Geological and technical uncertainty of a project

Quick refresher

Uncertainty is inexact knowledge. Risk is the possible loss (and gain) that comes from it. Technical uncertainty is created by the subsurface, the wells and the facilities. Unlike prices and laws, it can be reduced by data, before development and during it.
It matters not only how much will be produced but also when. If the wells give a third less than the planned plateau (the steady maximum rate), the same 12 million bbl are produced over a year longer. Discounted production falls by 6.1%, and NPV falls by 16.8%.
The estimates P90, P50 and P10 carry over to NPV only if there is really one uncertainty and NPV changes with it in one direction. With several parameters, the case “all at P90” is much worse than the P90 of NPV. The mean does not carry over like this if the dependence is not linear: we calculate it from the distribution, and from three cases with the weights 0.3, 0.4 and 0.3.
The NPV of the average case is not equal to the average NPV if the dependence is not linear. Upper limits (the capacity of a plant) and thresholds cut off the gain, so the average NPV is lower. The possibility to walk away cuts off the loss, so it is higher. In the example of the lesson, a plant built for the base plateau takes $3.1 million from the average NPV.
A link between parameters changes the spread of NPV. A positive link between parameters that act on NPV in opposite directions (larger reserves, more wells) makes the spread narrower. A link between parameters that act in the same direction makes it wider. In the example of the lesson, a link of 0.7 narrows the spread from 21.1 to 13.1 million dollars. We take the sign of the link from its physical cause.
Appraisal work narrows the range of what is measured. A well that clarifies the rock volume makes the P10/P90 ratio of the oil in place 1.22 instead of 1.39. For the recoverable oil it makes the ratio only 1.42 instead of 1.55, because the recovery factor stays unknown. Which way the estimate will move is not known in advance: on average, nowhere.

Key relationships

Q_\text{disc} = \sum_t \dfrac{Q_t}{(1+i)^{t}}

discounted production: the reserves with the time of production taken into account

\bar x \approx 0.3\,x_{90} + 0.4\,x_{50} + 0.3\,x_{10}

the mean from three estimates (the 0.3–0.4–0.3 rule)

\mathrm{E}[f(X)] \le f(\mathrm{E}[X])

for a dependence $f$ that flattens as it rises (a concave curve); for one that steepens as it rises (a convex curve), the reverse (Jensen's inequality)

\sigma^2_\text{NPV} = a^2\sigma_X^2 + b^2\sigma_Y^2 - 2ab\,\rho\,\sigma_X\sigma_Y

the spread of NPV ≈ $aX - bY$ with a correlation $\rho$ between $X$ and $Y$

Common pitfalls

Taking the NPV of the base case, or of the case with average values, for the expected NPV.
Building the P90 case of a project by putting all the parameters at their P90.
Drawing linked parameters independently, or setting a link without understanding its cause.
Assessing uncertainty by the volume of reserves alone, and forgetting the flow rates, the timing, the costs and the reliability of the facilities.
Expecting appraisal work to increase the reserves: data make the estimate more exact in both directions.

After this lesson you can

list the sources of technical uncertainty (volume, productivity, recovery, fluids, wells, facilities and timing), describe how each one shows up in the cash flow, and how it changes over the stages of a project
explain why the same reserves, produced more slowly, are worth less, and estimate the loss through discounted production
build the economic cases P90, P50 and P10 and estimate the expected NPV from them by the 0.3–0.4–0.3 rule, understanding when the P90–P10 estimates carry over to NPV and when they do not
explain why the NPV of the average case is not equal to the average NPV, and find from the shape of the dependence which way these numbers differ
estimate how a link between parameters changes the spread of NPV, and choose the sign of the link from its physical cause
estimate how far an appraisal well will narrow the range of volumes, and explain why the average estimate does not change, on average, with the data

Uncertainty and risk: what is known inexactly

Every estimate of a project rests on numbers that nobody has measured before development. Examples are the volume of rock in the trap, the permeability between the wells (how easily the rock lets fluid through), the share of oil that we can displace, and the time to drill through formations that have not been opened yet. Uncertainty is the inexactness of our knowledge of such quantities. We call it risk when money depends on it. Risk is usually linked with possible losses (the downside). But uncertainty has another side too, a possible gain (the upside): the reserves may turn out larger than expected, and the wells more productive.

It helps to divide the uncertainty of a project by its sources. Technical uncertainty is created by the subsurface, the wells and the surface facilities. Market, financial and regulatory uncertainty is created by prices, exchange rates, the cost of money, taxes and laws, and by relations with partners (Market, financial and regulatory uncertainty). The difference is practical: we can reduce technical uncertainty by data (appraisal wells, studies, experience of development), but no well will tell us the oil price ten years ahead.

Inside technical uncertainty there are two kinds of ignorance. The first is ranges: a quantity will certainly have some value, and the question is which one (a volume, a flow rate, a cost estimate). We describe it by a distribution (Probabilistic volumetrics by Monte Carlo simulation). The second is events: something may happen or may not (a dry well, an accident, a failure of a plant). We describe it by a probability and its consequences (Geological risk and chance of success). One cannot replace the other: a distribution of reserves says nothing about the chance of an empty trap, and the probability of an accident says nothing about its cost.

