Probabilistic volumetrics by Monte Carlo simulation

Quick refresher

The method does not change the formula for hydrocarbons in place. What changes is what we put into it: instead of one value of each parameter, we take its distribution. The result is a distribution of the volume in place.
We get the value of a parameter from a random number $u$ between 0 and 1. We take it as a cumulative probability and find the value with that cumulative probability, $x = F^{-1}(u)$ . Frequent values are chosen often, and rare ones rarely.
P90 is the value that is exceeded with a probability of 90% (a cautious estimate), P10 is exceeded with a probability of 10% (an optimistic estimate), and P50 is the median. In the example of the lesson they are 11.6, 13.8 and 16.2 million m³, against 7.4, 13.8 and 23.9 million m³ for the deterministic cases.
When the parameters are independent, the results are spread much more narrowly than the gap between the product of all the worst values and the product of all the best values. It is rare for all the parameters to deviate in one direction at once. Dependence between parameters widens the distribution. With full dependence, the P90 of the result equals the volume calculated from the cautious values of all the parameters, that is, the values for which each parameter is worse with a probability of 10%.
For independent parameters, the mean volume equals the product of the means. The quantity $1 + C_v^2$ of the result is the product of the same quantities of the parameters ( $C_v$ is the coefficient of variation: the standard deviation divided by the mean). This is a check on the simulation, and a way to see which parameter gives the main uncertainty.
The error of the calculation itself falls as the square root of the number of trials: for the mean it is $s/\sqrt{n}$ . A thousand trials are enough for the mean and the P50, but the “tails” need more. We report the result together with its assumptions: the distributions, the dependences, the number of trials, and the meaning of the labels P90 and P10.

Key relationships

N = \dfrac{\mathrm{GRV}\cdot\mathrm{NTG}\cdot\phi\cdot(1 - S_w)}{B_{oi}}

the formula for the volume in place is the same; each parameter is a distribution

x = F^{-1}(u)

the value of a parameter from a random number $u$ between 0 and 1

\mu = \dfrac{a + m + b}{3}

mean of the triangular distribution with the smallest value $a$ , the most likely value $m$ and the largest value $b$

1 + C_v^2(N) = \prod_j \left[1 + C_v^2(x_j)\right]

spread of a product of independent factors $x_j$

SE = \dfrac{s}{\sqrt{n}}

standard error of the mean from $n$ trials (how much the mean would change with another set of random numbers)

Common pitfalls

Calling the product of all the pessimistic values a P90 estimate.
Not finding out what the ends of a parameter's range mean: the limiting values, or the values with a probability of 10 and 90%.
Drawing linked parameters independently, or, the other way round, drawing all the parameters with one random number.
Taking P50 for the mean: for a distribution skewed to the right, the mean is larger than the median.
Adding up the P90 values of single accumulations and calling the sum the P90 of the field.
Taking the number of trials as a measure of reliability: the calculation is no more reliable than the distributions put into it.

After this lesson you can

explain how a probabilistic estimate differs from a set of deterministic cases, and what P90, P50, P10 and the mean tell us
describe the steps of a Monte Carlo calculation, and get the value of a parameter from a random number with the inverse distribution function
choose the shape of the distribution of an input parameter, and find out what the ends of its range mean
explain how dependence between parameters changes the result, and how we specify it
judge whether the number of trials is enough, from the standard error and from the stability of the estimates
check the simulation against the product of the means and the coefficients of variation, and present the result correctly

From three cases to a distribution

The volumetric method gives the volume in place as the product of five parameters, and none of them is known exactly (Volumetric estimation of hydrocarbons in place). The simplest way to show the uncertainty is to calculate three cases: we put all the pessimistic values into the formula, then all the base values, then all the optimistic ones. Such a calculation is called deterministic: each parameter is given one definite value. For the accumulation from the lesson on the volumetric method we get 7.4, 13.8 and 23.9 million m³.

This calculation has two weaknesses. It does not say how likely any value is: what should we take as the cautious estimate, 7 million m³ or 11? And its extreme cases are too extreme. To get 7.4 million m³, all five parameters must be at their worst values at the same moment. If the parameters are not linked to one another, such a coincidence is very rare.

Monte Carlo simulation (in Russian-language literature it is also called the method of statistical trials) replaces the three sets of numbers with thousands. Each set is one possible combination of the values of the parameters, chosen at random but in such a way that likely values are chosen often and unlikely ones rarely. The results show the whole distribution of the volume in place: its middle, its ends, and the probability of any value. The method helps most at the exploration stage, when we are still looking for oil and gas. Little is known about the trap (the structure that may hold oil or gas) then, and the ranges of the parameters are wide. The distributions are strongly skewed (stretched towards large values), above all that of the rock volume. Three deterministic cases describe such a picture worst of all.

How the calculation works

The order of the calculation

1
Model
A formula that links the result to the parameters. For oil in place it is the volumetric formula. For the recoverable volume we add to it the oil recovery factor (RF).
2
Distributions
We give each uncertain parameter a distribution: which values are possible and how likely each one is.
3
Dependences
We find out which parameters are linked to one another, and we specify the link.
4
Draw
For each parameter we take a random number between 0 and 1, and from the distribution function we find the value that corresponds to it.
5
Trial
We put the chosen values into the formula and write down the result. One such calculation is called a trial, or a realisation.
6
Repetition
We repeat the draw and the calculation thousands of times.
7
Result
We put the results in order, build their distribution, and read from it the mean and the estimates P90, P50 and P10.

The main step is the draw. The cumulative distribution function $F(x)$ shows the probability that the parameter does not exceed $x$ . It rises from 0 to 1 (from the histogram to the distribution). We take a random number between 0 and 1 as the value of this probability, and go along the curve in the opposite direction: from the probability to the value of the parameter.

Value of a parameter from a random number

x = F^{-1}(u)

where:

$x$ value of the parameter in this trial, units of the parameter
$u$ random number, uniformly distributed between 0 and 1, —
$F(x)$ distribution function of the parameter: the probability that it does not exceed $x$ , —
$F^{-1}(u)$ inverse function: the value below which the share $u$ of the distribution lies, units of the parameter

Units: $u$ and $F$ are dimensionless; $x$ is in the units of the parameter

Holds when

$u$ is uniformly distributed: any value between 0 and 1 is equally likely.
$F(x)$ is continuous and increasing, so that each $u$ corresponds to one value of $x$ .
For independent parameters, the random numbers are independent of one another.

