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In statistical mechanics, a semi-classical derivation of the entropy that does not take into account the indistinguishability of particles, yields an expression for the entropy which is not extensive (is not proportional to the amount of substance in question). This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox". If one takes the perspective that the definition of entropy must be changed so as to ignore particle permutation, the paradox is averted.

## Illustration of the problem

Gibbs himself considered the following problem that arises if the ideal gas entropy is not extensive.[1] Two identical containers of an ideal gas sit side-by-side. There is a certain amount of entropy S associated with each container of gas, and this depends on the volume of each container. Now a door in the container walls is opened to allow the gas particles to mix between the containers. No macroscopic changes occur, as the system is in equilibrium. The entropy of the gas in the two-container system can be easily calculated, but if the equation is not extensive, the entropy would not be 2S. In fact, the non-extensive entropy quantity defined and studied by Gibbs would predict additional entropy. Closing the door then reduces the entropy again to 2S, in supposed violation of the Second Law of Thermodynamics.

As understood by Gibbs,[2] and reemphasized more recently,[3][4] this is a misapplication of Gibb's non-extensive entropy quantity. If the gas particles are distinguishable, closing the doors will not return the system to its original state - many of the particles will have switched containers. There is a freedom in what is defined as ordered, and it would be a mistake to conclude the entropy had not increased. In particular, Gibbs' non-extensive entropy quantity for an ideal gas was not intended for varying numbers of particles.

The paradox is averted by concluding the indistinguishability (at least effective indistinguishability) of the particles in the volume. This results in the extensive Sackur–Tetrode equation for entropy, as derived next.

## Calculating the entropy of ideal gas, and making it extensive

In classical mechanics, the state of an ideal gas of energy U, volume V and with N particles, each particle having mass m, is represented by specifying the momentum vector p and the position vector x for each particle. This can be thought of as specifying a point in a 6N-dimensional phase space, where each of the axes corresponds to one of the momentum or position coordinates of one of the particles. The set of points in phase space that the gas could occupy is specified by the constraint that the gas will have a particular energy:

$U=\frac{1}{2m}\sum_{i=1}^{N} p_{ix}^2 + p_{iy}^2 + p_{iz}^2$

and be contained inside of the volume V (let's say V is a box of side X so that V=X3):

$0 \le x_{ij} \le X$

for :$i = 1...N$ and $j = 1, 2, 3$

The first constraint defines the surface of a 3N-dimensional hypersphere of radius (2mU)1/2 and the second is a 3N-dimensional hypercube of volume VN. These combine to form a 6N-dimensional hypercylinder. Just as the area of the wall of a cylinder is the circumference of the base times the height, so the area φ of the wall of this hypercylinder is:

$\phi(U,V,N) = V^N \left(\frac{2\pi^{\frac{3N}{2}}(2mU)^{\frac{3N-1}{2}}}{\Gamma(3N/2)}\right)~~~~~~~~~~~(1)$

The entropy is proportional to the logarithm of the number of states that the gas could have while satisfying these constraints. In classical physics, the number of states is infinitely large, but according to quantum mechanics it is finite. Before the advent of quantum mechanics, this infinity was regularized by making phase space discrete. Phase space was divided up in blocks of volume $h^{3 N}$. The constant h thus appeared as a result of a mathematical trick and thought to have no physical significance. However, using quantum mechanics one recovers the same formalism in the semi-classical limit, but now with h being Planck's constant. One can qualitatively see this from Heisenberg's uncertainty principle; a volume in N phase space smaller than h3N (h is Planck's constant) cannot be specified.

To compute the number of states we must compute the volume in phase space in which the system can be found and divide that by $h^{3 N}$. This leads us to another problem: The volume seems to approach zero, as the region in phase space in which the system can be is an area of zero thickness. This problem is an artifact of having specified the energy U with infinite accuracy. In a generic system without symmetries, a full quantum treatment would yield a discrete non-degenerate set of energy eigenstates. An exact specification of the energy would then fix the precise state the system is in, so the number of states available to the system would be one, the entropy would thus be zero.

When we specify the internal energy to be U, what we really mean is that the total energy of the gas lies somewhere in an interval of length $\delta U$ around U. Here $\delta U$ is taken to be very small, it turns out that the entropy doesn't depend strongly on the choice of $\delta U$ for large N. This means that the above "area" $\phi$ must be extended to a shell of a thickness equal to an uncertainty in momentum $\delta p = \delta\left(\sqrt{2 m U}\right) = \sqrt{\frac{m}{2 U}}\delta U$, so the entropy is given by:

$\left.\right. S=k\,\ln(\phi \delta p/h^{3N})$

where the constant of proportionality is k, Boltzmann's constant. Using Stirling's approximation for the Gamma function which omits terms of less than order N, the entropy for large N becomes:

$S = k N \ln \left[ V \left(\frac UN \right)^{\frac 32}\right]+ {\frac 32}kN\left( 1+ \ln\frac{4\pi m}{3h^2}\right)$

This quantity is not extensive as can be seen by considering two identical volumes with the same particle number and the same energy. Suppose the two volumes are separated by a barrier in the beginning. Removing or reinserting the wall is reversible, but the entropy increases when the barrier is removed by the amount

$\delta S = k \left[ 2N \ln(2V) - N\ln V - N \ln V \right] = 2 k N \ln 2 > 0$

which is in contradiction to thermodynamics if you re-insert the barrier. This is the Gibbs paradox.

The paradox is resolved by postulating that the gas particles are in fact indistinguishable. This means that all states that differ only by a permutation of particles should be considered as the same state. For example, if we have a 2-particle gas and we specify AB as a state of the gas where the first particle (A) has momentum p1 and the second particle (B) has momentum p2, then this state as well as the BA state where the B particle has momentum p1 and the A particle has momentum p2 should be counted as the same state.

For an N-particle gas, there are N! states which are identical in this sense, if one assumes that each particle is in a different single particle state. One can safely make this assumption provided the gas isn't at an extremely high density. Under normal conditions, one can thus calculate the volume of phase space occupied by the gas, by dividing Equation 1 by N!. Using the Stirling approximation again for large N, ln(N!) ≈ N ln(N) - N, the entropy for large N is:

$S = k N \ln \left[ \left(\frac VN\right) \left(\frac UN \right)^{\frac 32}\right]+kN\left( {\frac 52}+ {\frac 32}\ln\frac{4\pi m}{3h^2}\right)$

which can be easily shown to be extensive. This is the Sackur–Tetrode equation.

A closely related paradox is the mixing paradox. Again take a box with a partition in it, with gas A on one side, gas B on the other side, and both gases are at the same temperature and pressure. If gas A and B are different gases, there is an entropy that arises due to the mixing. If the gases are the same, no additional entropy is calculated. The additional entropy from mixing does not depend on the character of the gases. The paradox is that the two gases can be arbitrarily similar, but the entropy from mixing does not disappear unless they are the same gas.

The resolution is provided by a careful understanding of entropy. In particular, as explained concisely by Jaynes,[2] there is an arbitrariness in the definition of entropy.

A central example in Jaynes' paper relies on the fact that, if one develops a theory based on the idea that the two different types of gas are indistinguishable, and one never carries out any measurement which contradicts this fact, then the theory will have no internal inconsistencies. In other words, if there are two gases A and B and we have not yet discovered that they are different, then assuming they are the same will cause no theoretical problems. If ever an experiment is performed with these gases that yields incorrect results, we will certainly have discovered a method of detecting their difference and recalculating the entropy increase when the partition is removed.

This insight suggests that the idea of thermodynamic state and entropy are somewhat subjective. The differential increase in entropy (dS), as a result of mixing dissimilar element sets (the gases), multiplied by the temperature (T) is equal to the minimum amount of work we must do to restore the gases to their original separated state. Suppose that the two different gases are separated by a partition, but that we cannot detect the difference between them. We remove the partition. How much work does it take to restore the original thermodynamic state? None – simply reinsert the partition. The fact that the different gases have mixed does not yield a detectable change in the state of the gas, if by state we mean a unique set of values for all parameters that we have available to us to distinguish states. As soon as we become able to distinguish the difference, the amount of work necessary to recover the original macroscopic configuration becomes non-zero, and the amount of work does not depend on the magnitude of that difference.

This line of reasoning is particularly informative when considering the concepts of indistinguishable particles and correct Boltzmann counting. Boltzmann's original expression for the number of states available to a gas assumed that a state could be expressed in terms of a number of energy "sublevels" each of which contain a particular number of particles. While the particles in a given sublevel were considered indistinguishable from each other, particles in different sublevels were considered distinguishable from particles in any other sublevel. This amounts to saying that the exchange of two particles in two different sublevels will result in a detectably different "exchange macrostate" of the gas. For example, if we consider a simple gas with N particles, at sufficiently low density that it is practically certain that each sublevel contains either one particle or none (i.e. a Maxwell–Boltzmann gas), this means that a simple container of gas will be in one of N! detectably different "exchange macrostates", one for each possible particle exchange. Just as the mixing paradox begins with two detectably different containers, and the extra entropy that results upon mixing is proportional to the average amount of work needed to restore that initial state after mixing, so the extra entropy in Boltzmann's original derivation is proportional to the average amount of work required to restore the simple gas from some "exchange macrostate" to its original "exchange macrostate". If we assume that there is in fact no experimentally detectable difference in these "exchange macrostates" available, then using the entropy which results from assuming the particles are indistinguishable will yield a consistent theory. This is "correct Boltzmann counting". It is often said that the resolution to the Gibbs paradox derives from the fact that, according to the quantum theory, like particles are indistinguishable in principle. By Jaynes' reasoning, if the particles are experimentally indistinguishable for whatever reason, Gibbs paradox is resolved, and quantum mechanics only provides an assurance that in the quantum realm, this indistinguishability will be true as a matter of principle, rather than being due to an insufficiently refined experimental capability.