Sources of technical uncertainty

Source: Volume
What is known inexactly: the rock volume, the position of the fluid contacts (the boundaries between oil, gas and water), the net thickness, the porosity (the share of void space in the rock), the saturation (the share of that space filled with oil)
How it shows up in money: how much oil will be produced: revenue, number of wells

Source: Productivity and connectivity
What is known inexactly: permeability (how easily the rock lets fluid through), faults and barriers inside the accumulation, heterogeneity
How it shows up in money: flow rates and the production profile, number of wells

Source: Recovery
What is known inexactly: how much of the reservoir the displacing fluid reaches, the activity of the edge water (the water around the accumulation), the injectivity of the injection wells (how much water they can take)
How it shows up in money: what share of the reserves is produced, and for how long

Source: Fluids
What is known inexactly: viscosity, wax, hydrogen sulfide and carbon dioxide, sand production, gas hydrates (ice-like solids that block pipes)
How it shows up in money: materials and equipment, costs, downtime

Source: Wells
What is known inexactly: drilling problems, the time per well, the quality of the well completion, repairs (workovers)
How it shows up in money: capital and operating expenditure, timing

Source: Facilities and timing
What is known inexactly: the cost estimate and the construction schedule, the capacity and reliability of the plants, infrastructure shared with neighbours
How it shows up in money: costs, delay of first oil, lost production during downtime

Source	What is known inexactly	How it shows up in money
Volume	the rock volume, the position of the fluid contacts (the boundaries between oil, gas and water), the net thickness, the porosity (the share of void space in the rock), the saturation (the share of that space filled with oil)	how much oil will be produced: revenue, number of wells
Productivity and connectivity	permeability (how easily the rock lets fluid through), faults and barriers inside the accumulation, heterogeneity	flow rates and the production profile, number of wells
Recovery	how much of the reservoir the displacing fluid reaches, the activity of the edge water (the water around the accumulation), the injectivity of the injection wells (how much water they can take)	what share of the reserves is produced, and for how long
Fluids	viscosity, wax, hydrogen sulfide and carbon dioxide, sand production, gas hydrates (ice-like solids that block pipes)	materials and equipment, costs, downtime
Wells	drilling problems, the time per well, the quality of the well completion, repairs (workovers)	capital and operating expenditure, timing
Facilities and timing	the cost estimate and the construction schedule, the capacity and reliability of the plants, infrastructure shared with neighbours	costs, delay of first oil, lost production during downtime

An incomplete list. In each project it is made anew, starting from the parameters to which NPV is most sensitive (the tornado chart).

An example from practice: hydrogen sulfide and pipelines (2013–2016)

The offshore Kashagan field in the Kazakhstan sector of the Caspian Sea started production in September 2013, years later than expected when the project was prepared. A few weeks later production was stopped: leaks were found on the pipeline that carried gas with a high content of hydrogen sulfide from the sea to the shore. The cause was sulfide stress cracking, which is the cracking of steel under stress in the presence of hydrogen sulfide. The pipelines had to be replaced, and production restarted only in the autumn of 2016. The three-year stop and the replacement of the pipelines were caused not by the amount of oil in the reservoir but by the behaviour of metal in an aggressive environment. This was an uncertainty linked with the fluid and the facilities.

How uncertainty changes over the stages of a project

From exploration to production, both the main question and the means of answering it change. At first we do not know whether there is an accumulation at all. Then we ask how large it is and how it will give up its oil. Then we ask what the wells and the facilities will cost. Finally we ask how much we will be able to recover.

Uncertainty by stage

Stage: Exploration
The main question: is there an accumulation; if there is, what is the range of its volume
What reduces it: seismic survey, exploration well (Geological risk and chance of success)

Stage: Appraisal (the work after a discovery)
The main question: what are the volume, the flow rates and the connectivity; whether to develop, and by which project
What reduces it: appraisal wells, tests, core samples, detailed seismic survey

Stage: Development
The main question: what the wells and the facilities will cost, and when the oil will flow
What reduces it: project design, contracts, experience of the first wells

Stage: Production
The main question: how the reservoir gives up its oil, how much we will recover
What reduces it: production history, matching the model to the history (History matching)

Stage	The main question	What reduces it
Exploration	is there an accumulation; if there is, what is the range of its volume	seismic survey, exploration well (Geological risk and chance of success)
Appraisal (the work after a discovery)	what are the volume, the flow rates and the connectivity; whether to develop, and by which project	appraisal wells, tests, core samples, detailed seismic survey
Development	what the wells and the facilities will cost, and when the oil will flow	project design, contracts, experience of the first wells
Production	how the reservoir gives up its oil, how much we will recover	production history, matching the model to the history (History matching)

The range narrows from stage to stage, but it does not disappear: even a mature field has uncertainty about its remaining reserves. The estimate does not have to grow, either. At some fields the reserves are revised upwards after production starts, and at others downwards. Faults may divide the accumulation into blocks. Water may break through along thin layers of very high permeability (thief zones). Injection wells may take less water than expected. Or, the other way, the reservoir may turn out better connected and more productive. A revision itself is not a mistake. The mistake is a systematic bias, and it is usually optimistic (checking estimates against the results). The exploration statistics change over time too. In a new play the large traps are found first, so the average size of the discoveries falls as the play is studied. And what counts as a success (traces of oil, a flow at a test, or an economic accumulation) each company defines in its own way.

Optimism has a statistical cause too. The company chooses for development the projects with the best estimates. An estimate looks the best not only because the project is good, but also because its random error happened to be positive. So even unbiased estimates, when they are selected by their size, disappoint on average. In English-language literature this is called the “optimizer's curse”. At auctions for licence areas, where the highest estimate wins, it is called the “winner's curse”. We protect against it by comparing projects by estimates from one method, and by checking past forecasts against the facts.

Appraisal turns a geological success into a commercial one. Its aim is not to add to the reserves but to narrow their range enough to decide, with good reason, whether to develop the accumulation and by which project. How much data to collect depends on how close the project is to the limit of profitability. A large field, profitable at any plausible volume, needs appraisal to design the development correctly. A small one needs it to decide at all whether to take it on. This leads to a paradox: a doubtful project needs data most, and it is the hardest for it to pay for them. The time for appraisal is also limited by the terms of the licence (Licences and contracts for subsoil use). We use a decision tree to find what an appraisal is worth and when it is worth doing (Decision trees and the value of information).

Not only how much but also when: flow rates, timing, costs

Reserves are the sum of production over all the years, and the value depends on when we get that production. The same volume produced faster brings more: the money comes earlier, and we pay the fixed costs for fewer years. So the uncertainty of flow rates is a separate source of risk, even when the reserves are known exactly. A useful measure is discounted production.

