. By: Marco Antonio Guimarães Dias Technical Consultant by Petrobras

Doctoral Research by PUC-Rio

5th Annual International Conference on Real Options Theory Meets Practice

UCLA - Los Angeles, USA - July 2001

Selection of Alternatives of Investment in Information for Oilfield Development Using

Evolutionary Real Options Approach

Undelineated Field: Option to Appraise

Appraisal Appraisal InvestmentInvestment

RevisedVolume = B’

Developed Reserves: Options to Expand, to Stop Temporally, and to Abandon.

Real Options Process in Exploration & Production Concession: Option to Drill the Wildcat

Exploratory (wildcat) Exploratory (wildcat) InvestmentInvestment

Oil/Gas SuccessProbability = p

Expected Volumeof Reserves = B

Delineated Undeveloped Reserves: Options to Invest in Additional Information and to Develop

Development Development InvestmentInvestment

Alternatives of Investment in Additional Information into a Dynamic Framework Motivation: Answer the following questions related to a

discovered and delineated oilfield, but with remaining technical uncertainties about the reserve size and quality Is better to invest in information, to develop, or to wait? What is the best alternative to invest in information?

The presented model considers two information costs: the cost of investment in information itself; the cost to delay the development (time to learn)

Benefit: adequation of development cost to the “true” reserve size and quality, and eluding to develop revealed non-economic oilfield. Optimization under uncertainty Each costly learning alternative has different revelation

potential, modeled as distributions of probabilities

The Oilfield Development Option and NPV Equation By incurring the cost of information with certainty, we have

the option to develop the oilfield, which depends: The revealed technical scenario given by the information, the

market (oil price), and the legal time to expiration

For the scenarios that the development option is exercised, we get the net present value (NPV) given by: NPV = V D = q P B D

q = economic quality of the reserve, which has technical uncertainty, revelation modeled with a triangular distribution

– depends of rock’s perm-porosity quality, oil quality, etc.P is the oil price ($/bbl) source of the market uncertainty,

modeled with two risk neutral stochastic processes (Geometric Brownian motion and a special Mean Reversion Process);

B = reserve size (million barrels), which has technical uncertainty, revelation modeled with a triangular distribution;

D = oilfield development cost, function of the reserve size B: D = Fixed Cost + Variable Cost x B D = FC + VC . B

Alternatives of Investment in Information (0, 1 and 2) Alternative 0: Not invest in information (wait or develop,

until the expiration) Alternative 1: long-term well testing

Cost = 25 days of rig x 120 M$/day = US$ 3 million Time to Learn (development delay) = 30 days Information revelation (reserve size B and quality q):

B ~ Triangular (B0 x 0.8 ; B0 ; B0 x 1.2) q ~ Triangular (q0 x 0.9 ; q0 ; q0 x 1.1)

Alternative 2: an additional appraisal well (slim) Cost = US$ 4 million Time to Learn (development delay) = 30 days Information revelation (investigate more the size B):

B ~ Triang (B0 x 0.5 ; B0 ; B0 x 1.5) q ~ Triang (q0 x 0.9 ; q0 ; q0 x 1.1)

Alternatives of Investment in Information (3 and 4) Alternative 3: development drilling plan focused in

information instead drilling cost, but without tests Cost = 4 mob/demob = 40 days x 120 M$/day = US$ 4.8 million Time to Learn (development delay) = 5 months Information revelation:

B ~ Triang (B0 x 0.4 ; B0 ; B0 x 1.6) q ~ Triang (q0 x 0.8 ; q0 ; q0 x 1.2)

Alternative 4: Idem altern. 3 but with well testing Cost = 4 mob/demob + tests = 4 x (10 + 10) days x 120 M$/day =

US$ 9.6 million Time to Learn (development delay) = 6 months Information revelation:

B ~ Triang (B0 x 0.3 ; B0 ; B0 x 1.7) q ~ Triang (q0 x 0.6 ; q0 ; q0 x 1.4)

Monte Carlo Simulation of Uncertainties Simulation will combine uncertainties (technical and market) for

the equation of option exercise: NPV(t)dyn = q . B . P(t) D(B)