## Non-extensive entropy of two ideal gases and how to fix it

In this section, we present in rough outline a purely classical derivation of the non-extensive entropy for an ideal gas considered by Gibbs before "correct counting" (indistinguishability of particles) is accounted for. This is followed by a brief discussion of two standard methods for making the entropy extensive. Finally, we present a third method, due to R. Swendsen, for an extensive (additive) result for the entropy of two systems if they are allowed to exchange particles with each other.[5][6]

### Setup

We will present a simplified version of the calculation. It differs from the full calculation in three ways:

1. The ideal gas consists of particles confined to one spatial dimension.
2. We keep only the terms of order $n \log(n)$, dropping all terms of size n or less, where n is the number of particles. For our purposes, this is enough, because this is where the Gibbs paradox shows up and where it must be resolved. The neglected terms play a role when the number of particles is not very large, such as in computer simulation and nanotechnology. Also, they are needed in deriving the Sackur–Tetrode equation.
3. The subdivision of phase space into units of Planck's constant (h) is omitted. Instead, the entropy is defined using an integral over the "accessible" portion of phase space. This serves to highlight the purely classical nature of the calculation.

We begin with a version of Boltzmann's entropy in which the integrand is all of accessible phase space:

$S = k_B \ln\Omega = k_B\oint\limits_{H(\vec p, \vec q)=E} (d\vec p\,)^n (d\vec q\,)^n$

The integral is restricted to a contour of available regions of phase space, subject to conservation of energy. In contrast to the one-dimensional line integrals encountered in elementary physics, the contour of constant energy possesses a vast number of dimensions. The justification for integrating over phase space using the canonical measure involves the assumption of equal probability. The assumption can be made by invoking the ergodic hypothesis as well as the Liouville's theorem of Hamiltonian systems.

(The ergodic hypothesis underlies the ability of a physical system to reach thermal equilibrium, but this may not always hold for computer simulations (see the Fermi–Pasta–Ulam problem) or in certain real-world systems such as non-thermal plasmas.)

Liouville's theorem assumes a fixed number of dimensions that the system 'explores'. In calculations of entropy, the number dimensions is proportional to the number of particles in the system, which forces phase space to abruptly change dimensionality when particles are added or subtracted. This may explain the difficulties in constructing a clear and simple derivation for the dependence of entropy on the number of particles.

For the ideal gas, the accessible phase space is an (n-1)-sphere (also called a hypersphere) in the $n$ dimensional $\vec v$ space:

$E=\sum_{j=1}^n \frac 1 2 mv_j^2\,,$

To recover the paradoxical result that entropy is not extensive, we integrate over phase space for a gas of $n$ monatomic particles confined to a single spatial dimension by $0. Since our only purpose is to illuminate a paradox, we simplify notation by taking the particle's mass and Boltzmann's constant equal to unity: $m = k_B = 1$. We represent points in phase-space and its x and v parts by n and 2n dimensional vectors:

$\vec\xi = [x_1,..., x_n, v_1,...,v_n]=[\vec x, \vec v\,]$   where   $\vec x = [x_1,..., x_n]$   and   $\vec v = [v_1,...,v_n]\,.$

To calculate entropy, we use the fact that the (n-1)-sphere, $\Sigma v_j^2=R^2\, ,$ has an (n-1)-dimensional "hypersurface volume" of,

$\tilde A_n(R)=\frac{n\pi^{n/2}}{(n/2)!} R^{n-1}\,.$

For example, if n = 2, the 1-sphere is the circle $\tilde A_2(R)=2\pi R$, a "hypersurface" in the plane. When the sphere is even-dimensional (n odd), it will be necessary to use the gamma function to give meaning to the factorial; see below.

### Gibb's paradox in a one-dimensional gas

Gibb's paradox arises when entropy is calculated using an $n$ dimensional phase space, where $n$ is also the number of particles in the gas. These particles are spatially confined to the one-dimensional interval $\ell^n$. The volume of the surface of fixed energy is

$\Omega_{E,\ell} = \left( \int dx_1 \int dx_2 ...\int dx_n\right)\left(\underbrace{\int dv_1 \int dv_2 ...\int dv_n}_{\Sigma v_i^2 = 2E}\right)$

The subscripts on $\Omega$ are used to define the 'state variables' and will be will be discussed later, when it is argued that the number of particles, $n$ lacks full status as a state variable in this calculation. The integral over configuration space is $\ell^n$. As indicated by the underbrace, the integral over velocity space is restricted to the "surface area" of the n-1 dimensional hypersphere of radius $\sqrt{2E}$, and is therefore equal to the "area" of that hypersurface. Thus

$\Omega_{E,\ell} = \ell^n \frac{n\pi^{n/2}}{(n/2)!} (2E)^{\frac{n-1}{2}}$

After approximating the factorial and dropping the small terms, we obtain

\begin{align} \ln\Omega_{E,\ell} &\approx n\ln\ell + n \ln\sqrt{\frac E n} + const.\\ &= \underbrace{n\ln\frac{\ell}{n} + n \ln\sqrt{\frac E n}}_{extensive} + \,n\ln n + const.\\ \end{align}

In the second expression, the term $n\ln n$ was subtracted and added, using the fact that $\ln\ell-\ln n = \ln (\ell /n)$. This was done to highlight exactly how the "entropy" defined here fails to be an extensive property of matter. The first two terms are extensive: if the volume of the system doubles, but gets filled with the same density of particles with the same energy, then each of these terms doubles. But the third term is neither extensive nor intensive and is therefore wrong.

The arbitrary constant has been added because entropy can usually be viewed as being defined up to an arbitrary additive constant. This is especially necessary when entropy is defined as the logarithm of a phase space volume measured in units of momentum-position. Any change in how these units are defined will add or subtract a constant from the value of the entropy.

### Two standards ways to make the classical entropy extensive

As discussed above, an extensive form of entropy is recovered if we divide the volume of phase space, $\Omega_{E,\ell}$, by n!. An alternative approach is to argue that the dependence on particle number cannot be trusted on the grounds that changing $n$ also changes the dimensionality of phase space. Such changes in dimensionality lie outside the scope of Hamiltonian mechanics and Liouville's theorem. For that reason it is plausible to allow the arbitrary constant to be a function of $n$.[7] Defining the function to be, $f(n)=-n\ln n$, we have:

\begin{align} S = \ln\Omega_{E,\ell} &\approx n\ln\ell + n \ln\sqrt{E} + const.\\ &= n\ln\ell + n \ln\sqrt{E} + f(n)\\ \ln\Omega_{E,\ell,n}&\approx n\ln\frac{\ell}{n} + n \ln\sqrt{\frac E n} + const.\\ \end{align},

which has extensive scaling:

$S(\alpha E,\alpha\ell,\alpha n) = \alpha\, S(E,\ell,n)$

### Swendsen's particle-exchange approach

Following Swendsen,[5][6] we allow two systems to exchange particles. This essentially 'makes room' in phase space for particles to enter or leave without requiring a change in the number of dimensions of phase space. The total number of particles is $N$:

• $n_A$ particles have coordinates $0.
The total energy of these particles is $E_A$
• $n_B$ particles have coordinates $0.
The total energy of these particles is $E_B$
• The system is subject to the constraints, $E_A+E_B=E$ and $n_A+n_B=N$

Taking the integral over phase space, we have:

$\Omega_{E,\ell,N} =$ $\left( \underbrace{\int dx_? ... \int dx_?}_{n_A\;terms} \underbrace{\int dx_?...\int dx_N}_{n_B\;terms}\right)\left(\underbrace{\int dv_1 \int dv_2 ...\int dv_N}_{\Sigma v^2 = 2E_A\;or\;\Sigma v^2 = 2E_B}\right)$

$=\left(\ell_A\right)^{n_A}\left(\ell_B\right)^{n_B} \left(\underbrace{\frac{N!}{n_A!n_B!}}_{combination}\right) \left(\underbrace{\frac{n_A\pi^{n_A/2}}{(n_A/2)!} (2E_A)^{\frac{n_A-1}{2}}}_{n_A-sphere}\right) \left(\underbrace{\frac{n_B\pi^{n_B/2}}{(n_B/2)!} (2E_B)^{\frac{n_B-1}{2}}}_{n_B-sphere}\right)$

The question marks (?) serve as a reminder that we may not assume that that the first nA particles (i.e. 1 though nA) are in system-A while the other particles (nB through N) are in system-B. (This is further discussed in the next section.)

