Zero-variance schemes for kinetic Monte Carlo simulations

Abstract

Solving time-dependent transport problems for neutrons and precursors in a nuclear reactor is a daunting task in a naive Monte Carlo framework, mainly because of the enormous difference between the time scale associated with the prompt fission chains and that associated with the decay of delayed neutron precursors. Recently, the development of variance reduction techniques specific for reactor kinetics and the rapidly increasing computer power have paved the way towards the possibility of obtaining reference solutions to the time-dependent transport problem. However, the application of time-dependent Monte Carlo to large systems (i.e. at the scale of a full reactor core) is still considerably hindered by the huge computational requirements. In this paper, we construct an ideal Monte Carlo game that results in a zero-variance estimator for specific observables in time-dependent transport. Our derivation follows the pattern of the existing schemes for stationary problems. To the best of our knowledge, zero-variance Monte Carlo schemes for time-dependent transport including delayed neutrons precursors have never been considered before. As a proof of principle, we verify our construction for a simplified benchmark configuration where analytical reference solutions for the transport problem can be explicitly obtained.

Appendices

Derivation of the moment equations

Consider a generic random walk characterized by the flight kernel ${\hat{T}}(P,P')$ and the branchless collision kernel ${\hat{C}}(P',P'')$, with ${\hat{w}}_b(P)=\int {\hat{C}}(P,P') \,\mathrm {d}P'$. Each walker starts with a statistical weight w. After a flight from P to $P'$, the statistical weight becomes $w'$, and after a collision from $P'$ to $P''$ the statistical weight becomes ${\hat{w}}_b w''$, where $w'$ and $w''$ are expressions to be further determined. Suppose that the random walk is endowed with an estimator $h(P,P')$ along a flight from P to $P'$. Let us now consider the probability $\pi (s|P, w)$ to observe a score s for the quantity of interest, for a particle starting from P with statistical weight w. From considerations of normalization, we have

$$\begin{aligned} \pi (s|P, 1) \hbox {d}s = \pi (w s|P, w) \hbox {d}(ws). \end{aligned}$$

It can be shown that $\pi (s|P, w)$ satisfies the probability balance equation [5]

$$\begin{aligned} \pi (s|P, w) = \int {\hat{T}}(P, P') \,\mathrm {d}P' \int {\hat{C}}_b(P', P'') \pi (s - h(P,P') w' |P'', {\hat{w}}_b(P') w'') \,\mathrm {d}P'' , \end{aligned}$$

(44)

where ${\hat{C}}_b(P', P'') ={\hat{C}}(P', P'') / {\hat{w}}_b(P')$ is the normalized collision kernel.

We define the moments ${\hat{M}}_k(P)$ of the scores associated with the underlying random walk starting from P with unit statistical weight,

$$\begin{aligned} {\hat{M}}_k(P)=\int s^k \pi (s|P, 1) \,\mathrm {d}s . \end{aligned}$$

By multiplying Eq. (44) by s and integrating over $\,\mathrm {d}s$ we obtain the equation for the first moment, namely

$$\begin{aligned} w {\hat{M}}_1(P) = \int h(P,P') w' {\hat{T}}(P,P') \,\mathrm {d}P' +\int {\hat{T}}(P, P') \,\mathrm {d}P' \int {\hat{C}}(P', P'') w'' {\hat{M}}_1(P'') \,\mathrm {d}P'' . \end{aligned}$$

Similarly, multiplying Eq. (44) by $s^2$ and integrating over $\,\mathrm {d}s$ we obtain the equation for the second moment, namely

$$\begin{aligned} w^2 {\hat{M}}_2(P) = \sum _{k=0}^2 \left( {\begin{array}{c}2\\ k\end{array}}\right) \int {\hat{T}}(P,P') \left( h(P,P') w' \right) ^{2-k} \,\mathrm {d}P' \int {\hat{C}}_b(P', P'') \left( {\hat{w}}_b(P') w''\right) ^{k} {\hat{M}}_k(P'') \,\mathrm {d}P'' . \end{aligned}$$

As a special case, for the natural game we have ${\hat{T}}(P,P')= T(P,P')$, ${\hat{C}}(P',P'')= C(P',P'')$ and the statistical weights are not modified, namely $w'' = w' = w$. We denote by $M_k(P)$ the corresponding moments of the scores. The weights $w'$ and $w''$ for the non-analog game are chosen by imposing that the resulting game is unbiased, i.e. ${\hat{M}}_1(P) = M_1(P)$ [5], which leads to the weight generation rules given by Eqs. (20) and (21): $w' = w_T(P,P') w$ and $w'' = w_C(P',P'') w'$.