Reserve Size (B) (only at t = trevelation)

(in million of barrels)

Minimum = 60Most Likely = 120Maximum = 180

Oil Price (P) ($/bbl) (from t = 0 until t = T)

Mean = 20 US$/bbl Standard-Deviation:

changes with the time

Parameter Distribution Values (example)

Economic Quality of the Developed Reserve (q) (only at t = trevelation)

Minimum = 15% Most Likely = 20% Maximum = 25%

In the case of oil price (P) is performed a risk-neutral simulation of its stochastic process, because P(t) fluctuates continually along the time

Stochastic Processes for Oil Prices: the GBM The oil price in a future time t is uncertain and follows a stochastic process The paper use two stochastic processes.

The first is the popular Geometric Brownian Motion (GBM) The second one is a special mean-reversion process (next slide) In development the case combining mean-reversion and jumps

Used a risk-neutral simulation, in order to use the risk free discount rate (r) as the correct discount rate for the option For the GBM, the price Pt in a future time t is simulated:

Pt = P0 exp{ (r ) t + t

is the convenience yield of P and is the volatility of P

By sampling the standard Normal N(0, 1) we get values for Pt

Writing this format, detaching N(0, 1), is better for modular design

Sampling the N(0, 1), we get the risk-neutral paths for x(t). Writing this format, detaching N(0, 1), is better for modular design

The variable x(t) reverts to a long run mean Prices reverts to a long run equilibrium level. I set = 20 $/bbl In order to get the desirable mean is necessary to subtract from x the half of variance Var[x(t)], which is a deterministic function of the time:

P(t) = exp{ x(t) (0.5 * Var[x(t)])} This is necessary due the log-normal properties

Using the previous equation relating P(t) with x(t), we get the risk-neutral mean-reversion sample paths for the oil prices.

Risk-Neutral Mean Reversion Process for P The risk-neutral process for the variable x(t), considering

the AR(1) exact discretization (valid even for large t) is:

Example of Computational Modular Design C++ is suitable for modular software design. The module of random generation for Monte Carlo simulation can be: (a) performed online as usual; (b) offline until “perfection” and saved in files; and (c) even by quasi-random numbers files.

Offline simulation with two separed table files is very flexible

File with therevealed technicalscenarios are different for eachalternative of investment in information

File with the N(0, 1) isused for moststochastic processes and for anyalternative ofinvestment ininformation

Real Option Evaluation for a Given Decision Rule The simulated sample paths are checked with the threshold

A

Option F(t = 1) = V DF(t = 0) == F(t=1) * exp (r*t)

Present Value (t = 0)

B

F(t = 2) = 0ExpiredWorthless

Simulation and Technical Scenarios The investment in information can reveal bad or good technical

scenarios for q and B. The reveled news affect the quantity of sample paths reaching the same development threshold.

The Practical Power of Normalization The combination of technical and economic uncertainties

had a practical problem: After a technical revelation, both project value V and development cost D

change. So, the development threshold (V* or P*) will be not the same, because D is not the same.

Solution: work with the normalized threshold (V/D)* After the revelation, even changing D, the normalized threshold

(V/D)* is the same! It depends only of r, , The mathematical trick is that (V/D)* is homogeneous of degree

zero in V and D, and the real option value F is homogeneous of degree one in V and DThanks for McDonald & Siegel (1986, p.713 & appendix). This was proved for the geometric Brownian motion (GBM)

Extension of the model: D with both market (factor MF) and technical uncertainties: D = MF.(FC + VC . B) MF follows a GBM, (V/D)* depends of T = f(P, D, )

The population of threshold curvesare compared with the simulated

project value/cost. Evaluate the rules

Evolutionary Real Options FlowchartGenetic Algorithms (GA)Genetic operators generate n

decision rules (n threshold curves)

Monte Carlo SimulationProject value V and cost D are simulated

along the time combining uncertainties

Go to the next generation?

