumented code generated by
a compiler for p.next = q


Which barrier to use for:
 Copying
GC
 Mark & Sweep GC
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A5 Ex2 – Copying collectors
Compacting and copying GC cause the
object address to change at each
collection step.

Show how to solve the movement
problem (for the 2 GC types).
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A5 Ex2 – Copying collectors
Mark & Sweep vs. Copying GCs:
 Give a rough implementation of the
collection and allocation for the Copying
GC
 Which collector has the fastest allocation?
 Give an estimate of the collection cycle
cost (M = heap size, R = live objects)
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A5 Ex3 - Mark & Sweep

Phase 1:
 mark

every reachable object
Phase 2:
 remove
non-reachable objects
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Pointer Rotation - Introduction

Recursive traversal is very expensive
heap: list with 10’000 elements
PROCEDURE Traverse(root: Node);
VAR cnt: INTEGER;
10’000 * 16 = 160’000 Bytes Stack Size
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Pointer Rotation - Generic Case
q
p
q
p
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Pointer rotation example
P
Q
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R
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Pointer Rotation
Deutsch-Schorr-Waite (1967)
 Stores information in the data structure

+
+
–
–
–
memory efficient
iterative
structures are temporary inconsistent
non-concurrent
non-incremental
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Input Grammar
EBNF:
Graph := noOfNodes { Node }.
Node := noOfEdges { destination }.
 Implicit: each node is numbered starting
from 0.
 Example:

8312334562061700012
node
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Example
8312334562061700012
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4 5 6
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Overview

Assignment 5: hints
 Garbage

collection
Assignment 4: solution
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A4 Ex.1 – Loading Page Tables

The whole process’ page table is loaded in
hardware when the process is scheduled
 Advantage:
During the process execution, no
more memory references are needed for the
page table.
 Disadvantage: If the page table is large,
loading the whole page table at every context
switch can also hurt performance, as shown
in our example.
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Ex.1 – Loading Page Tables

Compute the fraction of the CPU time devoted to
loading the page tables if
 32-bit
address space, 8 KB pages
 each process runs for 100 msec
8KB pages  13 bits for the offset  219 entries in
the page table
TLoad = 219 · 100nsec = 52.4288msec
TLoad / T = 0.52
52% of the CPU time is devoted to loading the
page tables.
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A4 Ex.2 – Using TLBs

The time to read a word from
 page
table is 50 nsec
 TLB is 10 nsec

What hit rate is needed to have a mean
access time of 20 nsec?
10nsec + (1 - p) · 50nsec = 20nsec
p = 4 / 5 = 0.80
TLB hit rate = 80%
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Ex.2 – Using TLBs (cont)

How does a TLB function in a system with
multiple processes?

Some systems have an instruction which clears
all the validity bits. Linux uses this machine
instruction to invalidate all TLB entries at a
context switch.
Extend the TLB entries with a process identifier
field, and add a register to hold the PID of the
current process.

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A4 Ex.3 – Memory Size



The time to execute an instruction is 1 µsec or 2001 µsec
if a page fault occurs
A program has 15.000 page faults and an execution time
of 60 sec
We double the memory size

the interval between the page faults is doubled
T = Ninstr · 1µsec + 15.000 · 2000µsec = 60sec
Ninstr · 1µsec = 60.000.000 - 30.000.000µsec =
30.000.000µsec
T0 = 30.000.000µsec + 7.500 · 2000µsec = 30.000.000 +
15.000.000µsec = 45.000.000µsec = 45sec
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A4 Ex.5 – The Aging Algorithm
Page0: 01101110
Page1: 01001001
Page2: 00110111
Page3: 10001011
Problems with this algorithm?
 Loose the ability to distinguish between references early
in the tick interval from those occurring later.
 Because the counters have a finite number of bits, it may
happen that two pages have a counter value of 0 and we
have no way of seeing which of these two pages was
last referenced.
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A4 Ex.6 – Program Run Time

Application
 TLB hit rate is 75%
 number of memory access
 Page fault rate 0.005

System performance for this application
 average
 average
 average

is 55.500.000
TLB miss penalty is 130 nsec
DRAM access time is 50 nsec
disk access time is 9 msec
Which is the application run time
 on
 on
this system?
a system with a better disk with an access time of
6 msec?
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Ex.6 – Program Run Time
T = pTLB · Nacc · TTLBmiss + Nacc · TDRAM + ·pPF · Nacc · TDisk
T = 4.578.750.000nsec + 2.497.500msec
T = 2.502.078,75msec = 2.507,07875sec = 41min
T0 = 4.578,75msec + 0.005 · 55.500.000 · 6msec
T0 = 4.578,75msec + 1.665.000msec
T0 = 1.669.578,75msec = 1.669,57875sec = 27,8min
An increase in disk performance of 33% results in a
performance increase of 35% (for this scenario).
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