CHAPTER
4: Estimation
So far we have used the basic concepts and assumptions of Geostatistics to
build ourselves a ‘model’ of the structure and continuity within the deposit.
We have also (in Chapter 3) seen how this can lead to the production of
‘theoretical’ grade/tonnage curves and the study of how mining block size can
influence final production Figures. It is now time we returned to our original
problem of the estimation of ore reserves. The discussion in this (and the
next) chapter will be confined to ‘local’ estimation, i.e. interest is confined
to one portion of the deposit at a time. However, it should be borne in mind
that the same techniques can be applied on a global scale, i.e. to the whole
deposit at once. It should also be remembered that block-by-block or
stope-by-stope estimates will lead inevitably to global estimates.
Let us,
then, define the situation which is of interest to us. There is a point or an
area or a volume of ground over which we do not know the grade (or value), but
we wish to estimate it. Let us call this ‘unknown’ grade T, and the area (or point, or volume) of interest A. In order to produce an estimator we must have some
information, usually in the form of samples. To be completely general, let us
suppose n samples with values of g1,g2,g3...gn. This set of samples is generally
denoted by S. From these samples we can form a ‘linear’
type of estimator --- that is, a weighted average. We must restrict ourselves
to this type of estimator at this stage. The estimator is denoted by T* and is equal to:
where the w1,
w2, w3...wn are the weights assigned to each
sample. Most currently used local estimation techniques use a weighted average
approach --- inverse distance techniques and so on. The simplest case of all is
when all of the weights are the same, and T* is just the arithmetic mean of the
sample values.
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Fig 4.1. Hypothetical
sampling and estimation situation --- a uranium deposit. |
Table 4.1 Positions
and values on hypothetical Uranium estimation problem |
Now
consider the setup of samples and ‘unknown’ which we originally discussed in the
first chapter. Figure 4.1 shows the point of interest which lies at position A, and we have five ‘point’ samples lying around this
position. The co-ordinates of these six points and the values of the samples
are given in Table 4.1. The hypothetical deposit is a low-grade, large-tonnage
uranium one, which is assumed to be isotropic. The semi-variogram model fitted
to this deposit is a spherical one with a range of influence of 100 ft, a sill
value (C) of 700 (p.p.m.)² and a nugget effect of 100 (p.p.m.)². Let us take the simplest possible estimation
procedure. Take the value at the closest sample position (1) and ‘extend’ this
to the unknown point. In doing so we incur an estimation error, e, which will be equal to the
difference between the actual value T and the estimated value T*, which in this case equals g1. That is:
It is not too difficult to show that if there is no trend
(at least locally), this estimator is unbiased. That is, if we make lots of similar
estimations the average error will be zero.
The ‘reliability’ of the estimation can be measured by
looking at the spread of the
errors. If the errors take values consistently close to zero, then the
estimator is a ‘good’ one. If the spread of values is large, then the estimator
will be unreliable. The simplest stable measure of spread (statistically) is
the standard deviation. The standard deviation of an estimation error --- or
standard error as it is referred to in ordinary statistics --- will therefore
measure the reliability of that estimator.
No matter
how many estimations we perform, we cannot calculate the standard deviation of
the errors since we do not know the value of the error made. Therefore we must
look at the ‘theoretical’ form of the variance of the estimation error, i.e.
