chapter 4 sp 2014 economic uncertainties part 2
TRANSCRIPT
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PE 3003/6413
PE 3003/6413 Spring 2014
Chapter 4
Part 2
Econom ic Uncertaint ies
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Outline
1. Range Analysis
2. Probability
3. Cumulative Probability and Density Functions
4. Expected Value
5. Decision Three Analysis6. Distribution Functions
7. Monte Carlo Simulation
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Distribution Functions
Density functions are commonly used in engineering
to describe physical systems
Density distribution of a loading in a beam
For any point x along the beam , the density c an bedescr ibed b y a funct ion (in grams /cm )
The total loading between points a and b is
determ ined as the integral of the density fu nc t ion
f rom a to b
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Distribution Functions
Common Density Functions
Uniform
Triangular Distribution
Normal Distribution
Log-Normal Distribution
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Uniform Distribution
A continuous random variable X with p robabi li ty densi ty
funct ion
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Uniform Distribution
Let the continuous random variable X denote the porosity
measured in core samples. Assume that the range of X is [0,
20 %], and assume that the probability density function of Xis
f(X)=0.05, 0X20
1. What is the probability that a measurement of porosity isbetween 5 and 10 %?
2. What is the mean and variance of the core samples?
E(X)= 10% and Var(X)=33.33 %
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Uniform Distribution
Random Number
A set of numbers that have nothing to do with the other
numbers in the sequence
Random numbers have no defined sequence or formulation.
Thus, for any n random numbers, each appears with equalprobability
All computer languages and spread sheet has a random
function that return a random number uniformly distributed
between 0 and 1
a b
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Uniform Distribution
VBA Random number initialization
Const Seed As Double = 1000
' Random numbers Seed set
Dim RndVal As Double
Randomize (Seed)RndVal = Rnd(-Seed)
' After this command random numbers can be generated as
follows
RndVal = Rnd()
We will use this in Monte Carlo section
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Uniform Distribution
Random number Generation between aand b
VBA
Function RandAB(a As Double, b As Double) As Double
RandAB = (b - a) * Rnd() + a
End Function
Excel
=rand()*(b-a)+a
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Uniform Distribution
Strengths
Easy to create
Easy to understand
Simple implementation and calculation
Weakness Almost always inappropriate, except as random number
generation
Use cautiously with low iterations (Monte Carlo)
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Triangular Distribution
a bc
f(x)
x
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Triangular Distribution
Cumulative Distribution
bx
bxccbab
xb
bxaacab
ax
xax
PxF
1
1
0
)(
2
2
1. a is the minimum outcome of the random
variable.
2. cis the mode (most likely) of the random
variable.3. bis the maximum outcome of the random
variable.
4. F(x) is the cumulative probability of all
possible outcomes of the random variable
between aandx.
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Triangular Distribution
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Triangular Distribution
Random numbers generation
generate a random number (P) uniformly distributed between
0 to 1 (refer to uniform distribution random numbers)
random number is calculated using the inverse of the
triangular distribution as follows:
See VBA template on WebCT
1
11
0
0
Pb
Pab
accbabPb
ab
acP
aacabP
Pa
x
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Triangular Distribution
Strengths
Easy to create
Easy to understand
Simple implementation
Weakness Likeliestoften confuse with mean or P50 and very difficult to
defend
End points are sometimes difficult to estimate
Poor approximation of Log-Normal
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Normal Distribution
The most widely used model for the distribution of a
random variable
Whenever a random experiment is replicated, the random variable
that equals the average (or total) result over the replicates tends
to have a normal distribution as the number of replicates becomes
large.
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Normal Distribution
Probability density function
Mean and Variance
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Normal Distribution
Let the continuous random variable X denote the porosity
measured in core samples. Assume that X is normally
distributed with 10% mean and 4% variance
1. What is the probability that a measurement of porosity
exceeds 13%?
Excel => =1-NORM.DIST(13,10,SQRT(4),TRUE) = 0.06681
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Normal Distribution
Probabilities associated with a normal distribution
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Normal Distribution
Random numbers generation
generate a random number (P) uniformly distributed between
0 to 1 (refer to uniform distribution random numbers)
random number is calculated using the inverse of the normal
distribution as follows:
Excel
=NORM.INV(RAND(), Mean,Sigma) where p=rand()
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Normal Distribution
Strengths
Easy to create
Easy to understand
Simple implementation
Weakness Not really true nature at the ends
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Log-Normal Distribution
Variables in a system sometimes follow an exponential
relationship as x=exp(w)
If the exponent is a random variable, say W, X=exp(W)is a random
variable
An important special case occurs when W has a normal
distribution. In that case, the distribution of X is called alognormal distribution.