Derivation of the zero-variance kernels

Let us consider the first-moment equation for the natural Monte Carlo game associated with the kernels $T(P',P)$ and $C(P',P'')$, with estimator $h(P,P')$ along a flight:

$$\begin{aligned} M_1(P) = \int h(P,P') T(P,P') \,\mathrm {d}P' + \int T(P, P') \,\mathrm {d}P' \int C(P', P'') M_1(P'') \,\mathrm {d}P'' . \end{aligned}$$

(45)

Let $\kappa (P,P') \ge 0$ be a sufficiently regular, non-negative, but otherwise undetermined function. Equation (45) can be equivalently rewritten as follows:

$$\begin{aligned} M_1(P)= & {} \int h(P,P') T(P,P') \,\mathrm {d}P' \nonumber \\&+\,\int T(P, P') \left[ 1- \kappa (P,P') \right] \,\mathrm {d}P' \int C(P', P'') M_1(P'') \,\mathrm {d}P'' \nonumber \\&+\,\int T(P, P') \kappa (P,P') \,\mathrm {d}P' \int C(P', P'') M_1(P'') \,\mathrm {d}P'' . \end{aligned}$$

(46)

Consider then a non-analog branchless Monte Carlo game associated with the kernels ${\hat{T}}(P',P)$ and ${\hat{C}}(P',P'')$, with estimator $h(P,P')$ along a flight. Imposing a zero-variance condition for a unit starter weight amounts to setting

$$\begin{aligned} \frac{{\hat{M}}_2(P)}{{\hat{M}}_1(P)} = {\hat{M}}_1(P). \end{aligned}$$

(47)

By using the weight generation rules

$$\begin{aligned} w_T(P,P')= \frac{T(P,P')}{{\hat{T}}(P,P')} \end{aligned}$$

and

$$\begin{aligned} w_C(P',P'')=\frac{C(P',P'')}{{\hat{C}}(P',P'')} , \end{aligned}$$

this game is partially unbiased, with ${\hat{M}}_1(P) = M_1(P)$ for a single starting particle. For a unit starter weight, the equation for the second moment ${\hat{M}}_2(P)$ of the non-analog game is

$$\begin{aligned} {\hat{M}}_2(P)= & {} \int {\hat{T}}(P,P')\, h^2(P,P')\, w_T^2(P,P') \,\mathrm {d}P'\nonumber \\&+\,2\int {\hat{T}}(P,P')\, h(P,P')\, w_T^2(P,P') \int {\hat{C}}(P',P'')\,M_1(P'')\, w_C(P',P'') \,\mathrm {d}P'' \,\mathrm {d}P'\nonumber \\&+\,\int {\hat{T}}(P,P')\, w_T^2(P,P') \int {\hat{C}}(P',P'')\,w_C^2(P',P'')\, {\hat{w}}_b(P')\,{\hat{M}}_2(P'') \,\mathrm {d}P'' \,\mathrm {d}P' ,\nonumber \\ \end{aligned}$$

(48)

where

$$\begin{aligned} {\hat{w}}_b(P')=\int {\hat{C}}(P',P) \,\mathrm {d}P \end{aligned}$$

is the branchless weight correction for the outgoing neutron of the kernel ${\hat{C}}(P',P'')$. Let $\theta (P)\ge 0$ be a sufficiently regular, non-negative, but otherwise undetermined function; Eq. (48) can be equivalently rewritten as follows:

$$\begin{aligned} {\hat{M}}_2(P)= & {} \int {\hat{T}}(P,P')\, h^2(P,P')\, w_T^2(P,P') \,\mathrm {d}P'\nonumber \\&{}+2\int {\hat{T}}(P,P')\, h(P,P')\, w_T^2(P,P') \int {\hat{C}}(P',P'')\,M_1(P'')\, w_C(P',P'') \,\mathrm {d}P'' \,\mathrm {d}P'\nonumber \\&+\,\int {\hat{T}}(P,P')\, w_T^2(P,P') \theta (P') \int {\hat{C}}(P',P'')\,w_C^2(P',P'')\, {\hat{w}}_b(P')\, \frac{M_1(P'')}{\theta (P')} \frac{{\hat{M}}_2(P'')}{M_1(P'')} \,\mathrm {d}P'' \,\mathrm {d}P' ,\nonumber \\ \end{aligned}$$