Taking the logarithm and keeping only the largest terms, we have:

$S =\ln \Omega_{E,\ell,N}$ $\approx n_A\ln\left(\frac{n_A}{\ell_A}\sqrt{\frac{E_A}{\ell_A}}\right)+ n_B\ln\left(\frac{n_B}{\ell_B}\sqrt{\frac{E_B}{\ell_B}}\right)+ N\ln N + const.$

This can be interpreted as the sum of the entropy of system-A and system-B, both extensive. And there is a term, $N\ln N$, that is not extensive.

### Visualizing the particle-exchange approach in three dimensions

A three particle ideal gas partitioned into two parts

The correct (extensive) formulas for systems A and B were obtained because we included all the possible ways that the two systems could exchange particles. The use of combinations (i.e. N particles choose NA) was used to ascertain the number of ways N particles can be divided into system-A containing nA particles and system_B containing nB particles. This counting is not justified on physical grounds, but on the need to integrate over phase space. As will be illustrated below, phase space contains not a single nA-sphere and a single nB-sphere, but instead

${N \choose n_A} = \frac{N!}{n_A!n_B!}$

pairs of n-spheres, all situated in the same N+1-dimensional velocity space. The integral over accessible phase space must include all of these n-spheres, as can be seen in the figure, which shows the actual velocity phase space associated a gas that consists of three particles. Moreover, this gas has been divided into two systems, A and B.

If we ignore the spatial variables, the phase space of a gas with three particles is three dimensional, which permits one to sketch the n-spheres over which the integral over phase space must be taken. If all three particles are together, the split between the two gasses is 3|0. Accessible phase space is delimited by an ordinary sphere (2-sphere) with a radius that is either $\sqrt{2E_1}$ or $\sqrt{2E_2}$ (depending which system has the particles).

If the split is 2|1, then phase space consists of circles and points. Each circle occupies two dimensions, and for each circle, two points lie on the third axis, equidistant from the center of the circle. In other words, if system-A has 2 particles, accessible phase space consists of 3 pairs of n-spheres, each pair being a 1-sphere and a 0-sphere:

$v_1^2+v_2^2=2E_A \qquad v_3^2 = 2E_B$
$v_2^2+v_3^2=2E_A \qquad v_1^2 = 2E_B$
$v_3^2+v_1^2=2E_A \qquad v_2^2 = 2E_B$

Note that

${3 \choose 2} = 3.$

## References

1. ^ Gibbs, J. Willard (1875–1878). On the Equilibrium of Heterogeneous Substances. Connecticut Acad. Sci. ISBN 0-8493-9685-9. Reprinted in Gibbs, J. Willard (October 1993). The Scientific Papers of J. Willard Gibbs (Vol. 1). Ox Bow Press. ISBN 0-918024-77-3. and in Gibbs, J. Willard (February 1994). The Scientific Papers of J. Willard Gibbs (Vol. 2). Ox Bow Press. ISBN 1-881987-06-X.
2. ^ a b Jaynes, E.T. (1996). "The Gibbs Paradox" (PDF). Retrieved November 8, 2005.
3. ^ Grad, Harold (1961). "The many faces of entropy". Communications on Pure and Applied Mathematics 14 (3): 323. doi:10.1002/cpa.3160140312.
4. ^ van Kampen, N. G. (1984). "The Gibbs Paradox". In W. E. Parry. Essays in Theoretical Physics in Honor of Dirk ter Haar. Oxford: Pergamon. ISBN 978-0080265230.
5. ^ a b Swendsen, Robert (March 2006). "Statistical mechanics of colloids and Boltzmann's definition of entropy". American Journal of Physics 74 (3): 187–190. doi:10.1119/1.2174962.
6. ^ a b Swendsen, Robert H. (June 2002). "Statistical Mechanics of Classical Systems with Distinguishable Particles". Journal of Statistical Physics 107 (5/6): 1143–1166. Bibcode:2002JSP...107.1143S. doi:10.1023/A:1015161825292.
7. ^ Jaynes, E.T. (1992). The Gibbs Paradox in Maximum Entropy and Bayesian Methods (edited by C.R. Smith, G.J. Erickson, & P.O. Neudorfere) (PDF). Dordrecht, Holland: Kluwer Academic Publishers. pp. 1–22. In particular, Gibbs failed to point out that an 'integration constant' was not an arbitrary constant, but an arbitrary function. But this has, as we shall see, nontrivial physical consequences. What is remarkable is not that Gibbs should have failed to stress a one mathematical point in almost the last words he wrote; but that for 80 years thereafter all textbook writers (except possibly Pauli) failed to see it.


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### Proposal for Syncing Channel Updates

Lightning network is a peer to peer network of nodes, each node keeps a graph of the current existing nodes on the network with the nodes as vertices and the channels as edges. The routing table includes information about the channels existing on the network like their fee parameters, channel participants and others.

This information sometimes needs updating and so the node should be in continuous sync with the rest of the network, this syncing needs to be as efficient as possible. The first implementation of this was that nodes would request a complete dump of the routing table of the nodes they are connected to, updating their table with the new information which is a highly impractical way to do things. Then came a different implementation, every node would send its peers a list of channel ids, and the peer would reply back with information about the ids it has but the other participant doesn’t, nodes can also send a timestamp-based filter to their peers and their peers would reply back with updates that match this filter. This works well when the node misses a channel announcement but not when it misses a channel updates.

An update was suggested to include channel update timestamps to the channel range queries, the reply_channel_range will now include 4 bytes of a timestamp and the query_short_channel_ids will include a flag that is set to 1 if the node should send back a channel announcement and the updates, and would otherwise send just the updates.

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Linear Regression Problem for Pearson product moment correlation coefficient (r).

Find the linear correlation coefficient (r) and determine whether the relationship between average temperature and ice cream sales is strong, weak, or neutral for the following dataset of ice cream sales.

Temperature (x) 4 4 7 8 12 15 16 17 14 11 7 5
Ice Cream Sale (y) 73 57 81 94 110 124 134 139 124 103 81 80
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## Linear Regression Problem for correlation coefficient (r)

The given data sample size = n = 12

Formulae -

$$Linear\ Correlation\ Coefficient = r = \frac {S_{XY}}{\sqrt {S_{XX} \times S_{YY}}}$$

Where,

$$S_{XY} = \Bigl(\sum xy\Bigr) - n. \overline x. \overline y$$

$$S_{XX} = \Bigl(\sum x^2 \Bigr) - n. \overline x^2$$

$$S_{YY} = \Bigl(\sum y^2\Bigr) - n. \overline y^2$$

# Step 1

Calculate the $\sum x$, $\sum y$, $\sum x^2$, $\sum y^2$, and $\sum xy$

x y x2 y2 xy
4 73 16 5329 292
4 57 16 3249 228
7 81 49 6561 567
8 94 64 8836 752
12 110 144 12100 1320
15 124 225 15376 1860
16 134 256 17956 2144
17 139 289 19321 2363
14 124 196 15376 1736
11 103 121 10609 1133
7 81 49 6561 567
5 80 25 6400 400
120 1200 1450 127674 13362

# Step 2

Calculate the sample means of temperature (x) and ice cream sales (y) as follows:

$$\overline x = \frac {\sum x}{n} = \frac {120}{12} = 10$$

$$\overline y = \frac {\sum y}{n} = \frac {1200}{12} = 100$$

# Step 3

Calculate the $S_{XY}$

$$S_{XY} = \Bigl(\sum xy\Bigr) - n. \overline x. \overline y$$

$$S_{XY} = 13362 - 12 \times 10 \times 100$$

$$S_{XY} = 13362 - 12000$$

$$S_{XY} = 1362$$

Calculate the $S_{XX}$

$$S_{XY} = 1362$$

$$S_{XX} = \Bigl(\sum x^2 \Bigr) - n. \overline x^2$$

$$S_{XX} = 1450 - 12 \times (10)^2$$

$$S_{XX} = 1450 - 1200$$

$$S_{XX} = 250$$

Calculate the $S_{YY}$

$$S_{YY} = \Bigl(\sum y^2\Bigr) - n. \overline y^2$$

$$S_{YY} = 127674 - 12 \times (100)^2$$

$$S_{YY} = 127674 - 120000$$

$$S_{YY} = 7674$$

# Step 4

Finally, calculate the Linear Correlation Coefficient (r)

$$r = \frac {S_{XY}}{\sqrt {S_{XX} \times S_{YY}}}$$

$$r = \frac {1362}{\sqrt {250 \times 7674}}$$

$$r = \frac {1362}{1385.099274}$$

$$r = 0.983$$

## Step 5

• Thus we get the Linear Correlation Coefficient r = 0.983

• This implies that there is a strong, POSITIVE linear relationship between average temperature and ice cream sales since the linear correlation coefficient (r) is very close to +1.


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# Bayesian Signed-Rank Test for comparing algorithms in machine learning

This post is about Bayesian nonparametric tests for comparing algorithms in ML. This time we will discuss about Python module signrank in bayesiantests (see our GitHub repository). It computes the Bayesian equivalent of the Wilcoxon signed-rank test. It return probabilities that, based on the measured performance, one model is better than another or vice versa or they are within the region of practical equivalence.

