on the limits of partial compaction

On the limits of partial compaction

Anna Bendersky & Erez PetrankTechnion

Fragmentation• When a program allocates and de-allocates, holes

appear in the heap. • These holes are called “fragmentation” and then

– Large objects cannot be allocated (even after GC),– The heap gets larger and locality deteriorates– Garbage collection work becomes tougher– Allocation gets complicated.

• The amount of fragmentation is hard to define or measure because it depends on future allocations.

How Bad can Fragmentation Be?• Consider a game:

– Program tries to consume as much space as possible, but:

– Program does not keep more than M bytes of live space at any point in time.

– Allocator tries to satisfy program demands within a given space

– How much space may the allocator need to satisfy the requests at worst case?

• [Robson 1971, 1974]: There exists a program that will make any allocator use ½Mlog(n) space, where n is the size of the largest object.

• [Robson 1971, 1974]: There is an allocator that can do with ½Mlog(n).

Compaction Kills Fragmentation• Compaction moves all objects to the beginning of

the heap (and updates the references). • A memory manager that applies compaction after

each deletion never needs more than M bytes. • But compaction is costly !• A common solution: partial compaction.

– “Once in a while” compact “some of the objects”.– (Typically, during GC, find sparse pages & evacuate

objects.)• Theoretical bounds have never been studied. • Our focus: the effectiveness of partial

compaction.

Setting a Limit on Compaction• How do we measure the amount of (partial)

compaction? • Compaction ratio 1/c: after the program

allocates B bytes, it is allowed to move B/c bytes.

• Now we can ask: how much fragmentation still exists when the partial compaction is limited by a budget 1/c?

Theorem 1• There exists a program, such that for all

allocators the heap space required is at least:1/10 M min{ c , log(n)/log(c) }

• Recall: 1/c is the compaction ratio, M is the overall space alive at any point in time, n is the size of the largest object.

Theorem 1 Digest• If a lot of compaction is allowed (a small c), then

the program needs space Mc/10. • If little compaction is allowed (a large c), then the

program needs space M log(n) / ( 10 log(c) )

• Recall: 1/c is the compaction ratio, M is the overall space alive at any point in time, n is the size of the largest object.

What Can a Memory Manager Achieve?

• Recall Theorem 1: There exists a program, such that for all allocators the heap space required is at least:

1/10 M min{ c , log(n)/log(c) }

• Theorem 2 (Matching upper bound): There exists an allocator that can do with

M min{ c+1 , ½log(n) }

• Parameters: 1/c is the compaction ratio, M is the overall space alive at any point in time, n is the size of the largest object.

Upper Bound Proof Idea• There exists a memory manager that can serve any

allocation and de-allocation sequence in space M min( c+1 , ½log(n) )

• Idea when c is small (a lot of compaction is allowed): allocate using first-fit until M(c+1) space is used, and then compact the entire heap.

• Idea when c is large (little compaction is allowed): use Robson’s allocator without compacting at all.

Proving the Lower Bound• Provide a program that behaves “terribly”• Show that it consumes a large space overhead

against any allocator. • Let’s start with Robson: the allocator cannot

move objects at all. – The bad program is provably bad for any allocator.– (Even if the allocator is designed specifically to handle

this program only…)

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Robson’s “Bad” Program (Simplified)

• Allocate objects in phases. • Phase i allocates objects of size 2i. • Remove objects selectively so that future

allocations cannot reuse space.

• For (i=0, i<=log(n), ++i)– Request allocations of objects of size 2i (as many as

possible). – Delete as many objects as possible so that an object of

size 2i+1 cannot be placed in the freed spaces.

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Bad Program Against First Fit• Assume (max live space) M=48. • Start by allocating 48 1-byte objects.

The heap:

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Bad Program Against First Fit• Phase 0: Start by allocating 48 1-byte objects.

The heap:

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Bad Program Against First Fit• Phase 0: Start by allocating 48 1-byte objects. • Next, delete so that 2-byte objects cannot be

placed.

The heap:

x24

Memory available for allocations

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placed.• Phase 1: allocate 12 2-byte objects.

The heap:

x24


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placed.• Phase 1: allocate 12 2-byte objects. • Next, delete so that 4-byte objects cannot be

placed.

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The heap:

x24


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Bad Program Against First Fit• Phase 1: allocate 12 2-byte objects. • Next, delete so that 4-byte objects cannot be

placed.• Phase 2: allocate 6 4-byte objects.

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x24


First Fit Example -- Observations• In each phase (after the first), we allocate ½M

bytes, and space reuse is not possible. • We have log(n) phases. • Thus, ½Mlog(n) space must be used.

• To be accurate (because we are being videotaped): – The proof for a general allocator is more complex. – This bad program only obtains 4/13log(n) M for a

general collector. The actual bad program is more complex.

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Is This Program Bad Also When Partial Compaction is Allowed?

• Observation:– Small objects are surrounded by large gaps. – We could move a few and make room for future

allocations. • Idea: a bad program in the presence of partial

compaction, monitors the density of objects in all areas.

The heap:

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Is This Program Bad Also When Partial Compaction is Allowed?

• Observation:– Small objects are surrounded by large gaps. – We could move a few and make room for future

allocations. • Idea: a bad program in the presence of partial

compaction, monitors the density of objects in all areas.