Discounted production and the present-volume factor

Q_\text{disc} = \sum_{t} \frac{Q_t}{(1+i)^{t}}, \qquad k_\text{disc} = \frac{Q_\text{disc}}{\sum_t Q_t}

where:

$Q_t$ production of the year that is $t$ years from the time origin, million bbl (or million t)
$i$ discount rate, fraction per year
$t$ time from the time origin to the middle of the production year, years
$Q_\text{disc}$ discounted production: the reserves brought to the time origin (in the lesson on sensitivity it is $PV_Q$ ), million bbl
$k_\text{disc}$ present-volume factor: the share of the reserves that “reaches” the time origin, fraction

Units: production in units of volume (or mass), the rate as a fraction per year

Holds when

the same conventions about the time origin and the moment of the flows as in NPV (the time origin of NPV): here it is the middle of project year 1, and the flows are in the middle of each year, $t = 0, 1, 2, \ldots$
the rate is real if the prices and costs are taken in constant prices

At a constant price, the present value of the revenue is proportional to $Q_\text{disc}$ , not to the reserves. The longer and flatter the profile, the smaller $k_\text{disc}$ : the same oil comes later and is worth less today. The multiplier of the price in the formula of the break-even price is also the discounted production (break-even values).

Worked exampleThe same reserves, but more slowly

Given (assumed data)

Base case: the library's example project: 12 million bbl over 7 years of production, a plateau of 2.4 million bbl a year; NPV at 10% is 100.13 million dollars
Deviation: the wells give no more than 1.6 million bbl a year, a third less than the plateau; the reserves are the same: what we did not manage to produce is produced later
The rest: the price, the costs, the taxes and the rate are the base ones; the fixed part of the operating expenditure is paid in every year of production

Solution

The profile: 1.6 million bbl a year from year 3 to year 9, and 0.8 million bbl in year 10. In all it is 12 million bbl, produced over a year longer than the base case.
Discounted production: the base case 7.90 million bbl ( $k_\text{disc} = 0.658$ ), the slow case 7.42 million bbl ( $k_\text{disc} = 0.618$ ), which is 6.1% less.
NPV by the project model: 83.26 million dollars instead of 100.13, a loss of 16.87 million, or 16.8% less.

Answer: The same 12 million bbl, produced over a year longer, are worth 16.87 million dollars (16.8%) less.

What it means. In per cent, NPV loses almost three times more than discounted production. The revenue has moved to later years, while the capital expenditure stayed, and one more year of fixed operating expenditure was added. This is the same “lever” as in the sensitivity analysis (why NPV changes more than the parameter). The example is still mild: low productivity often reduces the recoverable reserves too, and does not only put off their production. The data in this example are assumed for illustration.

CalculatedProduction profiles of the example project: the base case (12 million bbl); reserves 30% larger with a plant built for the base plateau of 2.4 million bbl a year (the plateau lasts longer); and the same reserves with wells that give no more than 1.6 million bbl a year (production lasts a year longer).

Timing works in the same way. A delay of first oil by a year, with the dates of the capital investment unchanged, costs the example project $22.6 million of NPV (NPV with a delay). The technical reasons for delays are drilling problems, late equipment, the redesign of the project after unexpected data, and a long appraisal.

Cost estimates and schedules are also not symmetric: it is easier to exceed them than to save. While we drill, the tool may get stuck in the hole, the drilling fluid may be lost into the rock, gas may flow into the well, or we may have to wait for the weather. Each of these adds days, but it is possible to drill faster than planned only a little. The first wells in a new area take longer than the later ones. Experience builds up, and the time per well falls, but not smoothly: some wells are far slower than the rest (outliers). Complex facilities, a new technology and a remote location strengthen the same asymmetry. So the distributions of costs and times are skewed, with a long tail towards larger values, and the base case with the most likely values turns out optimistic (why the base case is not the average).

Failures, downtime and accidents

Part of technical uncertainty is not a range but possible events: the failure of a pump or a compressor, plugging with sand, hydrates in a pipeline, corrosion, a leak, a fire. Frequent small failures are allowed for by an average. We put into the production profile an uptime factor, the share of calendar time when the wells and the facilities work (for example, 0.95). Rare large accidents cannot be described by an average. Their probability is small, but their consequences may exceed the cost of the project.

The cost of downtime depends on whether the oil is lost or only deferred. If the plant sets the limit and the reservoir can still give oil, the oil not produced during the downtime will be produced later, so the loss is only in time. If the deferred oil cannot be produced later, the volume is lost. This happens when the licence ends before the reservoir is depleted, or when the downtime damaged the wells.

Worked exampleThree months of downtime: deferred or lost oil

Given (assumed data)

Project: the library's example project; the plant works at its full capacity of 2.4 million bbl a year
Downtime: in project year 4 the plant stands for three months: during the year we produce 1.8 million bbl instead of 2.4

Solution

The oil is deferred: the 0.6 million bbl not produced are produced in the following years, while the plant has spare capacity (years 6 and 7). NPV is 97.27 million dollars, a loss of 2.86 million.
The oil is lost: NPV is 85.71 million dollars, a loss of 14.43 million. A check: from a barrel, after royalty, variable costs and income tax, $(50\cdot0.9 - 5)\cdot0.8 = 32$ dollars remain; $0.6\cdot32/1.1^{3} = 14.43$ million dollars.

Answer: The same downtime costs 2.86 million dollars if the oil is only deferred, and 14.43 million, which is 5.0 times more, if it is lost.

What it means. That is why, when we assess reliability, we ask not only “how often and for how long” but also “what happens to the oil that was not produced”. Spare capacity, spare units and a second way to deliver the product cost money, and we compare them with the expected losses from downtime. A field that delivers oil into someone else's pipeline or to someone else's treatment plant depends on their reliability too: an accident at a neighbour stops everyone who is connected to it. The data in this example are assumed for illustration.

Chains of failures. Large accidents rarely have one cause. Usually several barriers fail one after another: permission to start work, maintenance, the protection automation. So reliability is built into the project and into the organisation of the work, and not only into the economic assessment.
Consequences outside the project. A spill or a blowout leads to clean-up costs, demands from the regulator, lawsuits and loss of trust. Part of such risks is passed to insurers (Risk reduction and diversification).
The human factor. Errors of design, construction and maintenance are a source of uncertainty of the same kind. It is reduced by training, discipline and the use of experience, not by extra data on the reservoir.

