6 ms·

Wow. This graph is certainly illuminating: https://public.brightside.io/v1/chart/321ef2c603e8d67080afcd92f48e1d69 https://public.brightside.io/v1/chart/321ef2c6

by oskarth 13y ago

Wow. This graph is certainly illuminating: https://public.brightside.io/v1/chart/321ef2c603e8d67080afcd92f48e1d69 https://public.brightside.io/v1/chart/321ef2c603e8d67080afcd... 450000000 is a lot of users. Metcalfe's law says that the value of a communication network is proportional to the square of the connected users. Then add to this the amazing growth they are seeing. I don't think people quite realize what this means. EDIT: In general, it baffles me how many armchair opinionators there are. Isn't it more interesting to try to understand why FB paid $16B, rather than saying why they shouldn't have?

TTPrograms 13y ago
This was exactly the chart I wanted to see after I saw the first one. ...but did they really put a LLS line through that dataset?
- rkuykendall-com 13y ago
  It's an API that puts an LLS line through everything. If it isn't meaningful, ignore it.
Retric 13y ago
Metcalfe's law is clearly wrong after a network get's overly large. I could in theory call a random person in China but if I have no way of communicating with them it's practically worthless to do so.
- nl 13y ago
  Cat gifs are pretty universal. But seriously: I've had people on conference calls when they couldn't understand the language, but someone wanted them to see slides. The fact they could join the call (in that case on Skype) definitely made the network more valuable. You might argue that the fact you can't speak to some of them drops the value per user, and that is true to some extent. But the fact they are on there and attract other people (some of whom you can speak to) adds value for you, too.
  - seanmcdirmid 13y ago
    Cat gifs aren't a thing in china, I think its mostly an American fetish.
    - lifeisstillgood 13y ago
      All in good time, all in good time
  - deleted 13y ago
    [deleted]
  - Retric 13y ago
    Orders of magnitude. Sure, adding someone else in china is in theory worth something. However adding one more friend that I regularly communicate with on a network is worth more to me than every person in China. PS: For large networks X Log X is probably much closer to reality than X ^2. Just compare the amount of internet bandwidth between NY to California vs the bandwidth between the US and China.
    - encoderer 13y ago
      Yeah but isn't that the point? Metcalf's law is not suggesting "The value of the network to any random node within the network." Because that measure is very subjective in the ways you say. Instead, it's "The total value of the network to all nodes". Because the more people in the network, the higher the probability that the nodes you DO care about are also connected.
- dredmorbius 13y ago
  See "A Refutation of Metcalfe's Law", by Odlyzko and Tilly: http://www.dtc.umn.edu/~odlyzko/doc/metcalfe.pdf http://www.dtc.umn.edu/~odlyzko/doc/metcalfe.pdf I'd extend the rule further and suggest that the value of a network scales as some Odlyzko and Tilly suggest, but with an additional negative function subtracting value (v) from the network: v = n(log(n) - f(n) That is: as the network grows, the added value of each additional member is reduced (log(n)). Further, each additional member of the network exacts a cost to the network as a whole as well. Doing some simple modeling, I suspect that this isn't a strictly linear factor, but itself grows with n, quite possibly as the original Metcalfe's law suggestion. That is: any given member is increasingly less likely to be a positive contribution to the network, but might well present an equal opportunity to be a net negative to the group as a whole: v = n(log(n)) - kn^2 Where 0 < k < 1 and n > 1. Moreover, let's look at some group sizes which might allow us to estimate for k in various contexts. For software team size, it's very typical that a core engineering group has a size of 5-10 members, more or less. This suggests that k is about 0.2 for software development: every added team member exacts a cost, within a single group, of about 20%. This sees value grow for 1 < n < 8, then fall with larger n, hitting negative values at about n=14. For an elementary school classroom where the ideal class size seems to be around 22-25 students, k would be around 0.08. For Dunbar's Number, the number of relationships people can manage (100 - 300, typically set at 150), k is between 0.028 (100), 0.2 (150), and 0.113 (300). For city sizes, it's likely that different cities offer different matches of positive and negative factors. k of 0.0005 gives value max at n ~ 10,000, k of 0.0006 is ~ 100,000, and 0.000007 is around 1 million. To scale to 1 billion users with net positive value means you have to keep k to less than 0.00000001. That is: any one member can have only a 1 in 10 million chance of being annoying to other members.
  - erichocean 13y ago
    > To scale to 1 billion users with net positive value means you have to keep k to less than 0.00000001. That is: any one member can have only a 1 in 10 million chance of being annoying to other members. Or you could just design a network where new users don't annoy existing users, and reduce it down all the way to zero.
    - 13y ago
logicallee 13y ago
isn't the obvious graph to ask for cost per user / revenue per user? (i.e. "acquisition price" per earnings per user.) After all, that's the main way that we can consider a per-user cost to be "high".
- tav 13y ago
  A bunch of ZeroDivisionErrors doesn't look so pretty :)
Shivetya 13y ago
450 million is a number that immediately brings up the question, how many are real? This is almost as if the site is trying to make up an excuse for the purchase. However to me, having watched different entities out there I am always curious as to what is real and what isn't

by oskarth 13y ago

TTPrograms 13y ago

This was exactly the chart I wanted to see after I saw the first one. ...but did they really put a LLS line through that dataset?

rkuykendall-com 13y ago
It's an API that puts an LLS line through everything. If it isn't meaningful, ignore it.

