Metcalfe's Law: Bitcoin Price and Adoption Analysis for the Future - page 3.

painlord2k

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Quote from: FattyMcButterpants on June 19, 2014, 01:19:23 PM

i think metcalfe's law is wrong. it assumes all nodes are equally valuable, which is pretty controversial. here is a good rundown of why:

http://spectrum.ieee.org/computing/networks/metcalfes-law-is-wrong

perhaps growth is not quadratic, but rather n log(n), where network size = n.

The nodes joining the network first are the nodes receiving the most benefit, others will follow.
The nodes joining first are not, usually, the largest.
Often , the traffic increase a lot when some subset of nodes is able to form closed loops where there is a positive feedback loop.

FattyMcButterpants

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i think metcalfe's law is wrong. it assumes all nodes are equally valuable, which is pretty controversial. here is a good rundown of why:

http://spectrum.ieee.org/computing/networks/metcalfes-law-is-wrong

perhaps growth is not quadratic, but rather n log(n), where network size = n.

Bitcoin_is_here_to_stay

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Quote from: chriswilmer on June 19, 2014, 12:56:49 PM

Quote from: Bitcoin_is_here_to_stay on June 19, 2014, 12:35:28 PM

I am afraid I do not understand what are "unique addresses" - I first thought these are bitcoin addresses, somehow filtered. But if so, how they can ever be declining?

I think it's unique addresses used in transactions PER DAY.

Thanks, that makes sense

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chriswilmer

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Quote from: Bitcoin_is_here_to_stay on June 19, 2014, 12:35:28 PM

I am afraid I do not understand what are "unique addresses" - I first thought these are bitcoin addresses, somehow filtered. But if so, how they can ever be declining?

I think it's unique addresses used in transactions PER DAY.

Bitcoin_is_here_to_stay

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I am afraid I do not understand what are "unique addresses" - I first thought these are bitcoin addresses, somehow filtered. But if so, how they can ever be declining?

saddambitcoin

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i believe we will go super exponential this year and hit $10,000.

just around the corner...

Raystonn

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I have updated the graph with the latest data. We continue to have higher lows on the Metcalfe data. I have also added a zoomed chart to the first post. We can clearly see we're working ourselves into a tighter and tighter range on adoption. We should see a large price move upward when we break above the Metcalfe high of June 1st.

Wary

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Quote from: Raystonn on June 18, 2014, 02:49:03 AM

I think you are just playing at definitions and missing the larger point. As rate of adoption increases beyond linear, rate of price appreciation will increase beyond the current exponential. Until we pass the 50% point on the S curve, the rate of price appreciation should therefore accelerate, not diminish.

If you don’t believe me or don’t get it, I don’t have time to try to convince you, sorry. (c)-you know who Wink

Raystonn

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Quote from: Wary on June 18, 2014, 02:08:56 AM

Quote from: Raystonn on June 18, 2014, 01:21:28 AM

N(y) = y^2, where N is the number of addresses and y the year number. N increases exponentially as time marches on in a linear fashion. As we approach the center of the S curve, adoption will begin increasing at an exponential rate. The Metcalfe value is this exponential rate, squared. So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number. This is simplified to N(y) = y^4. N increases much faster here.

OK. I think I get the root of our misunderstanding. You seem to think that y^2 or y^4 is an exponent. It is not. If you don't trust me, ask somebody whom you trust.

Quote

I will stop you here. What you describe is not an S curve. This is an S curve:

rogerwilco already posted reference to the the S-curve aka logistic function: http://en.wikipedia.org/wiki/Logistic_function. In particular, this article says: "the logistic curve shows early exponential growth for negative argument, which slows to linear growth of slope 1/4 for an argument near zero, then approaches one with an exponentially decaying gap.". Which means exactly what I've said: exponential curve that slows down to horizontal line. Again, sorry for bothering you with calculus 101.

I think you are just playing at definitions and missing the larger point. As rate of adoption increases beyond linear, rate of price appreciation will increase beyond the current exponential. Until we pass the 50% point on the S curve, the rate of price appreciation should therefore accelerate, not diminish.