Ranges and the estimates P90, P50, P10 in an economic assessment

At the stages of appraisal and design, we usually calculate the economics for three technical cases: low, base and high. Each case is not simply a different volume but its own consistent project: the production profile, the number of wells, the capacity of the facilities, the costs (consistent scenarios). The cases are linked with probabilities in this way. The low case corresponds to the estimate P90 of the main technical parameter (the reserves will be at least this large with a probability of 90%), the base case to P50, and the high case to P10 (how to read P90, P50 and P10).

Do these probabilities carry over to NPV? Only on two conditions: there is really one uncertainty, and NPV changes with it in one direction, for example it grows with the reserves. Then the NPV of the low case will indeed be exceeded with a probability of 90%. But if there are several uncertain parameters, the case “all the parameters at their P90” is not the P90 of the project. It is a much rarer coincidence of failures, and we get the P90 of NPV by simulation (simulation of NPV).

The mean NPV, by which we compare decisions (Expected monetary value and preference theory), does not carry over like this either, unless NPV depends on the parameter linearly: we calculate it from the distribution. When there are only three cases, we weigh them. The usual weights are 0.3, 0.4 and 0.3. This is the so-called Swanson's rule, or the 0.3–0.4–0.3 rule.

The mean from three estimates

\bar x \approx 0.3\,x_{90} + 0.4\,x_{50} + 0.3\,x_{10}

where:

$\bar x$ the mean (the expected value) of the quantity, units of the quantity
$x_{90},\ x_{50},\ x_{10}$ its estimates P90, P50 and P10: the values exceeded with a probability of 90, 50 and 10%, units of the quantity

Units: the units of the quantity itself (million bbl, US$ million …)

Holds when

the three numbers are the estimates P90, P50 and P10 of one distribution
the distribution is not too skewed: for a log-normal distribution with a P10/P90 ratio of 4 the error of the mean is less than 1%, with a ratio of 10 it is about 3.6%, and with a ratio of several tens the rule clearly understates the mean
for NPV the rule is used as an approximate averaging over the distribution of one main parameter: the three cases are its P90, P50 and P10. If NPV changes with this parameter in one direction, the NPVs of the cases are also the P90, P50 and P10 of NPV

Three points with these weights are a crude replacement for the whole distribution. For a normal distribution they reproduce the mean exactly and the variance almost exactly, with an error of about 1.5%. For a symmetric distribution the rule gives a mean equal to P50. For a distribution skewed to the right it gives more than P50, as it should.

Worked exampleThree cases of reserves and the expected NPV

Given (assumed data)

Reserves: P90 8.4 million bbl (0.7 of the base), P50 12, P10 15.6 (1.3 of the base); the distribution is symmetric
Project: the library's example project; the plant and the wells are built for the base plateau of 2.4 million bbl a year. With smaller reserves the profile is proportionally lower, with larger ones the plateau lasts longer; the costs are the same

Solution

NPV at 10%: low case 24.31, base case 100.13, high case 165.68 million dollars. NPV grows with the reserves, so these are also the P90, P50 and P10 of NPV.
Mean NPV by the 0.3–0.4–0.3 rule: $0.3\cdot24.31 + 0.4\cdot100.13 + 0.3\cdot165.68$ $= 97.05$ million dollars.
Mean reserves: $0.3\cdot8.4 + 0.4\cdot12 + 0.3\cdot15.6 = 12$ million bbl. The NPV of the case with the mean reserves is 100.13 million dollars, which is 3.08 million more than the mean NPV.
If the plant did not limit production, the high case would give 175.96 million dollars, and the mean NPV would equal the NPV of the average case: $0.3\cdot24.31 + 0.4\cdot100.13 + 0.3\cdot175.96 = 100.13$ .

Answer: The expected NPV is about 97.05 million dollars, which is 3.08 million less than the NPV of the case with the mean reserves.

What it means. The difference is not a rounding error but a result of the shape of the dependence. With reserves smaller than the base, NPV falls in proportion, and with larger ones it grows more slowly, because the plant lets us produce the extra oil only later. The loss of the low case is not balanced by the gain of the high one. So the rule is: we calculate the expected NPV from the cases or by simulation, not from one case with mean parameters. The example simplifies two things. With larger reserves, the plant could be expanded, and part of the gain would return, so the loss of the mean NPV would be smaller. And in a real project each case has its own cost estimate (see the section on the links between parameters). The data in this example are assumed for illustration.

Why the NPV of the average case is not the average NPV

In the example above, a calculation by the mean reserves overstated the expected NPV. This is a special case of a general rule that in mathematics is called Jensen's inequality, and in practice the “flaw of averages”: the average value of a function equals the function of the average only if the function is linear.

Jensen's inequality

\mathrm{E}[f(X)] \le f(\mathrm{E}[X])

where:

$X$ an uncertain parameter: the reserves, a flow rate, the costs, units of the parameter
$f$ the dependence of NPV on this parameter, other things equal, US$ million
$\mathrm{E}[\cdot]$ the mean over the distribution (the expected value), —

Units: units of NPV

Holds when

$f$ flattens as $X$ grows (it is concave) on the whole range of $X$ : each next increase of $X$ adds less than the previous one; for a function that steepens as $X$ grows (a convex one) the sign of the inequality is reversed
equality holds only for a linear $f$ or when there is no spread

The difference grows with the spread of $X$ and with the curvature of $f$ . For a smooth $f$ and a small spread, $\mathrm{E}[f(X)] \approx f(\bar X) + \tfrac12 f''(\bar X)\,\sigma_X^2$ : the correction is proportional to the variance, and it is negative when $f$ flattens as it rises (is concave).

The direction of the difference depends on what makes the dependence non-linear. In an oil and gas project, two kinds of causes are typical.