This notebook demonstrates the use of the module. You can download the notebook from GitHub as well.

We will load the classification accuracies of the naive Bayesian classifier and AODE on 54 UCI datasets from the file Data/accuracy_nbc_aode.csv. For simplicity, we will skip the header row and the column with data set names.

In [11]:
import numpy as np
scores = np.loadtxt('Data/accuracy_nbc_aode.csv', delimiter=',', skiprows=1, usecols=(1, 2))
names = ("NBC", "AODE")

Functions in the module accept the following arguments.

• x: a 2-d array with scores of two models (each row corresponding to a data set) or a vector of differences.
• rope: the region of practical equivalence. We consider two classifiers equivalent if the difference in their performance is smaller than rope.
• prior_strength: the prior strength for the Dirichlet distribution. Default is 0.6.
• prior_place: the region into which the prior is placed. Default is bayesiantests.ROPE, the other options are bayesiantests.LEFT and bayesiantests.RIGHT.
• nsamples: the number of Monte Carlo samples used to approximate the posterior.
• names: the names of the two classifiers; if x is a vector of differences, positive values mean that the second (right) model had a higher score.

### Summarizing probabilities

Function signrank(x, rope, prior_strength=0.6, prior_place=ROPE, nsamples=50000, verbose=False, names=('C1', 'C2')) computes the Bayesian signed-rank test and returns the probabilities that the difference (the score of the first classifier minus the score of the first) is negative, within rope or positive.

In [2]:
import bayesiantests as bt
left, within, right = bt.signrank(scores, rope=0.01)
print(left, within, right)

0.0 0.12244 0.87756

The first value (left) is the probability that the first classifier (the left column of x) has a higher score than the second (or that the differences are negative, if x is given as a vector).

In the above case, the right (AODE) performs worse than naive Bayes with a probability of 0.88, and they are practically equivalent with a probability of 0.12.

If we add arguments verbose and names, the function also prints out the probabilities.

In [12]:
left, within, right = bt.signrank(scores, rope=0.01, verbose=True, names=names)

P(NBC > AODE) = 0.0, P(rope) = 0.1232, P(AODE > NBC) = 0.8768

The posterior distribution can be plotted out:

1. using the function signrank_MC(x, rope, prior_strength=1, prior_place=ROPE, nsamples=50000) we generate the samples of the posterior
2. using the function plot_posterior(samples,names=('C1', 'C2')) we then plot the posterior in the probability simplex
In [13]:
%matplotlib inline
import matplotlib.pyplot as plt

samples = bt.signrank_MC(scores, rope=0.01)

fig = bt.plot_posterior(samples,names)

plt.show()

### Checking sensitivity to the prior

To check the effect of the prior, let us a put a greater prior on the left.

In [19]:
samples = bt.signrank_MC(scores, rope=0.01,  prior_strength=0.6, prior_place=bt.LEFT)
fig = bt.plot_posterior(samples,names)
plt.show()

… and on the right

In [20]:
samples = bt.signrank_MC(scores, rope=0.01,  prior_strength=0.6, prior_place=bt.RIGHT)
fig = bt.plot_posterior(samples,names)
plt.show()

The prior with a strength of 1 has negligible effect. Only a much stronger prior on the left would shift the probabilities toward NBC:

In [18]:
samples = bt.signrank_MC(scores, rope=0.01,  prior_strength=6, prior_place=bt.LEFT)
fig = bt.plot_posterior(samples,names)
plt.show()

## Auxiliary functions

The function signrank_MC(x, rope, prior_strength=0.6, prior_place=ROPE, nsamples=50000) computes the posterior for the given input parameters. The result is returned as a 2d-array with nsamples rows and three columns representing the probabilities $p(-\infty, rope), p[-rope, rope], p(rope, \infty)$. Call signrank_MC directly to obtain a sample of the posterior.

The posterior is plotted by plot_simplex(points, names=('C1', 'C2')), where points is a sample returned by signrank_MC.

## References

@ARTICLE{bayesiantests2016, author = {{Benavoli}, A. and {Corani}, G. and {Demsar}, J. and {Zaffalon}, M.}, title = "{Time for a change: a tutorial for comparing multiple classifiers through Bayesian analysis}", journal = {ArXiv e-prints}, archivePrefix = "arXiv", eprint = {1606.04316}, url={https://arxiv.org/abs/1606.04316}, year = 2016, month = jun }

@inproceedings{benavoli2014a, title = {A {B}ayesian {W}ilcoxon signed-rank test based on the {D}irichlet process}, booktitle = {Proceedings of the 30th International Conference on Machine Learning ({ICML} 2014)}, author = {Benavoli, A. and Mangili, F. and Corani, G. and Zaffalon, M. and Ruggeri, F.}, pages = {1--9}, year = {2014}, url = {http://www.idsia.ch/~alessio/benavoli2014a.pdf} }

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text:
## Thursday, 10 August 2017

### Understanding the definition of limits of sequences

The definition of limit of a sequence (in the real numbers) is well known:

Let $(u_n)_{n \in \mathbb{N}}$ be a sequence of real numbers. Then:

$$\lim_{n \to \infty} u_n = L \iff \forall \epsilon >0: \exists N \in \mathbb{N}: \forall n > N: |u_n - L| < \epsilon$$

and we say that the sequence converges to its limit $L$.

While this definition may seem abstract, there is a good reason it looks like this. When we speak about convergence, we mean that when we look far, for very great indices $n$, the terms of the sequence come arbitrarily close to the limit. So, the distance between the limit value, and the value of the sequence for this large $n$, can become as small as possible.

So for a certain (small) $\epsilon > 0$, we have that $|u_n - L| < \epsilon$, if we look far enough, which translates by saying that we can always find a positive integer $N$ such that the condition $|u_n - L| < \epsilon$ is met whenever $n > N$.

## Sunday, 16 April 2017

### $(p-1)! \equiv -1 \bmod p$

A group theoretic proof:

Obviously, for $p = 2$, the statement is true.
For the rest of the proof, let's assume $p \neq 2$

Now, if we have a finite abelian group $G = \{x_1, \dots , x_n\}$, we know that $x_1*x_2* \dots x_n = \prod\limits_{ord(x_i)=2}x_i$, because every element that does not have order 2 cancels with its inverse (if $ord(x) = 2$, then $x^2 = e$, then $x = x^{-1}$)

Now, let's apply this to the group $(\mathbb{Z}/p\mathbb{Z}-\{0\},.)$:

(Denote the elements of $\mathbb{Z}/p\mathbb{Z}-\{0\}$  as $\{ [1], \dots [p-1]\}$)

We find $$[1][2][3]\dots[p-1] = [(p-1)!] =\prod\limits_{ord(x_i)=2}x_i$$

So, we only have to find the elements of order 2 in the group:

Hence, let's solve the equation $[a]^2 = [1]$:

$[a]^2 = [1] \iff [a^2] = [1] \iff a^2 \equiv 1 \bmod p \iff p|(a-1)(a+1)$

and because $p$ is prime (that's important here), it follows that either $p|(a-1)$ or $p|(a+1)$. Hence, $a \equiv 1 \bmod p$ or $a \equiv -1 \bmod p$

and thus:

$[a] = [1]$ or $[a] = [-1]$

Because $ord([1]) = 1$, it follows that $[-1]$ is the only element of order $2$ in the group. Hence:

$[(p-1)!] = [-1]$

and thus:

$(p - 1)! \equiv -1 \bmod p \quad \triangle$


text:
Conference paper Open Access

Synthesizing Invariants by Solving Solvable Loops

de Oliveira, Steven; Bensalem, Saddek; Prevosto, Virgile

Dublin Core Export

<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>de Oliveira, Steven</dc:creator>
<dc:creator>Bensalem, Saddek</dc:creator>
<dc:creator>Prevosto, Virgile</dc:creator>
<dc:date>2017-09-15</dc:date>
<dc:description>Formal program verification faces two problems. The first
problem is related to the necessity of having automated solvers that are
powerful enough to decide whether a formula holds for a set of proof obli-
gations as large as possible, whereas the second manifests in the need of
finding sufficiently strong invariants to obtain correct proof obligations.
This paper focuses on the second problem and describes a new method
for the automatic generation of loop invariants that handles polynomial
and non deterministic assignments. This technique is based on the eigen-
vector generation for a given linear transformation and on the polynomial
optimization problem, which we implemented on top of the open-source
tool Pilat.
</dc:description>
<dc:identifier>https://zenodo.org/record/893067</dc:identifier>
<dc:identifier>10.5281/zenodo.893067</dc:identifier>
<dc:identifier>oai:zenodo.org:893067</dc:identifier>
<dc:relation>info:eu-repo/grantAgreement/EC/H2020/731453/</dc:relation>
<dc:relation>doi:10.5281/zenodo.893066</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:rights>http://creativecommons.org/licenses/by-sa/4.0/legalcode</dc:rights>
<dc:subject>invariant generation</dc:subject>
<dc:subject>linearization</dc:subject>
<dc:subject>polynomial mapping</dc:subject>
<dc:title>Synthesizing Invariants by Solving Solvable Loops</dc:title>
<dc:type>info:eu-repo/semantics/conferencePaper</dc:type>
<dc:type>publication-conferencepaper</dc:type>
</oai_dc:dc>

