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Overview

Overview. Assignment 5: hints Garbage collection Assignment 4: solution. 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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Overview

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  1. Overview • Assignment 5: hints • Garbage collection • Assignment 4: solution

  2. 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

  3. 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).

  4. 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)

  5. A5 Ex3 - Mark & Sweep • Phase 1: • mark every reachable object • Phase 2: • remove non-reachable objects 4 1 5 o 2 6 3 7

  6. 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

  7. q q p p Pointer Rotation - Generic Case

  8. P Q R Pointer rotation example 0 1 2 3

  9. Pointer Rotation • Deutsch-Schorr-Waite (1967) • Stores information in the data structure • memory efficient • iterative • structures are temporary inconsistent • non-concurrent • non-incremental

  10. Input Grammar • EBNF:Graph := noOfNodes { Node }.Node := noOfEdges { destination }. • Implicit: each node is numbered starting from 0. • Example: 8 3 1 2 3 3 4 5 6 2 0 6 1 7 0 0 0 1 2 node

  11. Example 8 3 1 2 3 3 4 5 6 2 0 6 1 7 0 0 0 1 2 0 1 2 3 4 5 6 7 4 1 5 o 2 6 3 7

  12. Overview • Assignment 5: hints • Garbage collection • Assignment 4: solution

  13. 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.

  14. 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.

  15. 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%

  16. 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.

  17. 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

  18. 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.

  19. A4 Ex.6 – Program Run Time • Application • TLB hit rate is 75% • number of memory access is 55.500.000 • Page fault rate 0.005 • System performance for this application • average TLB miss penalty is 130 nsec • average DRAM access time is 50 nsec • average disk access time is 9 msec • Which is the application run time • on this system? • on a system with a better disk with an access time of 6 msec?

  20. 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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