(49)

Equation (47) involves the first moment of the score, Eq. (46), and the second moment, Eq. (49). Both sides of Eq. (47) involve integrals over intermediate points in phase space. In order for Eq. (47) to hold true, an inspired sufficient condition is that the integrands are equal. Since ${\hat{M}}_1(P) = M_1(P)$, we thus equate the integrand in the term

$$\begin{aligned} \int T(P, P') \kappa (P,P') \,\mathrm {d}P' \int C(P', P'') M_1(P'') \,\mathrm {d}P'' \end{aligned}$$

appearing at the right-hand side of Eq. (46) (for $M_1(P)$) to the integrand in the term

$$\begin{aligned} \frac{1}{M_1(P)}\int {\hat{T}}(P,P')\, w_T^2(P,P') \theta (P') \int {\hat{C}}(P',P'')\,w_C^2(P',P'')\, {\hat{w}}_b(P')\, \frac{M_1(P'')}{\theta (P')} \frac{{\hat{M}}_2(P'')}{M_1(P'')} \,\mathrm {d}P'' \,\mathrm {d}P' \end{aligned}$$

appearing at the right-hand side of Eq. (49) (for $M_2(P)$), after dividing by $M_1(P)$. We therefore have

$$\begin{aligned}&T(P, P') \kappa (P,P') C(P', P'') M_1(P'') \\&\quad = \frac{1}{M_1(P)}{\hat{T}}(P,P')\, w_T^2(P,P') \theta (P') {\hat{C}}(P',P'')\,w_C^2(P',P'')\, {\hat{w}}_b(P')\, \frac{M_1(P'')}{\theta (P')}M_1(P'') . \end{aligned}$$

By using the definitions of $w_T(P',P'')$ and $w_C(P'',P)$, we obtain

$$\begin{aligned} \kappa (P,P') = \frac{1}{M_1(P)} \frac{ \theta (P') }{ {\hat{T}}(P,P') {\hat{C}}(P',P'')} {\hat{w}}_b(P')\, \frac{M_1(P'')}{\theta (P')} T(P,P') C(P',P'') , \end{aligned}$$

which is satisfied by setting

$$\begin{aligned} {\hat{T}}(P,P') = T(P,P') \frac{ \theta (P') }{M_1(P) \kappa (P,P') } , \end{aligned}$$

and

$$\begin{aligned} {\hat{C}}(P',P'') = C(P',P'') \frac{M_1(P'') {\hat{w}}_b(P')}{\theta (P')} . \end{aligned}$$

By imposing the normalization of ${\hat{T}}(P,P')$, we have

$$\begin{aligned} \int {\hat{T}}(P,P') \,\mathrm {d}P' = \int T(P,P') \frac{ \theta (P') }{M_1(P) \kappa (P,P') } \,\mathrm {d}P' = 1 , \end{aligned}$$

whence

$$\begin{aligned} \int T(P,P') \theta (P') \frac{ \theta (P') }{\kappa (P,P') } \,\mathrm {d}P' = M_1(P) . \end{aligned}$$

From Eq. (45), by imposing the equality of the integrands we obtain

$$\begin{aligned} \frac{ \theta (P') }{\kappa (P,P') } = h(P,P') + \int C(P', P'') M_1(P'') \,\mathrm {d}P'' . \end{aligned}$$

Finally,

$$\begin{aligned} {\hat{T}}(P,P') = T(P,P') \frac{ h(P,P') + \int C(P', P'') M_1(P'') \,\mathrm {d}P'' }{M_1(P)}. \end{aligned}$$

(50)