the estimation variance:
The average would be made (theoretically) over the whole
deposit. That is, the same estimation situation would be repeated over the
whole deposit and the variance found. This cannot be done in practice, of
course, so let us look closer at the form of this variance. It is found by
taking the grade at point A, subtracting the grade at point 1,
squaring the result, repeating the process over all possible pairs of such
points and then averaging the values. This sounds exactly like the definition
of a variogram. In fact, it is the variogram between the
two points A and (1). Given the distance between them (h) we
can evaluate this estimation variance simply by reading a value from the
semi-variogram model (g ) and
multiplying it by 2. This is one of the reasons why it is good policy to avoid
confusing the variogram and the semi-variogram. Thus:
In the case of our particular example given in Fig. 4.1:
Given our knowledge about this deposit, i.e. the
semi-variogram model, we can state (without too much fear of error) that the
estimator used has a standard error of 25.4 p.p.m. Turning this standard error
into a confidence interval, however, requires the assumption of some kind of
probability distribution for the deposit. For instance if we hope that the
Central Limit Theorem holds, we can say that a 95% confidence interval for T would be given by T*± 1.96se, i.e. (350 p.p.m., 450 p.p.m.). On
the other hand, if we were to assume a log-normal distribution for the errors,
the 95% confidence interval would be given by (354 p.p.m., 453 p.p.m.).
Fig. 4.2. More realistic estimation --- the
value of the block is required (uranium deposit).
Now, let us
complicate the procedure a little. Instead of estimating the value at the point
A, in a more realistic situation (at least
in mining) we would be interested in the average grade over an area or block or
some mining unit. In Fig. 4.2, a ‘panel’ 60 ft by 30 ft has been centred on the
original point A. The estimation procedure then becomes:
The same arguments as previously still hold. The average
error can be shown to be zero if there is no local trend. The estimation
variance is still a variogram, but it is now the variogram between the grade at
sample point (1) and the average grade over the panel A. We saw in Chapter 3 that we could
cope with average grades over samples if we wanted the semi-variogram between
samples of the same size, but so far we have not considered the possibility of
having two different sizes to compare. The model semi-variogram supplies us
with the difference in grades between two points. We could find the value of
the semi-variogram between the sample point and every point within the panel A, and we could average those values. Let us define
this quantity as 
(S,A), read as ‘gamma-bar between the sample
and every point in the panel’. The ‘bar’ notation is the standard one for
arithmetic mean. This gamma-bar term will take the place of the g(h) in our previous relationship. However,
what we really need is the semi-variogram between the average grade of panel
and the sample, not between all the individual points within
the panel and the sample. 2(S,A) would be the variance of the error made
if we tried to estimate every point within the panel. To correct for this
difference in emphasis we need to take into account the variation of the grades
at points within the panel.
This was
discussed in Chapter 3, and we evaluated it using the auxiliary function F(l,b). This was the average semi-variogram
between all possible pairs of points within the panel. We can rewrite this in a
more general way using the gamma-bar notation. That is, 
(A,A) will be the average semi-variogram value
between every point in the panel and every point in the panel. In the case
shown in Fig. 4.2, then, when using the value at sample point (1) to estimate
the average grade of the panel, the estimation variance becomes:
The calculation of these gamma-bar terms will be discussed
more fully later.
Now, let us complicate the mathematics still further. We
actually have more than one sample available to us, so why not use them in the
estimation procedure. Suppose we use the arithmetic mean of the samples as our T*. This gives us the simplest form of the
weighted average type of estimator. That is:
In this case the term (S,A) is the average semi-variogram value between each
point in the ‘sample set’ S and each point in the panel A. The term (A,A) is still the average semi-variogram
between each point in the panel and each point in the panel. However, now we
have yet another source of spurious variation. We only consider the average
grade of the samples as the estimator, but (S,A) takes the individual grades into account. Thus we have
also to subtract a (S,S) term from the variance, where this is the
average semi-variogram value between each point in the sample set and each
point in the sample set (i.e. 25 ‘pairs’ of samples). The final version of the
estimation variance then becomes:
The arithmetic mean is often known in Geostatistics as an extension estimator, and the above
variance is referred to as the extension variance. To distinguish this variance
from the more general estimation variance for a weighted average, the subscript
e is used rather than the general e.
CALCULATION
OF GAMMA-BAR TERMS
Having produced a formula for the extension variance, it only remains to
explain how to calculate such terms as (S,A) in practice. For the sake of our
(too) simplistic approach, we will consider for the moment only simple
idealistic cases, and these only in one or two dimensions. Generalisation will
be discussed later.
