Spatial sampling density variance¶
Import packages¶
import numpy as np
import pandas as pd
import sys
import os
import matplotlib.pyplot as plt
import gstlearn as gl
import gstlearn.plot as gp
import gstlearn.document as gdoc
gdoc.setNoScroll()
Introduction¶
This document is meant to give some introductory information on various tools derived from the geostatistical etimation variance, which have been developed to measusre the sampling density, either for a regular sampling design or an irregular one. It follows the paper entitled From the Spatial Sampling of a Deposit to Mineral Resources Classification, from Jacques Rivoirard, Didier Renard, Felipe Celhay, David Benado, Celeste Queiroz, Leandro Jose Oliveira, and Diniz Ribeiro, presented in Geostatistics Valencia 2016.
The concepts¶
In this paper, $Z(x)$ denotes the regionalized (additive) variable. Considering a domain V, the estimation of $Z(V)$ by inner samples of $V$ is denoted $Z(V)^*$ and the corresponding estimation variance $\sigma_E^2 (V)$ which is calculated from the variogram as follows: $$\sigma_E^2 (V)={2 \over N} \sum_i \bar \gamma(x_i,V) - {1 \over N^2} \sum_{ij} \gamma(x_i,x_j) - \bar \gamma(V,V)$$
The spatial sampling density variance (ssdv) corresponds is $$\chi(V) = {\sigma_E^2 (V) \times |V|}$$.
The specific volume measures the density of a spatial sampling of the variable, relatively to its mean: the smaller the specific volume, the more precise the sampling pattern: $$V_0 = { {\sigma_E^2 (V) \times |V|} \over M^2 }$$
The coefficient of variation for a given Volume is: $$CoV = \sqrt{ V_0 \over V}$$
The specific area corresponding to a given CoV (area over which the mean variable has a given coefficient of variation) is: $$V = { V_0 \over CoV^2 }$$
Application to regular sampling¶
The previous concepts are be illustrated using the following example.
Consider a large deposit in two dimensions, sampled with regular grid. the variable of interest is the thickness (with a mean of 10m), a sample point variance of 10, and a variogram equal to: $$1 nug(h) + 9sph(h/200m)$$
mean_10 = 10
model1 = gl.Model.createFromParam(gl.ECov.NUGGET,sill=1)
model1.addCovFromParam(gl.ECov.SPHERICAL,range=200,sill=9)
model1
Model characteristics ===================== Space dimension = 2 Number of variable(s) = 1 Number of basic structure(s) = 2 Number of drift function(s) = 0 Number of drift equation(s) = 0 Covariance Part --------------- Nugget Effect - Sill = 1.000 Spherical - Sill = 9.000 - Range = 200.000 Total Sill = 10.000 Known Mean(s) 0.000
Most of the subsequent calculations need the determination of quantities which involve extension variance, the latter being based on calulation of average variogram over a block. This operation is performed by discretization which needs to be specified explicitely:
ndisc = [30,30]
The extension variance of a $100m \times 100m$ by its center is:
target = gl.Db.createFromOnePoint([0,0])
block_100 = [100, 100]
Sigma_E = model1.extensionVariance(target, block_100, ndisc)
print("Extension Variance = %1.3f" %Sigma_E)
Extension Variance = 2.697
Therefore, for a $100m \times 100m$ grid, the spatial sampling density variance is:
ssdv = model1.samplingDensityVariance(target, block_100, ndisc)
print("Spatial Sampling Density Variance = %1.2f m².m²" %ssdv)
Spatial Sampling Density Variance = 26972.85 m².m²
The specific area (i.e. the specific volume in two dimensions) is:
V_100 = model1.specificVolume(target, mean_10, block_100, ndisc)
print("Specific Area (case #1) = %1.2f m²" %V_100)
Specific Area (case #1) = 269.73 m²
If the grid is now a $50m \times 50m$, the specific volume becomes:
block_50 = [50, 50]
V_50 = model1.specificVolume(target, mean_10, block_50, ndisc)
print("Specific Area (small block) = %1.2f m²" %V_50)
Specific Area (small block) = 45.77 m²
Another issue is whether the initial $100m \times 100m$ grid is better, or not, than a $70m \times 70m$ sampling in a similar deposit, where the thickness has a mean of 8 and a variogram equal to $$6 nug(h) + 6sph(h/200m)$$
mean_8 = 8
model2 = gl.Model.createFromParam(gl.ECov.NUGGET,sill=6)
model2.addCovFromParam(gl.ECov.SPHERICAL,range=200,sill=6)
model2
Model characteristics ===================== Space dimension = 2 Number of variable(s) = 1 Number of basic structure(s) = 2 Number of drift function(s) = 0 Number of drift equation(s) = 0 Covariance Part --------------- Nugget Effect - Sill = 6.000 Spherical - Sill = 6.000 - Range = 200.000 Total Sill = 12.000 Known Mean(s) 0.000
block_70 = [70, 70]
V_70 = model2.specificVolume(target, mean_8, block_70, ndisc)
print("Specific Area (case #2) = %1.2f m²" %V_70)
Specific Area (case #2) = 518.70 m²
We can finally characterize resources given an assumption on the production volume (or more exactly the ore + waste volume to be extracted). For instance, a specific volume (3) of 6000 m3 gives a CoV of:
Volume = 6000
CoV = 100 * model1.coefficientOfVariation(target, Volume, mean_10, block_100, ndisc)
print("Coefficient of Variation = %1.2f" %CoV)
Coefficient of Variation = 21.20
Conversely we can define the specific volume to be mined in order to ensure a given CoV. For instance we can deduce that the 5% CoV area (area over which the mean thickness has a coefficient of variation of 5%) must practically reach 200,000 m2, while it is only108,000 m2 in the first case
cov_5 = 0.05
Vol1 = model1.specificVolumeFromCoV(target, cov_5, mean_10, block_100, ndisc)
print("Specific area (case #1) = %1.2f m²" %Vol1)
Vol2 = model2.specificVolumeFromCoV(target, cov_5, mean_8, block_70, ndisc)
print("Specific area (case #2) = %1.2f m²" %Vol2)
Specific area (case #1) = 107891.40 m² Specific area (case #2) = 207479.07 m²