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Log-Normal Distribution
Probability density function
Mean and Variance
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Log-Normal Distribution
Random numbers generation
generate a random number (P) uniformly distributed between
0 to 1 (refer to uniform distribution random numbers)
random number is calculated using the inverse of the normal
distribution as follows:
Excel
=LOGNORM.INV(rand(),,)
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Log-Normal Distribution
Strengths
Easy to create
Easy to understand
Logical starting points for many inputs
Extremely common in nature Weakness
Not really true nature at the ends
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Binomial Distribution
Bernoulli Trial or Experiment:
Flip a coin 10 times. Let X=number of heads obtained.
A worn machine tool produces 1% defective parts. Let X
=number of defective parts in the next 25 parts
produced.
The random variable in each case is a count of the
number of trials that meet a specified criterion
Assumed that the trials that constitute the random
experiment are independent
probability of a success in each trial is constant
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Binomial Distribution
A random experiment consists of n Bernoulli trialssuch that
1. The trials are independent
2. Each trial results in only two possible outcomes,
labeled as success and failure3. The probability of a success in each trial, denoted as p,
remains constant
The random variable Xthat equals the number of
success has a binomial random variable with
parameters n(number of trials or experiments) andsuccess probability p
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Binomial Distribution
We are drilling 5 wildcats in a new basin where thechance of a discovery (wildcat success ratio) is 0.15
on each well. Assuming each well is a Bernoulli trial,
what is the probability of only one discovery in the
fives wells drilled?
n=5
p=0.15
S=1
P(S=1)=0.3915
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Outline
1. Range Analysis
2. Probability
3. Cumulative Probability and Density Functions
4. Expected Value
5. Decision Three Analysis
6. Distribution Functions
7. Monte Carlo Simulation
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Gas Metering Problem
wq Oq
Gq GP GT
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Gas Metering Problem
The gas flow rate is measured a certain pressure andtemperature
Gas flow rate need to be reported at standard
conditions (Deterministic Model)
)67.459(0283.0)(
G
GGSCG
Tz
Pqq
qG(SC): gas flow rate at standard conditions [MSCF/D]qG: gas flow rate at in-situ conditions [MCF/D]
PG: gas pressure [psia]
TG: gas temperature [F]
z: gas compressibility factor [-]. For simplicity z=0.9
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Gas Metering Problem
For a given time
][19.203
)67.459120(9.00283.0
30010)( MSCF/D
SCGq
qG: 10 [MCF/D]
PG: 300 [psia]
TG: 120 [F]z: 0.9 (for simplicity)
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Gas Metering Problem
Unfortunately qG, PGand TGhas associateduncertainties
0
0.1
0.2
0.3
0.4
0.5
0 10 20
f(qG
)
qG[MCFD/D]
0
0.02
0.04
0.06
0.080.1
0.12
0.14
280 300 320
f(PG
)
PG[psia]
0
0.05
0.1
0.150.2
0.25
100 120 140
f(TG
)
TG[F]
qGis normally distributed
with mean of 10MCFD
and 1 MCFD of standard
deviation
PGis normally
distributed with mean of
300 psia and 3 psia of
standard deviation
TGis normally
distributed with
mean of 120 F and
2 F of standard
deviation
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Gas Metering Problem
Uncertainties need to be propagated to get theuncertainty in qG,
)67.459(0283.0)(
G
GGSCG
Tz
Pqq
0
0.1
0.2
0.3
0.4
0.5
0 10 20
f(qG
)
qG[MCFD/D]
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
280 300 320
f(PG
)
PG[psia]
0
0.05
0.1
0.15
0.2
0.25
100 120 140
f(TG
)
TG[F]
0
0.1
0.2
0.3
0.4
0.5
0 10 20
f(qG
)
qG[MCFD/D]
Monte Carlo is used to
propagates this uncertainties
Though the Deterministic
Model
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Gas Metering Problem
Assume qG, PGand TGare not correlated
Assume
qGis no rmal ly dis tr ibuted w ith mean of 10MCFD and 1
MCFD of s tandard deviat ion
PGis normally dis tr ibuted with mean of 300 psia and 3 psia
of standard deviat ion