Upper limits cut off the gain: The capacity of a plant or a pipeline, quotas, progressive taxes, and production sharing with a growing share of the state. If the outcome is good, the project gets less than it “should” by proportion, and if it is bad, it loses in full. The dependence flattens as it rises, and the mean NPV is lower than the NPV of the average case.
The possibility to walk away cuts off the loss: We may not develop a small discovery, stop production at the economic limit (economic limit), or end a programme after failures. A bad outcome costs less than it would by proportion. The dependence steepens as it rises (it is convex), and the mean result is higher than the result of the average case. This is what we saw in the lesson on EMV, where a small discovery was not developed (discoveries of different sizes).

There is also a third cause. It is not a non-linearity, but it acts in the same way: asymmetric ranges. If a cost estimate is more likely to be exceeded than saved, then even a linear model with the most likely values gives an NPV above the mean (simulation of NPV). In practice these causes add up.

The width of the range of reserves and the loss of the mean NPV, US$ million

Reserves P90 and P10: 0.9 and 1.1 of the base
NPV P90, P50, P10: 74.9, 100.1, 123.4
Mean NPV: 99.54
Below the NPV of the average case by: 0.60

Reserves P90 and P10: 0.8 and 1.2 of the base
NPV P90, P50, P10: 49.6, 100.1, 145.2
Mean NPV: 98.49
Below the NPV of the average case by: 1.64

Reserves P90 and P10: 0.7 and 1.3 of the base
NPV P90, P50, P10: 24.3, 100.1, 165.7
Mean NPV: 97.05
Below the NPV of the average case by: 3.08

Reserves P90 and P10: 0.6 and 1.4 of the base
NPV P90, P50, P10: −1.0, 100.1, 183.6
Mean NPV: 94.86
Below the NPV of the average case by: 5.28

Reserves P90 and P10	NPV P90, P50, P10	Mean NPV	Below the NPV of the average case by
0.9 and 1.1 of the base	74.9, 100.1, 123.4	99.54	0.60
0.8 and 1.2 of the base	49.6, 100.1, 145.2	98.49	1.64
0.7 and 1.3 of the base	24.3, 100.1, 165.7	97.05	3.08
0.6 and 1.4 of the base	−1.0, 100.1, 183.6	94.86	5.28

The 0.3–0.4–0.3 rule; the plant is built for the base plateau; the NPV of the average case is $100.13 million. In the bottom row the NPV of the low case is slightly below zero, and it would not be developed. With zero instead of −1.0, the mean NPV would grow by 0.3 million dollars. The data in this example are assumed for illustration.

CalculatedNPV of the example project against the reserves: with the plant built for the base plateau (solid line) and without a capacity limit (dashed line). The points are the cases P90, P50 and P10. The mean NPV from them ($97.05 million) is below the NPV of the average case by $3.08 million. There is a small step between 16.6 and 16.8 million bbl. The remainder that did not fit into the plateau is produced as a separate year only when it pays for the fixed costs of that year. Until then it is left in the reservoir.

The loss grows faster than the width of the range: doubling the range from ±10% to ±20% makes it 2.8 times larger, and from ±20% to ±40% it makes it 3.2 times larger. The wider the uncertainty, the more a calculation by averages misleads.

Links between parameters

Technical parameters are rarely independent, because they have common causes. If an accumulation is larger because its area is larger, we need more wells to drain it, so both the reserves and the capital expenditure grow. If it is larger because the rock is better, the reserves and the flow rates grow, and we need no more wells. A cost overrun usually comes together with a delay of start-up. The porosity and the water saturation of one reservoir are linked through the size of the pores (dependence between parameters).

The strength of a link is measured by the correlation coefficient $\rho$ , from −1 to +1. It is easiest to see how a link acts on NPV when NPV depends on two parameters almost linearly.

The spread of NPV with two linked parameters

\sigma^2_\text{NPV} = a^2\sigma_X^2 + b^2\sigma_Y^2 - 2ab\,\rho\,\sigma_X\sigma_Y

where:

$\sigma_\text{NPV}$ standard deviation of NPV (a measure of its spread), US$ million
$X,\ Y$ uncertain parameters; here, the fractions of the base production and of the base capital expenditure, units of the parameters
$a$ how much NPV grows when $X$ grows by one unit, US$ million per unit of $X$
$b$ how much NPV falls when $Y$ grows by one unit, US$ million per unit of $Y$
$\sigma_X,\ \sigma_Y$ standard deviations of the parameters, units of the parameters
$\rho$ correlation coefficient between $X$ and $Y$ , —

Units: $\sigma_\text{NPV}$ is in US$ million; $a\sigma_X$ and $b\sigma_Y$ are also in US$ million

Holds when

within the ranges, NPV ≈ $\mathrm{NPV}_0 + a\,\Delta X - b\,\Delta Y$ , a linear approximation
$X$ raises NPV and $Y$ lowers it, hence the minus before the last term; if both act in the same direction, the sign is plus
$\rho$ is the linear correlation; a link set through ranks or normal values gives a number close to it

A positive link between parameters that act on NPV in opposite directions reduces the spread: the good and the bad deviations partly compensate each other. The same link between parameters that act in one direction increases the spread. In a linear model the link does not affect the mean NPV. It changes only the width of the distribution, and so the P90 and P10.

Worked exampleLarger reserves, more wells

Given (assumed data)

Uncertain parameters: the fraction of the base production $X$ and the fraction of the base capital expenditure $Y$ ; the triangular distributions from the lesson on sensitivity: $X$ from 0.8 to 1.15, $Y$ from 0.9 to 1.3, and the most likely value of both is 1; the rest is at the base
Link: the reserves are larger because of a larger area, and we need more wells for them: correlation $\rho = 0.7$
Slopes of the model: $a = 252.758$ million dollars per unit of the production fraction, $b = 126.292$ million dollars per unit of the cost fraction

Solution

Standard deviations of the triangular distributions: $\sigma_X = 0.071686$ , $\sigma_Y = 0.084984$ . The spread of NPV from each parameter taken alone: $a\sigma_X = 18.119$ and $b\sigma_Y = 10.733$ million dollars.
Independent parameters: $18.119^2 + 10.733^2 = 443.5$ , $\sigma_\text{NPV} = 21.1$ million dollars. This is less than the sum 28.9, because the deviations partly cancel each other.
With a correlation of 0.7: the correction is $2\cdot0.7\cdot18.119\cdot10.733 = 272.3$ ; $\sigma^2_\text{NPV} = 443.5 - 272.3 = 171.2$ , $\sigma_\text{NPV} = 13.1$ million dollars.
A check by simulation of the full model (10,000 trials, the link set through normal values): without a link, the standard deviation is 21.0 million dollars, P90 = 59.8, P10 = 114.9. With a link (actual correlation 0.68): 13.2, P90 = 69.8, P10 = 104.5 million dollars. The mean in both cases is about 88 million dollars.