Raystonn

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Quote from: zimmah on June 18, 2014, 02:10:02 AM

Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

Funny how human psychology makes most humans buy when the price is rising , but no one dares buying when the price is low after a bubble. Even when it's pretty obvious a new bubble will come.

I mean right now we are sitting on 600s but people will only really start buying once we past 1000 or so. And then once we peaked at about 5000 or so, people will stop buying until we are once again past the previous ATH rather than just buying during the downtrend.

Without buyers near the top, we'd have no one to sell to. I plan to keep no more than 90% of my liquid wealth in Bitcoin. I will need to sell at regular price intervals to keep that ratio.

zimmah

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Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

Funny how human psychology makes most humans buy when the price is rising , but no one dares buying when the price is low after a bubble. Even when it's pretty obvious a new bubble will come.

I mean right now we are sitting on 600s but people will only really start buying once we past 1000 or so. And then once we peaked at about 5000 or so, people will stop buying until we are once again past the previous ATH rather than just buying during the downtrend.

Wary

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Quote from: Raystonn on June 18, 2014, 01:21:28 AM

N(y) = y^2, where N is the number of addresses and y the year number. N increases exponentially as time marches on in a linear fashion. As we approach the center of the S curve, adoption will begin increasing at an exponential rate. The Metcalfe value is this exponential rate, squared. So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number. This is simplified to N(y) = y^4. N increases much faster here.

OK. I think I get the root of our misunderstanding. You seem to think that y^2 or y^4 is an exponent. It is not. If you don't trust me, ask somebody whom you trust.

Quote

I will stop you here. What you describe is not an S curve. This is an S curve:

rogerwilco already posted reference to the the S-curve aka logistic function: http://en.wikipedia.org/wiki/Logistic_function. In particular, this article says: "the logistic curve shows early exponential growth for negative argument, which slows to linear growth of slope 1/4 for an argument near zero, then approaches one with an exponentially decaying gap.". Which means exactly what I've said: exponential curve that slows down to horizontal line. Again, sorry for bothering you with calculus 101.

Swordsoffreedom

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Thanks for the chart I was wondering what the chart looks like normalized for Price Volatility
I assume that as the market price moves upward or downward that the price affects the users and unique transactions that occur in the network
That said more data is always a good thing

Raystonn

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Quote from: Wary on June 18, 2014, 12:36:34 AM

That's why I'm surprised. On the chart we can see a curve that fluctuates around a straight line. Correct? The vertical scale is log. Correct? Therefore the straight line is an exponential line. Because no other line can be straight in the log scale. Correct? Therefore rate of adoption (squared) fluctuates around an exponential rate. Correct? Exponential line squared is still an exponential line. Correct? Therefore rate of adoption grows at an exponential rate. Correct?

Right now adoption is increasing at a linear rate. This means we gain more users at some average constant rate of new addresses per day. The Metcalfe value is this value squared. So, to approximate, N(y) = y^2, where N is the number of addresses and y the year number. N increases exponentially as time marches on in a linear fashion.

As we approach the center of the S curve, adoption will begin increasing at an exponential rate. The Metcalfe value is this exponential rate, squared. So, to approximate, N(y) = (y^2)^2, where N is the number of addresses and y the year number. This is simplified to N(y) = y^4. N increases much faster here.

Quote from: Wary on June 18, 2014, 12:36:34 AM

Now the S-curve. It is basically an exponential curve that at it's last third slows down to horizontal line. Until that stage it's rate of growth is constant.

I will stop you here. What you describe is not an S curve. This is an S curve:

Wary

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Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

Sorry, one more correction: Super-exponential curve is a curve that looks exponential on log scale. Something like y=exp(exp(x)). While exponential curve, squared, is still just an exponential: y = (exp(x))^2 = exp(2x). In log scale it will still looks like straight line, just more steep, than non-squared one.

P.S. Sorry for nitpicking, nothing personal (and thanks for the OP), I just wanted to clarify some math details.

kashish948

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the price will keep increasing with new merchants especially the big ones!

