Replicated scheme - data format

Question

Hi all
     I was doing memory size experiment on coherence. I was wondering if someone has answer to the following case:
     I am doing this investigation using Replicated cache scheme.
     I want to load 3 million id/value pairs, into a named cache (key:id, data:value)
     I can put these 3 million entries by calling put(key,data), or putAll(collection). At the end, the size of the application is around 900MB.
     I can put these 3 million entries by putting these ids into smaller hashtable (~100K items) and put these hashtables into named cache. put(somekey, hashtable). So, if I need to access an id, i find the respective hashmap, and check the value from that hashmap. The total size of the application is ~150MB for this case.
     I know there is some redundant data that goes into each items in the namedcache. For the first case, named cache has 3 million entries, for the second case it has 30 entries (100K hashmaps).
     The difference is 750MB, some of which is due to ~3million more indexes for the first case and some due to extra overhead for each items in the named cache.
     The main reason should not be the more index, i would expect that to be ~100MB. Is that the case that memory size is a lot higher since..
     3 million x EXTRA >> 30 x EXTRA
     If so, can someone tell me what this extra part consists of, when we put some data into named cache when we are using Replicated-scheme.

Answer 1

There are a few things to consider:
     1) The data is stored in deserialized form until it is read. The first read will deserialize the data.
     2) There is some overhead per-entry. 750MB/(3,000,000-30) = ~250 bytes per entry. Some portion of this is attributable to the per-entry overhead.
     3) How did you measure memory usage? Usually invoking <tt>System.gc()</tt> before measuring heap usage will clear garbage but is not guaranteed.
     Jon Purdy
     Oracle