Originally posted by starfleetX
Wrong.
Older G4's have a four-stage pipeline.
Apple (well, Motorola) changed with the 733 and up.
Hmm, really? Didn't know that. I guess you learn something every day...
About those nasty percentages, I'm still iffy on how to understand them. I guess 100% faster is twice as fast, but it's still really weird.
Help me out here -- if 100% faster is twice as fast, which means effectively where the benchmark system is at 100% speed (meaning it's as fast as the benchmark, which is itself), then the faster system is at 200% speed, right? So if, let's say, the Pentium that Apple is using as a benchmark is set at 100%, and it performs a certain task in 100 seconds. The Dual GHz G4 is supposedly 72% faster, so it is 172% the speed of the benchmark Pentium. So how long would it take to perform the same task on the Dual GHz G4?
UPDATE: OK, I think I actually get it now. So if 100% faster is effectively 200% of the benchmark system, then this is 2x as fast. So you take the time it takes for the benchmark system to do a certain job and divide it by the factor, in this case 2. So something 100% faster than the base Pentium Apple is using would take 50 seconds to do a task that the Pentium took 100 seconds to do.
So, using that, if the Pentium took 100 seconds to do a task, then the 800 MHz G4 would take 79.37 seconds, the 933 would take 66.23 seconds, and the dual 1 GHz would take 58.14 seconds. And using the same reasoning, I can use the 933 as a benchmark for the dual 1 GHz, and use the reverse of the method described above to get that the dual 1 GHz machine is 13.91% faster than the 933 MHz machine.
Whew.
So now, if you say 933 MHz is the base clock speed, then 1 GHz is 107.3% of that clock speed, which means the clock speed is 7.3% faster, contrasted with the real-world estimate (by Apple, anyway) that the dual 1 GHz is actually 13.91% faster. So it seems maybe it does add up, eh?