The share of random numbers smaller than $u$ is equal to $u$ itself, so the share of the values smaller than $x$ is equal to $F(x)$ : the sample reproduces the given distribution. Where the curve $F(x)$ is steep, an equal step in $u$ corresponds to a small step in $x$ , and such values are chosen more often.

CalculatedDrawing a value of the rock volume (GRV) from a triangular distribution: the probability density (how likely each value is) at the top, the distribution function at the bottom. Three random numbers, 0.10, 0.57, 0.90, give the values 159, 186, 209 million m³.

The computer supplies the random numbers. Their sequence is set by a starting number (the seed), and with the same seed the calculation repeats exactly. The advanced part shows how this works.

Every distribution has its own inverse function. For the uniform distribution from $a$ to $b$ it is simply $x = a + u\,(b - a)$ . For the normal distribution it is $x = \mu + z_u\sigma$ , and for the log-normal distribution it is $x = \e^{\mu + z_u\sigma}$ . Here $z_u$ is the quantile of the standard normal distribution for the probability $u$ (the value below which the share $u$ of that distribution lies). $\mu$ and $\sigma$ are the mean and the standard deviation: of the quantity itself for the normal distribution, and of its logarithm for the log-normal one (the normal distribution). For the triangular distribution (specified by the smallest, the most likely and the largest value), the inverse function consists of two branches: before the peak and after it.

Distribution function of the triangular distribution

F(x) = \begin{cases} \dfrac{(x - a)^2}{(b - a)(m - a)}, & a \le x \le m \\[10pt] 1 - \dfrac{(b - x)^2}{(b - a)(b - m)}, & m < x \le b \end{cases}

where:

$F(x)$ probability that the parameter does not exceed $x$ , —
$x$ value of the parameter, units of the parameter
$a$ smallest possible value, units of the parameter
$m$ most likely value (the mode), units of the parameter
$b$ largest possible value, units of the parameter

Units: $x$ , $a$ , $m$ and $b$ are in the units of the parameter; $F$ is dimensionless

Holds when

The probability density rises linearly from zero at $a$ to its largest value at $m$ , and falls linearly to zero at $b$ .
Values smaller than $a$ and larger than $b$ are impossible: there $F = 0$ and $F = 1$ .
$a < m < b$ . If $m$ is in the middle, the distribution is symmetrical; otherwise it is skewed.

The density is a triangle with its base from $a$ to $b$ and its peak at $m$ . Its area equals one, so the height of the peak is $2/(b - a)$ . The distribution function is the area of this triangle to the left of $x$ : two branches of a parabola that meet at the peak, at $F(m) = (m - a)/(b - a)$ . The mean of the triangular distribution is $(a + m + b)/3$ . It coincides with the most likely value only for a symmetrical triangle.

Drawing from the triangular distribution

x = \begin{cases} a + \sqrt{u\,(b - a)(m - a)}, & u \le F(m) \\[4pt] b - \sqrt{(1 - u)(b - a)(b - m)}, & u > F(m) \end{cases} \qquad F(m) = \frac{m - a}{b - a}

where:

$x$ value of the parameter that corresponds to the random number $u$ , units of the parameter
$a$ smallest possible value, units of the parameter
$m$ most likely value (the mode), units of the parameter
$b$ largest possible value, units of the parameter
$u$ random number between 0 and 1, —
$F(m)$ cumulative probability of the most likely value: the share of the distribution to the left of the peak, —

Units: $x$ , $a$ , $m$ and $b$ are in the units of the parameter

Holds when

The same as for the distribution function above.
$u$ is uniformly distributed between 0 and 1.

The formula follows if we solve the equation $F(x) = u$ for $x$ separately for each branch. First we compare $u$ with $F(m)$ : smaller numbers give values to the left of the peak, and larger ones to the right.

Distributions of the input parameters

The distribution of a parameter is a record of what is known and what is unknown about it. We choose its shape from the data and from the meaning of the quantity (Distributions and sample size).

Distributions used to describe the parameters of the calculation

Distribution: Uniform
Specified by: the smallest and the largest value
When it fits: only the limits are known, and there is no reason to prefer any value between them

Distribution: Triangular
Specified by: the smallest, the most likely and the largest value
When it fits: there are three estimates from a specialist; it can be symmetrical or skewed

Distribution: Normal
Specified by: the mean and the standard deviation
When it fits: a symmetrical spread around the mean: mean porosity, measurement errors; it is cut off at the physical limits

Distribution: Log-normal
Specified by: the mean and the standard deviation of the logarithm
When it fits: the quantity is formed by multiplying factors and is skewed to the right: the area and volume of a trap, the sizes of accumulations

Distribution: From data (empirical)
Specified by: a histogram of measurements or of analogues
When it fits: there are enough data, and there is no need to fit a formula

Distribution	Specified by	When it fits
Uniform	the smallest and the largest value	only the limits are known, and there is no reason to prefer any value between them
Triangular	the smallest, the most likely and the largest value	there are three estimates from a specialist; it can be symmetrical or skewed
Normal	the mean and the standard deviation	a symmetrical spread around the mean: mean porosity, measurement errors; it is cut off at the physical limits
Log-normal	the mean and the standard deviation of the logarithm	the quantity is formed by multiplying factors and is skewed to the right: the area and volume of a trap, the sizes of accumulations
From data (empirical)	a histogram of measurements or of analogues	there are enough data, and there is no need to fit a formula

The shape must not claim more than the data show. The peak of a triangle is a statement that one value is more likely than the others. If there is no basis for it and only the limits are known, the uniform distribution is more honest.
Respect the physical limits. The net-to-gross ratio and the saturation lie between 0 and 1, and the porosity of a reservoir lies above the cut-off value.
The distribution refers to the mean over the accumulation. The porosity of individual plugs (small cylinders cut from core) may vary from 12 to 25%, but the formula contains the mean porosity of the accumulation, and its uncertainty is much narrower than the scatter of the plugs. Its measure is the standard error of the mean, not the standard deviation of the data (how precisely the mean is known).
Include all the sources of uncertainty in a parameter. For the rock volume these are the conversion of seismic times to depths, the position of the contact, and the thickness of the layer.

Example: the volume in place of the accumulation from the lesson on the volumetric method

Let us take the accumulation for which the lesson on the volumetric method calculated three deterministic cases (uncertainty of the estimate). We take the three values of each parameter, the pessimistic, the base and the optimistic one, as the smallest, the most likely and the largest, and we describe each parameter by a triangular distribution. There are five parameters: the gross rock volume (GRV), the net-to-gross ratio (NTG), the porosity $\phi$ , the water saturation $S_w$ and the oil formation volume factor $B_{oi}$ . For now we take them to be independent.