The EndThe Best

Decision Rule

No

Stopping rule can be by computational time or by the user

Yes

New

gen

erat

ion

: G

A r

epro

du

ctio

n

GA operators: crossover,mutation, andvariations

Risk-neutral

stochasticprocesses + technical uncertaintiesExercise theoption if theV/D reaches the threshold

Why Evolutionary Real Options? Why genetic algorithms (GA) as optimization tool?

Use with Monte Carlo simulation, which works better with the curse of dimensionality and curse of modeling Alternative in Monte Carlo for American options problemsMonte Carlo is forward looking but dynamic programming is a backward

looking: GA is not backward optimization

Complex real options: only near optimal is possible Schwartz & Zozaya-Gorostiza (May, 2000): not known

optimal exercise for path dependant and random expiration Genetic algorithm has a high implicit parallelism

Schema theorem proved that processing each generation a population of only n chromosomes, GA processes usefully something like n3 schemata! (see Goldberg, 1989, pp.40-45)

Genetic algorithms drawback: computational time First paper version: 29.38 min (P-III, 800 MHz, population = 40,

generations = 31) to get the optimal with 2% from the theoretical. New version: this time is reduced to less than 2 min.! Forthcoming!

Threshold Approximation with Function and Free Points The genetic algorithm approximates the threshold curve using 4 genes: two free points (a, b) and two coefficients of a logarithmic function, c + d ln(T - t)

Software in C++ The interface (input windows) for the mean-reversion case:

Results for the Geometric Brownian Case For the Geometric Brownian case, the threshold solution is

known (finite differences) so is possible to compare with the evolutionary approach setting a limit of ~30 minutes

The solution was close to theoretical for both option value and threshold

Real Options Results for Mean Reversion Case The table shows the evaluation of the alternatives of investment in

information and some output details

Vector X AlternativeCase

gen/pop_sizeBest Chromosome

ValueGeneration of the best

1.14562201 0 60/40 49.044 401.152721171.254576560.071844731.03630149 1 25/20 46.128 251.040269021.228392840.0424125

1.13232374 2 30/40 48.771 301.168233871.229260680.043729921.08463931 3 30/40 50.02 301.1159513

1.261269210.072263961.10089219 4 30/40 51.711 301.158927441.2680186

0.07771666

Even with different absolute values, again the best alternative was the number 4 (the alternatives ranking was the same)

Evolution of the Best Threshold (Chromosome) The chart shows the progress along generations of best

chromosome (development threshold) for mean-reversion. The points only appear if a new real options maximum

occurs. Sometimes there is a jump as in generation 6.

Conclusion and Extensions Genetic algorithm optimization is a good option for Monte Carlo

simulation of complex American options Monte Carlo in real options problems is a flexible way to face the

problems of curse of dimensionality and curse of modeling. In addition is easier to “sell” to managers.

The evolutionary real options approach performed better than expected in terms of computational time (now 2 min.)

The threshold normalization (V/D)* eases a lot the combination of technical with market uncertainties

In the case study, the costlier alternative of investment in information reached the maximum real options value due the larger information revelation potential

There are several extensions in execution or consideration: Stochastic process: jumps, random equilibrium level, random D Optimal timing search also for investment in information

Anexos

APPENDIXSUPPORT SLIDES

Paper New Contributions and Limitations The main new contributions of this paper are:

Use of evolutionary approach (genetic algorithms + data structures) for optimization under uncertainty in E&P Flexibility: non-backward optimization and choice of stochastic process Paper used a non-conventional mean-reversion process

Combination of market uncertainty (oil price) with technical uncertainties made simple using a normalized threshold curve Revealed technical uncertainty affects both the underlying asset V and the development

cost D. Normalization V/D uses the same threshold! Modular computational design, including offline simulation

The main limitations (for future improvement): Considered only the development timing, not the learning timing: there is

additional value for cases when is not optimal the costly learning now, neither the immediate development

Considered only alternatives with single learning stage, not sequential investment in information Focus is additional information decision, not the whole appraisal phase

Case Study and the Model Equations Consider an undeveloped oilfield

The expected reserve size before revelation is estimated in:E[B] = 120 million barrels

By looking the portfolio of projects of a basin we can set a cost function for the development investment D = Fixed Cost + Variable Cost x B

FC = 240, VC = 2 D = 240 + 2 x 120 = US$ 480 million We can set a linear equation for the net present value (NPV) as

function of the price and the reserve size B NPV = V D = q P B D

q = economic quality of the developed reserve, with expected value of E[q] = 20%

P is the current oil price (P0 = 20 $/boe) and is the current source of market uncertainty.