Fig.4.3. Example of using a peripheral point to
estimate the average value of the line segment.
Consider,
as an example, the setup in Fig. 4.3. There is a length of, say, drive, l m long, whose grade is unknown.
We have
at our disposal a single sample, perhaps at a development heading, whose value is
known. In our previous notation T is the average grade over l, T* is the grade at the sample
position, A is the length and S is the single sample point. The reliability of this
estimator is given by:
(S,S) is the semi-variogram between the sample point
and itself, which is zero because the sample is a ‘point’. (A,A) is none other than the F(l) function encountered in Chapter 3. Our
problem arises with (S,A) which has been defined as the average
semi-variogram between the sample point and every point in the line. That is,
we take M as a fixed point (the sample) and M’ can be anywhere on the line. We
take all such pairs that are possible, calculate the value of the
semi-variogram for each pair, sum these (using an integration), and average
this sum. Because the ‘sum’ is being performed over a continuous length, we
cannot divide it by the ‘number of points’ in the sum. Instead we divide by the
length of the line itself, l. This produces another auxiliary function
which is called c(l) and deals with the specific case
of points on the end of lines. Thus our extension variance becomes:
It remains only to determine the function c(l) for the particular model in use and the
standard error is immediately available. The one-dimensional auxiliary
functions are given below for the three ‘common’ models. Semi-variograms
comprising more than one component model are easily handled. The auxiliary
function for each component is evaluated and then the component auxiliary
functions added together.
Auxiliary functions
Linear
model for the semi-variogram;
Exponential model for the semi-variogram;
Spherical model for the semi-variogram:
Thus in our example above, if we have a linear
semi-variogram, the extension variance for the setup in Fig. 4.3 becomes:
For any specific problem, we need specify only the length
of the line l and the slope of the semi-variogram, p.
Let us now consider a slightly more interesting example,
such as that shown in Fig. 4.4.
Fig 4.4. Example of using a central point to estimate the average value of the line segment.
Here the
point sample is in the middle of the line, but otherwise the
situation remains the same. In:
only the first term (S,A) has changed. Rather than invent a new
auxiliary function, or have to do the integration all over again, we can use
the existing c(l) function to produce the required
term.
The term
we require is as follows:
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= the average semi-variogram value between the sample point and every point along the line = (sum of all semi-variogram values between the sample point and every
point along the line)/l = (sum of all the semi-variogram values between the sample point and
every point in the left hand half of the line + sum of all the values between
the sample and the right hand half of the line)/l |
Fig 4.5. Simplifying the central point problem
to allow the use of auxiliary functions.
Figure 4.5 illustrates the ‘splitting’ of the
line so as to put the sample point at the end of two shorter lines. Now, c(l/2) would give us the average of all the
semi-variogram values between M (the sample point) and the M’ on the left hand
half of the line. Returning to the definition of the c function, it can easily be seen
that the sum of all the semi-variogram values between M and M’
will be the average multiplied by the length of line under
consideration.
Thus:
so that
In a particular case the user may substitute his own model for
the semi-variogram, and hence the appropriate auxiliary functions. Before
moving on let us compare this result with the previous situation, where the
sample lay at the end of the
line. In the former case the extension variance was:
By definition c(l) must be greater than (or at least equal
to) c(l/2). The conclusion? If you can only
take one sample, it is better to take it in the middle of what you are trying
to estimate. It is reassuring to find that so-called common sense has a sound
mathematical background.
Fig. 4.6 Generalisation of the ‘central’ point problem.
Using the
same sort of logic on Fig. 4.6, you should be able to deduce that:
so that
Fig. 4.7. Extrapolation of the peripheral point
problem.
Figure 4.7 at first sight seems to be a different kettle of
fish. However, let us follow the same procedure and see where it leads.
