On the limits of partial compaction Anna Bendersky & Erez Petrank Technion

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 ½M  log(n) space, where n is the size of the largest object. [Robson 1971, 1974]: There is an allocator that can do with ½M  log(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 M  c/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, ½  log(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…)

Robson’s “Bad” Program (Simplified) Allocate objects in phases. Phase i allocates objects of size 2 i. Remove objects selectively so that future allocations cannot reuse space. 11 For (i=0, i<=log(n), ++i) – Request allocations of objects of size 2 i (as many as possible). – Delete as many objects as possible so that an object of size 2 i+1 cannot be placed in the freed spaces.

Bad Program Against First Fit 12 Assume (max live space) M=48. Start by allocating 48 1-byte objects. The heap:

Bad Program Against First Fit 13 Phase 0: Start by allocating 48 1-byte objects. The heap:

Bad Program Against First Fit 14 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

Bad Program Against First Fit 15 Phase 0: Start by allocating 48 1-byte objects. Next, delete so that 2-byte objects cannot be placed. Phase 1: allocate 12 2-byte objects. The heap: x24 Memory available for allocations

Bad Program Against First Fit 16 Phase 0: Start by allocating 48 1-byte objects. Next, delete so that 2-byte objects cannot be placed. Phase 1: allocate 12 2-byte objects. Next, delete so that 4-byte objects cannot be placed. 16 The heap: x24 Memory available for allocations

Bad Program Against First Fit 17 Phase 1: allocate 12 2-byte objects. Next, delete so that 4-byte objects cannot be placed. Phase 2: allocate 6 4-byte objects. 17 x24 Memory available for allocations

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, ½M  log(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 4/13  log(n) M for a general collector. The actual bad program is more complex.

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. 19

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

Simplifications for the Presentation Objects are aligned. The compaction budget c is a power of Aligned allocation: an object of size 2 i, must be placed at a location k*2 i, for some integer k

The Adversarial Program For (i=0, i<=log(n), ++i) – Request allocations of as many as possible objects of size 2 i, not exceeding M. – Partition the memory into consecutive aligned areas of size 2 i+1 – Delete as many object as possible, so that each area remains at least 2/c full. 22

An Execution Example 23 The heap: Assume compaction budget C=4 Compaction quota: 48/4 = 12 i=0, allocate M=48 objects of size 1.

An Execution Example 24 The heap: x23 Memory available for allocations 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 Example Assume compaction budget C=4 Compaction Quota: = 12 i=1, allocate 11 objects of size 2. The heap: x1 Memory available for allocations Compaction Quota: = 12+22/4=17.5

An Execution Example Assume compaction budget C=4 i=1, allocate 11 objects of size 2. Memory manager compacts. Compaction Quota: = 17.5 The heap: x1 Memory available for allocations Compaction Quota: = =9.5

An Execution Example Assume compaction budget C=4 i=1, allocate 11 objects of size 2. Memory manager compacts. i=1, deletion. Lowest density 2/c=1/2. Compaction Quota: = 9.5 The heap: x1 x20 Memory available for allocations

An Execution Example Assume compaction budget C=4 i=2, allocate 5 objects of size 4. Compaction Quota: = 9.5 The heap: Memory available for allocations 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 List Segregated 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 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. 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. …

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?

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.) n 1248n

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.) n 1248n

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 s i – 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 s i – Delete as many objects as possible while leaving at least 2b/c bytes in each block. // where b is the size of the block. 37

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] 38

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. 39