Input parameters: triangular distributions

Parameter: GRV, million m³
Smallest: 140
Most likely: 180
Largest: 230
Mean: 183.3

Parameter: NTG
Smallest: 0.60
Most likely: 0.70
Largest: 0.78
Mean: 0.693

Parameter: $\phi$
Smallest: 0.17
Most likely: 0.19
Largest: 0.21
Mean: 0.190

Parameter: $S_w$
Smallest: 0.24
Most likely: 0.28
Largest: 0.33
Mean: 0.283

Parameter: $B_{oi}$ , m³/m³
Smallest: 1.20
Most likely: 1.25
Largest: 1.30
Mean: 1.250

Parameter	Smallest	Most likely	Largest	Mean
GRV, million m³	140	180	230	183.3
NTG	0.60	0.70	0.78	0.693
$\phi$	0.17	0.19	0.21	0.190
$S_w$	0.24	0.28	0.33	0.283
$B_{oi}$ , m³/m³	1.20	1.25	1.30	1.250

For $S_w$ and $B_{oi}$ the pessimistic value is the largest: more water and a greater shrinkage of the oil mean a smaller volume in place. The data in this example are assumed for illustration.

Worked exampleOne trial step by step

Given (assumed data)

Random numbers of the first trial (GRV, NTG, $\phi$ , $S_w$ , $B_{oi}$ ): 0.5701, 0.5202, 0.6998, 0.6989, 0.5955
Distributions: triangular, from the table above

Solution

GRV. The share of the distribution to the left of the peak is $F(m) = (180 - 140)/(230 - 140)$ $= 0.444$ . The number 0.5701 is larger, so the second branch of the formula applies.
Under the root: $(1 - 0.5701)\cdot90\cdot50 = 1934.6$ ; the root is 43.98. $\mathrm{GRV} = 230 - 43.98 = 186.02$ million m³.
NTG. $F(m) = 0.10/0.18 = 0.556$ ; the number 0.5202 is smaller, so the first branch applies. Under the root: $0.5202\cdot0.18\cdot0.10 = 0.00936$ ; the root is 0.0968. $\mathrm{NTG} = 0.60 + 0.0968 = 0.6968$ .
In the same way we get $\phi = 0.1945$ , $S_w = 0.2932$ and $B_{oi} = 1.255$ .
The volume in place in this trial is $N = 186.02\cdot0.6968\cdot0.1945\cdot(1 - 0.2932)/1.255$ $= 14.20$ million m³.
The second and third trials, with new random numbers, give 12.75 and 14.93 million m³.

Answer: First trial:

N = 14.20

million m³. This is one possible accumulation that does not contradict the input data.

What it means. A single trial is not a forecast and not a “case”: it is one of thousands of equally valid sets of values. In the first trial the rock volume turned out a little above the most likely value, and the water saturation also came out higher, that is, worse. The deviations of different parameters point in different directions. Only the whole set of trials speaks about probabilities. The data in this example are assumed for illustration.

Worked exampleTen thousand trials

Given (assumed data)

Distributions of the parameters: from the table above; the parameters are independent
Number of trials: 10,000
Deterministic cases: 7.4, 13.8 and 23.9 million m³

Solution

The mean of the trials is 13.83 million m³, the standard deviation is 1.74 million m³, and the coefficient of variation is 0.126.
Let us put the results in ascending order. One tenth of the trials gave less than 11.6 million m³. This is the P90: with a probability of 90% the volume in place is not smaller. Half of the trials gave less than 13.8, which is the P50. Nine tenths gave less than 16.2 million m³, which is the P10.
The smallest and the largest results are 8.9 and 20.3 million m³. The extreme deterministic cases (7.4 and 23.9) did not occur even once.
The ratio of P10 to P90 is 1.39; the ratio of the optimistic case to the pessimistic one is 3.2.
In other units, the same three estimates are 73, 87 and 102 million barrels (a cubic metre holds 6.2898 barrels) or, at an oil density of 850 kg/m³, 9.9, 11.7 and 13.7 million t. Converting units multiplies all the results by one number and does not change the probabilities.
A check with the mean values of the parameters from the table (the mean of the oil saturation is $1 - 0.2833 = 0.7167$ ): $183.33\cdot0.6933\cdot0.19\cdot0.7167/1.25$ $= 13.85$ million m³. The difference from the mean of the trials is 0.1%.

Answer: P90 = 11.6, P50 = 13.8, P10 = 16.2 million m³; the mean is 13.83 million m³.

What it means. The distance between the cautious and the optimistic estimates turned out much smaller than between the extreme deterministic cases: the P90 is one and a half times the pessimistic case. The mean and the P50 here are close to the base case, because the skews of the parameters almost cancel out: the rock volume is skewed to the high side (it has a long tail of large values), and the net-to-gross ratio to the low side. If all five distributions were symmetrical, the mean volume in place, for independent parameters, would practically coincide with the base case. Usually it is different: at the exploration stage the rock volume is strongly skewed to the high side, and the mean volume in place clearly exceeds the base case. The difference between them then tells us about the shape of the input distributions. The data in this example are assumed for illustration.

CalculatedResult of 10,000 trials. At the top is a histogram of the volume in place, and at the bottom the exceedance probability curve with the estimates P90, P50 and P10. The triangles on the lower axis are the pessimistic and the optimistic deterministic cases.

Worked exampleThe same three numbers, another reading

Given (assumed data)

Input numbers: the same pessimistic, base and optimistic values of the parameters
New assumption: the pessimistic and optimistic values are not limits: each parameter will turn out worse than the first, or better than the second, with a probability of 10%
Distributions: triangular, with the same peak; the limits are chosen so that 10% of the probability is left beyond each of the two values

Solution

The limits move apart. For a symmetrical triangle, the condition “10% to the left of the value” gives a half-width 1.81 times larger than the distance from the peak to that value. For porosity this is from 0.154 to 0.226 instead of 0.17–0.21.
The other parameters: GRV from 105 to 268 million m³, NTG from 0.52 to 0.85, $S_w$ from 0.205 to 0.368, $B_{oi}$ from 1.16 to 1.34.
10,000 trials with the new distributions: P90 = 9.9, P50 = 13.6, P10 = 18.1 million m³; the mean is 13.8 million m³.
With the first reading we had P90 = 11.6 and P10 = 16.2 million m³. The P90–P10 range has widened from 4.6 to 8.2 million m³.
Where is the product of all the pessimistic values, 7.4 million m³, now? Among 50,000 trials, a smaller result came out in 0.5%. It is exceeded with a probability of about 99.5%, not 90%.