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text:
# OPTIMAL CLONING OF PURE STATES

@article{Werner1998OPTIMALCO,
title={OPTIMAL CLONING OF PURE STATES},
author={Reinhard F. Werner},
journal={Physical Review A},
year={1998},
volume={58},
pages={1827-1832}
}
• R. Werner
• Published 1 April 1998
• Mathematics
• Physical Review A
We construct the unique optimal quantum device for turning a finite number of d-level quantum systems in the same unknown pure state \sigma into M systems of the same kind, in an approximation of the M-fold tensor product of the state \sigma.
36 Citations
Bounds on the efficiency of cloning for two-state quantum systems
We analyse the efficiency of quantum cloning for qubits. We investigate, in particular, the role played by the form of allowed input states and show how the fidelity of cloning changes by restricting
Quantum cloning machine of a state in a belt of Bloch sphere
• Physics
• 2009
AbstractWe analyze the problem of approximate quantum cloning when the quantum state is between two latitudes on the Bloch’s sphere. We present an analytical formula for the optimized 1-to-2 cloning.
A quantum network for implementation of the optimal quantum cloning
• Physics
• 2009
This paper presents a quantum network to implement the optimal 1 → 2 quantum cloning in 2 dimensions, including the optimal asymmetric universal, the optimal symmetric phase-covariant, and the
Optimal cloning for two pairs of orthogonal states
• Mathematics
• 2001
We study the optimal cloning transformation for two pairs of orthogonal states of two-dimensional quantum systems, and derive the corresponding optimal fidelities.
On the Entanglement Structure in Quantum Cloning
• Physics
• 2002
We study the entanglement properties of the output state of a universal cloning machine. We analyse in particular bipartite and tripartite entanglement of the clones, and discuss the “classical
Quantum Cloning Optimal for Joint Measurements
• Mathematics
• 2002
We show that universally covariant cloning is not optimal for achieving joint measurements of noncommuting observables with minimum added noise. For such a purpose a cloning transformation that is
Optimal cloning of pure states, testing single clones
• Mathematics
• 1999
We consider quantum devices for turning a finite number N of d-level quantum systems in the same unknown pure state σ into M>N systems of the same kind, in an approximation of the M-fold tensor

## References

SHOWING 1-10 OF 33 REFERENCES
Universal Algorithm for Optimal Estimation of Quantum States from Finite Ensembles via Realizable Ge
• Computer Science
• 1998
A universal algorithm for the optimal quantum state estimation of an arbitrary finite dimensional system that specifies a physically realizable positive operator valued measurement on a finite number of identically prepared systems is presented.
Optimal extraction of information from finite quantum ensembles.
• Computer Science
Physical review letters
• 1995
It is proved that optimal measurement procedures must necessarily view the ensemble as a single composite system rather than as the sum of its components, i.e., optimal measurements cannot be realized by separate measurements on each particle.
Optimal Quantum Cloning Machines
• Computer Science
• 1997
Quantum Cloning Machines (QCM) are universal devices to translate quantum information into classical information and it is proved that the fidelity (quality) of these copies is optimal.
Quantum copying: Beyond the no-cloning theorem.
• Physics
Physical review. A, Atomic, molecular, and optical physics
• 1996
It is shown that there exists a ‘‘universal quantum-copying machine’’ ~i.e., transformation! which approximately copies quantum-mechanical states such that the quality of its output does not depend on the input.
Two non-orthogonal states can be cloned by a unitary-reduction process
• Mathematics
• 1997
We show that, there are physical means for cloning two non-orthogonal pure states which are secretly chosen from a certain set $% \$={ | \Psi_0 > , | \Psi_1 > }\$. The states are cloned through a
QUANTUM COPYING : FUNDAMENTAL INEQUALITIES
• Physics
• 1997
How well can one copy an arbitrary qubit? To answer this question we consider two arbitrary vectors in a two-dimensional state space and an abstract copying transformation which will copy these two
Probabilistic and Statistical Aspects of Quantum Theory
• Physics
• 1982
Foreword to 2nd English edition.- Foreword to 2nd Russian edition.- Preface.- Chapters: I. Statistical Models.- II. Mathematics of Quantum Theory.- III. Symmetry Groups in Quantum Mechanics.- IV.
A single quantum cannot be cloned
• Physics
Nature
• 1982
If a photon of definite polarization encounters an excited atom, there is typically some nonvanishing probability that the atom will emit a second photon by stimulated emission. Such a photon is


text:
# New Journal of Physics (IOPP/DPG)

Latest additions:
 2016-09-2713:19 Effects of scalar leptoquark on semileptonic Λb decays / Sahoo, Suchismita ; Mohanta, Rukmani We study the scalar leptoquark effects on the rare semileptonic decays of Λb baryon, governed by the quark level transition . [...] Published in New J. Phys. 18 (2016) 093051 10.1088/1367-2630/18/9/093051 arXiv:1607.04449 Fulltext: PDF XML; 2016-09-0717:34 Search for scalar leptoquarks in pp collisions at  = 13 TeV with the ATLAS experiment / , ; Aaboud, M ; Aad, G ; Abbott, B ; et al An inclusive search for a new-physics signature of lepton-jet resonances has been performed by the ATLAS experiment. [...] Published in New J. Phys. 18 (2016) 093016 10.1088/1367-2630/18/9/093016 arXiv:1605.06035 Fulltext: PDF XML; 2016-07-1215:23 A search for an excited muon decaying to a muon and two jets in pp collisions at with the ATLAS detector / Aad, G ; Abbott, B ; Abdallah, J ; Abdinov, O ; et al A new search signature for excited leptons is explored. [...] Published in New J. Phys. 18 (2016) 073021 10.1088/1367-2630/18/7/073021 arXiv:1601.05627 Fulltext: PDF XML; 2016-07-1118:06 The evolution of σγP with coherence length / Caldwell, Allen Assuming the form at fixed Q2 for the behavior of the virtual-photon proton scattering cross section, where l is the coherence length of the photon fluctuations, it is seen that the extrapolated values of for different Q2 cross for fm. [...] Published in New J. Phys. 18 (2016) 073019 10.1088/1367-2630/18/7/073019 arXiv:1601.04472 Fulltext: PDF XML; 2016-06-1520:28 Neutrino oscillations with the MINOS, MINOS+, T2K, and NOvA experiments / Nakaya, Tsuyoshi ; Plunkett, Robert K This paper discusses recent results and near-term prospects of the long-baseline neutrino experiments MINOS, MINOS+, T2K and NOvA. [...] Published in New J. Phys. 18 (2016) 015009 10.1088/1367-2630/18/1/015009 arXiv:1507.08134 Fulltext: XML PDF; 2016-06-1520:28 Electroweak breaking and neutrino mass: ‘invisible’ Higgs decays at the LHC (type II seesaw) / Bonilla, Cesar ; Romão, Jorge C ; Valle, José W F Neutrino mass generation through the Higgs mechanism not only suggests the need to reconsider the physics of electroweak symmetry breaking from a new perspective, but also provides a new theoretically consistent and experimentally viable paradigm. [...] Published in New J. Phys. 18 (2016) 033033 10.1088/1367-2630/18/3/033033 arXiv:1511.07351 Fulltext: PDF XML; 2016-06-1520:28 Leptoquark effects on and decay processes / Sahoo, Suchismita ; Mohanta, Rukmani We study the rare semileptonic decays of B mesons induced by as well as transitions in the scalar leptoquark (LQ) model, where the LQs have the representation (3, 2, 7/6) and (3, 2, 1/6) under the standard model gauge group. [...] Published in New J. Phys. 18 (2016) 013032 10.1088/1367-2630/18/1/013032 arXiv:1509.06248 Fulltext: PDF XML; 2016-03-0118:16 Lefschetz-thimble analysis of the sign problem in one-site fermion model / Tanizaki, Yuya ; Hidaka, Yoshimasa ; Hayata, Tomoya The Lefschetz-thimble approach to path integrals is applied to a one-site model of electrons, i.e., the one-site Hubbard model. [...] Published in New J. Phys. 18 (2016) 033002 10.1088/1367-2630/18/3/033002 arXiv:1509.07146 Fulltext: PDF XML; 2015-11-1114:12 Effective field theory out of equilibrium: Brownian quantum fields / Boyanovsky, D The emergence of an effective field theory out of equilibrium is studied in the case in which a light field—the system—interacts with very heavy fields in a finite temperature bath. [...] Published in New J. Phys. 17 (2015) 063017 10.1088/1367-2630/17/6/063017 arXiv:1503.00156 Fulltext: XML PDF; 2015-10-2814:17 Hydrodynamic transport in strongly coupled disordered quantum field theories / Lucas, Andrew We compute direct current thermoelectric transport coefficients in strongly coupled quantum field theories without long lived quasiparticles, at finite temperature and charge density, and disordered on long wavelengths compared to the length scale of local thermalization. [...] Published in New J. Phys. 17 (2015) 113007 10.1088/1367-2630/17/11/113007 Fulltext: PDF XML;


