Friday, November 14, 2008

 

Fiber: Thinking it through

Doc Searls has an essay about bringing fiber optics to every home in America. It is aimed in the right direction, but makes a couple of mistakes on the numbers and falls to ground way short of its target. It troubles me that I appear to be the sole source for Doc's numbers (on the basis of some informal conversation and my Telecom Day speech in Wellington NZ last May).

This post is an attempt to correct the record, and to create one where my previous thinking has been private.

Doc writes:
A typical fiber trunk fits an 864-fiber cable inside a 1.5-inch conduit. Each fiber can carry 10 gigabits of data. The total comes to 1.6 terabits. Here’s how David Isenberg puts that into perspective:

If all 6.5 billion people on earth had a telephone, and if they were all off-hook, generating 64 kilobits a second, and all those conversations were routed to this cable, there would be 100 fibers still dark.
Almost. Each fiber, in fact, can carry 10 gigabits per second **on each of 160 wavelengths** using today's off-the-shelf technology. In other words, each fiber, lit like this, carries 1.6 terabits. The arithmetic shows that 10 gigabits times 864 fibers is 8.64 terabits -- wrong. The actual computed capacity of an 864 fiber cable, lit as described above, is 864 times 1.6 terabits, or 1.4 petabits.

It is very bad to trust me on simple arithmetic; I have been known to think about dividing by six while I divided by four. I have been known to divide when I should be multiplying. And I've made plenty of oversight mistakes. But I've worked this particular problem a few times from scratch, but please check my work . . .

I'll leave it to the reader to calculate whether 6.5 billion people on earth, each generating 64 kbit/s at the same time, with all that traffic going to a fiber cable in which each fiber is carrying 10 gigabits per second on each of 160 wavelengths, whether all that traffic could squeeze into 764 fibers.

Another quibble. Doc writes that a *typical* fiber cable is 864 fibers. Typical? The fiber in front of my house is a 144 fiber Sumitomo cable. I saw it installed. My understanding is that 864 fiber cables are mostly used for long-haul these days. The example of 10 gigabits by 160 lambdas by 864 fibers is only *typical* of what's possible. You can light more than 160 wavelengths, and jam more than 10 gigabits down each one. I saw a Sumitomo fiber cable with 2000 fibers in it in 2004. You can have more **or less** of any of these quantities.

Here's another item from Doc that needs a bit of post-hoc sharpening: Doc writes,
Bringing fiber to homes and offices costs between $1000 to $7000 per “drop.”
Yes and no. I've been using $2000 per urban/suburban home and $6000 per rural home as a rule of thumb, a first approximation average cost. This includes homes hooked up for 50% of the homes passed. It is an impressionistic amalgam of lots of stuff I've been told in private conversation with fiber mavens, and it has been verified as reasonable by several people who have actually built fiber networks. [ By the way, others, telco types and Washington DC policy people, tend to cite lower numbers, closer to $1000 per home, but when I dig, this is only the cost of passing the home, not hooking it up.] But it is also worth noting that the costs can be very much higher than $6000 in individual cases, urban or rural.

So, if you assume 100 million urban and dense-suburban residences in the United States at $2000, and 15 million rural homes at $6000, that's $290 billion. (Heck, the U.S. found $700 Billion in a few days last month cause we had to have it now . . . but we still don't know what we're going to use it for.)

Doc didn't get anything overtly wrong when he wrote:
David Isenberg tells me a good sum to invest is $300 billion. That would be for “every home passed with more density than about four per road mile and a 50% take rate.”
But he sure ignored a lot of relevant detail.

I'm nitpicking, right? Well, how do we know whether Doc missed another factor of 160 (or its conceptual equivalent) when he concludes (perhaps based on another unvetted morsel from another sole source) that a public-private partnership can build all the bandwith for any purpose its users want to use it for, even as it
". . . leave[s] old arguments — about Net Neutrality, bandwidth hogs, and who is to blame for what — outside the door."
Sorry, Doc. We can't pretend there's no incumbent, no installed base or that the market is always right. The intellectual heavy lifting in the Fiber to Every Citizen effort will be in getting the details right. This includes how the new infrastructure is regulated. Yes, regulated with laws that say, "No snooping, no blocking, no discriminatory pricing or treatment, and what goes in is what comes out." We don't need any more Enrons, Blackwaters or AIGs. We need the private companies to partner with the public, not the reverse. We need to get the incentives right. We might even consider whether public-private-partnership is the right model at all. But, in any case, we need to do the math.