Answer: With the second reading, P90 = 9.9 and P10 = 18.1 million m³, against 11.6 and 16.2 with the first.

What it means. The same three numbers per parameter gave two clearly different distributions of the volume in place, and neither is “more correct” by itself: everything depends on what the ends of the range meant to the person who named them. So we report this assumption together with the result. This also confirms the conclusion of the lesson on the volumetric method. Suppose each parameter turns out worse than its pessimistic value with a probability of 10%. Then a calculation with all the pessimistic values is not the P90 of the volume in place but an estimate at the level of P99 and beyond. The event “all five parameters are worse than their pessimistic values” is even rarer, one case in a hundred thousand. The volume in place falls short of 7.4 million m³ more often, because it is enough for two or three parameters to deviate strongly. The data in this example are assumed for illustration.

Dependence between parameters

So far the parameters were drawn independently: the random number for porosity said nothing about the number for water saturation. In a reservoir this is not always so. In a more porous rock the pores are usually larger and hold less water, so a high mean porosity more often goes with a low water saturation (Fluid saturation, Capillary pressure and the transition zone). Both quantities are in the formula, and in a lucky trial both work to increase the volume in place: the deviations stop cancelling each other.

We specify the link between parameters in one of three ways.

Full dependence. Both parameters get one random number: $u$ for porosity and $1 - u$ for water saturation if the link is inverse. A high porosity always goes with a low water saturation.
Partial dependence. We specify a correlation coefficient (a number from −1 to 1 that shows how strongly two quantities move together) and draw a pair of linked random numbers (exactly how is shown in the advanced part). We estimate the strength of the link from data: from the mean values of the parameters in wells or in analogue accumulations.
A calculated link. We draw one parameter, calculate the second from it through a relation found from data, and add a random deviation from that relation.

Worked exampleFour calculations with the same distributions

Given (assumed data)

Distributions of the parameters: triangular, as in the example above; 10,000 trials, the same random numbers
Independent parameters: P90 = 11.6, P10 = 16.2 million m³; standard deviation 1.74 million m³

Solution

Porosity and water saturation linked with a correlation of −0.7: the standard deviation is 1.83 million m³, P90 = 11.5, P10 = 16.3 million m³.
A full inverse link between porosity and water saturation: the standard deviation is 1.87 million m³, P90 = 11.5, P10 = 16.4 million m³.
All five parameters drawn with one random number, each in its favourable direction: the standard deviation is 3.37 million m³, P90 = 9.8, P10 = 18.9 million m³.
For comparison, a calculation without simulation. The value for which a parameter will be worse with a probability of 10% is different for each parameter: for water saturation and the formation volume factor it is larger than the most likely value, and for the others it is smaller. The volume in place calculated from these five values is 9.9 million m³. From the values for which a parameter will be better with a probability of 10%, it is 18.9 million m³.

Answer: From independent to fully dependent parameters, the P90 falls from 11.6 to 9.8 million m³, and the P10 rises from 16.2 to 18.9 million m³.

What it means. The link between porosity and water saturation widened the distribution by only a few per cent: together these two parameters give a small part of the total uncertainty. A link that involves the main parameter acts much more strongly. In the limit, when all the parameters change in step, the distribution is twice as wide, and its P90 and P10 coincide, to the accuracy of the simulation, with the volumes calculated from such values of the parameters themselves. So the rule is this: to multiply the cautious values of the parameters is to assume silently that they are all tightly linked. The data in this example are assumed for illustration.

CalculatedExceedance probability curves for three calculations of the example: independent parameters, a full inverse link between porosity and water saturation, and all parameters changing in step.

It is just as easy to make a mistake in either direction. Drawing linked parameters independently narrows the distribution and makes the P90 too high. Drawing all the parameters with one number takes us back to the deterministic cases and loses what the calculation was started for. We specify links where there is a physical reason and data for them.

How many trials are enough

The result of a simulation is itself random: another set of random numbers will give a slightly different mean and P90. This error is a property of the calculation, not of the accumulation. It falls as the number of trials grows, and we can estimate it in advance. We denote the number of trials by $n$ , because the letter $N$ is taken in this lesson by the volume in place.

Error of the mean and of a probability from a simulation

SE = \frac{s}{\sqrt{n}}, \qquad SE_p = \sqrt{\frac{p\,(1 - p)}{n}}

where:

$SE$ standard error of the mean over the trials, units of the result
$s$ standard deviation of the results of the trials, units of the result
$n$ number of trials, —
$p$ a probability estimated as a share of the trials, for example the share of results below a given value, fraction
$SE_p$ standard error of this share, fraction

Units: $SE$ is in the units of the result (million m³); $SE_p$ and $p$ are fractions

Holds when

The trials are independent of one another: a simple random draw.
$n$ is large enough for the mean and the share to follow a distribution close to normal (hundreds of trials or more).
The formulas describe the error of the calculation for the given distributions, not the uncertainty of the volume in place.

Both errors fall as $1/\sqrt{n}$ : to reduce them by half, we need four times as many trials (the same formula for the mean of plugs). For an estimate of P90 ( $p = 0.1$ ) with a thousand trials, $SE_p \approx 0.0095$ : the value found in fact corresponds to an exceedance probability somewhere between 88 and 92%.

Result of the simulation for different numbers of trials, million m³

Number of trials: 100
Mean ± $SE$: 13.72 ± 0.17
P90: 11.42
P50: 13.59
P10: 16.25

Number of trials: 1000
Mean ± $SE$: 13.78 ± 0.05
P90: 11.57
P50: 13.69
P10: 16.10

Number of trials: 10,000
Mean ± $SE$: 13.83 ± 0.02
P90: 11.61
P50: 13.76
P10: 16.16

Number of trials: 50,000
Mean ± $SE$: 13.84 ± 0.01
P90: 11.64
P50: 13.75
P10: 16.18

Number of trials	Mean ± $SE$	P90	P50	P10
100	13.72 ± 0.17	11.42	13.59	16.25
1000	13.78 ± 0.05	11.57	13.69	16.10
10,000	13.83 ± 0.02	11.61	13.76	16.16
50,000	13.84 ± 0.01	11.64	13.75	16.18

One sequence of random numbers: each row is its first $n$ trials. A calculation without simulation gives a mean of 13.85 million m³ (the advanced part). The data in this example are assumed for illustration.

CalculatedHow the estimates settle as the number of trials grows: the mean, P90 and P10 over the first

n

trials of one sequence. The band around the mean is two standard errors in each direction.