The point lies on the end of a ‘line’ of length l+b. The expression (l+b) c(l+b) would give us the sum of all the semi-variogram
values between the sample and the length l+b. However we do not require the
points corresponding to M’ within the length b, so we may subtract those in the
form c(b). That is:
so that
For the linear model, for example, this would be:
This is obviously larger than the expression when the point
was on the end of the line, as would be expected.
One last example before we abandon one-dimensional
examples: Fig. 4.8 shows the ‘same’ line, which now contains three samples.
Fig. 4.8. More complex problem when three samples are available to estimate the line segment.
We shall
use the arithmetic mean of the three grades to estimate the length, i.e. T*=(g1+g2+g3)÷3. Then our extension variance is
where S is now a set of three points. (A,A) remains unchanged, equal to F(l) since we have not changed the length to
be estimated at all. However, (S,A) is now the average semi-variogram value
between each of the three points and the line, so that
where S1 represents sample 1 and so on. Now
(S1,A) is simply c(l), as is (S3,A). The term (S2,A) is the same situation as that in Fig.
4.4, so this equals c(l/2). Thus,
The middle term of the variance (S,S) requires us to take each point in the
sample set with each point in the sample set. Since there are three points in
the set, there are nine such pairs of points:
Each of the individual terms is simply the semi-variogram
between a pair of points. Three of the terms, g(S1,S1), g(S2,S2) and g(S3,S3)
are automatically zero since the samples are points. The terms g(S1,S2), g( S2,S1), g(S2,S3)
and g(S3,S2)
are all equal to g(l/2), whilst g(S1,S3)
and g( S3,S1)
are equal to g(l). Thus:
so that
For our linear example, this reduces to:
This is considerably less than the other evaluations, as
seems only sensible with three times as many samples.
TWO-DIMENSIONAL
EXAMPLES
In two dimensions we have again a set of auxiliary functions which are mostly
generalisations of the one-dimensional ones. These are shown in Fig. 4.9.
Fig. 4.9. The two-dimensional auxiliary
functions.
g(l;b) is the two-dimensional
analogue of g(l) --- sometimes read as ‘g(l) stretched through length b’. It is the average semi-variogram value between all
points on one length, b, and all points on another
parallel to it and of the same length. This function is useful for parallel
boreholes, channel samples, drives, raises and so on. The generalisation of c(l) becomes c(l;b) and is the average semi-variogram
between a length b (drive, raise, core etc.) and an
adjacent panel l by b. The function F(l,b) has been introduced in Chapter 3
for such terms as (A,A) for rectangular panels. We also
introduce a ‘new’ function H(l,b) which does duty for two
rather different situations. It represents the average semi-variogram value
between a panel and a point on one corner of it; it also
represents the average semi-variogram value between two lengths l and b at right angles to one another.
This is simply a mathematical accident.
A few examples will be given here to show how to
‘manoeuvre’ situations into the required form. Figure 4.10 shows a stoping
panel in a cassiterite vein, 100 ft by 125 ft, with a development drive passing
through it parallel to the
sides. The drive is 25 ft from the ‘bottom’ of the panel.
Fig. 4.10. Estimation of the panel
average from the drive average.
The drive
is so closely sampled that we can assume we know its average grade. We wish to
use that average grade as an estimate of the average grade of the panel, so
that:
(A,A) is the average semi-variogram value
between every point in A and every point in A, which is the definition of the two-dimensional
function F(l,b). Thus if we have a model for the
semi-variogram we can evaluate this. Let us suppose that in the above example
the semi-variogram is a spherical model with a range of influence of 80 ft and
a sill of 450 (lb/ton)². The table given in Chapter 3 for the F(l,b) function (Table 3.4) is for the ‘standardised’
spherical model with a=1 and C=1. We require F(125,100) for a
model in which a=80 and C=450. To convert to a=1 we divide the measurements l and b by the range a (80 ft). Thus we require F(125/80,100/80) =F(1.54,1.20) for a=1 and C=450. From the table (interpolating
linearly) we find that F(1.54,1.20)=0.7807 when C=1. Therefore, if C=450, our F value must be 0.7807\450=351.3 (lb/ton)². This finally is (A,A).
