4 ms·
It's always a good idea to do a quick back-of-the-envelope calculation before generalizing like that: About 2 million servers were sold world-wide in 2014Q2 A
by phkamp 12y ago
It's always a good idea to do a quick back-of-the-envelope calculation before generalizing like that:
About 2 million servers were sold world-wide in 2014Q2
Assume 25% runs UNIX and NTPD -> half a million servers
Assume NTPD uses 0.1% of machine resources -> 500 fully loaded servers.
Assume 100W/server -> 50 kW
50kW for half a year -> 220,000 kWh
Assume 500g CO2/kWh -> 110 tons of CO2
QED: I think CPU overhead is a very relevant concern for time synching programs.
- lazyjones 12y ago> Assume NTPD uses 0.1% of machine resources -> 500 fully loaded servers. Bold assumption. It uses 0.0006% on my slow 6 years old box (over 13 days uptime). Let's be generous and say it's 0.001%, then you get 1.1 ton of CO2, which is the equivalent of the production of about 32.8 Kg of beef. I don't think such dramatic conclusions can be made from those facts.
- phkamp 12y agoFirst, I doubt your kernels statistics actually account correctly at that level of precision. (If it does, that alone would waste a lot of CPU cycles!) Second, have you checked the memory footprint of NTPD ? "machine resources" is more than CPU cycles. And don't forget: That number was one quarters purchase of servers, running for half a year. All the servers bought in 2010, 2011, 2012, 2013 and the other half of the year were not included. Things add up.
- lazyjones 12y ago> Second, have you checked the memory footprint of NTPD ? "machine resources" is more than CPU cycles. 284K, which is a whopping 0.04% of available memory. I don't see how this increases CO2 output though. > Things add up To a still very insignificant number, dwarfed by the beef consumption of the people reading this. I'd say there are better arguments in favour of using dated languages than this.
- phkamp 12y agoI can't change how much beef people eat. I can change how their computers synchronize time.
- zzzcpan 12y agoI think you missed something in your calculations. Removing 0.1% load from the machine will not result in 0.1% of energy savings, very very far from it if in any at all. And that's not even what we want to calculate. We want the difference with hypothetical alternative in a safer language, which is hard to guess, but is unlikely to be in the same order or two with your calculations.