A hundred trials give the mean with an error of about 0.17 million m³, a thousand give 0.05, and ten thousand give 0.02 million m³. A thousand trials are usually enough for the mean and the P50. The estimates P90 and P10 settle more slowly, and probabilities of a fraction of a per cent need tens or hundreds of thousands of trials.

Increase the number of trials until the estimates you need stop changing in the digits that you are going to report.
Repeat the calculation with another seed of the generator. The difference between the two calculations is a clear measure of the error.
Do not aim for a precision of calculation that is an order of magnitude higher than the precision of the input data. The limits of the distributions are known to within per cent or tens of per cent, and the error of ten thousand trials is a fraction of a per cent.
Layered (stratified) sampling, called the Latin hypercube, gives the same precision of the mean with far fewer trials (the advanced part).

How to read and present the result

We show the result with an exceedance probability curve (also called an expectation curve): the horizontal axis is the volume in place, and the vertical axis is the probability that it is at least this value. The curve falls from 100% to zero. The steeper it is, the narrower the uncertainty. We read the three standard estimates from it, and we give the mean next to them.

The estimates that we report from a simulation

Estimate: P90
What it means: with a probability of 90% the volume is not smaller; the cautious (“low”) estimate
In the example, million m³: 11.6

Estimate: P50
What it means: the median: the volume is equally likely to be larger or smaller
In the example, million m³: 13.8

Estimate: P10
What it means: with a probability of 10% the volume is not smaller; the optimistic (“high”) estimate
In the example, million m³: 16.2

Estimate: Mean
What it means: the mean over all outcomes, taking their probabilities into account; calculations of expected value are built on it
In the example, million m³: 13.8

Estimate	What it means	In the example, million m³
P90	with a probability of 90% the volume is not smaller; the cautious (“low”) estimate	11.6
P50	the median: the volume is equally likely to be larger or smaller	13.8
P10	with a probability of 10% the volume is not smaller; the optimistic (“high”) estimate	16.2
Mean	the mean over all outcomes, taking their probabilities into account; calculations of expected value are built on it	13.8

The labels are as in the estimation of resources and reserves: the number after P is the exceedance probability. In mathematical statistics and in many programs it is the opposite: P90 is a large value. We always state the convention (two readings of P10 and P90).

The mean and the P50 are different estimates. The volume in place is a product, so its distribution is usually skewed to the right, and the mean is larger than the median. For wide distributions the difference reaches tens of per cent. In the example it is small, because the spread is small.

We go from the volume in place to the recoverable volume by adding the recovery factor, with its own distribution, to the product (volume in place and recoverable volume).

Worked exampleRecoverable volumes and three levels of confidence

Given (assumed data)

Hydrocarbons in place: the 10,000 trials of the example above
Recovery factor: triangular distribution: 0.22, 0.30, 0.38; it does not depend on the other parameters

Solution

In each trial the volume in place is multiplied by its own value of the recovery factor. The mean recoverable volume is 4.15 million m³.
P90 = 3.3, P50 = 4.1, P10 = 5.1 million m³. The ratio of P10 to P90 is 1.55; for the volume in place it was 1.39.
Three levels of confidence: the low estimate is 3.3 million m³; the middle one is 4.1, which is 0.8 million m³ more; the high one is 5.1, another 1.0 million m³ more.

Answer: Recoverable volume: P90 = 3.3, P50 = 4.1, P10 = 5.1 million m³.

What it means. At an early stage the recovery factor is poorly known, and the uncertainty of the recoverable volume is clearly wider than that of the volume in place. In the probabilistic approach to classification, the three estimates correspond to the levels 1P, 2P and 3P. Proved reserves are the volume that will be produced with a probability of at least 90%; proved plus probable, at least 50%; and with the possible added, at least 10%. The probable and possible reserves are the additions between the levels: 0.8 and 1.0 million m³. But a probability alone is not enough for the word “reserves”: we need a development project and its economic justification (Classification of reserves and resources). The data in this example are assumed for illustration.

As the accumulation is studied and developed, the curve narrows. New wells reduce the uncertainty of the rock volume and the properties, the production history refines the recovery factor, and the oil that has already been produced stops being an estimate. For a mature field, the P90 and P10 of the remaining volume are close to each other.

For an undrilled trap, the distribution describes the volume on the condition that the accumulation exists. The probability of the discovery itself is a separate number (geological chance of success). The estimates P90–P10 do not contain the risk of failure. The “risked” volume (the mean multiplied by the chance of success) is used when we compare prospects (possible drilling targets) (Geological risk and chance of success, Expected monetary value and preference theory).

The contribution of the parameters to the uncertainty of the volume in place

Factor: GRV
$C_v$: 0.100
$1 + C_v^2$: 1.0101
Share of the spread: 64%

Factor: NTG
$C_v$: 0.053
$1 + C_v^2$: 1.0028
Share of the spread: 18%

Factor: $\phi$
$C_v$: 0.043
$1 + C_v^2$: 1.0018
Share of the spread: 12%

Factor: $1 - S_w$
$C_v$: 0.026
$1 + C_v^2$: 1.0007
Share of the spread: 4%

Factor: $1/B_{oi}$
$C_v$: 0.016
$1 + C_v^2$: 1.0003
Share of the spread: 2%

Factor: Volume in place $N$
$C_v$: 0.125
$1 + C_v^2$: 1.0157
Share of the spread: 100%

Factor	$C_v$	$1 + C_v^2$	Share of the spread
GRV	0.100	1.0101	64%
NTG	0.053	1.0028	18%
$\phi$	0.043	1.0018	12%
$1 - S_w$	0.026	1.0007	4%
$1/B_{oi}$	0.016	1.0003	2%
Volume in place $N$	0.125	1.0157	100%

The coefficients of variation are from the formulas of the triangular distribution; the bottom row is the product of the quantities $1 + C_v^2$ (the advanced part). The share of the spread is the share of $\ln(1 + C_v^2)$ of the factor in their sum. The data in this example are assumed for illustration.

This kind of breakdown is called the parametric method. Here almost two thirds of the uncertainty (64%) comes from the rock volume. So what we need to improve is the structure and the position of the contact, with seismic and appraisal wells, and not the porosity: even an exact knowledge of the porosity would narrow the range of the volume in place only a little.

Together with the result, we report everything without which it cannot be checked or compared with another.

what has been estimated: hydrocarbons in place or a recoverable volume, in what units and at what standard conditions
the distributions of the parameters: the type, the values, and the meaning of the ends of the range
which parameters were treated as dependent, and how the link was specified
the number of trials, the method of drawing, and the seed of the generator
the estimates P90, P50, P10 and the mean, with the convention for the labels stated
the deterministic base case next to the probabilistic result

The last item is not a formality. The deterministic and the probabilistic calculations check each other: the base case must lie near the middle of the distribution. A large difference is a reason to look for the cause: a strongly skewed distribution, a missed dependence, or an error in the input data.

