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I did read part II, and I think his analogy is off. Memory isn't linearly limited (it happens to be in the processor, but this isn't a given), I don't even unde

by hacknat 10y ago

I did read part II, and I think his analogy is off. Memory isn't linearly limited (it happens to be in the processor, but this isn't a given), I don't even understand why he brings circles into his conception of memory. Memory isn't linearly bound to the previous local cache. RAM isn't an order of magnitude larger than the L3 cache, it's MANY of order of magnitude larger! His analogy is just wrong. A better analogy would be library A takes X amount of time to access its books and can store 100 of them, and library B takes 10X amount of time to access its books, but it can store 1,000,000 books! Yes as we increase in memory size latency goes up, but he never proves that this is sqrt(N) (he correlates that it is). Each jump in up can be explained by cache misses in each successive local cache, but RAM can scale more than a few orders of magnitude beyond the L3 cache. He needed to keep his chart going past 1GB to see that there is actually a plateau to be hit. If he had scaled to 10GB and then to 100GB he would have seen access times be about the same that they were for 1GB.

zeroer 10y ago
Re-read the "The theoretical limit" section in part II. His arguments depend on the physics in this particular universe, not how computers happen to be built.
- hacknat 10y ago
  My response was about his math and physics: If you scale the radius of a sphere by one unit then you indeed get an order of magnitude increase in volume (bits of info), but that's not the correct model! We don't increase by on unit! RAM isn't a one unit increase in radius! It's an order of magnitude increase! If you increase the radius by an order of magnitude you get a 3 fold order of magnitude increase in memory. So you jump up in latency by one order of magnitude and you get 3 orders of magnitude of memory in return. His analysis and math are wrong.
  - Jweb_Guru 10y ago
    You clearly did not understand the linked article at all. He is referring to theoretical results on the information density of a black hole, which indeed collapses to area. In practice, the reason RAM doesn't appear to increase by three orders of magnitude is not that (it's that we don't actually build RAM in a sphere) but his point is that whether you are looking at theory or practice the square root is the correct model.
  - zeroer 10y ago
    You're totally missing his argument. If you scale RAM cubicly at any positive density, as you are suggesting, eventually you will achieve the matter density sufficient for your RAM to gravitationally collapse into a black hole. If the density of your RAM is d, then the volume that a mass of M RAM takes up is M/d. If it's arranged in a sphere, the radius is ((3M)/(4dpi))^(1/3). Notice how this radius is a constant times the cube root of M. Whereas the Schwarzschild radius scales proportionally to mass. https://en.wikipedia.org/wiki/Schwarzschild_radius https://en.wikipedia.org/wiki/Schwarzschild_radius Thus, if you put enough mass of constant postive density together (no matter how small that density is) eventually you get a black hole. N.B. - The author's argument is a little more subtle because he's talking about information density via the Berkenstein Bound, and he gets a square root instead of a cube root. But the argument is the same flavor.

by hacknat 10y ago

zeroer 10y ago

Re-read the "The theoretical limit" section in part II. His arguments depend on the physics in this particular universe, not how computers happen to be built.

hacknat 10y ago
My response was about his math and physics: If you scale the radius of a sphere by one unit then you indeed get an order of magnitude increase in volume (bits of info), but that's not the correct model! We don't increase by on unit! RAM isn't a one unit increase in radius! It's an order of magnitude increase! If you increase the radius by an order of magnitude you get a 3 fold order of magnitude increase in memory. So you jump up in latency by one order of magnitude and you get 3 orders of magnitude of memory in return. His analysis and math are wrong.
- Jweb_Guru 10y ago
  You clearly did not understand the linked article at all. He is referring to theoretical results on the information density of a black hole, which indeed collapses to area. In practice, the reason RAM doesn't appear to increase by three orders of magnitude is not that (it's that we don't actually build RAM in a sphere) but his point is that whether you are looking at theory or practice the square root is the correct model.
- zeroer 10y ago
  You're totally missing his argument. If you scale RAM cubicly at any positive density, as you are suggesting, eventually you will achieve the matter density sufficient for your RAM to gravitationally collapse into a black hole. If the density of your RAM is d, then the volume that a mass of M RAM takes up is M/d. If it's arranged in a sphere, the radius is ((3M)/(4dpi))^(1/3). Notice how this radius is a constant times the cube root of M. Whereas the Schwarzschild radius scales proportionally to mass. https://en.wikipedia.org/wiki/Schwarzschild_radius https://en.wikipedia.org/wiki/Schwarzschild_radius Thus, if you put enough mass of constant postive density together (no matter how small that density is) eventually you get a black hole. N.B. - The author's argument is a little more subtle because he's talking about information density via the Berkenstein Bound, and he gets a square root instead of a cube root. But the argument is the same flavor.