Report

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 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…) 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. 11 Bad Program Against First Fit • Assume (max live space) M=48. • Start by allocating 48 1-byte objects. The heap: 12 Bad Program Against First Fit • Phase 0: Start by allocating 48 1-byte objects. The heap: 13 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 14 Bad Program Against First Fit • 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 15 Bad Program Against First Fit • • • • 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. The heap: x24 Memory available for allocations 16 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. x24 Memory available for allocations 17 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. 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: 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. Guideline for adversarial program: remove as much space as possible, but maintain a minimal “density” ! 20 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 21 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. 22 An Execution Example • i=0, allocate M=48 objects of size 1. Assume compaction budget C=4 The heap: Compaction quota: 48/4 = 12 23 An Execution Example • i=0, allocate M=48 objects of size 1. • i=1, memory manager does not compact. Assume compaction budget C=4 • i=1, deletion step. The heap: Compaction Quota: = 12 x23 Memory available for allocations 24 An Execution Example • i=1, allocate 11 objects of size 2. Assume compaction budget C=4 The heap: Compaction Quota: = 12+22/4=17.5 12 x1 Memory available for allocations An Execution Example • i=1, allocate 11 objects of size 2. • Memory manager compacts. The heap: Assume compaction budget C=4 Compaction Quota: = 17.5 17.5-8=9.5 x1 Memory available for allocations An Execution Example • i=1, allocate 11 objects of size 2. • Memory manager compacts. • i=1, deletion. Lowest density 2/c=1/2. The heap: Assume compaction budget C=4 Compaction Quota: = 9.5 x1 x20 Memory available for allocations An Execution Example • i=2, allocate 5 objects of size 4. Assume compaction budget C=4 The heap: Compaction Quota: = 9.5 x20 Memory available for allocations 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 1 2 4 8 2 Different object sizes n … Each object is allocated in a block matching its size. The first free chunk in33 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: 1 2 3 4 5 n 1 n – If no compaction allowed: space used ≥ 4/5 M√n • For exponentially increasing sizes: – With no compaction: space used ≥ M log(n) n 1 2 4 8 2 • (Should be interpreted as: there exists a bad program that will make the segregated free list allocator use at least … space.) 35 n Lower Bound Differ by Possible Sizes • For all possible sizes: 1 2 3 4 5 n 1 n – 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: n 1 2 4 8 – With no compaction: space used ≥ M log(n) – With partial compaction: space used ≥ ¼M min(log(n),c/2) 2 • (Should be interpreted as: there exists a bad program that will make the segregated free list allocator use at least … space.) 36 n 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. 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