Terms and symbols

Trial (realisation): One calculation of the result with a randomly chosen set of values of the parameters.
$u$ , $F^{-1}(u)$: A random number between 0 and 1, and the value of the parameter that has this cumulative probability.
Triangular distribution: A distribution specified by the smallest ( $a$ ), the most likely ( $m$ ) and the largest ( $b$ ) value. Its mean is $(a + m + b)/3$ .
P90, P50, P10: The values that are exceeded with a probability of 90, 50 and 10%. In estimating reserves, P90 is the cautious estimate.
Exceedance probability curve: The probability that the result is at least a given value. It falls from 100% to zero.
Dependence of parameters: A link in which the deviation of one parameter makes a deviation of another more likely. We specify it with a common random number, a correlation, or a calculation formula.
$n$ , $SE$: The number of trials and the standard error of their mean, $s/\sqrt{n}$ .
Seed of the generator: The number from which the sequence of pseudo-random numbers starts. With the same seed the calculation repeats exactly.

Common mistakes and misconceptions

Wrong

“The P90 of the volume in place is what we get if we put the cautious values (P90) of all the parameters into the formula.”

In fact

That would be so only if all the parameters changed in step. For independent parameters the P90 of the volume in place is clearly higher: in the example of the lesson, 11.6 million m³ against 9.9 million m³.

Wrong

“P50 is the mean expected volume in place.”

In fact

P50 is the median. The distribution of the volume in place is skewed to the right, and the mean is larger than the median; for wide distributions, much larger. Calculations of expected value use the mean.

Wrong

“A hundred thousand trials give an exact estimate of the volume in place.”

In fact

The number of trials reduces only the error of the calculation. The reliability of the result is set by the distributions of the parameters and the links between them. If they are specified wrongly, a large number of trials will reproduce the wrong answer exactly.

Wrong

“The distribution of porosity for the calculation is the histogram of the porosity of the plugs.”

In fact

The formula contains the mean porosity of the accumulation. Its uncertainty is much narrower than the scatter of the plugs, and it falls as the number of wells and measurements grows. A histogram of plugs describes the variability of the rock, not how precisely the mean is known.

Wrong

“The P90 of a field is the sum of the P90 values of its accumulations.”

In fact

The means add up. The sum of the P90 estimates of independent accumulations is smaller than the P90 of the sum: it is unlikely that all the accumulations will be at their cautious values at the same time. We must combine the distributions, not the finished estimates.

In depth: advanced detail

A check without simulation: the mean and spread of a product

For a simple model, a product of independent factors, the mean and the spread of the result can be found without a single random number. This is both a check of the simulation and an explanation of where its result comes from.

Mean and variance of the triangular distribution

\mu = \frac{a + m + b}{3}, \qquad \sigma^2 = \frac{a^2 + m^2 + b^2 - am - ab - mb}{18}

where:

$\mu$ mean, units of the parameter
$\sigma^2$ variance; the standard deviation $\sigma$ is its square root, square of the units of the parameter
$a,\ m,\ b$ the smallest, the most likely and the largest values, units of the parameter

Units: in the units of the parameter; the coefficient of variation $C_v = \sigma/\mu$ is dimensionless

Holds when

A triangular distribution with $a < m < b$ .

For a symmetrical triangle with a half-width $w$ , the formula gives $\sigma = w/\sqrt{6} \approx 0.41\,w$ : the standard deviation is clearly smaller than half the range, because values near the ends are rare.

Mean and coefficient of variation of a product of independent quantities

\mu_N = \prod_j \mu_j, \qquad 1 + C_v^2(N) = \prod_j \left[1 + C_v^2(x_j)\right]

where:

$N$ the result, a product of factors $x_j$ ; for the volume in place these are GRV, NTG, $\phi$ , $1 - S_w$ and $1/B_{oi}$ , million m³
$\mu_N$ mean of the result, million m³
$\mu_j$ mean of the factor $x_j$ , units of the factor
$C_v(x_j)$ coefficient of variation of the factor: the standard deviation divided by the mean, —
$C_v(N)$ coefficient of variation of the result, —

Units: the means are in the units of the quantities; the coefficients of variation are dimensionless

Holds when

The factors are independent. With dependence, the mean of the product is no longer equal to the product of the means.
The factor is exactly what enters the product: $1 - S_w$ , not $S_w$ , and $1/B_{oi}$ , not $B_{oi}$ .
The type of the distributions does not matter: the formulas are exact for any independent factors.

The mean of a product of independent quantities is equal to the product of their means, and the mean square is the product of the mean squares. The second formula follows from this. For small spreads it turns into the familiar rule for adding relative errors: $C_v^2(N) \approx \sum_j C_v^2(x_j)$ . The factor with the largest $C_v$ decides the result, and factors with a small $C_v$ add almost nothing.

The distribution of a product of several factors is close to log-normal (log-normal distribution). If we know the mean and the coefficient of variation, we can also estimate the percentiles. Here $\mu$ and $\sigma$ are the mean and the standard deviation of the natural logarithm of the volume in place, $\ln N$ , not of the volume itself: $\sigma^2 = \ln[1 + C_v^2(N)]$ , $\mu = \ln\mu_N - \sigma^2/2$ , and the values of P90 and P10 are $\e^{\mu \mp 1.282\sigma}$ .

Worked exampleThe formulas against ten thousand trials

Given (assumed data)

Means of the factors: GRV 183.3 million m³; NTG 0.6933; $\phi$ 0.19; $1 - S_w$ 0.7167; $1/B_{oi}$ 0.8002
Coefficients of variation: 0.100, 0.053, 0.043, 0.026, 0.016

Solution

Mean: $183.3\cdot0.6933\cdot0.19\cdot0.7167\cdot0.8002$ $= 13.85$ million m³. From the trials it is 13.83 million m³.
Spread: the product of the quantities $1 + C_v^2$ is 1.0157, so $C_v(N) = 0.1255$ and the standard deviation is 1.74 million m³. From the trials it is 1.74 million m³.
The log-normal approximation: $\sigma = 0.1250$ , $\mu = 2.6205$ ; P90 = 11.71, P50 = 13.74, P10 = 16.13 million m³. From the trials: 11.61, 13.76 and 16.16 million m³.
The mean of $1/B_{oi}$ is 0.8002, while one divided by the mean of $B_{oi}$ is 0.8000: a difference of 0.03%. For a narrow distribution we can ignore it; for a wide one we cannot.
Linked porosity and water saturation (a correlation of −0.7) add to the sum of squares a term $2\cdot0.7\cdot0.043\cdot0.026 = 0.0015$ : $C_v(N) \approx 0.131$ , and the standard deviation is about 1.82 million m³. From the trials it is 1.83 million m³.