Wary

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Quote from: Raystonn on June 18, 2014, 12:18:28 AM

Quote from: Wary on June 18, 2014, 12:15:07 AM

Quote from: Raystonn on June 18, 2014, 12:10:29 AM

Quote from: Wary on June 18, 2014, 12:06:28 AM

Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No. The rate of adoption is currently linear. The Metcalfe value is exponential. That is the number of unique addresses used per day, squared. Vertical is an expression. The middle of an S curve looks nearly vertical.

Look at left side of the OP chart: it is in log scale. Curve that looks linear in log scale is exponential.

I *am* the OP. The Metcalfe value displayed on the chart is the rate of adoption, *squared*. That is the definition of the value from Metcalfe's Law.

That's why I'm surprised. On the chart we can see a curve that fluctuates around a straight line. Correct? The vertical scale is log. Correct? Therefore the straight line is an exponential line. Because no other line can be straight in the log scale. Correct? Therefore rate of adoption (squared) fluctuates around an exponential rate. Correct? Exponential line squared is still an exponential line. Correct? Therefore rate of adoption grows at an exponential rate. Correct?

Now the S-curve. It is basically an exponential curve that at it's last third slows down to horizontal line. Until that stage it's rate of growth is constant. Say, it grows 10x each year. It was growing this way at the beginning, it grows that way now, it will keep growing this way till it starts slowing down. The rate of growth in relative number, percentage-wise will never increase. That's why I said that S-curve has no vertical stage. The area that you are referring to as "vertical stage" is the area somewhere around 50% where the growth in absolute numbers, will reach the maximum. But the growth rate in relative numbers, in %, will be decreasing by then and will be lower than it is now.

Edit: typo

rogerwilco

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Quote from: Wary on June 18, 2014, 12:06:28 AM

Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

Actually, the S-curve is the colloquial term for the Logistic function, which can be intuitively understood as a system initially taking on exponential growth but eventually running into limiting factors.

The fastest rate of expansion ("going vertical") is halfway between the lower and upper limits. This inflection point represents a paradigm shift, because whereas before the growth could seemingly expand anywhere in the system, now it's become the norm so there are fewer places left to expand into. As an example, in the outside chance Bitcoin becomes the world reserve currency, this would be the point at which people would no longer value a bitcoin in dollars, but rather value a dollar in terms of bitcoins.

Raystonn

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Quote from: Wary on June 18, 2014, 12:15:07 AM

Quote from: Raystonn on June 18, 2014, 12:10:29 AM

Quote from: Wary on June 18, 2014, 12:06:28 AM

Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No. The rate of adoption is currently linear. The Metcalfe value is exponential. That is the number of unique addresses used per day, squared. Vertical is an expression. The middle of an S curve looks nearly vertical.

Look at left side of the OP chart: it is in log scale. Curve that looks linear in log scale is exponential.

I *am* the OP. The Metcalfe value displayed on the chart is the rate of adoption, *squared*. That is the definition of the value from Metcalfe's Law.

Wary

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Quote from: Raystonn on June 18, 2014, 12:10:29 AM

Quote from: Wary on June 18, 2014, 12:06:28 AM

Quote from: Raystonn on June 17, 2014, 04:48:20 PM

I've been analyzing further. When we hit the vertical stage of the adoption S curve, rate of adoption will go exponential. Metcalfe will go super-exponential at that point, along with price.

It's an illusion. S curve has no vertical stage, just like an exponential curve doesn't have it. The rate of adoption is already (and had always been) exponential, that's why it looks like straight line on the log-scale chart.

No. The rate of adoption is currently linear. The Metcalfe value is exponential. That is the number of unique addresses used per day, squared. Vertical is an expression. The middle of an S curve looks nearly vertical.

Look at left side of the OP chart: it is in log scale. Curve that looks linear in log scale is exponential.

Topic: Metcalfe's Law: Bitcoin Price and Adoption Analysis for the Future - page 3. (Read 14808 times)