| Bonita Montero <Bonita.Montero@gmail.com>: Jun 03 09:16PM +0200 > "In one point of comparison, the RISC-V Rocket core is twice as energy efficient as the most similar ARM implementation, the Cortex-A5," the foundation said in a Q&A. That's depending on the efficiency. If you would desin a CPU with a comparable IPC and clocking like a current x86 there won't be a big difference. > Is Open Source RISC-V Ready to Take on Intel, AMD, and ARM in the Data Center? If even ARM doesn't make it for years, RISC-V is even more unlikely. |
| aminer68@gmail.com: Jun 06 01:00PM -0700 Hello, Read again, i correct a typo.. I was just reading the following webpage: PageRank Algorithm - The Mathematics of Google Search http://pi.math.cornell.edu/~mec/Winter2009/RalucaRemus/Lecture3/lecture3.html I think this webpage above is not rigorous mathematics, because you have to speak about the conditions required for the Markov chain to find its way to an equilibrium distribution, so read my following tutorial to understand: My name is Amine Moulay Ramdane, i will talk about Markov chains in mathematics.. In mathematics, many Markov chains automatically find their own way to an equilibrium distribution as the chain wanders through time. This happens for many Markov chains, but not all. I will talk about the conditions required for the chain to find its way to an equilibrium distribution. If in mathematics we give a Markov chain on a finite state space and asks if it converges to an equilibrium distribution as t goes to infinity, an equilibrium distribution will always exist for a finite state space, but you need to check whether the chain is irreducible and aperiodic. If so, it will converge to equilibrium. If the chain is irreducible but periodic, it cannot converge to an equilibrium distribution that is independent of start state. If the chain is reducible, it may or may not converge. So i will give an example: Suppose that for the course you are currently taking there are two volumes on the market and represent them by A and B. Suppose further that the probability that a teacher using volume A keeps the same volume next year is 0.4 and the probability that it will change for volume B is 0.6. Furthermore the probability that a professor using B this year changes to next year for A is 0.2 and the probability that it again uses volume B is 0.8. We notice that the matrix of transition is: 0.4 0.6 0.2 0.8 The interesting question for any businessman is whether his market share will stabilize over time. In other words, does it exist a probability vector (t1, t2) such that: (t1, t2) * (transition matrix above) = (t1, t2) [1] So notice that the transition matrix above is irreducible and aperiodic, so it will converge to an equilibrum distribution that is (t1, t2) that i will mathematically find, so the system of equations of [1] above is: 0.4 * t1 + 0.2 * t2 = t1 0.6 * t1 + 0.8 * t2 = t2 this gives: -0.6 * t1 + 0.2 * t2 = 0 0.6 * t1 - 0.2 * t2 = 0 But we know that (t1, t2) is a vector of probability, so we have: t1 + t2 = 1 So we have to solve the following system of equations: t1 + t2 = 1 0.6 * t1 - 0.2 * t2 = 0 So i have just solved it with R, and this gives the vector: (0.25,0.75) Which means that in the long term, volume A will grab 25% of the market while volume B will grab 75% of the market unless the advertising campaign does change the probabilities of transition. Thank you, |
| aminer68@gmail.com: Jun 06 12:51PM -0700 Hello, I was just reading the following webpage: PageRank Algorithm - The Mathematics of Google Search http://pi.math.cornell.edu/~mec/Winter2009/RalucaRemus/Lecture3/lecture3.html I think this webpage above is not rigorous mathematics, because you have to speak about the conditions required for the Markov chain to find its way to an equilibrium distribution, so read my following tutorial to understand: My name is Amine Moulay Ramdane, i will talk about Markov chains in mathematics.. In mathematics, many Markov chains automatically find their own way to an equilibrium distribution as the chain wanders through time. This happens for many Markov chains, but not all. I will talk about the conditions required for the chain to find its way to an equilibrium distribution. If in mathematics we give a Markov chain on a finite state space and asks if it converges to an equilibrium distribution as t goes to infinity, an equilibrium distribution will always exist for a finite state space, but you need to check whether the chain is irreducible and aperiodic. If so, it will converge to equilibrium. If the chain is irreducible but periodic, it cannot converge to an equilibrium distribution that is independent of start state. If the chain is reducible, it may or may not converge. So i will give an example: Suppose that for the course you are currently taking there are two volumes on the market and represent them by A and B. Suppose further that the probability that a teacher using volume A keeps the same volume next year is 0.4 and the probability that it will change for volume B is 0.6. Furthermore the probability that a professor using B this year changes to next year for A is 0.2 and the probability that it again uses volume B is 0.8. We notice that the matrix of transition is: 0.4 0.6 0.2 0.8 The interesting question for any businessman is whether his market share will stabilize over time. In other words, does it exist a probability vector (t1, t2) such that: (t1, t2) * (transition matrix above) = (t1, t2) [1] So notice that the transition matrix above is ??irreducible and aperiodic, so it will converge to an equilibrum distribution that is (t1, t2) that i will mathematically find, so the system of equations of [1] above is: 0.4 * t1 + 0.2 * t2 = t1 0.6 * t1 + 0.8 * t2 = t2 this gives: -0.6 * t1 + 0.2 * t2 = 0 0.6 * t1 - 0.2 * t2 = 0 But we know that (t1, t2) is a vector of probability, so we have: t1 + t2 = 1 So we have to solve the following system of equations: t1 + t2 = 1 0.6 * t1 - 0.2 * t2 = 0 So i have just solved it with R(look inside test.pl example), and this gives the vector: (0.25,0.75) Which means that in the long term, volume A will grab 25% of the market while volume B will grab 75% of the market unless the advertising campaign does change the probabilities of transition. Thank you, |
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