Now let
us consider the term (S,S). Our sample is a ‘line’ of length 125 ft,
and the (S,S) is the average semi-variogram value
between every point in the line and every point in the line, i.e. the
one-dimensional function F(l). For a spherical semi-variogram
model, where the length is greater than the range of influence, we know that:
Substituting the values l=125 ft, a=80 ft and C=450 (lb/ton)², gives F(125) =270.9 (lb/ton)², and this is the value used for (S,S).
Finally, we
must turn to (S,A) which presents a problem, since the
auxiliary function c(l;b) is only for situations where the
‘line’ is on one side of the panel. We must employ the same sort of manoeuvre
as before:
|
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= average semi-variogram value
between the line and a panel 125 ft × 100 ft. = (sum of the semi-variogram
values between the line and the panel) ÷ (100
× 125) =(sum of the semi-variogram
values between the line and the points in the ‘upper’ panel + sum of the
semi-variogram values between the line and the points in the ‘lower’ panel) ÷ (100 × 125) |
However,
the average semi-variogram value between the line and the ‘upper’ panel (125 ×
75 ft) is given by c(75;125), so that the sum of the semi-variogram values
between the line and every point in the upper panel will be 75×125×c(75;125. Similarly, the sum of the
semi-variogram values between the line and every point in the ‘lower’ panel
(125 ft × 25 ft) will be given by 25×125×c(25;125). This gives:
Table 4.2 shows the values of the c(l;b)
function for the ‘standardised’ spherical model with a=1 and C=1. This table is used in the same
way as the F(l,b) table. It should be noted that the order
of the arguments is important. The value of c(25;125) is obviously different from the value of c(125;25). Standardising the
measurements of the panel in the same way as before, we require the values of c(0.94;1.54) and c(0.31;1.54) from the table. Interpolating linearly gives
the values 0.8196 and 0.6538 respectively. Multiplying by the sill of 450
(lb/ton)² gives 368.8 (lb/ton)² and 294.2 (lb/ton)² respectively. Putting these
together in the expression for (S,A) gives a value of
350.2(lb/ton)².
Having
evaluated all of the individual terms, we can now calculate the extension
variance for the setup in Fig. 4.10, so that:
This gives an estimation standard deviation or ‘standard error’ of 8.8 lb/ton
for the estimation of the panel grade. If we are willing to assume Normality
for the errors (sic) we
could be 95% certain that the ‘true’ grade of the panel lay between T* ± 2se equal to T*± 16.6 lb/ton. This standard
error would be correct for any panel having the same sampling setup, anywhere
within this deposit, since the estimation variance depends not on
the actual grades involved but on the geometric positioning of the sample and
the panel.
Fig. 4.11. Estimation of the panel
average from two drive averages.
As a second example,
consider a disseminated nickel deposit in the late stages of development. On a
particular underground level, we have a stoping block 40m by 30m to be
estimated. If we consider only the one level we can treat the problem as a
two-dimensional one. Suppose this ‘panel’ has been developed along two sides,
and the information available consists of (i) the average grade along the 40m
drive, g1 and (ii) the average grade along
the 30m drive, g2. Suppose we use the average of
these two grades to estimate the value inside the panel. Then:
For this particular deposit we have a spherical
semi-variogram with a range of influence of 60m, a sill of 0.75(%)² and a nugget effect of 0.10(%)². This implies a standard deviation for the ‘point’
sample values of 0.92%. The extension variance, as always, is given by:
(A,A)is, as before, the two-dimensional function F(l,b), but now we have two components to the
evaluation of this term. We can calculate the F(40,30) for the spherical part of the model, with a=60m and C=0.75(%)². This turns out to be 0.4336 × 0.75=0.325(%)². To this we must add the nugget effect of 0.10(%)², so that (A,A) =0.425(%)².