TGis norm all y d is tr ib u ted w ith mean of 120 F and 2 F o f
standard deviat ion
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Gas Metering Problem
Follow the steps below:
1. Calculate the average qG(SC) using the average values for qG,PGand TG
2. Generate a random number following qGdistribution
3. Generate a random number following PGdistribution4. Generate a random number following TGdistribution
5. Calculate qG(SC)using the deterministic model
6. Repeat from steps 2 to 5 until maximum number of iterations is
reached or the average value of all calculated qG(SC)is close
enough to the qG(SC)
calculated in step 1
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Gas Metering Problem
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Gas Metering Problem
192
194
196
198
200
202
204
0 200 400 600 800 1000 1200
Average
Iterations
Convergence Plot
Simulation Deterministic Model
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Gas Metering Problem
Histogram
Min 141.0443
Max 271.6897
N of classes 10
Step 13.06453288
Left Right Interval Freq Freq [frac]
141.04 154.10 [141.04-154.1] 6 0.006006006
154.10 167.16 (154.1-167.16) 22 0.022022022
167.16 180.22 (167.16-180.22) 102 0.102102102
180.22 193.28 (180.22-193.28) 185 0.185185185
193.28 206.34 (193.28-206.34) 269 0.269269269206.34 219.40 (206.34-219.4) 222 0.222222222
219.40 232.46 (219.4-232.46) 128 0.128128128
232.46 245.52 (232.46-245.52) 52 0.052052052
245.52 258.58 (245.52-258.58) 11 0.011011011
258.58 271.64 (258.58-271.64) 2 0.002002002
Total 999 1
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Gas Metering Problem
Histogram
0
0.05
0.1
0.15
0.2
0.25
0.3
Histogram
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Gas Metering Problem
Average and Standard Deviation and Normal
Approximation
Average 203.28
St Deviation 20.38
Pessimistic Mean Optimistic162.53 203.28 244.03
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0.00%
10.00%
20.00%
30.00%
40.00%
50.00%
60.00%
70.00%
80.00%
90.00%
100.00%
140 160 180 200 220 240 260
F(qG(SC)
qG(SC) [MSCFD]
Gas Metering Problem
Cumulative Freq DistributionPercentile qG(SC)
2.50% 161.89
5.00% 169.61
10.00% 176.58
15.00% 182.73
20.00% 187.41
25.00% 190.97
30.00% 193.8035.00% 196.52
40.00% 200.47
45.00% 202.97
50.00% 205.85
55.00% 208.28
60.00% 210.93
65.00% 213.48
70.00% 216.58
75.00% 219.57
80.00% 223.08
85.00% 227.71
90.00% 232.09
95.00% 239.09
97.50% 247.22
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Monte Carlo Process
1. Define all the variables
Identify all the measure variables of interest (i.e. NPV, oil price,
etc.) or all variables that affect the evaluation
2. Define the variables relationships in the deterministic
projection model Equations or other numerical calculations
3. Sort the input variables into two groups
Variables that we known with certain (represented by single
point value)
Significant unknowns will be represent by random variables(exact value can not be specified)
4. Define random variable distributions
Usually, the best available expert is assigned to judge each
random variable
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Monte Carlo Process
5. Perform the respective simulation trials1. Calculate the deterministic average using the average values all
random variables into the deterministic equation
2. Generate a random number following the distribution of each
random variable
3. Calculate the outcome using the deterministic model4. Repeat from steps 2 to 5 until maximum number of iterations is
reached or the average value of all calculated values in step 3 is
close enough to the calculated in step 1
6. Calculate
Plot the histogram Expected and standard deviation value of the outcomes
Plot the cumulative frequency distribution
P90
P50
P10
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P90, P50 and P10
Other disciplines
Mao-Jones, J. (2012). Decision & Risk Analysis. Merrick & Company Technical Paper.
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P90, P50 and P10
Other disciplines
Holm, S. Suleymanov, V. Vanvik, N.
(2011). Shtokman flow assurance
challenges a
Systematic approach to analyzeuncertaintiesPart 2.15thMultiphase
Production Technology Conference,
Canes (2012)
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P90, P50 and P10
Other disciplines
Holm, S. Suleymanov, V. Vanvik, N.
