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Paul Kent
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In this appendix we derive the relative statistical efficiency of
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twist averaging with an irreducible (weighted) set of k-points
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versus using uniform weights over an unreduced set of k-points
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(\emph{e.g.} a full Monkhorst-Pack mesh).
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(e.g., a full Monkhorst-Pack mesh).
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Consider the weighted average of a set of statistical variables
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$\{x_m\}$ with weights $\{w_m\}$:
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\begin{align}
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x_{TA} = \frac{\sum_mw_mx_m}{\sum_mw_m}
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x_{TA} = \frac{\sum_mw_mx_m}{\sum_mw_m}\:.
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\end{align}
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If produced by a finite quantum Monte Carlo run at a set of
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If produced by a finite QMC run at a set of
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twist angles/k-points $\{k_m\}$, each variable mean $\mean{x_m}$
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has a statistical error bar $\sigma_m$ and we can also obtain
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the statistical error bar of the mean of the twist averaged
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has a statistical error bar $\sigma_m$, and we can also obtain
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the statistical error bar of the mean of the twist-averaged
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quantity $\mean{x_{TA}}$:
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\begin{align}
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\sigma_{TA} = \frac{\left(\sum_mw_m^2\sigma_m^2\right)^{1/2}}{\sum_mw_m}
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\sigma_{TA} = \frac{\left(\sum_mw_m^2\sigma_m^2\right)^{1/2}}{\sum_mw_m}\:.
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\end{align}
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The error bar of each individual twist $\sigma_m$, is related to the
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autocorrelation time $\kappa_m$, intrinsic variance $v_m$, and number
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of post-equilibration Monte Carlo steps $N_{step}$ in the following way:
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The error bar of each individual twist $\sigma_m$ is related to the
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autocorrelation time $\kappa_m$, intrinsic variance $v_m$, and the number
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of postequilibration MC steps $N_{step}$ in the following way:
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\begin{align}
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\sigma_m^2=\frac{\kappa_mv_m}{N_{step}}
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\sigma_m^2=\frac{\kappa_mv_m}{N_{step}}\:.
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\end{align}
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In the setting of twist averaging, the autocorrelation time and
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variance for different twist angles are often very similar across
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twists and we have
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twists, and we have
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\begin{align}
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\sigma_m^2=\sigma^2=\frac{\kappa v}{N_{step}}
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\sigma_m^2=\sigma^2=\frac{\kappa v}{N_{step}}\:.
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\end{align}
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If we define the total weight as $W$, \emph{i.e.} $W\equiv\sum_{m=1}^Mw_m$,
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for the weighted case with $M$ irreducible twists the error bar is
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If we define the total weight as $W$, that is, $W\equiv\sum_{m=1}^Mw_m$,
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for the weighted case with $M$ irreducible twists, the error bar is
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\begin{align}
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\sigma_{TA}^{weighted}=\frac{\left(\sum_{m=1}^Mw_m^2\right)^{1/2}}{W}\sigma
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\sigma_{TA}^{weighted}=\frac{\left(\sum_{m=1}^Mw_m^2\right)^{1/2}}{W}\sigma\:.
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\end{align}
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For uniform weighting with $w_m=1$ the number of twists is $W$ and
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For uniform weighting with $w_m=1$, the number of twists is $W$ and
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we have
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\begin{align}
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\sigma_{TA}^{uniform}=\frac{1}{\sqrt{W}}\sigma
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\sigma_{TA}^{uniform}=\frac{1}{\sqrt{W}}\sigma\:.
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\end{align}
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We are interested in comparing the efficiency of choosing weights
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uniformly or based on the irreducible multiplicity of each twist angle
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for a given target error bar $\sigma_{target}$. The number of Monte Carlo
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for a given target error bar $\sigma_{target}$. The number of MC
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steps required to reach this target for uniform weighting is
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\begin{align}
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N_{step}^{uniform} = \frac{1}{W}\frac{\kappa v}{\sigma_{target}^2}
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N_{step}^{uniform} = \frac{1}{W}\frac{\kappa v}{\sigma_{target}^2}\:,
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\end{align}
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while for non-uniform weighting we have
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while for nonuniform weighting we have
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\begin{align}\label{eq:weighted_step}
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N_{step}^{weighted} &= \frac{\sum_{m=1}^Mw_m^2}{W^2}\frac{\kappa v}{\sigma_{target}^2} \nonumber\\
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&=\frac{\sum_{m=1}^Mw_m^2}{W}N_{step}^{uniform}
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N_{step}^{weighted} &= \frac{\sum_{m=1}^Mw_m^2}{W^2}\frac{\kappa v}{\sigma_{target}^2} \nonumber\:,\\
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&=\frac{\sum_{m=1}^Mw_m^2}{W}N_{step}^{uniform}\:.