The average semi-variogram value between each sample and
each sample now contains four terms which have to be averaged, i.e.:
It is generally easier to look at each term separately and
then to combine them to produce the final answer. The first term (drive1,drive1) is the average semi-variogram between
all points within a length of 40m. That is, F(40).
For a
spherical model, when the length of the ‘line’ is shorter than the range of
influence, F(l) is given by:
Substituting l=40 and a=60, C=0.75(%)² gives a value of 0.239 (%)². Adding on the nugget effect produces (drive1,drive1) =0.339(%)². Similarly, (drive2,drive2) =0.283(%)². The two terms (drive1,drive2)
and ( drive2,drive1)
are identical and have been defined as the auxiliary function H(l,b). Using Table 4.3 for the H(l,b) function in the same way as the other tables
and adding in the nugget effect gives:
Putting all the individual terms together, we find that (S,S) =0.428(%)².
Finally, to calculate the estimation variance, we also
require the term (S,A). This is the average semi-variogram value between each
sample and the panel, so that:
The extension variance for the problem illustrated in Fig.
4.11 is thus:
giving a standard error for the prediction of the panel
average of 0.376% nickel.
A third
example is shown in Fig. 4.12.
Fig. 4.12. Estimation of the panel
average from two raise averages.
This time
the metal is zinc, the semi-variogram is spherical with a range of 20m and a
sill of 49(%)². The value to be estimated is the average
over the 30m by 15m panel, and the information available is the average grade of
each of the two raises through the ‘stope panel’. In the idealised situation
chosen, the raises are both 7.5m from the edges of the panel. Very briefly, the
extension variance is produced as follows:
The term (A,A) =F(30,15) when a=20 and C=49(%)², which is 0.7021×49 = 34.4(%)².
The g(l;b) function is given for the ‘standardised’
spherical model in Table 4.4. The last term to be evaluated is:
Since the setup is symmetrical, (raise1,A)= (raise2,A), so that
This gives an extension variance of:
This gives a standard error for the estimation of the panel
average of 1.14% Zn. Thus although the original sample standard deviation for
‘point’ samples was 7% Zn, two ‘samples’ within the panel produce a standard
error one-sixth of this Figure.
One last
brief example: a porphyry copper deposit has been explored by means of vertical
boreholes. To simplify the problem, each ‘bench’ in the deposit is considered
separately as a plane, and the borehole intersections as points within the
plane. Take a typical small block, 25m by 25m, with a borehole passing through
it as shown in Fig. 4.13.
Fig. 4.13. Estimation of panel average from one ‘point’ sample.
Suppose
we ‘extend’ the value of the core intersecting the block to the whole block. Then:
Then
From previous investigation we know that the semi-variogram
has a spherical form with a range of influence of 90m and a sill of 0.6(%)² Cu. (A,A) is yet another F(25,25) function, whose value can be calculated to be
0.2145 × 0.6 = 0.129(%)² Cu. (S,S) =0 since the sample is supposed to be a
point. (S,A) must be calculated with the now familiar
(?) manoeuvring, as follows:
So, finally, the extension variance becomes:
so that the standard error for the estimation is 0.288 %Cu.
The same exercise can be repeated for a sample outside the panel, using the same logic as that applied to the
one-dimensional problem in Fig. 4.7.
SUMMARY OF MAJOR POINTS
1.
When an estimation is performed, an error is made in
the prediction.
2. The magnitude of that error is
dictated by the structure and type of the deposit, and by the mineral itself.
Different minerals within the same deposit may have different structures.
3. The structure can (probably) be
described by a semi-variogram model, in the absence of significant trend on the
local level.
4. The estimation error variance can
be calculated if the semi-variogram model is known. Tables for the spherical
model have been supplied. These may also be used as approximations for the
linear model.
5. If we use the extension type of
estimator, i.e. the arithmetic mean of the samples, then the extension variance
may be written:
That is, the ‘reliability’ of the
estimator depends on three quantities: the relationship of the samples to the
area to be estimated; the relationship amongst the samples; and the variation
of grades within the area being estimated.