Answer: The formulas give a mean of 13.85 and a standard deviation of 1.74 million m³. These are the same as the simulation gives, within its error.

What it means. For such a simple model the simulation added nothing to the formulas except the shape of the distribution, and this is a good check on both calculations. The method becomes irreplaceable when there are no formulas: with dependent parameters, truncated distributions (cut off at a limit), cut-offs, several accumulations with common parameters, or when the formula for the volume in place is followed by a production model and an economic model. The data in this example are assumed for illustration.

Where random numbers come from

When such calculations were done by hand, random numbers were taken from printed tables: columns of two-digit numbers from 00 to 99, drawn up in advance. The scale of cumulative probability was divided into a hundred parts, and each number in the table pointed to its own part. A computer gets random numbers differently: it calculates them. Such numbers are called pseudo-random. The sequence is fully set by a rule and by the seed (the starting number), but in its statistical properties it looks random.

The pseudo-random number generator used in the lesson

x_{k+1} = 16{,}807\cdot x_k \bmod \left(2^{31} - 1\right), \qquad u_k = \frac{x_k}{2^{31} - 1}

where:

$x_k$ an integer, the state of the generator at step $k$ ; $x_0$ is the seed, —
$\bmod$ remainder of division, —
$u_k$ the random number between 0 and 1 that goes into the calculation, —

Units: dimensionless; $2^{31} - 1 = 2{,}147{,}483{,}647$

Holds when

The seed is any integer from 1 to $2^{31} - 2$ .
The sequence repeats after $2^{31} - 2$ steps. For the calculations of the lesson this is enough with a large margin.
This is one of the simplest generators, chosen because it is easy to repeat. In working calculations people use the tested generators of calculation packages.

Each next number is the remainder of dividing the previous one, multiplied by 16,807, by the prime number $2^{31} - 1$ (a number that has no divisors except 1 and itself). Such a generator is called multiplicative congruential. Neighbouring numbers are linked by the rule, but in the calculation they look and behave as independent.

Worked exampleThe first random numbers of the lesson

Given (assumed data)

Seed $x_0$: 20,261,003

Solution

$x_0\cdot16{,}807 = 340{,}526{,}677{,}421$ . We divide by 2,147,483,647: the integer part is 158, and the remainder is $x_1 = 1{,}224{,}261{,}195$ .
We divide $x_1$ by $2^{31} - 1$ : $u_1 = 0.5701$ . This number set the rock volume in the first trial.
By the same rule, $x_2 = 1{,}117{,}082{,}458$ and $u_2 = 0.5202$ ; $x_3 = 1{,}502{,}829{,}532$ and $u_3 = 0.6998$ .

Answer: The first three random numbers: 0.5701, 0.5202, 0.6998.

What it means. We can repeat the calculation in a spreadsheet: one cell with the formula for the remainder of a division, copied downwards. All the products are smaller than $2^{53}$ , so a spreadsheet calculates them exactly. Reproducibility is a practical requirement: a calculation that gives different digits at each run can be neither checked nor compared with last year's. So we record the seed together with the result.

How partial dependence is specified

Full dependence comes from a common random number, and independence from two different ones. To get a link of intermediate strength, we use the fact that normally distributed quantities are easy to link with a correlation coefficient.

A pair of linked random numbers

z_2 = \rho\,z_1 + \sqrt{1 - \rho^2}\;z', \qquad u_2 = \Phi(z_2)

where:

$z_1,\ z'$ quantiles of the standard normal distribution for two independent random numbers: $z_1 = \Phi^{-1}(u_1)$ , $z' = \Phi^{-1}(u')$ , —
$\rho$ the specified correlation coefficient, from −1 to 1, —
$z_2$ a normal quantity linked with $z_1$ by the correlation $\rho$ , —
$\Phi$ the standard normal distribution function, —
$u_2$ the random number for the second parameter; for the first we take $u_1$ itself, —

Units: all quantities are dimensionless

Holds when

$u_1$ and $u'$ are independent and uniformly distributed between 0 and 1.
$\rho$ is the correlation of the normal quantities $z_1$ and $z_2$ . The correlation of the parameters themselves is close to it, but does not coincide with it exactly.
The distributions of both parameters stay the same: only the way their values combine changes.

At $\rho = 0$ we get $u_2 = u'$ , an independent draw. At $\rho = -1$ we get $u_2 = 1 - u_1$ , a full inverse link. Intermediate values give a link of intermediate strength. We find the values of the parameters as usual: $x_1 = F_1^{-1}(u_1)$ , $x_2 = F_2^{-1}(u_2)$ (the normal distribution and its quantiles).

For several linked parameters we specify a table of pairwise correlations. It cannot be filled in arbitrarily: if the first parameter is strongly linked to the second, and the second to the third, the first and the third cannot be independent. Calculation packages check this themselves, but we should not assign correlations “by eye”: behind each one there must be data or a physical reason.

Layered sampling: the Latin hypercube

With simple random drawing, random numbers fall unevenly: in a hundred trials some parts of the probability scale get more numbers and others fewer. Latin hypercube sampling (LHS), a kind of layered (stratified) sampling, removes this unevenness. We divide the probability scale of each parameter into $n$ equal layers, as many as there will be trials, and take exactly one value from each layer: at a random point of the layer or, as in the example below, in its middle. We combine the layers of different parameters in a random order. Each parameter is then represented by all parts of its distribution, from the smallest values to the largest.

Worked exampleA hundred trials: simple drawing and the Latin hypercube

Given (assumed data)

Distributions of the parameters: as in the main example; the parameters are independent
Number of trials in a calculation: 100
Number of repetitions: 200 calculations by each method with different seeds

Solution

Simple drawing. The means of the two hundred calculations are scattered with a standard deviation of 0.17 million m³, as the formula says: $1.74/\sqrt{100} = 0.17$ .
Latin hypercube. The scatter of the means is 0.011 million m³, which is 16 times smaller.
The error of simple drawing falls as the square root of the number of trials, so to get the same precision of the mean it would need about 250 times more trials.
The P90 estimate gains less: the scatter is 0.26 million m³ with simple drawing and 0.16 million m³ with layered sampling.

Answer: With a hundred trials, the Latin hypercube finds the mean about 16 times more precisely than simple drawing.

