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Case 1a: A DC is using 60kW for equipment and 40kW for cooling Case 1b: Code efficiency of 5% improves the equipment to 57kW for equipment; I'd exp
by sokoloff 3y ago
Case 1a: A DC is using 60kW for equipment and 40kW for cooling
Case 1b: Code efficiency of 5% improves the equipment to 57kW for equipment;
I'd expect cooling to drop to around 38kW rather than remain at 40kW to remove that 57kW of heat load.
If that's the case, a 5% improvement in equipment power draw is a 5% improvement in overall DC power draw, not a 3% (5% improvement on 60% of it).
- packetlost 3y agoA code efficiency improvement of 5% (represented as a 5% improvement in CPU utilization) does not equate to a 5% improvement in equipment power draw though. You at least need to factor in the percentage of an individual server's power draw that goes to the CPU, which is about 30% (again, on average). So a 5% equipment power draw improvement would need at least a 15% code efficiency improvement (0.05 / 0.3 = 0.1666), assuming you're CPU bound on whatever your code is doing. I factored in the average CPU load (50%) to account for this, but it's heavily application dependent.
- sokoloff 3y agoWhatever the percentage is for equipment applies closer to 1:1 than 0:1 to the cooling is my main point. I agree that it's not 1:1 from CPU to equipment.
- packetlost 3y agoYeah, that would make sense
- deleted 3y ago[deleted]