Hence, the static NPV is zero (= 0.2 x 20 x 120 480)

Alternatives of Investment in Information under Uncertainties

The alternatives considered in this study are: Alternative 0: Not invest in information (wait or develop) Alternative 1: long-term well testing Alternative 2: an additional appraisal well (slim) Alternative 3: development drilling plan focused in

information instead drilling cost, but without tests Alternative 4: Idem alternative 3 but with well testing

Previous job must eliminate alternatives of investment in information which are dominated by the others:

For more expensive alternatives, is useful to analyze only if they reveal more information than the cheaper one or present other economic benefit.

Alternative 5: Future Case Study Not included in the paper, an extension. Alternative 5: Pilot Production System

Pilot of NPV = zero Cost = Benefit of early revenue + salvage value

Impact over the development phase: Pilot production decrease the reserve in 20 million barrelsInvestment cost D is reduced in US$ 40 million

Time to Learn (development delay) = 4 years Information revelation:

B ~ Triang [(B0 20) x 0.25 ; B0 20 ; (B0 20) x 1.75] q ~ Triang [q0 x 0.4 ; q0 ; q0 x 1.6]

Investment Cost (in million) D = CF 40 + CV x B

Technical Uncertainty Technical uncertainty decreases as the investment in information is performed (step-by-step approach). Geological uncertainty is reduced by the investment of the whole industry in a basin

The “cone of uncertainty” (Amram & Kulatilaka) can be adapted to the technical uncertainty understanding:

Risk reduction by the investment in information of all firms in the basin(driver is the investment, not directly by the passage of time)

Project evaluation with additionalinformation(t = T)

Lower Risk

ExpectedValue

Current project evaluation(t=0)

HigherRisk

ExpectedValue

con

fide

nce

inte

rval

Lack of Knowledge Trunk of Cone

Technical Uncertainty But in addition to the risk reduction process, there is

another important issue: revelation of the true value The expected value after the investment in information

can be very different of the initial estimative Investment in information reveals good or bad news

Value withgood revelation

Value withbad revelationCurrent

project evaluation(t=0)

Investment inInformation Project value

after investment

t = T

Value withneutral revelation

Technical Uncertainty The number of possible scenarios to be revealed is proportional to the cumulative

investment in information Information can be costly (our investment) and/or free, from the other firms investment (free-rider)

.

Investmentin information(wildcat drilling, etc.)

Investment in information(costly and free-rider)

Todaytechnicaland economicvaluation

t = 0 t = 1

Possible scenariosafter the informationarrived during the first year of option term

t = T

Possible scenariosafter the informationarrived during the option leaseterm

The arrival of information process leverage the option value of a tract

Investment in Information and Revealed Scenarios Suppose the following stylized delineation of an oilfiled

(appraisal phase of a discovered field). MM = million The drilled well in the area A proved 100 MM bbl

A B

DC

Area A: provedBA = 100 MM bbl

Area B: possible50% chances of

BB = 100 MM bbl& 50% of nothing

Area D: possible50% chances of

BD = 100 MM bbl& 50% of nothing

Area C: possible50% chances of

BC = 100 MM bbl& 50% of nothing

Suppose there are three alternatives of investment in information: (1) drill one well (B); (2) drill two wells (B + C); (3) drill three wells (B + C + D)

Alternatives: Revealed Scenarios and Uncertainty Alternative Zero: Not invest in information

This case there is only a single scenario, the current situation So, we work with the expected value for the reserve B:

E(B) = 100 + (0.5 x 100) + (0.5 x 100) + (0.5 x 100) E(B) = 250 MM bbl

But this value can be as low as 100 and as higher as 400 MM bbl. Hence, there is large uncertainty Without learn, after the development you find one of the values:

100 MM bbl with 12.5 % chances200 MM bbl with 37,5 % chances300 MM bbl with 37,5 % chances400 MM bbl with 12,5 % chances

The standard deviation (uncertainty) is 86.6 MM bbl

Alternative 1: Invest in Information with Only One Well Suppose that we drill the well in the area B.