Answer: The link “larger reserves, more wells” narrows the spread of NPV from 21.1 to 13.1 million dollars. P90 rises by 10 million, P10 falls by 10 million, and the mean hardly changes.

What it means. An independent draw would exaggerate the risk here. The combinations “large reserves and cheap development” and “small reserves and expensive development” are in fact rare, because the number of wells follows the size of the accumulation. With the opposite cause (the reserves are larger because the rock is better, and fewer wells are needed per barrel) the link would be negative, and the spread would grow to 26.8 million dollars. So we take the sign of the link from its physical cause, and its strength from analogues or from consistent development cases. The mean NPV of 88 million dollars is below the base 100.13 for another reason: the asymmetric ranges, as in the lesson on sensitivity. The data in this example are assumed for illustration.

How appraisal work narrows the range

Each kind of data clarifies its own part of the uncertainty. An appraisal well that reaches the oil–water contact away from the first well clarifies the rock volume. Core samples (rock brought up from the well) and well logs (measurements made along the well) clarify the porosity and the saturation. Well tests clarify the productivity and the connectivity of the reservoir (Well testing: purposes and types of test). A seismic survey clarifies the shape of the trap and the faults. The recovery factor is truly clarified only by production: the experience of an extended well test and the history of development.

We can represent the reserves as a product of factors (the mean and spread of a product). Data that clarify one factor remove its share from the spread, but do not touch the other factors: the relative uncertainty after them is the uncertainty of the remaining factors.

Worked exampleAn appraisal well clarifies the rock volume

Given (assumed data)

Before the well: the accumulation from the lesson on the Monte Carlo method: oil in place P90 = 11.6, P50 = 13.8, P10 = 16.2 million m³, mean 13.83; recoverable oil P90 = 3.28, P10 = 5.08 million m³
The well: it will reach the contact and clarify the rock volume to ±5%; the other parameters and the recovery factor keep their earlier distributions
Possible results: the rock volume 159, 183 or 209 million m³: the estimates P90, P50 and P10 of the earlier distribution

Solution

Oil in place after the well (10,000 trials for each result): P90–P10 from 10.8 to 13.2, from 12.4 to 15.2, or from 14.2 to 17.4 million m³. The P10/P90 ratio is 1.22 instead of 1.39. The difference P10 − P90 for the middle result is 2.77 million m³ instead of 4.55, which is 39% less.
Recoverable oil: the P10/P90 ratio is 1.42 instead of 1.55, and the difference P10 − P90 is 1.44 million m³ instead of 1.80, only 20% less. The well did not clarify the recovery factor.
The mean oil in place after the well: 12.00, 13.78 or 15.76 million m³. With the weights 0.3, 0.4 and 0.3 they give $0.3\cdot12.00 + 0.4\cdot13.78 + 0.3\cdot15.76 = 13.84$ , almost the same as the mean before the well, 13.83 million m³.

Answer: The well narrows the range of oil in place by 39% and that of recoverable oil by 20%, and it does not change the mean expected estimate.

What it means. Which way the estimate will move is not known in advance: the result “less than expected” is as possible as “more”. The mean estimate before the well is the average of the possible estimates after it. To expect an “increase of reserves” from data means to admit that the earlier estimate was too low. The value of the well is that after it, the decisions on development (the number of wells, the capacity of the plant, the development itself) are taken for a narrower range (Decision trees and the value of information). The uncertainty of recovery remains the main one, and only production reduces it. The weights 0.3, 0.4 and 0.3 are an approximation: the means are exactly equal when we average over all the possible results of the well. The data in this example are assumed for illustration.

CalculatedOil in place before the appraisal well (dashed curve) and after it, for three possible results: a rock volume of 159, 183 and 209 million m³. Each curve is the probability that the oil in place is not less than a given value.

From this come practical rules of appraisal. Collect data on the parameter that gives the main share of the uncertainty and can change the decision. Before drilling, ask what will change in the project for each possible result. Remember that the time for appraisal also costs money: a delay puts off all the outcomes of the project (when information is worth buying).

Terms and symbols

Uncertainty and risk: Uncertainty is inexact knowledge of a quantity. Risk is the possible loss (the downside) and gain (the upside) that come from it.
Discounted production $Q_\text{disc}$: The sum of the production of the years multiplied by the discount factors. $k_\text{disc}$ is its share of the reserves.
Cases P90, P50, P10: The low, base and high technical cases of a project. They correspond to the estimates of the main parameter with a probability of being exceeded of 90, 50 and 10%.
The 0.3–0.4–0.3 rule: An approximate mean from the estimates P90, P50 and P10 (Swanson's rule).
Jensen's inequality: For a dependence that flattens as it rises (a concave one), the average value of the function is not larger than the function of the average. For a dependence that steepens as it rises (a convex one), it is not smaller.
Correlation coefficient $\rho$: A measure of the linear link between two quantities, from −1 to +1.
Uptime factor: The share of calendar time when the wells and the facilities work.
Appraisal: The work between a discovery and the decision on development that narrows the range of the volumes and properties of the accumulation.

Common mistakes and misconceptions

Wrong

“The NPV calculated by the mean reserves is the expected NPV of the project.”

In fact

Only if NPV depends on the reserves linearly. In this lesson's example the plant built for the base plateau cuts off the gain of larger reserves, and the expected NPV is $97.05 million, against $100.13 million for the case with the mean reserves. The possibility to walk away acts in the opposite direction.