Retric 13y ago

Metcalfe's law is clearly wrong after a network get's overly large. I could in theory call a random person in China but if I have no way of communicating with them it's practically worthless to do so.

nl 13y ago
Cat gifs are pretty universal. But seriously: I've had people on conference calls when they couldn't understand the language, but someone wanted them to see slides. The fact they could join the call (in that case on Skype) definitely made the network more valuable. You might argue that the fact you can't speak to some of them drops the value per user, and that is true to some extent. But the fact they are on there and attract other people (some of whom you can speak to) adds value for you, too.
- seanmcdirmid 13y ago
  Cat gifs aren't a thing in china, I think its mostly an American fetish.
  - lifeisstillgood 13y ago
    All in good time, all in good time
- deleted 13y ago
  [deleted]
- Retric 13y ago
  Orders of magnitude. Sure, adding someone else in china is in theory worth something. However adding one more friend that I regularly communicate with on a network is worth more to me than every person in China. PS: For large networks X Log X is probably much closer to reality than X ^2. Just compare the amount of internet bandwidth between NY to California vs the bandwidth between the US and China.
  - encoderer 13y ago
    Yeah but isn't that the point? Metcalf's law is not suggesting "The value of the network to any random node within the network." Because that measure is very subjective in the ways you say. Instead, it's "The total value of the network to all nodes". Because the more people in the network, the higher the probability that the nodes you DO care about are also connected.
dredmorbius 13y ago
See "A Refutation of Metcalfe's Law", by Odlyzko and Tilly: http://www.dtc.umn.edu/~odlyzko/doc/metcalfe.pdf http://www.dtc.umn.edu/~odlyzko/doc/metcalfe.pdf I'd extend the rule further and suggest that the value of a network scales as some Odlyzko and Tilly suggest, but with an additional negative function subtracting value (v) from the network: v = n(log(n) - f(n) That is: as the network grows, the added value of each additional member is reduced (log(n)). Further, each additional member of the network exacts a cost to the network as a whole as well. Doing some simple modeling, I suspect that this isn't a strictly linear factor, but itself grows with n, quite possibly as the original Metcalfe's law suggestion. That is: any given member is increasingly less likely to be a positive contribution to the network, but might well present an equal opportunity to be a net negative to the group as a whole: v = n(log(n)) - kn^2 Where 0 < k < 1 and n > 1. Moreover, let's look at some group sizes which might allow us to estimate for k in various contexts. For software team size, it's very typical that a core engineering group has a size of 5-10 members, more or less. This suggests that k is about 0.2 for software development: every added team member exacts a cost, within a single group, of about 20%. This sees value grow for 1 < n < 8, then fall with larger n, hitting negative values at about n=14. For an elementary school classroom where the ideal class size seems to be around 22-25 students, k would be around 0.08. For Dunbar's Number, the number of relationships people can manage (100 - 300, typically set at 150), k is between 0.028 (100), 0.2 (150), and 0.113 (300). For city sizes, it's likely that different cities offer different matches of positive and negative factors. k of 0.0005 gives value max at n ~ 10,000, k of 0.0006 is ~ 100,000, and 0.000007 is around 1 million. To scale to 1 billion users with net positive value means you have to keep k to less than 0.00000001. That is: any one member can have only a 1 in 10 million chance of being annoying to other members.
- erichocean 13y ago
  > To scale to 1 billion users with net positive value means you have to keep k to less than 0.00000001. That is: any one member can have only a 1 in 10 million chance of being annoying to other members. Or you could just design a network where new users don't annoy existing users, and reduce it down all the way to zero.
  - 13y ago

logicallee 13y ago

isn't the obvious graph to ask for cost per user / revenue per user? (i.e. "acquisition price" per earnings per user.) After all, that's the main way that we can consider a per-user cost to be "high".

tav 13y ago
A bunch of ZeroDivisionErrors doesn't look so pretty :)

Shivetya 13y ago

450 million is a number that immediately brings up the question, how many are real? This is almost as if the site is trying to make up an excuse for the purchase. However to me, having watched different entities out there I am always curious as to what is real and what isn't