This case generated 2 scenarios, because the well B result can be dry (50% chances) or success proving more 100 MM bbl

In case of positive revelation the expected value is:

E1(B) = 100 + 100 + (0.5 x 100) + (0.5 x 100) = 300 MM bbl In case of negative revelation the expected value is:

E2(B) = 100 + 0 + (0.5 x 100) + (0.5 x 100) = 200 MM bbl The uncertainty remains. In case of positive revelation, can occur:

200 MM bbl with 25 % chances 300 MM bbl with 50 % chances 400 MM bbl with 25 % chances

The negative revelation case is analog: can occur 100 MM bbl (25% chances); 200 MM bbl (50%); and 300 MM bbl (25%)

The remaining standard-deviation in both cases of revelation is 70.7 MM bbl (decrease when compared with alternative zero)

Alternative 2: Invest in Information with Two Wells Suppose that we drill the wells in the areas B and C.

In this case 3 scenarios are generated because the results after the information (drilling) are:Positive revelation: both wells proving oil (25%); Neutral revelation: one dry hole + one with oil (50%); andNegative revelation: two dry holes (25% chance).

The new expected values given the information are:Positive revelation: E1(B) = 350 MM bbl; Neutral revelation: E2(B) = 250 MM bbl; eNegative revelation: E3(B) = 150 MM bbl.

Uncertainty still remains (a little bit). In any of revealed scenarios, the reserve can vary +/- 50 MM bbl

Eg.: a new positive revelation, B can reaches 400 MM bbl The remaining standard-deviation in any of the three revealed scenarios is 50

MM bbl (decreases even more compared with alternative 1, but the number of revealed scenarios is larger)

The Full Revelation Alternative 3: Invest in Three Wells Suppose that we drill the wells in the areas B, C and D.

In this case 4 scenarios (the a priori scenarios) are revealed because the results after the information (drilling) are: Positive revelation: all three wells proving oil (12.5%); Neutral-positive revelation: two wells with oil and one dry hole (37.5 %

chance for this scenario);Neutral-negative revelation: one wells with oil and two dry holes (37.5

%); andNegative revelation: all 3 wells are dry hole (12.5% chance).

The expected values given the information are, respectively: E1(B) = 400 MM bbl; E2(B) = 300 MM bbl; E3(B) = 200 MM bbl; e

E4(B) = 100 MM bbl. In this case all the uncertainty is revealed. In any of the revealed scenarios, the

reserve (100, 200, 300 or 400) become deterministic The standard-deviation in any of the four revealed scenarios is zero (total

reduction of the uncertainty on the reserve size)

Alternatives: Revealed Scenarios and Uncertainty The simple and stylized example shows: As higher is the (non-redundant) investment in information, as

higher is the number of revealed scenariosIdea is to associate the investment in information with the

distribution of probability of a revelationFor any scenario, the development investment (if happen) will take

place in a more adjusted way (benefit) As higher is the (non-redundant) investment in information, as

lower is the remaining uncertainty in each revealed scenarioInverted cone of uncertainty

In this example, all the information were totally non-redundant and independent (no correlation between the wells) In the real life can happen some degree of redundancy and/or

correlation. In addition to the volume, the well can reveal the fluid and reservoir rock qualities.