Wrong

“The low case of a project is all the parameters at their P90.”

In fact

With several independent parameters, all of them landing at P90 at the same time happens much more rarely than in 10% of cases. Such a case is a stress test, not the P90 of the project. We get the P90 of NPV with several uncertainties by simulation.

Wrong

“The reserves are the same, so low flow rates only stretch production out and will hardly affect the value.”

In fact

In this lesson's example, wells that give a third less than the planned plateau produce the same 12 million bbl over a year longer, and NPV falls by 17%.

Wrong

“If the link between the parameters is not known, it is safer to take them as independent.”

In fact

Independence is also an assumption, and we can be wrong in both directions: for parameters with a common cause, it understates or overstates the spread, depending on the sign of the link. We find the sign from the physical cause, and the strength from analogues.

Wrong

“An appraisal well is needed to add to the reserves.”

In fact

Its task is to narrow the range. The estimate after it can grow or fall; on average it equals the earlier one. The value of the well lies in the decisions that are taken after it for a narrower range.

In depth: advanced detail

How much the curvature takes away: the calculation from three cases

For three symmetric cases we can find the loss of the mean NPV without trial and error. It is proportional to the “second difference”: how much the NPV of the middle case is above the midpoint between the extreme ones.

The loss of the mean NPV from three symmetric cases

\mathrm{NPV}_{50} - \overline{\mathrm{NPV}} = 0.3\,(2\,\mathrm{NPV}_{50} - \mathrm{NPV}_{90} - \mathrm{NPV}_{10})

where:

$\mathrm{NPV}_{90},\ \mathrm{NPV}_{50},\ \mathrm{NPV}_{10}$ NPV of the cases P90, P50 and P10 of the parameter (the index is the case, not the discount rate), US$ million
$\overline{\mathrm{NPV}}$ the mean NPV by the 0.3–0.4–0.3 rule, US$ million

Units: US$ million

Holds when

the cases P90 and P10 are at the same distance from P50, and the middle case is P50
the weights are 0.3, 0.4 and 0.3

It follows directly from the rule: $\mathrm{NPV}_{50} - (0.3\,\mathrm{NPV}_{90} + 0.4\,\mathrm{NPV}_{50} + 0.3\,\mathrm{NPV}_{10})$ . The bracket on the right is the second difference with the opposite sign, a finite-difference form (made from the differences between the three points) of $-f''\Delta^2$ , where $\Delta$ is the step between the cases. The variance of the three points is $0.6\,\Delta^2$ , so the loss is $-\tfrac12 f''\sigma^2$ , which is the correction from Jensen's inequality with the opposite sign. For a linear dependence the bracket is zero.

Worked exampleA check on the example with the plant

Given (assumed data)

NPV of the cases: 24.306, 100.134 and 165.680 million dollars

Solution

$0.3\cdot(2\cdot100.134 - 24.306 - 165.680) = 0.3\cdot10.28 = 3.08$ million dollars.

Answer: 3.08 million dollars: the same loss as in the main example.

What it means. The sign of the bracket shows the direction at once. A positive one means that the dependence flattens as it rises (it is concave) and the mean NPV is below the NPV of the average case. A negative one means that it steepens as it rises (it is convex), and the reverse holds. The data in this example are assumed for illustration.

Where the formula for the spread of two parameters comes from

If $\mathrm{NPV} \approx \mathrm{NPV}_0 + a\,\Delta X - b\,\Delta Y$ , the deviation of NPV from its mean is $a\,\delta X - b\,\delta Y$ , where $\delta$ are the deviations of the parameters from their means. Squaring and averaging, we get $a^2\sigma_X^2 + b^2\sigma_Y^2 - 2ab\,\mathrm{cov}(X, Y)$ . The covariance is the mean of the product of the deviations, $\mathrm{cov}(X, Y) = \rho\,\sigma_X\sigma_Y$ , and this gives the formula of the lesson.

The spread of NPV with several parameters (linear approximation)

\sigma^2_\text{NPV} = \sum_{j}\sum_{k} s_j\,s_k\,\rho_{jk}\,\sigma_j\,\sigma_k

where:

$s_j$ slope: the change of NPV when the $j$ -th parameter grows by one unit (with its own sign), US$ million per unit of the parameter
$\sigma_j$ standard deviation of the $j$ -th parameter, units of the parameter
$\rho_{jk}$ correlation of the $j$ -th and $k$ -th parameters; $\rho_{jj} = 1$ , —

Units: (US$ million)²

Holds when

NPV depends on the parameters linearly within their ranges
for strongly non-linear models and skewed distributions the result is approximate, and then we calculate by simulation

With independent parameters, the squares of the contributions $s_j\sigma_j$ add up. The spread grows more slowly than the sum of the contributions, and the largest contributions decide everything. The contributions $s_j\sigma_j$ are the halves of the bars of the tornado chart, only with a range of one standard deviation.

In a simulation we set a link not by the values themselves but by their normal values. Two random numbers are turned into normal variables $z_1$ and $z_2$ , the second is mixed with the first as $\rho z_1 + \sqrt{1 - \rho^2}\,z_2$ , and the result is turned back into probabilities (how partial dependence is set). The correlation of the triangular variables themselves comes out a little smaller than the one that was set: in the example of the lesson, about 0.68 at 0.7.

What “narrowing the range” means: the law of total variance

The uncertainty before the data is made of two parts: what the data will clarify, and what will remain. This is the law of total variance.

The law of total variance

\mathrm{Var}(X) = \mathrm{E}\left[\mathrm{Var}(X\mid D)\right] + \mathrm{Var}\left(\mathrm{E}[X\mid D]\right)

where:

$X$ the quantity being assessed: the reserves, million m³
$D$ the result of the data, for example the measured rock volume, —
$\mathrm{Var}(X\mid D),\ \mathrm{E}[X\mid D]$ variance and mean of the reserves after the data $D$ , (million m³)², million m³

Units: (million m³)²

Holds when

the probabilities of the results $D$ are those that follow from the distributions before the data.