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Comments:
David

When you want to work out data rates and capacity just remember Google is your friend :) go there and enter

6500000000*64Kb=?Tb
and the answer you'll get is

6,500,000 000 * 64 kilobits = 387.430191 terabits or 396,728.516 gigabits so with each fiber, using DWDM, getting 10Gbps per wavelength then you need 396728.516/10= 6,539,672.85 lambdas to carry that traffic. At 160 lambdas per fiber that means you need 40,872.9553 fibers to carry that much traffic. Divide that by 764 for the number of fibers in a cable and you find you would actually need 53.4986326 cables to carry those conversations simultaneously.
 
Thank you for running these numbers!

Doug, if I understand you correctly, you say there is even more space than is suggested by official usage calculations?
 
Wizkid

If I understand your question correctly the answer is no, there is insufficient capacity not excess capacity on a single cable.

Doc Searls was saying that David claimed a single 764 strand cable, using DWDM producing 160 lambdas per strand of fiber, would be capable of carrying the phone calls of the entire world's population simultaneously. What my calculations indicate is that this is not correct. It would require 54 cables, not one, to carry that much traffic.

Or did I misunderstand your question?
 
OK, here goes my own rusty arithmetic:

If 6.5 billion people generate 64 kbit/s at once, in round numbers, that's about 400 terabits, or 400,000,000,000,000 bits/s. So far, that agrees with Doug Alder's math.
So the aggregate maximum world traffic is 400 terabits. OK so far?

If you light each of 160 wavelengths, at 10 gigabits a second, that's 1.6 terabits, or 1,600,000,000,000 bits/s, per fiber.

When I divide 400 terabits by 1.6 terabits to get the number of fibers needed to carry 400 terabits, I get 250 fibers.

So it looks like I misoverestimated by more than half, by my calculation. This is probably because previously when I've done the calculation, I've rounded several items up to the next whole number.

I could still be wrong. Doug Alder says I'd need 50-some cables but I don't understand his math. For example, he gets
"396728.516/10= 6,539,672.85 lambdas."
By my calculation, 396728/10 is 3967 lambdas, or (dividing by 160) 25 fibers.

So far, I stand by a more general claim that if an 864 fiber cable, lit as described above, were carrying 6.5 billion 64 kilobit streams, there'd be LOTS of spare capacity.
 
Looks I got careless in my math in my second calculation David. 396,728.516 Gb/10 is 39672.8516 lambdas, my apolgies - no idea how I came up with almost double that amount - and that of course changes everything.

So 396,728.516 divided by 160 lambdas per fiber gets you 2,479.553225 fibers required and divide that by the number of fibers in a cable 2,479.553225/764 and you get 3.245488514 cables

In your calculation you made a division error as well

"By my calculation, 396728/10 is 3967 lambdas, or (dividing by 160) 25 fibers."

No it's 39,672.8 - you're of by a factor of 10 - don't know about you but I'm blaming my oversight on oldsheimers ;)

Does that add up now?
 
So there you have it -- with all of my innumeracy (and Alder's) hanging out in public for all to see.

Embarrassing, but I think we finally got the arithmetic right.

The concepts are harder to prove obviously wrong. It is easy to say, "Hey let's leave that old Net Neutrality argument behind," and it sounds very breezy and post-partisan, but who knows the option value of the next great invention you're not inventing when you do it?
 
I am very proud of myself for not even sharpening a pencil to get into that arithmetic! If I understand what just happened on the whiteboard, we came down to the assertion that four 864 fiber cables could handle 6.5 billion users all on the telephone at the same time.

Now for the hard part... if four cables will do it, how long do they have to be to pass near every house, igloo, hogan, and yurt on the planet?
 
So here is my whack at it:

864 fibers, 160 wavelengths per, 10 gbps
per wavelength.
6.5 billion (6.5e9) people, 64 kbps (64e2) if all offhook.

Using octave (a matlab clone):

octave:13> (6.5e9*64e3)/1e12
ans = 416

So 416 petabits for all 6.5 billion folks to go offhook (talking to whom?).

Then, what can that bundle carry:

octave:15> (864*160*10e9)/1e12
ans = 1382.4

So the bundle has a capacity of
1382.4 petabit/sec, before framing,
encap/etc. So one bundle can handle
all the voice traffic.
 
. . . to finish RickS' thinking, 864 fibers * 416/1382.4 is 260 -- 260 fibers to carry the world's conventional telephony traffic. I think we're limping towards consensus here.
 
Oh, the shame of it - I've dated myself, when I went to school peta anything was something discussed in chemistry class or such :) Meant to say "tera" in some of that previous posting.. But you premise seems correct - get enough fibers together and the worlds voice traffic can be passed in a single bundle.
 
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