Visualization of Revealed Scenarios: Revelation Distribution

.t = 0

Information (1 well)

E(B) = 250(MM boe)

E1(B) = 300

E2(B) = 200Alternative 1

.t = 0

Information (2 wells)

E(B) = 250(MM boe)

E1(B) = 350

E2(B) = 250

E3(B) = 150

Alternative 2

.t = 0

Information (3 wells)

E(B) = 250(MM boe)

E1(B) = 400

E2(B) = 300

E3(B) = 200

E4(B) = 100

Alternative 3

Mean Reversion Process Consider the arithmetic mean reversion process

The solution is given by the equation with stochastic integral:

Where is the reversion speed. The variable x(t) has normal distribution with mean and variance given by:

We want a mean reversion process for the oil prices P with lognormal distribution with mean E[P(T)] = exp{E[x(T)]}

Sampling the N(0, 1), we get the risk-neutral paths for x(t). Writing this format, detaching N(0, 1), is better for modular design

It is similar but not equal the known Schwartz’s model 1 (1997). Schwartz used x = ln P, but with a more complex relation between and here is the long-run equilibrium (mean) level, which the oil price reverts

I prefer a slight different mean-reversion model for the oil price letting E[P(T)] = exp{E[x(T)]} and . I set = 20 $/bbl In order to get the desirable mean is necessary to subtract from x the half of variance, which is a deterministic function of the time:

P(t) = exp{ x(t) (0.5 * Var[x(t)])}

Using the previous equation relating P(t) with x(t), we get the risk-neutral mean-reversion sample paths for the oil prices.

Risk-Neutral Mean Reversion Process for P The risk-neutral process for the variable x(t), considering

the AR(1) exact discretization (valid even for large t) is:

Mean Reversion Forecasting

Genetic Algorithm Chromosome Adopted a simple chromosome: heuristic studying American

options thresholds. The chromosome has two free points (a, b) with linear

interpolation + logarithmic function with 2 coefficient (c, d). Vector (a, b, c, d)T (V/D)t through a generation function of (V/D)t

For less known cases is possible to use only (and many) free points

2 linear constrains to make sure the threshold is monotonic in = T t : b a c + d * ln(0.2 t) b

c dThreshold a at = 0.1 year

Function = c + d ln()Free Points a, b

Threshold b at = 0.2 year

Genetic Algorithm Flowchart

The EndBest

Chromosome

NoGo to the newGeneration?

Yes

Evaluate the Chromosomes The most complex job

Ranking for Fitness Linear normalization(1, 2, ……. N)

Parent Selection for Reproduction Roulette wheel method

Genetic Operators Crossover & mutation

Steady-State Generations with high GAP n chromosomes are createdGeneration change: n = N

Initialize the Population P(g = 0)Using N chromosomes (eg. N = 40)

Random with seed

Genetic Operators The program uses 5 genetic operators, being two for

mutation and 3 for crossover (see Michalewicz, p.127) Genocop III has 10 genetic operators, but with expensive

fitness function (by MC simulation), it is necessary to eliminate operators that generate sub-optimal offsprings:“Boundary mutation” was eliminated because here is

known that values at the limit of constrains are sub-optimal

Operator 1: Uniform Mutation All genes have the same probability of mutation Mutation for gene k: uniform distribution in the selection

of the mutant gene inside the valid domain for k.

5.2 6.3 7.4 8.5

k

Mutation 5.2 6.3 2.1 8.5

k

Genetic Operators Operator 2: Non-Uniform Mutation

Operator responsible for the fine tuning capabilities Mutation of gene k: mutant gene value xk has a variation

of , in function of the relation (1 g/ G)b , where: g = actual generation, G = maximum generation (to be set) and b

is a parameter that gives the non-uniformity degreeThe value decreases along the generation (g getting close

of G). For g G 0 In words, the operator search uniformly the search space

initially, and very locally at later generationsMutation 5.25.2 6.3 7.4 8.5 6.3 2.6 8.5Initially:

Mutation 5.25.2 6.3 7.4 8.5 6.3 7.3 8.5Later:

Genetic Operators Operator 3: Arithmetical Crossover

Given two selected chromosomes (parents) J1 e J2 the resultant offspring J’1 e J’2 have genes values that are linear combination of parents genes:

J’1 = a . J1 + (1 a) . J2 J’2 = a . J2 + (1 a) . J1 a is random and chosen on the interval [0, 1]

Ex.: for a = 0.5 (“average crossover”), twin offsprings:

J111.01.0 2.0 3.0 4.0 12.0 13.0 14.0J2 +

6.0 7.0 8.0 9.0J’1J’2 6.0 7.0 8.0 9.0

Genetic Operators Operator 4: Simple Crossover

The cut position k is selected for the selected parents J1 e J2 resulting in the offspring J’1 e J’2

Example: in case of k = 3 (after the third gene)

J111.01.0 2.0 3.0 4.0 12.0 13.0 14.0J2 +

1.0 2.0 3.0 14.0J’1J’2

11.0 12.0 13.0 4.0

Michalewicz uses a linear combination for the new genes in order to guarantee the viability in terms of constrains.