Guideline for adversarial program: remove as much space as possible, but maintain a minimal “density” !

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Simplifications for the Presentation

• Objects are aligned. • The compaction budget c is a power of 2.

Aligned allocation: an object of size 2i, must be placed at a location k*2i, for some integer k.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

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The Adversarial Program• For (i=0, i<=log(n), ++i)

– Request allocations of as many as possible objects of size 2i, not exceeding M.

– Partition the memory into consecutive aligned areas of size 2i+1

– Delete as many object as possible, so that each area remains at least 2/c full.

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An Execution Example

The heap:

Assume compaction budget C=4

Compaction quota: 48/4 = 12

• i=0, allocate M=48 objects of size 1.

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An Execution Example

The heap:

x23


Assume compaction budget C=4

Compaction Quota: = 12

• i=0, allocate M=48 objects of size 1. • i=1, memory manager does not compact.• i=1, deletion step.

An Execution ExampleAssume compaction budget C=4

Compaction Quota: = 12

• i=1, allocate 11 objects of size 2.

The heap:

x1


Compaction Quota: = 12+22/4=17.5


• i=1, allocate 11 objects of size 2.• Memory manager compacts.

Compaction Quota: = 17.5

The heap:

x1


Compaction Quota: = 17.5-8=9.5


• i=1, allocate 11 objects of size 2.• Memory manager compacts. • i=1, deletion. Lowest density 2/c=1/2.


The heap:

x1x20



• i=2, allocate 5 objects of size 4.


The heap:


x20

Bad Program Intuition• The goal is to minimize reuse of space. • If we leave an area with density 1/c, then the

memory manager can: – Move allocated space out (lose budget area-size/c)– Allocate on space (win budget area-size/c)

• Therefore, we maintain density 2/c in each area.

Proof Skeleton• The following holds for all memory managers. • Fact: space used ≥ space allocated – space reused.• Lemma 1: space reused < compaction c/2• By definition: compaction < ( space allocated ) / c• Conclusion:

space used ≥ space allocated – compaction c/2

≥ space allocated ( 1 - ½ )• Lemma 2: either allocation is sparse (and a lot of

space is thus used) or there is a lot of space allocated during the run.

• And the lower bound follows. The full proof works with non-aligned objects...

The Worst-Case for Specific Systems

• Real memory managers use specific allocation techniques.

• The space overhead for a specific allocator may be large. – Until now, we considered a program that is bad for all

allocators. • A malicious programs can take advantage of

knowing the allocation technique. • One popular allocation method is segregated free

list.

Block-Oriented Segregated Free ListSegregated free list: • Keep an array for different object sizes (say, a

power of 2). • Each entry has an associated range of sizes and it

points to a free list of chunks with sizes in the range.

Block Oriented: • Partition the heap to blocks (typically of page

size).• Each block only holds objects of the same size.

Segregated Free List

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1 2 4 8 n

Different object sizes

Each object is allocated in a block matching its size.The first free chunk in the list is typically used. If there is no free chunk, a new block is allocated and split into chunks.

2n

…

What’s the Worst Space Consumption?

• No specific allocators were investigated in previous work (except for obvious observations).

• Even with no compaction: what’s the worst space usage for a segregated free list allocator?

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Lower Bound Differ by Possible Sizes

• For all possible sizes:– If no compaction allowed: space used ≥ 4/5 M√n

• For exponentially increasing sizes:– With no compaction: space used ≥ M log(n)

• (Should be interpreted as: there exists a bad program that will make the segregated free list allocator use at least … space.)

1 2 3 4 5 n1n

1 2 4 8 n2n

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Lower Bound Differ by Possible Sizes

• For all possible sizes:– If no compaction allowed: space used ≥ 4/5 M√n – With partial compaction: space used ≥ ⅙ M min(√n,c/2)

• For exponentially increasing sizes:– With no compaction: space used ≥ M log(n)– With partial compaction: space used ≥ ¼M

min(log(n),c/2)

• (Should be interpreted as: there exists a bad program that will make the segregated free list allocator use at least … space.)

1 2 3 4 5 n1n

1 2 4 8 n2n

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The Adversarial Program for SFL• When no compaction is allowed: For (i=0; i<k; ++i) // k is the number of different

sizes. – Allocate as many objects as possible of size si – Delete as many objects as possible while leaving a

single object in each block.• In the presence of partial compaction: For (i=0; i<k; ++i)

– Allocate as many objects as possible of size si – Delete as many objects as possible while leaving at least

2b/c bytes in each block. // where b is the size of the block.

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Related Work• Theoretical Works:

– Robson’s work [1971, 1974]– Luby-Naor-Orda [1994,1996]

• Various memory managers employ partial compaction. For example:– Ben Yitzhak et.al [2003]– Metronome by Bacon et al. [2003]– Pauless collector by Click et al. [2005]– Concurrent Real-Time Garbage Collectors by Pizlo et al.

[2007, 2008]

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Conclusion• Partial compaction is used to ameliorate the

pauses imposed by full compaction. • We studied the efficacy of partial compaction in

reducing fragmentation.• Compaction is budgeted as a fraction of the

allocated space. • We have shown a lower bound on fragmentation

for any given compaction ratio. • We have shown a specific lower bound for

segregated free list.

on the limits of partial compaction

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