The first term is the uncertainty that will remain, on average, after the data. The second is the spread of the future estimates themselves: how far the result of the data will shift the mean. The data move variance from the first term to the second, but the sum stays the same. The uncertainty does not disappear but turns into knowledge of which part of the range we are in. And the mean of the future estimates equals the present estimate: $\mathrm{E}[\mathrm{E}[X\mid D]] = \mathrm{E}[X]$ .

For a product of independent factors, the answer comes without simulation. The relative spread of the reserves, the coefficient of variation $C_v$ , is linked with the spread of the factors in this way: $1 + C_v^2 = \prod_j (1 + C_{v,j}^2)$ . If the data have measured one factor, its multiplier leaves the product.

Worked exampleCoefficients of variation before and after the well

Given (assumed data)

Factors: $C_v$ of the rock volume 0.100; of the other factors of the oil in place 0.075; of the recovery factor 0.109
After the well: the rock volume is known to ±5%: the $C_v$ of the triangular distribution is $0.05/\sqrt6 = 0.020$

Solution

Oil in place: $C_v$ before the well 0.125, after it 0.078; by simulation, for the middle result of the well, 0.078.
Recoverable oil: $C_v$ before the well 0.167, after it 0.134. The recovery factor, with a $C_v$ of 0.109, is now the main source of spread.

Answer: The well lowers the

C_v

of the oil in place from 0.125 to 0.078, and that of the recoverable oil only from 0.167 to 0.134.

What it means. The calculation without simulation agrees with the simulation, and it shows in advance what an appraisal job will give. Before drilling, it is useful to ask what share of the spread it can remove at all. The data in this example are assumed for illustration.

Connections to other topics

Probabilistic volumetrics by Monte Carlo simulation — the distribution of volumes, the estimates P90, P50, P10 and the dependence of parameters.
Sensitivity analysis and scenarios — sensitivity, scenarios, simulation of NPV and asymmetric ranges.
Expected monetary value and preference theory — the expected value of decisions, and the upward curve (convexity) that the possibility to walk away creates.
Decision trees and the value of information — the value of appraisal work and of sequential decisions.
Geological risk and chance of success — the risk of an empty trap, and checking estimates against the results of drilling.
Cash flow model of an oil and gas project — the model through which uncertainty turns into a cash flow.
Market, financial and regulatory uncertainty — the uncertainty of prices, exchange rates, taxes and laws.
Risk reduction and diversification — how companies reduce risk: partnership, insurance, the portfolio.

Check yourself

1
The reserve cases P90, P50 and P10 give NPVs of 20, 60 and 130 million dollars; NPV grows with the reserves, and the other uncertainties are small. What is the expected NPV by the 0.3–0.4–0.3 rule?
1. A.70 million dollars, the average of the three numbers
2. B.75 million dollars, the midpoint between P90 and P10
3. C.69 million dollars
4. D.60 million dollars, the NPV of the P50 case
Show the answer
Correct answer: C. 69 million dollars
$0.3\cdot20 + 0.4\cdot60 + 0.3\cdot130 = 6 + 24 + 39 = 69$ million dollars. It is larger than the NPV of the P50 case: the distribution of NPV is skewed to the right, and the high case is farther from P50 than the low one.
2
Production is limited by the capacity of a plant built for the base case, and the reserves are, with equal chances, 25% larger or 25% smaller than the base. How do the expected NPV and the NPV of the base case compare?
1. A.They cannot be compared without simulation
2. B.The expected NPV is smaller
3. C.The expected NPV is larger
4. D.They are equal: the reserves are on average the base ones
Show the answer
Correct answer: B. The expected NPV is smaller
The dependence of NPV on the reserves flattens as the reserves grow (it is concave). With smaller reserves NPV falls in proportion. With larger ones it grows more slowly, because the plant lets us produce the extra oil only later. By Jensen's inequality, the mean NPV is smaller than the NPV at the mean reserves.
3
Why does the possibility of not developing a small discovery make the expected value of a discovery, averaged over its possible sizes, larger than the value of a discovery of the average size?
Show the answer
The possibility to walk away cuts off the loss: a small discovery is worth not the negative NPV of its development but zero. The dependence of value on size becomes convex (it steepens as the size grows), and by Jensen's inequality the average over the sizes is larger than the value of the average size. A calculation by one average discovery loses this possibility and understates the expected value.
4
The spread of NPV from the uncertainty of the reserves is 30 million dollars, and from the uncertainty of the capital expenditure it is 20 million (each taken alone, as standard deviations). The reserves and the costs are linked by a correlation of +0.5: larger reserves, more wells. What is the spread of NPV?
1. A.36.1 million dollars
2. B.26.5 million dollars
3. C.50 million dollars
4. D.10 million dollars
Show the answer
Correct answer: B. 26.5 million dollars
The reserves raise NPV and the costs lower it, so $\sigma^2 = 30^2 + 20^2 - 2\cdot0.5\cdot30\cdot20 = 900 + 400 - 600 = 700$ , $\sigma \approx 26.5$ million dollars. Without a link it would be $\sqrt{1300} \approx 36.1$ , and with a full link, $30 - 20 = 10$ . The plain sum of 50 is right only for a full link of parameters that act in one direction.
5
The same reserves will be produced with the same profile but two years later, because of a delay in development. By what factor will discounted production fall at a rate of 10%, and why will NPV fall more?
Show the answer
All the production years move by two years, and the discounted production is multiplied by $1/1.1^2 \approx 0.83$ , which is a fall of 17%. NPV falls more because the capital expenditure made before the delay does not move: only the arrival of the money is put off, and the investments stay in place (NPV with a delay of first oil).
6
After an appraisal well, the mean estimate of the reserves fell by 10%. Does this mean that the well was drilled for nothing?
Show the answer
No. It was not known in advance which way the estimate would move; on average it does not change, and a fall is as possible as a rise. The well narrowed the range, and so made it possible to fit the project to the real accumulation, for example not to build a plant for reserves that do not exist. The value of data is assessed before getting them, over all the possible results, not by the one result received.

Updated: 7 October 2026