From owner-freebsd-advocacy Thu May 13 17:49:43 1999 Delivered-To: freebsd-advocacy@freebsd.org Received: from smtp2.vnet.net (smtp2.vnet.net [166.82.1.32]) by hub.freebsd.org (Postfix) with ESMTP id C2AFE14FD5 for ; Thu, 13 May 1999 17:49:40 -0700 (PDT) (envelope-from rivers@dignus.com) Received: from dignus.com (ponds.vnet.net [166.82.177.48]) by smtp2.vnet.net (8.9.1a/8.9.1) with ESMTP id UAA13534; Thu, 13 May 1999 20:50:41 -0400 (EDT) Received: from lakes.dignus.com (lakes.dignus.com [10.0.0.3]) by dignus.com (8.9.2/8.8.5) with ESMTP id UAA04614; Thu, 13 May 1999 20:49:35 -0400 (EDT) Received: (from rivers@localhost) by lakes.dignus.com (8.9.2/8.6.9) id UAA45243; Thu, 13 May 1999 20:49:35 -0400 (EDT) Date: Thu, 13 May 1999 20:49:35 -0400 (EDT) From: Thomas David Rivers Message-Id: <199905140049.UAA45243@lakes.dignus.com> To: advocacy@freebsd.org, jesus.monroy@usa.net, licia@o-o.org, rivers@dignus.com Subject: Re: [Re: BUDS Coming to you soon.] In-Reply-To: Sender: owner-freebsd-advocacy@FreeBSD.ORG Precedence: bulk X-Loop: FreeBSD.ORG > > Uhm, not to be a nitpicker, but 2^313 isn't -that- huge, according to bc, > it's > > 166873987181321100187111070794496258953336290809113497652112625611110\ > 91607661254297054391304192 > > Which is a lot of kernels, but I could print it out on a 3x5 note card pretty > easily ;) > > Are you sure that's right? Seems kinda small to me :-) But, I'm wrong more than I care to admit :-) So, 2**64 is 18446744073709551615 (according to "%qu") - 20 digits, seems like 2**313 is going to be a lot longer... let's say we gain a 10's place digit for every 4 powers of two (which is, of course, way conservative, as you actually gain more than one 10s digit.) 313-64 is 249; 249/4 is 62... Hmm... seems like you're right - you should have about 20+62 with my quick conservative math - and you have about 100 digits up there (I didn't count them.. just guessed) So - yep - you could print it on a 3x5 card :-) But - it still is an awful lot of kernels... How long does it take to build a single kernel - say, a minute, or better, say, 30 seconds... let's make that a power of two and say, uh, 32 seconds (2**5). So, we need 2**5 * 2**313 total seconds... that's 2**318 seconds. Again, fudging a little on the math; let's pretend that a minute is 64 seconds instead of 60 - which is 2**6. So, we need 2**312 minutes. Again, fudging and saying an hour is 64 minutes, we'd need 2**306 hours. (bc could likely figure this out much more precisely - but I'm "thinking with my fingers") Fudging again, and saying there are 32 hours in a day, that's 2**301 days. If we make a week 8 days long, that's 2**298 weeks. Making a year 64 weeks long (we are *really* stretching things here, but I think we're stretching in the "right" way), that makes 2**293 years. Make a millenium 1024 years, and we have 2**283 millenia. If I've kept everything straight - that's an *awfully* long time :-) Maybe I've messed up along the way... so, just lop off a whole chunk of years and say 2**250 millenia... Now, how old is the universe? Of course, there are other ways to tackle the problem: If you had 8 billion people (2**33) (the world has about 6 billion people on it, right now) and each did, say 2*280 kernel builds a night, you could get it in one night. Let's give each of these 8 billion people 4 computers. That's 2**35 computers - so, each computer only needs to do 2**278 kernel builds per night... :-) :-) :-) - Dave Rivers - To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-advocacy" in the body of the message