text:
# 30 inch bathroom mirror

And they talked about the role that TeamQuest capacity management solutions play at HelpSystems. Amazon addresses this strategic decision area through a finished goods inventory using just-in-time inventory management in some areas. Operations management 1. If your business uses standardized operations throughout production, and if there is little or no variation in the setup time, run time, consumption calculation, cost calculation, and so on, you might benefit from creating default operation relations for all operations. Operational Risk Management (ORM) is simply a formalized way of thinking about these things. PDF; Size: 1 MB. 08/16/2019; 2 minutes to read ; j; D; K; H; In this article. lib.mdp.ac.id. January 28, 2011 | SCRC SME. Operation costing is a mix of job costing and process costing, and is used in either of the following situations:. ECONOMIC ORDER QUANTITY (EOQ) MODEL. In Asset Management, you can calculate capacity load on: maintenance schedule lines; work orders that have not yet been scheduled; scheduled work orders; This is useful if you want to get an overview of expected capacity load for a specific period. Even if … Data from 5 samples are shown in the table below: Construct an x chart. Details. Operations Management Examples 2. The economic order quantity (EOQ) is the order quantity that minimizes total holding and ordering costs for the year. In this video from KS2 Maths, the order of operations is used to solve a long calculation. Operations Management Solvers Multiple Server Model Calculator Instructions: You can use this Multiple Server Model Calculator, by providing the arrival rate per time period $$(\lambda)$$, the service rate per time period $$(\mu)$$, and the number of servers $$(s)$$ using the form below: PDF; Size: 3 MB. In general, terms capacity is referred as maximum production capacity, which can be attained within a normal working schedule. Account Details Login Options Account Management Settings Subscription Logout No new notifications. Pages 1. File Format. Find out why ETAP has … Load more. Introduction. ETAP’s integrated digital twin platform offers the best comprehensive suite of enterprise solutions. Compute allowed time for the entire job by adding elemental standard times considering frequency of occurrence of each element. Capacity measures the rate that the operation can transform inputs into outputs. A measure of how efficiently inputs are converted into outputs is called productivity. The Microsoft Operations Management Suite is the IT management solution for the era of the cloud; a single solution, built to manage your entire enterprise environment regardless of location, operating system, hypervisors, or cloud provider choices. to give standard time for each element. This is the total amount of time a worker is engaged in productive activity. LCL = 10.27 - (0.729)(0.21) = 10.12. He presented his list of 10 principles of operations management at an APICS conference in 2007, saying the violation of these … Calculation groups that are assigned to items let you specify how the cost or sales price of components, as outlined by the calculation group, is sourced for the calculation. The typical company carries out various functions as a part of its operation. Get help with your Operations management homework. Operations Management. PERT Network with Slack Time . Production and operations management talks about applying business organization and management concepts in creation of goods and services. Add the suitable allowances to compensate for fatigue, personal needs, contingencies etc. Multiply Step 1 and Step 2. Comment on the process average and the process variability. To calculate the EMV in project risk management, you need to: Assign a probability of occurrence for the risk. Production. 1) Operations management refers to the direction and control of inputs that transform processes into products and services. The diameter is a critical factor. The calculation for slack time is conducted for every event in the network as shown in the following figure by recognizing the latest starting date and the earliest expected date. ORM is a simple six-step process, which identifies operational hazards and takes reasonable measures to reduce risk to … Per Bauer is the Director of International Services for capacity management at HelpSystems. vtechworks.lib.vt.edu. The study of component and process reliability is the basis of many efficiency evaluations in Operations Management discipline. The management of steel bar manufacturer is concerned about the production of it's steel bars. Recall that operations management is responsible for managing the transformation of numerous inputs into a range of outputs, such as goods or services. Watch the full webinar below. Product design involves creating a product that will be sold to the end consumer. recap . Pre Algebra. In this case, avoid creating operation relations that are specific to any route or released product. Production is a scientific process which involves transformation of raw material (input) into desired product or service (output) by adding economic value. By learning the key decision-making skills involved in operations management, you will make handling all the activities required to create and deliver an organisation's goods or services to its customers on a day-to-day basis much easier. This preview shows page 1 out of 1 page. Unit 34 Operation Management in Business Assignment. Step-by-Step › Algebra › Basic Operations. Chapter 1 Using Operations to Compete. All FAA operations in the United States, and indeed even our personal daily activities involve risk, and require decisions that include risk ass essment and risk management. Capacity is about the quantity of a product or service that can be made within a given time period. Download. When you work out a calculation with more than one operation (eg 8 + 2 × 3) follow the BIDMAS rule. Operations Management Plan Guide. School Southern New Hampshire University; Course Title QSO 300; Uploaded By srosenelson33. Operations management case studies helps to come out with various operational and project planning challenges in an organization. Calculate the normal time for Calculate the actual work hours for each employee. Randall Schaeffer is an experienced manufacturing and operations management professional, an industrial philosopher, and regular speaker at conferences organized by APICS, the leading US association of supply chain and operations management. These case study deals with project management leadership, operational challenges and opportunities, inventory management and planning, strategic network optimization, different forecasting dilemmas etc. Operations Management Suite manages your Windows and Linux workloads across your datacenter, Azure, Amazon Web Services (AWS), … This free online Diploma in Operations Management course will help you understand the role of an operations manager. Some of the functions include: 1. Cycle time is the time taken by the longest activity (here it is 7 min).Throughput time is 46 min. Capacity management affects all areas of an operation. 1. Assign monetary value of the impact of the risk when it occurs. Example of Operations Management Plan. Operations management is a multi-disciplinary field that focuses on managing all aspects of an organization's operations. (1) UCL = 10.27 + (0.729)(0.21) = 10.42. Or keep scrolling for the short recap. Operations Management 4-2 Milestone 2 Weighted Calculations.xlsx. By Mary Ann Anderson, MSE, Edward J. Anderson, Geoffrey Parker . There are countless ways for an operations manager to design a process. Sorting of the calculation methods System of cost management What was our topic at the last time? The value you get after performing Step 3 is the Expected Monetary Value. This value is positive for opportunities (positive risks) and negative for threats (negative risks). TL – TE = 0 for event 1. You define calculation groups under Predetermined cost policies setup in Cost management. A product initially has identical processing for a group of products, and is then finished using more product-specific procedures. Sales and Operations Planning; Supply Market Intelligence; Sustainability; Transportation Planning; ECONOMIC ORDER QUANTITY (EOQ) MODEL: Inventory Management Models : A Tutorial . This value does not include any profit earned from the firm's investments, such as … Operation Management Assignment Help, Calculate the standard time for the job, An industrial engineer has obtained the following data in connection with a time study on the lines of the Hawthorne studies taking four different elements and five cycle of the job. Business; Operations Management; How to Design a Process in Operations Management; How to Design a Process in Operations Management . ETAP ® is an analytical engineering solution company specializing in the simulation, design, monitoring, control, operator training, optimizing, and automating power systems. In operations, management capacity is referred as an amount of the input resources available to produce relative output over period of time. Energy Management Solutions to Design, Operate, and Automate Power Systems . Operations management is a key function in any organization. For example, in the calculation of the Overall Equipment Effectiveness (OEE) introduced by Nakajima , it is necessary to estimate a crucial parameter called availability. File Format. In inventory management, operations management’s focus is on maintaining optimal inventory ordering and holding. Calculate capacity load. Why ETAP. Predecessor ComponentsSequencing the operations 4. Operations Management: Processes and Supply Chains, 10e (Krajewski et al.) Product Design. They tackled the role of capacity management in operations management today (including the cloud). Agenda• Takira Motors: Creating Assembly and Process Chart• Grouping Activities• Bakery Example 3. Download. Assemblychart 5. How to Manage Bottlenecks in Operations Management By Mary Ann Anderson, MSE, Edward J. Anderson, Geoffrey Parker If, as an operations manager, you’re lucky enough to be in a situation where demand for your product or service exceeds your ability to make the products or deliver the service, then you want to find ways to increase your production so you can sell more. It is computed as a ratio of … Details. Project Management Professional (PMP)® Earned Value Management : Cost Control | iZenBridge - Duration: 36:58. iZenBridge Consultancy Pvt Ltd 144,135 views 36:58 Any time a worker is unable or unwilling to work on her regular tasks due to factors such as a lack of raw materials or machines that don't work, these hours must be deducted from her standard work hours to arrive at her actual work hours. A product initially uses different raw materials, and is then finished using a common process that is the same for a group of products; or. Slack Time Formula & Calculation in Project Management. But how do we know whether this transformation process is efficient? Construct an R chart. Productivity measures how well resources are used. What constitutes a good or bad … It means that the individual in charge of the department will be required to perform various strategic functions. Operations management is a field of business that involves managing the operations of a business to ensure efficiency in the execution of projects. The Components of an Operations Management Aggregate Plan. Point Out What Needs to be Done. This is strictly related to reliability. en English Español Português 中文(简体) עברית العربية. Operating profit is the profit earned from a firm's normal core business operations. Account management Settings Subscription Logout No new notifications a firm 's normal core business.... Processing for a group of products, and is used to solve a long calculation shows page 1 of. About these things new Hampshire University ; Course Title QSO 300 ; by. Specific to any route or released product output over period of time the suitable to... Assign a probability of occurrence of each element the operations of a product initially has processing! An organization out a calculation with more than one operation ( eg 8 + 2 × 3 follow! In either of the input resources available to produce relative output over period of time of job costing process! The economic order quantity ( EOQ ) is the time taken by the longest activity here! That TeamQuest capacity management solutions to Design, Operate operation management calculation and is then finished more! The total amount of time costing and process Chart• Grouping Activities• Bakery Example.! Concerned about the role of operation management calculation management at HelpSystems is engaged in productive.. Are countless ways for an operations manager to Design a process in operations management Course will help you the! Produce relative output over period of time a worker is engaged in productive activity calculate the in. In an organization management today ( including the cloud ) for capacity management at HelpSystems decision area a... Business that involves managing the operations of a business to ensure efficiency in the execution projects! … you define calculation groups under Predetermined cost policies setup in cost management for! Management discipline, terms capacity is about the quantity of a business to ensure in... Ensure efficiency in the table below: Construct an x chart ) time... Account Details Login Options account management Settings Subscription Logout No new notifications product or service that can made! Uploaded by srosenelson33 עברית العربية How do we know whether this transformation process is?. Long calculation data from 5 samples are shown in the table below: Construct an x chart transformation process efficient. Costing, and Automate Power Systems ’ s integrated digital twin platform offers the comprehensive! Of capacity management at HelpSystems policies setup in cost management ( EOQ is. + 2 × 3 ) follow the BIDMAS rule management Settings Subscription Logout No new.... Needs, contingencies etc creation of goods and services transform inputs into outputs the total amount of the will... They talked about the quantity of a product that will be required to perform various strategic functions J. Anderson Geoffrey. Longest activity ( here it is 7 min ).Throughput time is the order operations. Page 1 out of 1 page calculation with more than one operation ( eg 8 + 2 × )! Taken by the longest activity ( here it is computed as a ratio of … define... A group of products, and Automate Power Systems that are specific to any route released. Be required to perform various strategic functions digital twin platform offers the best comprehensive suite of enterprise solutions inputs... No new notifications process Chart• Grouping Activities• Bakery Example 3 and is then using... The end consumer the department will be sold to the end consumer the BIDMAS rule value positive. Amount of the following situations: total amount of the risk when it.! Cost management: creating Assembly and process reliability is the profit earned from a firm 's normal business! An organization inventory management in some areas operation management calculation, you need to Assign... Monetary value of the department will be required to perform various strategic functions route or released.! But How do we know whether this transformation process is efficient value of the input resources available to produce output... Management ( ORM ) is simply a formalized way of thinking about these.! 中文 ( 简体 ) עברית العربية of projects comment on the process average and the process variability How! Enterprise solutions capacity management in operations management ’ s focus is on maintaining optimal inventory ordering and.. 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text:
# Linear algebra area with vectors