(2011). Shtokman flow assurance
challenges a
Systematic approach to analyzeuncertaintiesPart 2.15thMultiphase
Production Technology Conference,
Canes (2012)
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P90, P50 and P10
Other disciplines
P10, P50 and P90 are defined as the percentile of the
outcome
This classical definition will apply if you are dealingwith
1. Facility engineers or any other engineer who is
designing an equipment
2. Flow assurance engineers
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P90, P50 and P10
Petroleum Resources Management System (PRMS)
(Society of Petroleum Engineers 2007)
f
Reserves [MMBBL or BSCF]
1P
10
th
2P
50th
3P
90th
Proved Provable Possible
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P90, P50 and P10
Petroleum Resources Management System (PRMS)
(Society of Petroleum Engineers 2007)
Low Est imate:
Conservative estimate of the quantity that will
actually be recovered from the accumulation bya project
There should be at least 90% probability (P90)
that the quantities actually recovered will equal
or exceed the low estimate
This is the definition that will be used in this
course owing to we are working directely with
the reserves
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P90, P50 and P10
Reserves [MMBBL or BSCF]1P
10th
Should be at least a 90%
probability that the quantities
actually recovered will equal or
exceed the estimate
P90
Low Estimate
This is the definition that will be used in this course
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P90, P50 and P10
Reserves [MMBBL or BSCF]
2P
P50
there should be at least a 50%
probability that the actual
quantities recovered will equal or
exceed the 2P estimate
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P90, P50 and P10
Petroleum Resources Management System (PRMS)
(Society of Petroleum Engineers 2007)
High Estimate:
Optimistic estimate of the quantity that will
actually be recovered from an accumulation by aproject
There should be at least a 10% probability (P10)
that the quantities actually recovered will equal
or exceed the high estimate
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P90, P50 and P10
Reserves [MMBBL or BSCF]3P
90th
there should be at least
a 10% probability that
the actual quantities
recovered will equal
or exceed the 3P
estimate.
P10This is the definition that will be used in this course
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P90 P50
P90, P50 and P10
Summary: we will use P10= 10% sure that the value (i.e. Reserve or NPV) is
higher than P10
P90= 90% sure that the value (i.e. Reserve or NPV) is
higher than P90
P10
NPV[MM$]
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Example 1
The production of a new well can be predicted using decline curve analysisand historical data from nearby reservoirs. The hyperbolic decline curve is
given by:
bi iDibqq1
0 1
Where
q= Oil flow rate [BPD]
q0= Initial oil flow rate (time zero) [BPD]Di= Initial Nominal Decline [Fraction / year]
b= Hyperbolic Exponent factor (some authors use the term n)
i = integervalue of the time [year].
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Example 1
The net revenue of the well in a particular yearican be estimated by:
DTOATvenueReNet iii 1)(
1. A(i) is the gross revenue in year igiven by
Wc
WcWaterCosticePrGasGORicePrOildaysqA ii
1365
Wcis the water cut
2. O(i)is the operational cost in yearI
3. D= is the depreciation in year i.
4
CapexD
where Capexis the initial investment and 4 years is the taxable life of the
well; Tis the income tax rate
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Example 1
Finally, the net present value of this well can be calculated as follows:
Capex
MROR
venueReNetesentValuePrNet
ii
i
1
)(
1
where MRORis the minimum rate of return. The well will be closed when the Net
Revenue becomes negative (NetRevenue(i) < 0). Thus the life of the well corresponds
to the year before the net revenue becomes less than zero. In other words the
summation will be carried out until Net Revenue becomes negative
Using this mathematical formulation, We created a user define function in excel that
can return the NPV for a well (Deterministic Model)
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Example 1
Lets consider the following random variables:1. Initial flow rate q0is normally distributed with mean =500
and standard deviation of 30 BPD
2. Oil price is uniform distributed between 85 and 120 $/BPD
3. Gas price in uniform distributed between 2 and 6 $/Mscf
The certain variables are:
GOR 0.3 [Mscf/BBL]
WaterCost 0.13 [$/BBL]
Wc 0.5
O(i) 301,000.00 [$]T 0.29
MROR 0.13
Di 0.7 [1/year]
b 1.5
Capex 2,000,000 [$]
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Example 1
Run a Monte Carlo simulation assuming that the random variablesare independent and not correlated
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Example 2
The recoverable oil using the volumetric approach in a givenprospect is given
rf
Bo
ShA,Nr W
17587
whereAis the drainage area [acres], his the net pay [ft], is porosity [-], Swis
the water saturation [-],Bois the formation volume factor [bbl/stb] and rf
is the recovery factor [-]. Nris in stb
Example 2 template contains a user define function that calculatesNrand
can be used to setup a Monte Carlo Simulation.
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Example 2
Consider the following random variables1. is porosity [-] is normally distributed with mean 0.14 and
standard deviation of 0.02
2. Swis the water saturation [-]. Uniform distributed between 0.2
and 0.4.
3. his the net pay [ft]. Normally distributed mean 15 and standarddeviation of 1.5 feet's
4. Ais the drainage area [acres]. Lognormal distributed with =4.35 and
=0.246 (mean 80 and std of 20 Acres)
5. rf is the recovery factor [-]. Normally distributed with mean of
0.34 and standard deviation of 0.05.6. Bois the formation volume. Uniform distributed between 1.15
and 1.25