On Mon, May 2, 2022 at 4:02 PM Tom Lane <t...@sss.pgh.pa.us> wrote: > David Rowley <dgrowle...@gmail.com> writes: > > On Mon, 2 May 2022 at 21:00, Pavel Stehule <pavel.steh...@gmail.com> > wrote: > >> I found a query that is significantly slower with more memory > > > If it was work_mem you increased, it seems strange that the plan would > > switch over to using a Nested Loop / Memoize plan. > > Yeah, there's something unexplained there. > > I spent some time studying cost_memoize_rescan, and the only > conclusions I arrived at were that the variable names are poorly > chosen and the comments are unhelpful. For instance, one would > think that est_entry_bytes is the expected size of one cache entry, > but it seems to actually be the total space that would be occupied > if the whole input relation were loaded into the cache.
And > the est_cache_entries computation seems nonsensical; if it does > make sense, the comment sure doesn't illuminate why. My take on this is that a cache entry is keyed by a parameterization and any given entry can have, at most, every tuple saved into it (hence the computation of tuples*per-tuple-size). So the maximum number of hash keys is the total available memory divided by input relation size. This upper bound is stored in est_cache_entries. If the number of unique parameterizations expected (at worst one-per-call) is less than this we use that value and never evict. If it is more we use the est_cache_entries and plan to evict. What I'm expecting to find but don't see is that by definition each unique parameterization must positively match a unique subset of the input relation tuples. If we remember only those tuples that matched then at no point should the total memory for the hash table exceed the size of the input relation. Now, I'm not completely confident the cache only holds matched tuples...but if so: >From that the mpath->est_entries should be "min(hash_mem_bytes / est_entry_bytes, 1.0) * ndistinct" i.e., all groups or a fractional subset based upon available hash memory Then: ndistinct = estimate_num_groups() || calls retention_ratio = min(hash_mem_bytes / est_entry_bytes, 1.0) est_entries = retention_ratio * ndistinct evict_ratio = 1.0 - retention_ratio hit_ratio = (est_entries / ndistinct) - (ndistinct / calls) || clamp to 0.0 I don't understand the adjustment factor ndistinct/calls evictions total cost adjustment also assumes we are evicting all tuples as opposed to tuples/est_entries This is a "rescan" so aside from cache management isn't the cost of originally populating the cache already accounted for elsewhere? David J.
