Metcalfe’s law says that a network of n people has a value of n2. This law is useful for networks like a phone network where value comes from the connections between two people.
Reed’s law says that a network of n people has a value of 2n. This law is useful for networks like Google Groups where value comes from communities of many people.
Reed’s law grows a lot faster than Metcalfe’s law. Umair Haque has written about how these laws effect the financial gains of companies like Yahoo, Google, and eBay.
Between Metcalfe’s law and Reed’s law are a class of networks that grow like n3, n4, n5, and so on. In general, these networks grow like nc where c is a constant.
I’ll explain why and how. But first, here is a table of the various laws. Sarnoff’s law grows the slowest as the network gets bigger and Haque’s law grows the fastest.
| Value | Who | Model | Example |
| n | Sarnoff | Broadcasting | TV |
| nlog(n) | Odlyzko | A practical Metcalfe’s Law | Telephone |
| n2 | Metcalfe | Networks | Telephone |
| nc | Me | A practical Reed’s Law | Google Groups |
| 2n | Reed | Communities | Google Groups |
| n! | Haque | ? | ? |
So where does my nc law come from?
Certain folks have observed that the value of a community can decrease as the network grows. After a certain point, you can’t remember everybody’s name and you start getting trolls, idiots, and spammers in the community. This limit is often called the Dunbar Number and empirical evidence indicates that it is less than 150.
After 150 people, the value of your community doesn’t grow as fast as it used to. The value may, in fact, decrease as your community grows.
If you assume that any community larger than 150 people has no value, you get an n150 law. If you assume that any community larger than c people has no value, you get an nc law.
So, between Metcalfe’s n2 law and Reed’s 2n law are n3, n4, n5, … laws that span the gap between Metcalfe and Reed.
Please send the Nobel Prize to nivi@alum.mit.edu.
Notes: Propers go to Dan Gould for bringing the Dunbar Number to my attention. And somebody please check my math, I haven’t done math since grade 3. The logic is that if you have a set of n elements, the number of subsets that have c elements or less is O(nc).
Interesting thoughts Nivi… You have triggered some thoughts on the measurability and effectiveness of social networks… Thanks.
ceo
You seem to be using n^150 weirdly here. Are you saying that the value of a 2 node network of this type has a value of 2^150? The notation also suggests that a 151 node network of this type has a value of 151^150, rather than zero.
Death,
n is a big number. n is much bigger than c. n is the total size of the network. For example, n could be the number of the people on the Internet.
If you have a 151 people in a network, all the subgroups with 150 people or less are valuable. The 151 person subgroup is not valuable (by assumption).
Aha. I see what you mean.
Aren’t you opening up yourself to Odlyzkoing? Clearly the cost of finding the right 150 people among the masses, and the diminishing value of overlapping groups wildly reduces the actual value of the network. Metcalfe’s case is at least plausible, since each interaction between two people is disjoint. But being a part of two subnetworks (cliques?) of 150 people that differ by only one person would be a minor annoyance, and being a part of n choose 150 such cliques would be maddening. You should head this off at the pass and figure out a tighter bound.
What’s the network’s value to an individual member, as opposed to as a whole? Is the value of the network as a whole bounded by the sum of these member valuations? What about leveraging synergies?
The number of subsets with (precisely) k elements is n!/(k!(n-k)!) . It is not a power law.
Metcalfe: each mmeber can make (n-1) connections, namely to all but himself. There are therefore n(n-1) ~ O(n^2) connections possible
Reed: There are 2^n possible subsets of a set of n (different) elements. For each element, decide if it is in or not. Two options chosen n times = 2^n possibilities.
Nivi,
I guess you are making your argument from the limits to the binomial coefficients:
http://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_for_binomial_coefficients
If
C(n,k) 2k, C(n,k) > C(n,k-1), so the number of subsets with no more than k elements is
S = C(n,k)+C(n,k-1)+…+C(n,1)
Death,
Those are good thoughts. If you come up with an interesting result, I would be glad to publish it here.