1. Sep 20, 2007

The figure below shows the tetrahedron determined by a,b,c E R^3. The area vector of a face is a vector perpendicular to the face, pointing outward, whose magnitude is the area of the face. Show that the sum of the four outward pointing area vectors of the faces equals the zero vector. I have attached the image.

Since the area of each face is identical can I jsut say it is some variable say c. So i am trying to look for the directional vector of each side.. but i am having trouble doing this. I think that vec b is one and vec (b-c) is another, but i am having trouble picturing the other two vectors especially the one pointing downwards.
Is this even the right approach to this problem?

#### Attached Files:

• ###### tetrahedron.JPG
File size:
3.6 KB
Views:
120
2. Sep 20, 2007

### HallsofIvy

Are you told that the area of each face is identical? That is NOT true for a general tetrahedron. (And, even if they were, it would be really bad idea to use the same symbol, c, for the area as for one of the vectors!)

Do you know the "cross product"? In particular, do you know that the area of a parallelogram having vectors $\vec{a}$ and $\vec{b}$ as adjacent sides is $|\vec{a}\times\vec{b}|$? If you know that then the vectors referred to are just the cross products of two vectors each. Using the notation and order of edges that you have in your picture, the outward 'area vectors' are:
bottom face: $\vec{c}\times\vec{b}$
front face: $\vec{b}\times\vec{a}$
back face: $\vec{a}\times\vec{c}$
right face: $(\vec{b}-\vec{c})\times\(\vec{b}-\vec{a})$
Using the distributive law and the fact that the cross product is anti-commutative will give you the result.

If you can't use the cross product- sorry!

3. Sep 20, 2007


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# six demon bag

Wind, fire, all that kind of thing!

2013-12-20

## Automatic Updates Consuming 100% CPU

For a couple months now I've been observing the Automatic Updates service (wuauserv) causing high CPU load on hosts running Windows Server 2003 or Windows XP after new updates had been released. The problem vanished after the updates had been installed.

Apparently this is a problem with the Windows Update Agent, caused by inefficient handling of superseded updates. Installing the latest cumulative Internet Explorer update should resolve the problem.

2013-12-11

## Some Useful DFS Commands

List DFS replication groups:

dfsradmin rg list

List replicated folders in a replication group:

dfsradmin rf list /rgname:<REPL_GROUP>

See more ...

2013-10-30

## Interactive VBScript Shell

Python's interactive mode is very convenient, because you can try simple stuff without having to write it to a script first. Since I have to do a lot of VBScript lately, I wanted to have something like that for VBScript, too.

I found this blog post that has almost exactly what I wanted, except for line continuation. Which is what I added (for my own convenience). You can download the modified script here.

Update: Added an Import() procedure for loading/executing additional code from other VBScript files.

Update: vbsh can now be customized with an optional init script %USERPROFILE%\init.vbs.

Update: Added a function to look up keywords in the VBScript documentation (requires that script56.chm is installed in the current working directory, the Windows help directory, or one of the directories in the %PATH%). Note that you'll need the English language version of script56.chm, since other language versions use different internal paths.

2013-10-09

## Process Explorer Error: .NET Performance Counters Are Corrupt

Today I encountered the following error message on one of my servers (running Windows Server 2008) when I started Process Explorer:

See more ...

2013-08-27

## Import other script files in VBScript

The VBScript language doesn't provide a feature for including other code files, so you can't easily build and import code libraries. However, the missing feature can be emulated using the ExecuteGlobal statement in a custom Import procedure.

See more ...

2013-06-20

## PrintBrm Error 0x8007007b

Running the following command for backing up a printserver configuration

"%SystemRoot%\System32\spool\tools\PrintBrm.exe" -B -S server -F "C:\path\to\backup.pbk"

gave me an error after listing the printer ports:

The following error occurred: 0x8007007b.
The filename, directory name, or volume label syntax is incorrect.

Check the eventlog for detailed information about the error which occurred.

Turns out, PrintBrm doesn't seem to like double quotes around the backup file name, at least not on Windows Server 2008 R2 SP1. I changed the command line to

"%SystemRoot%\System32\spool\tools\PrintBrm.exe" -B -S server -F C:\path\to\backup.pbk

and the backup ran without issues.

2013-04-03

## Attach to Internet Explorer

Internet Explorer exposes a COM object that can be controlled programmatically e.g. from a VBScript. The usual way is to create a new Internet Explorer instance and work with that:

 Set ie = CreateObject("InternetExplorer.Application")

However, sometimes you may want to use an already running instance instead of creating a new one.

See more ...

2013-03-12

## TSM BA Client 6.4 on Debian

Although IBM doesn't officially support its TSM client on Linux distributions other than SLES and RHEL the client works quite nicely on Debian. With version 6.4 you need at least the following 4 packages on AMD64 systems:

• TIVsm-BA
• TIVsm-API64
• gskcrypt64
• gskssl64

See more ...

2013-03-09

## Features on Demand in Server 2012

Back in the days of Windows 2000 Server and Windows Server 2003 it was common practice to copy the i386 folder from the CD to the server and change the SourcePath registry value, so you wouldn't have to shuffle installation media around for adding/removing Windows features. With the advent of Windows Vista and Server 2008 Microsoft introduced the component store. Now all Windows features were put on disk by default and simply had to be activated/deactivated. That increased the size of a Windows installation by factor 3-4, but on the other hand disk space had become rather inexpensive, so it wasn't that big a deal. And although the component store can be a little fragile at times it usually works quite well.

So far, so good. But what happened now with the release of Windows 8 and Server 2012? Someone at Microsoft decided that the component store took up too much space, so they introduced a new concept called "Features on Demand", which simply means that administrators can free disk space by removing components from the store. The .Net Framework 3.5 for instance was removed by default.

This is not a bad thing per se. However, the removed framework still shows up as an installable feature (e.g. in Server Manager), only now you're prompted for an alternate source path when you choose to install the framework. So what we have now is that the component store still eats up disk space (I didn't notice any significant reduction in used disk space) and we're back to either shuffling installation media around or copying sources to the local harddisk (eating up even more disk space).

Srsly, Microsoft?

2013-02-07

## Auditing Windows File ACLs

Sometimes you run into a situation where you need to determine the permissions on some directory tree. Be it to document or clean up permissions on the subdirectories of a share, to troubleshoot permission issues due to deleted accounts or groups, or whatever. Manually analyzing permissions is quite tedious, even when using standard tools like cacls or xcacls. The output of XCACLS.vbs isn't any better, but since it's a script, I considered modifying it to suit my needs … until I took an actual look at the script. I abondoned the thought afterwards.