By imposing then the normalization of ${\hat{C}}(P',P'')$, we have ${\hat{w}}_b(P')=1$ and

$$\begin{aligned} \int {\hat{C}}(P',P'') \,\mathrm {d}P'' = \int C(P',P'') \frac{M_1(P'') }{\theta (P')} = 1 , \end{aligned}$$

whence

$$\begin{aligned} \theta (P') = \int C(P',P'') M_1(P'') \,\mathrm {d}P'' . \end{aligned}$$

Finally,

$$\begin{aligned} {\hat{C}}(P',P'') = C(P',P'') \frac{ M_1(P'') }{\int C(P',P'') M_1(P'') \,\mathrm {d}P''} . \end{aligned}$$

(51)

We have now to verify that this choice for the kernels ${\hat{T}}(P,P')$ and ${\hat{C}}(P',P'')$ actually satisfies Eq. (47) (remember that we have determined $\kappa (P,P')$ and $\theta (P)$ by equating the integrands of only one of the terms). For the integrands, we must therefore have

$$\begin{aligned}&T(P,P') h(P,P') +T(P, P') \left[ 1- \kappa (P,P') \right] \int C(P', P'') M_1(P'') \,\mathrm {d}P''\\&\quad = \frac{1}{M_1(P)} {\hat{T}}(P,P')\, h^2(P,P')\, w_T^2(P,P')\\&\qquad +\frac{2}{M_1(P)} {\hat{T}}(P,P')\, h(P,P')\, w_T^2(P,P') \int {\hat{C}}(P',P'')\,M_1(P'')\, w_C(P',P'') \,\mathrm {d}P'' . \end{aligned}$$

By using the expressions of the weight correction formulas and the definition of $\theta (P')$, we obtain

$$\begin{aligned} h(P,P') + \left[ 1- \kappa (P,P') \right] \theta (P') =\frac{h(P,P')}{M_1(P)} \frac{T(P, P')}{{\hat{T}}(P,P')} \left[ h(P,P') + 2 \theta (P')\right] . \end{aligned}$$

Now,

$$\begin{aligned} \frac{T(P, P')}{{\hat{T}}(P,P')} = \frac{M_1(P)}{h(P,P') + \theta (P')} , \end{aligned}$$

so that we have

$$\begin{aligned} h(P,P') + \left[ 1- \kappa (P,P') \right] \theta (P') =\frac{h(P,P') }{h(P,P') + \theta (P')} \left[ h(P,P') + 2 \theta (P')\right] . \end{aligned}$$

(52)

Since

$$\begin{aligned} h(P,P') + \theta (P') = \frac{\theta (P')}{\kappa (P,P')} , \end{aligned}$$

the equality of the two terms in Eq. (52) is satisfied, which proves the identity leading to the zero-variance condition in Eq. (47).

We have thus shown that a non-analog Monte Carlo game with kernels given by Eqs. (50) and (51) yields a zero-variance estimate due to a unit-weight starter from any source point P. However, the zero-variance property of the total score

$$\begin{aligned} \int {\hat{Q}}(P) w_Q(P) {\hat{M}}_1(P) \,\mathrm {d}P = \int Q(P) M_1(P) \,\mathrm {d}P = R \end{aligned}$$

is not ensured, unless the non-analog source ${\hat{Q}}(P)$ is properly chosen. The variance of the total score is given by

$$\begin{aligned} {\hat{V}} = {\hat{m}}_2 - {\hat{m}}^2_1= \int {\hat{Q}}(P) w_Q^2(P) {\hat{M}}_2(P) \,\mathrm {d}P - \left[ \int {\hat{Q}}(P) w_Q(P) {\hat{M}}_1(P) \,\mathrm {d}P \right] ^2 . \end{aligned}$$

Since ${\hat{M}}_1(P) =M_1(P)$ and ${\hat{M}}_2(P) = M^2_1(P)$, in order to ensure a zero-variance for the total score we equate the integrands

$$\begin{aligned} w_Q(P) M^2_1(P) = R Q(P) M_1(P) , \end{aligned}$$

whence the condition

$$\begin{aligned} {\hat{Q}}(P)=\frac{Q(P) M_1(P)}{R} , \end{aligned}$$

which completes the requirements for a non-analog partially unbiased Monte Carlo game resulting in a zero-variance estimate for the total score R with the estimator $h(P,P')$.