SOME MORE COMPLEX PROBLEMS
Let us
return to the original problem posed in Figs. 4.1 and 4.2. We showed that if we
used the grade of sample 1 to estimate the point A, we obtained a standard
error of 25.4 p.p.m. We then introduced the problem of estimating a 60 ft by 30
ft block centred at A. After the examples above, it should be easy enough to
calculate that the extension standard error will now be 19.2 p.p.m. [(S,A)=356, (A,A)=344, (S,S)=0] , when estimating the average grade of the panel from the
single sample point 1. This is over 20% lower than when trying to estimate the
central point. The real conclusion is simply that it is easier to estimate the
average grade over a block than to specify the grade at a single point. Now, if
we consider the arithmetic mean of the 5 point-samples, we obtain the
following:
If we use
this estimator to predict the central point of the block, the extension
standard deviation is 21.8 p.p.m. However, if we estimate the panel, the
extension standard deviation reduces to 12.8 p.p.m. Notice that both of these
Figures are lower than when only considering sample 1. It would appear that,
even though the other samples are a lot further away from the centre of the block, they are
contributing a fair amount of information about the block grade.
To conclude this chapter, an example on a slightly grander scale, on which the
reader can exercise his new-found knowledge. For this deposit a simulation has
been used, since in that case we know the
semi-variogram and the value at every point within the deposit. This enables us
to compare estimates with ‘actual values --- a situation which is rare to the
point of extinction in the real world. It also enables us to produce a set of
samples on any given sampling scheme proposed. The simulated deposit
is a low grade sedimentary iron ore, with an overall average of about 35% Fe, a
standard deviation of 5% Fe, a range of influence of 100m and a sill of 25(%)² Fe---obviously. The semi variogram is spherical (yet
again) with no nugget effect. An area 400 metres square has been simulated and
a set of 50 samples taken from it at random. The positions and values of these
are shown in Fig. 4.14 and Table 4.5.
Fig 4.14. A set of
random samples taken from a simulated iron ore deposit
The initial estimation, at the pre-feasibility
stage, is to be done on 50m by 50m blocks. For this first example, each block
has been allocated the average grade of all interior samples. Where a sample
falls on the edge it has been allocated to both blocks. Figure 4.15 shows the
estimated value for each block.
Fig 4.15. Estimates
of block values formed by averaging all interior samples in the iron ore
deposit, and the corresponding extension standard deviations.
Blocks without internal sampling have been
shaded in. The upper Figure shown in each block is the estimator T*, and the lower is the extension standard
deviation. Since this deposit is actually Normally
distributed, 95% confidence limits would be given (approximately) by T*± 2se. For comparison, the ‘true average grade
for each block is shown in Fig. 4.16.
Fig 4.16. ‘Actual’
average values within each block in the simulated iron ore deposit.
It can be seen that in most of the 37
estimated blocks, the true value is within the 95% confidence interval. Four or
five blocks lie just outside the interval, and three or four are considerably
outside. This is a little higher than would be expected, since we would only
expect about two blocks to lie outside a 95% interval. However, if we consider
the 99% confidence interval (3 standard deviations) only one block is
significantly outside the interval, i.e. the lower left-hand block of the area
(south-west corner).
For comparison, Fig. 4.17 shows a set of
samples taken on a regular grid from the same ‘deposit’. In this case, each 50m
block has two samples in opposing corners.
Fig 4.17. A set of
samples taken on a regular grid from the simulated iron ore deposit.
Using the average of these two samples to
estimate the block results in an extension standard deviation of 2.5% Fe, Fig.
4.18 shows the estimated values in each block.
Fig 4.18 Estimated
block values from regular samples.
It can be seen that although five or six
blocks lie outside the 95% confidence interval, not one lies more than 2.25
standard deviations from the true value.
Having
exhausted the possibilities of extension in idealised circumstances, let us
move on to some more interesting situations.