See more ...


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## Sunday, February 3, 2013

### Molecular Weight Distributions: The Log-Normal Distribution

A log-normal distribution in molecular weights often arises as a consequence of anionic polymer synthesis. You can see pictures of this asymmetric distribution on the linked wikipedia page.

Let us review some properties of this distribution. The number distribution of segments with molecular weight $M$ has two parameters
$N(M;m, s) = \frac{1}{M s \sqrt{2 \pi}} \exp \left( {-\frac{(\ln M - m)^2}{2 s^2}} \right).$
The quantity $\ln M$ is normally distributed around $m$ with variance $s$. You can verify that:
$\int_0^\infty N(M) dM = 1$
$M_n = \int_0^\infty M N(M) dM = \exp\left(m + \frac{s^2}{2} \right)$
$M_w = \int_0^\infty M W(M) dM =\exp\left(m + \frac{3 s^2}{2} \right),$
where I tacitly used the relation for the weight distribution $W(M) = N(M)/M_n$ in the last line. The polydispersity index $p = M_w/M_n = \exp(s^2)$
It is usual to report the $p$ and $M_w$ of an empirically obtained distribution. One can express the parameters of the log-normal distribution in terms of these numbers as:
$s^2 = \ln p$
$m = \ln M_w - \frac{3}{2} \ln p$
One could use these to directly express the number distribution in the following somewhat ugly form:
$N(M) = \frac{1}{\sqrt{2 \pi } M \sqrt{\ln (p)}} \exp \left[- \frac{\left( \ln M-\ln \left(\frac{2 M_n}{p} \right) \right)^2}{2 \ln (p)}\right]$
Note: The cumulative distribution function is given by:
$C(M; m,s) = \frac{1}{2} + \frac{1}{2} \text{erf} \left(\frac{\log M-\mu}{\sqrt{2}\sigma} \right)$


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# What is the range of the function 2(x-3)^2+1?

Jan 12, 2018

Range: $f \left(x\right) \in \left[1 , + \infty\right)$
Note that ${\left(x - 3\right)}^{2}$ has a minimum value of $\textcolor{b l u e}{0}$
in which case $2 {\left(x - 3\right)}^{2} + 1$ has a value of $\textcolor{red}{1}$
Also note that ${\left(x - 3\right)}^{2}$ has no upper limit
so as the value of ${\left(x - 3\right)}^{2}$ increases $2 {\left(x - 3\right)}^{2} + 1 \rightarrow + \infty$


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## Realtime global illumination

### April 2, 2011

I’ve had an idea floating around in my head for several months now, but evening classes, a hard drive failure, then a GPU failure prevented me from doing much about it until this weekend. First, a couple of screenshots of what I’ll be talking about.

Cornell box, I choose you!

The above screenshots are of real-time single-bounce GI in a static scene with fully dynamic lighting. There are 7182 patches, and the lighting calculations take 36ms per frame on one thread of an Intel Core 2 @2.4 GHz. The code is not optimized.

The basic idea is simple and is split into two phases.

A one-time scene compilation phase:

• Split all surfaces in the scene into roughly equal sized patches.
• For each patch i, build a list of all other patches visible from i along with the form factor between patches i and j:
$F_{ij} = \frac{cos \phi_i \, cos \phi_j}{\pi \, |r|^2} \, A_j$
where:
Fij is the form factor between patches i and j
Aj is the area of patch j
r is the vector from i to j
Φi is the angle between r and the normal of patch i
Φj is the angle between r and the normal of patch j

And a per-frame lighting phase:

• For each patch i, calculate the direct illumination.
• For each patch i, calculate single-bounce indirect illumination from all visible patches:
• $I_i = \sum_j D_j \, F_{ij}$
where:
Ij is the single-bounce indirect illumination for patch i
Dj is the direct illumination for patch j

So far, so radiosity. If I understand Michal Iwanicki’s GDC presentation correctly, this is similar to the lighting tech on Milo and Kate, only they additionally project the bounce lighting into SH.

The problem with this approach is that the running time is O(N2) with the number of patches. We could work around this by making the patches quite large, running on the GPU, or both. Alternatively, we can bring the running time down to O(N.log(N)) by borrowing from Michael Bunnell’s work on dynamic AO and cluster patches into a hierarchy. I chose to perform bottom-up patch clustering similarly to the method that Miles Macklin describes in his (Almost) realtime GI blog post.

Scene compilation is now:

• Split all surfaces in the scene into roughly equal sized patches.
• Build a hierarchy of patches using k-means clustering.
• For each patch i, build a list of all other patches visible from i along with the form factor between patches i and j. If a visible patch j is too far from patch i look further up the hierarchy.

And the lighting phase:

• For each leaf patch i in the hierarchy, calculate the direct illumination.
• Propagate the direct lighting up the hierarchy.
• For each patch i, calculate single-bounce indirect illumination from all visible patches clusters.

Although this technique is really simple, it supports a feature set similar to that of Enlighten:

• Global illumination reacts to changes in surface albedo.
• Static area light sources that cast soft shadows.

That’s basically about it. There are a few of other areas I’m tempted to look into once I’ve cleaned the code up a bit:

• Calculate directly and indirect illumination at different frequencies. This would allow scaling to much larger scenes.
• Perform the last two lighting steps multiple times to approximate more light bounces.
• Project the indirect illumination into SH, HL2 or the Half-Life basis.
• Light probes for dynamic objects.

You can grab the source code from here. Expect a mess, since it’s a C++ port of a C# proof of concept with liberal use of vector and hash_map. Scene construction is particularly bad and may take a minute to run. You can fly around using WASD and left-click dragging with the mouse.

## Fighting fireflies

### December 23, 2010

I’ve been playing around with path tracing on and off for longer than I care to admit. Although my dalliances never produced anything earth-shattering (and certainly nothing I’d be willing to post about on ompf), I’ve found it to be an endlessly fascinating subject. No matter how much I read about it, I only ever seem to be scratching the surface.

One of the biggest headaches I encountered were caused by “fireflies”: those bright pixels that can occur when a sampling a strong response combined with a small PDF somewhere along the path. For a long time, I was “fixing” these by hand painting over the offending pixels and pretending like nothing ever happened. Eventually though, the guilt of this gnawed away at me long enough to motivate finding some kind of better solution.

My first thought was to write a filter that estimated variance in an image and replace any “bad” pixels it found with a weighted average of their neighbours. Luckily, my second thought was of shadow maps, the only other context in which I’d read about variance before. Based on the ideas in that paper, I accumulate two separate per-pixel buffers: one storing the running sum of the samples and the other storing the sum of their squares. Having these two buffers then makes it trivial to compute the sample variance of each pixel in the image.

My path tracer already had support for progressive refinement, so it was straightforward to add a separate “variance reduction” pass that would run at the touch of a button. During this pass, the N pixels with the highest variance are identified and oversampled a few hundred times, which hopefully reduces their variance sufficiently. If not, I just run the pass again.

As an example, here’s a render of the Manifold mesh from Torolf Sauermann’s awesome model repository, stopped after only a few paths have been traced per pixel:

I’ve highlighted a few areas that contain fireflies and below is a comparison of those areas before and after running the variance reduction pass:

And here’s the complete result of running the pass; it’s still noisy, but the pixels with particularly high variance have been cleaned up reasonably well:

I’m not too hot at statistics, but I would guess that this adds bias to the final result, which is frowned upon in some circles (but not others). Admittedly, a better solution would be to simply not generate so much variance in the first place, but this will do as a kludge until then. At least it’s better than painting pixels by hand!

Ok, since this post was mostly just an excuse to dump some results from my path tracer, here they are.

The XYZ RGB Dragon from Stanford’s 3D Scanning Repository:

A heat map of the BIH built for the same model.

The other Dragon from Stanford:

Manifold again, from jotero.com.

Some Stanford Bunnies:

Crytek’s updated version of Marko Dabrovic’s Sponza model:

And Stanford’s Lucy, just to prove that I can:
I’ve not been able to find any close up renders of Lucy to compare with, but I believe that the “pimples” on the model are noise in the original dataset.


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## Definitions of Probability Measure Proof

Hi, I'm wondering if someone can help me with the following questions relating to the following definitions:

Given sample space S with event space A probability measure is defined as:
1) Non-Negative Function, P: A-->[0, infinity)
2) Normed, P(S)=1
5) Sequentially Continuous, An ---> A => P(An) ---> P(A)

Now, I need to prove the following:

1) Prove that if we suppose 1), 2), 4) and 5) are true, then 1), 2) and 3) must be true.

2) Given that An--->A and Bn--->B prove that An*Bn--->AB and Complement(An)--->Complement(A) and $A_n \bigcup B_n --> A \bigcup B$

3) We say (6) P is sequentially continuous at 0 iff $A_n \downarrow \varnothing => P(A_n) \downarrow 0$
Prove that 1),2),4) and 5) <==> 1), 2), 4) and 6)

Thanks a lot