Zero k-eff uncertainty in infinite homogeneous medium 2-group problem

Code version
Version | 0.13.0
Git SHA1 | cff247e35785e7236d67ccf64a3401f0fc50a469

I was working out a simple infinite box problem to learn about multigroup cross section generation and running in MG mode. I came across an odd result. In continuous energy mode, all three criticality estimators returned an uncertainty, per below:
k-effective (Collision) = 1.00091 +/- 0.00102
k-effective (Track-length) = 1.00083 +/- 0.00104
k-effective (Absorption) = 0.99895 +/- 0.00112
Combined k-effective = 1.00006 +/- 0.00075

However, when running the problem in 2-group multigroup mode, the absorption estimator returns a 0-valued uncertainty, as:

k-effective (Collision) = 1.00084 +/- 0.00099
k-effective (Track-length) = 1.00084 +/- 0.00101
k-effective (Absorption) = 0.99970 +/- 0.00000
Combined k-effective = 0.99970 +/- 0.00000

I have compared the fast/thermal fluxes in both problem variants and they are in excellent agreement.

Problem statement
The geometry is a simple cube with reflective boundaries. There is a single material, composed of a mix of H-1 and U-235 in the ratio of 2058:1. The model temperature is 600K. The cross section library is the NNDC, and the WMP method is used to handle temperatures.

Two models are used. The runs the continuous energy problem, with tallies setup to create a simple two-group library for the fuel “gas”, The resulting mgxs.h5 file is read by the problem, which is set to run in multigroup.
mgxs.h5 (13.1 KB) (955 Bytes) (4.3 KB)

@AlexandreTrottier Interesting question. The absorption estimate of keff works by scoring w\Sigma_f/\Sigma_a each time a neutron is absorbed (where w is the particle weight). The fundamental reason that the uncertainty is so small (it’s actually not zero but when rounded to 5 digits appears as 0) is that a vast majority of the neutrons are absorbed in a single energy group. Thus, most of the keff estimates are exactly the same. There is a small probability of absorption in the other group, which means there is some non-zero variance in the estimate. If that probability were zero or if you were running a one-group problem, you would indeed get an uncertainty of exactly zero.