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Unfortunately, I can't find any freely available text with an estimation of worst case (external) fragmentation overhead in (binary) buddy memory system. I've found only something like M(1+lg2 m) , without any proof. This expression estimates(?) a buddy heap size that guarantees to allocate total memory of size M (m is the longest allocated block). This estimation seems too rough at least for m=2; also the proof would be interesting.

I would be grateful for any explanations or links concerning this matter.

user396672
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This seems to be covered by exercise 41 in section 2.5 of Volume 1 of The Art of Computer Programming, by Knuth. Knuth considers blocks of size 1, 2, .. 2^k with total store allocated of n - Quote follows

We can prove by induction on k that the total storage consumed by such split blocks never exceeds kn; for after every request to split a block of size 2^(k + 1), we are using at most kn locations in split 2^k blocks and at most n locations in unsplit ones.

(end quote) So I think you can view this as if you had a memory allocator for blocks of size up to 2^(k+1) which produced a guarantee of using at most n(k+1) store by using at most n store to deliver blocks of size 2^(k+1) and deferring requests for smaller blocks to an inductively magic allocator which could service those requests using at most nk store. If you start off with a pool of (k+1)n blocks, the magic allocator never needs more than kn store, so the allocator for blocks of size 2^(k+1) will always have at least n store of untouched unfragmented blocks to serve requests for size 2^(k+1) from.

mcdowella
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  • Thank you for detailed answer, I understand now the expression M(1+lg2 m) and it source :). It seems, however, too pessimistic, I'm almost sure that for k=2 the bound is about n*(3/2) but I don't know the precise answer for k>2 ... – user396672 Jun 10 '11 at 14:06