When I reopen a large, multi-layered image in CS4, the efficiency pop-up displays 100%. Is this display dynamic, meaning, will the displayed efficiency reduce to a number less than 100 ONLY during execution of a process; eg, applying a filter, if insufficient RAM in available requiring access to virtual memory? Thank you, Kenneth Schang
Thanks Neil. My question is somewhat academic. Everything is well with image processing fine on my system. My question has nothing to do with my ability to process and my question is not images specific. I am only questioning how the efficiency pop-up works functions. Does it always show 100%, except when a task being processes requires the use of virtual memory in which case the displayed number would fall below 100? Thank you, Ken
Thanks Ramdon. I did check the help files before posting my question. I don’t believe my question is answered. If you can direct me to the section of the help files that discusses that states that it is either a dynamic (displaying processing efficiency only during execution of a process) or if it is static display, I would appreciate that input. Again, I haven’t found an answer to my question in the help files. Thanks again, Ken
K.W.- Yes, the Efficiency reading will display 100% until you start taxing RAM and other resources. The reading is "dynamic" in that it shows the amount for the last performed task. If the last task shows 70%, it will remain until the next action.
Never thought Ken was asking such an obvious, elementary question with a self-explanatory answer. I thought he wanted to know what the percentage change meant and didn’t want to go into explanations of what happens in multiprocessor machines.
I’ve been using Photoshop seriously for close to six years and have never had any need to look at the efficiency figures, which I regard as useful only as a troubleshooting aid when a user encounters performance problems, which I haven’t had even on my older box.
Thank you PShock. I appreciate you taking your time to read, digest and then answer my obvious, elementary question that has a self explanatory answer. It is amusing to me how some are unable to understand such an obvious, elementary question when asked. Kenneth Schang
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