Not (right). If not told otherwise (specifying an explicit storage pool for every object), OS/400 itself chooses physicall space for any objetct, usually trying to keep sotrage usage at a level for all disks.

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Wrong. All data is spread over all disks evenly, to make optimal use of all disk-arms.

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Also, in a related performance impact: If you seperate the objects into 10 libraries, the as/400 can place them on seperate disk packs, reducing the risk of (entire) loss, and spreading the disk head activity among several physical devices. (right?) :)

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In each library there is an independent index (similar to the index of a logical file) which the system can use to find 'normal' objects. That means it will take approximately 5 times as long (averagely) to find one object with 10 libraries than with one library, given that they are evenly distributed in the 10 libraries. But a lookup in an index is fast, so you really don't need to bother.

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I'm afraid that your statement is true if you have qualified the objects ( I mean if you have stated the library/object of the call/open ) if not there is a search. As was stated in a later message that penalty is nowadays not that significant ( I recall having read somewhere in the old S/38 to put the data libraries before the pgms libraries to diminish the search )

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The is absolutley no difference between the two. I runs on a pointer system and does not need to search libraries. Once the program is compiled , os400 knows exactly where it is.

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This is an interesting question, and I'd love to hear what people say. But, I really don't think it's a big performance worry. My guess is this. It searches each library for the object, but uses a good search technique. Kind of like the way we would search through a dictionary for a word. We wouldn't start with the A's and scan until we reach our desired word (in this case let's say "STEAK"). We would go to the Ss and then the STs and so forth. I can't for the life rember the name of the search method. But it's the one where you go to the middle of the list. If the value is equal, your done. If it is greater than the desired value, you go to middle of the top half, and do the same until a value is or is not found. If the value is less than the desired value, you use the bottom half of the list and do the same until a match is or isn't found... Darn mental blocks! Binary search? Someone help me out…

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