What it means. The gain is large because the volume in place almost “decomposes” into the contributions of single parameters, and layered sampling represents each parameter perfectly. The estimates of the ends of the distribution depend on combinations of values, and here the gain is more modest. When each trial takes a few seconds, this does not matter. When each trial needs a run of a reservoir simulation model, the saving decides. Simple formulas for the error do not suit layered sampling: its precision is checked by repeated calculations, as here. The data in this example are assumed for illustration.

Adding the estimates of several accumulations

A field often consists of several accumulations or blocks, and their estimates have to be combined. We can always add the means. We cannot add the P90 and P10 estimates, except when the accumulations are fully dependent.

Worked exampleTwo identical independent accumulations

Given (assumed data)

Each accumulation: as in the main example: P90 = 11.6, mean 13.8 million m³
Link between the accumulations: none: each has its own random numbers

Solution

The second accumulation is drawn with its own random numbers: the distribution is the same, the sample is different, and its mean over 10,000 trials is 13.9 million m³. The sum of the means is 27.7 million m³. The mean of the sum is the same number: means always add up, whatever the link between the accumulations.
The sum of the P90 estimates of the two accumulations is 23.2 million m³.
The P90 of the sum is the value that the sum of the volumes exceeds in 90% of the trials: 24.5 million m³, which is 1.3 million m³ more.
For P10 it is the opposite: the sum of the estimates is 32.3 million m³, and the P10 of the sum is 30.9 million m³.

Answer: The P90 of the two accumulations together is 24.5 million m³, not 23.2; the sum of the separate P90 values underestimates the cautious estimate.

What it means. A failure in one accumulation is usually partly made up by the other, so the spread of the sum, as a share of the mean, is smaller than for each accumulation. The arithmetic sum of the P90 estimates would correspond to full dependence of the accumulations. In reality the accumulations of one field are linked partly, by a common contact and common reservoir properties, and we must combine the distributions with this link, not the finished numbers. The data in this example are assumed for illustration.

Connections to other topics

Volumetric estimation of hydrocarbons in place — the volumetric formula, its parameters, and the three deterministic cases that this lesson starts from.
Distributions and sample size — density and distribution function, the normal and log-normal distributions, the standard error of the mean.
Descriptive statistics in geology — percentiles, two readings of the labels P10 and P90, the coefficient of variation.
Classification of reserves and resources — the categories of reserves and resources and their link with the estimates 1P, 2P and 3P.
Geological risk and chance of success — the geological chance of success, which the distribution of the volume does not contain.
Sensitivity analysis and scenarios — the same method in project economics: a distribution of NPV instead of one base case.
Expected monetary value and preference theory — how mean volumes and probabilities become decisions.
Gross rock volume from maps — where the rock volume comes from: the main source of uncertainty in the volume in place.

Check yourself

1
A parameter is described by a triangular distribution with the smallest value 10, the most likely 20 and the largest 40. What value does the random number $u = 0.25$ give?
1. A.17.5
2. B.25.0
3. C.18.7
4. D.20.0
Show the answer
Correct answer: C. 18.7
The share of the distribution to the left of the peak is $F(m) = (20 - 10)/(40 - 10) = 0.333$ . The number 0.25 is smaller, so we take the first branch: $x = 10 + \sqrt{0.25\cdot30\cdot10}$ $= 10 + 8.66 = 18.7$ . The answer 17.5 is a linear interpolation over the range, and it suits only a uniform distribution.
2
Why is the P90 of the volume in place, found by simulation with independent parameters, higher than the volume calculated from the cautious values (P90) of all the parameters?
Show the answer
A calculation from the cautious values gives the result on the condition that all the parameters are at these values at the same time. For independent parameters such a coincidence is far less likely than 10%: the deviations of different parameters mostly cancel out. So the value below which the volume falls in 10% of the trials lies higher. These numbers coincide only when the parameters are fully dependent.
3
From 400 trials, the standard deviation of the result is 2.0 million m³. What is the standard error of the mean, and how many trials are needed to halve it?
1. A.2.0 million m³; any number: it does not change
2. B.0.1 million m³; 800 trials
3. C.0.005 million m³; 800 trials
4. D.0.1 million m³; 1600 trials
Show the answer
Correct answer: D. 0.1 million m³; 1600 trials
$SE = 2.0/\sqrt{400} = 0.1$ million m³. The error falls as the square root of the number of trials, so to halve it we need four times as many trials, 1600. The standard deviation of the result does not fall: it describes the uncertainty of the volume in place, not the error of the calculation.
4
A report gives: P90 = 40, P50 = 60, P10 = 95 million t. What can we say about the mean?
1. A.Nothing: the mean has no link with percentiles
2. B.It is larger than 60 million t
3. C.It equals 60 million t
4. D.It is smaller than 60 million t
Show the answer
Correct answer: B. It is larger than 60 million t
The distance from P50 to P10 (35 million t) is larger than from P90 to P50 (20 million t): the distribution is skewed to the right. In such a distribution the rare large values pull the mean up, and it is larger than the median. For volumes in place this is the usual picture.
5
The mean values of the factors are: rock volume 100 million m³, net-to-gross ratio 0.8, porosity 0.2, oil saturation 0.7, formation volume factor 1.25. The coefficients of variation of the first three are 0.20, 0.10 and 0.05, and the others can be neglected. Estimate the mean volume in place and its coefficient of variation for independent parameters.
Show the answer
Mean: $100\cdot0.8\cdot0.2\cdot0.7/1.25 = 8.96$ million m³. Spread: $1.04\cdot1.01\cdot1.0025 = 1.0530$ , so $C_v = \sqrt{0.0530} \approx 0.23$ . Almost all the spread comes from the rock volume: without the two other factors it would be 0.20.
6
In a calculation of the volume in place, porosity and water saturation were drawn independently, although the wells show that the higher the mean porosity, the lower the water saturation. How did this affect the result?
1. A.The distribution came out too wide
2. B.It did not: the distributions of the parameters are specified correctly
3. C.The mean volume in place is overestimated twofold
4. D.The distribution of the volume in place came out too narrow: the P90 is too high and the P10 is too low
Show the answer
Correct answer: D. The distribution of the volume in place came out too narrow: the P90 is too high and the P10 is too low
With an inverse link, a high porosity comes together with a high oil saturation: both factors deviate in the same direction and reinforce each other. An independent draw makes these combinations rare, and the distribution comes out narrower than the true one. How much narrower depends on the share of these parameters in the total uncertainty. In the example of the lesson the standard deviation grew by a few per cent.

Updated: 3 October 2026