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\end{align}
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The Monte Carlo efficiency is defined as
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The MC efficiency is defined as
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\begin{align}
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\xi = \frac{1}{\sigma^2t}
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\xi = \frac{1}{\sigma^2t}\:,
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\end{align}
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where $\sigma$ is the error bar and $t$ is the total cpu time required
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for the Monte Carlo run.
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where $\sigma$ is the error bar and $t$ is the total CPU time required
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for the MC run.
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The main advantage made possible by irreducible twist weighting is to
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reduce the equilibration time overhead by having fewer twists, and
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hence fewer Monte Carlo runs to equilibrate. In the context of twist
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averaging, the total cpu time for a run can be considered to be
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reduce the equilibration time overhead by having fewer twists and,
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hence, fewer MC runs to equilibrate. In the context of twist
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averaging, the total CPU time for a run can be considered to be
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\begin{align}
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t=N_{twist}(N_{eq}+N_{step})t_{step}
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t=N_{twist}(N_{eq}+N_{step})t_{step}\:,
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\end{align}
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where $N_{twist}$ is the number of twists, $N_{eq}$ is the number of Monte
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Carlo steps required to reach equilibrium, $N_{step}$ is the number
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of Monte Carlo steps included in the statisical averaging as before,
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where $N_{twist}$ is the number of twists, $N_{eq}$ is the number of MC steps required to reach equilibrium, $N_{step}$ is the number
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of MC steps included in the statistical averaging as before,
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and $t_{step}$ is the wall clock time required to complete a single
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Monte Carlo step. For uniform weighting $N_{twist}=W$ while for irreducible
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MC step. For uniform weighting $N_{twist}=W$; while for irreducible
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weighting $N_{twist}=M$.
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We can now calculate the relative efficiency ($\eta$) of irreducible vs.
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uniform twist weighting with the aim of obtaining a target error bar
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$\sigma_{target}$:
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\begin{align}
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\eta &= \frac{\xi_{TA}^{weighted}}{\xi_{TA}^{uniform}} \nonumber \\
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&= \frac{\sigma_{target}^2t_{TA}^{uniform}}{\sigma_{target}^2t_{TA}^{weighted}} \nonumber \\
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&= \frac{W(N_{eq}+N_{step}^{uniform})}{M(N_{eq}+N_{step}^{weighted})} \nonumber \\
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&= \frac{W(N_{eq}+N_{step}^{uniform})}{M(N_{eq}+\frac{\sum_{m=1}^Mw_m^2}{W}N_{step}^{uniform})} \nonumber \\
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&= \frac{W}{M}\frac{1+f}{1+\frac{\sum_{m=1}^Mw_m^2}{W}f}
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\eta &= \frac{\xi_{TA}^{weighted}}{\xi_{TA}^{uniform}} \nonumber\:, \\
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&= \frac{\sigma_{target}^2t_{TA}^{uniform}}{\sigma_{target}^2t_{TA}^{weighted}} \nonumber\:, \\
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&= \frac{W(N_{eq}+N_{step}^{uniform})}{M(N_{eq}+N_{step}^{weighted})} \nonumber\:, \\
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&= \frac{W(N_{eq}+N_{step}^{uniform})}{M(N_{eq}+\frac{\sum_{m=1}^Mw_m^2}{W}N_{step}^{uniform})} \nonumber\:, \\
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&= \frac{W}{M}\frac{1+f}{1+\frac{\sum_{m=1}^Mw_m^2}{W}f}\:.
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\end{align}
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In this last expression, $f$ is the ratio of the number of usable
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Monte Carlo steps to the number that must be discarded during equilibration
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($f=N_{step}^{uniform}/N_{eq}$) and as before $W=\sum_mw_m$, which is the number of
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MC steps to the number that must be discarded during equilibration
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($f=N_{step}^{uniform}/N_{eq}$); and as before, $W=\sum_mw_m$, which is the number of
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twist angles in the uniform weighting case. It is important to recall
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that $N_{step}^{uniform}$ in $f$ is defined relative to uniform weighting; it is
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the number of Monte Carlo steps required to reach a target accuracy in the
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that $N_{step}^{uniform}$ in $f$ is defined relative to uniform weighting and is
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the number of MC steps required to reach a target accuracy in the
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case of uniform twist weights.
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The formula for $\eta$ above can be easily changed with the help of
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Eq. \ref{eq:weighted_step} to reflect the
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number of Monte Carlo steps obtained in an irreducibly weighted run
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instead. A good exercise is to consider runs one has already completed
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The formula for $\eta$ in the preceding can be easily changed with the help of
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Equation~\ref{eq:weighted_step} to reflect the
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number of MC steps obtained in an irreducibly weighted run
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instead. A good exercise is to consider runs that have already completed
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with either uniform or irreducible weighting and calculate the
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expected efficiency change had the opposite type of weighting been used.
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The break even point $(\eta=1)$ can be found at a usable step fraction of
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\begin{align}
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f=\frac{W-M}{M\frac{\sum_{m=1}^Mw_m^2}{W}-W}
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f=\frac{W-M}{M\frac{\sum_{m=1}^Mw_m^2}{W}-W}\:.
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\end{align}
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The relative efficiency $(\eta)$ above is useful to consider in view of certain
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The relative efficiency $(\eta)$ is useful to consider in view of certain
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scenarios. An important case is where the number of required sampling
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steps is no larger than the number of equilibration steps, \emph{i.e.}
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$f\approx 1$. For a very simple case with 8 uniform twists with
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steps is no larger than the number of equilibration steps (i.e.,
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$f\approx 1$). For a very simple case with eight uniform twists with
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irreducible multiplicities of $w_m\in\{1,3,3,1\}$ ($W=8$, $M=4$), the
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relative efficiency of irreducible vs. uniform weighting is
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$\eta=\frac{8}{4}\frac{2}{1+20/8}\approx 1.14$. In this case,
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irreducible weighting is about $14$\% more efficient than uniform weighting.
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Another interesting case is where the number of sampling steps you can
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Another interesting case is one in which the number of sampling steps you can
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reach with uniform twists before wall clock time runs out is small
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relative to the number of equilibration steps ($f\rightarrow 0$).
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In this limit $\eta\approx W/M$. For our 8 uniform twist example, this would
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result in a relative efficiency of $\eta=8/4=2$ making irreducible
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In this limit, $\eta\approx W/M$. For our eight-uniform-twist example, this would
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result in a relative efficiency of $\eta=8/4=2$, making irreducible
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weighting twice as efficient.
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A final case of interest is where the equilibration time is short
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A final case of interest is one in which the equilibration time is short
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relative to the available sampling time $(f\rightarrow\infty)$,
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giving $\eta\approx W^2/(M\sum_{m=1}^Mw_m^2)$. Again for our simple example
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giving $\eta\approx W^2/(M\sum_{m=1}^Mw_m^2)$. Again, for our simple example
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we find $\eta=8^2/(4\times 20)\approx 0.8$, with uniform weighting being
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$25$\% more efficient than irreducible.
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For this simple example, the crossover point for irreducible weighting being
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more efficient than uniform weighting is $f<2$, \emph{i.e.} when the
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$25$\% more efficient than irreducible weighting. For this example, the crossover point for irreducible weighting being
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more efficient than uniform weighting is $f<2$, that is, when the
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available sampling period is less than twice the length of the equilibration
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period. The expected efficiency ratio and crossover point should be checked
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for the particular case under consideration to inform the choice between
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@ -5,11 +5,10 @@
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The following is a list of all papers, theses, and book chapters
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known to use QMCPACK. Please let the developers know if your paper
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is missing, if you know of other works, or an entry is incorrect. We
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list papers whether they cite QMCPACK directly or not. This list
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is missing, you know of other works, or an entry is incorrect. We
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list papers whether or not they cite QMCPACK directly. This list
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will be placed on the \url{http://www.qmcpack.org} website.
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\btPrintAll
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\end{btSect}
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