The case above is for a = 1 in the linear combination. If it is not viable, the algorithm reduces this value until reach 0

Genetic Operators Operator 5: Heuristic Crossover

It uses values of the objective function in determining the direction of the search. Best evaluated parents have more weight in terms of value to be passed to offspring genes.

Fine tuning operator and to search the + promising direction Maximization: if f(J2) > f(J1) J’ = J2 + a . (J2 J1) a is a random number on the interval (0, 1) Only one child is generate, but no child is possible (constrains)

Example: f(J2) > f(J1) and a = 0.5:

J111.01.0 2.0 3.0 4.0 12.0 13.0 14.0J2 +

J’ 16.0 17.0 18.0 19.0

Genetic Algorithm under Uncertainty The GA application must consider some specific issues for

the case of uncertainties The most expensive and complex item is the chromosomes

evaluation, performed under uncertainty (MC simulation) The evaluation (fitness) function is a noisy type

Davis (Handbook of GAs): steady-state reproduction does not work well for noisy evaluation function

The RiskOptimizer (Evolver +@Risk, an Excel add-in) is very flexible, but has the disadvantage of using steady state and with the additional problems of no elimination of duplicates and uses GAP = 1 (no option to increase GAP) GAP = number of chromosomes replaced in each generation

Adopted the language C++, adapting the source of the software “Genocop”, described in Michalewicz’s book. GAP is controled by the frequency of operators utilization

Adaptation of Genocop Program After the decision to build the software in C++, was adopted the

Genocop I (version 3.0), because: C source available, but needing to adapt to this application It works with floating point (real) not binary as other It works with linear constrains in reliable and efficient form

Creator of Genocop (Michalewicz) is the greatest expert in terms of genetic algorithms under constrains (linear and non-linear)

Why not to use the latest Genocop version, Genocop III? Genocop III is more complex because works with non-linear constrains,

but my application demands only linear constrains Costlier adaptation for version III, without advantage here

Problems in Genocop adaptation Insufficient documentation: read me file and book gives an idea Integration with the complex evaluation program is non-trivial Input file replaced by a minimum interface mínima, fixing immutable

Adaptation of Genocop and C++ Software Difficulty of integration with the complex evaluation system,

reading two files, using pointers, and performing 2 million calculation by chromosome Problems when passing from C to C++

Eg.: “new” and “try” are reservated words for C++ but not for C and Genocop was using as variable Genocop code adaptation

The Genocop’s chromosome evaluation method was changed. Genocop values individually. The software evaluate in parallel all the population (thanks to Prof. Pacheco)

Performance Pentium III, 800 Mhz: Total time for a basic experiment (31 generations and population of 40

chromosomes): ~ 29 min. Current program improvement: less than 2 min. for GBM

Evolution with average technical scenario (alternative 0) and only one complete (technical + market) simulation for the last generation: normalized threshold (V/D)* is the same

Hybrid Approch for Simulation Simulate scenarios and oil prices in C++ or only read a file with a

previously simulated distributions (eg. @Risk)? Simulation in C++: without a good variance reduction technique requires

several iterations (high computation cost) Simulation in C++ only with good variance reduction (what?) Using @Risk for offline simulation: it has a good method of variance

reduction (Latin Hypercubic) Problem with @Risk to save all the simulations in the same file: giant file,

poor C++ performance, loose design flexibility Hybrid solution is more feasible and more flexible:

@Risk only supply the very basic input distributions, that is, the triangular distributions and the Normal(0, 1) Independentes distributions can be divided into several files

C++ program uses module (function) to generate the output of the simulation, especially oil price simulation and the V/D path

Flexibility: with N(0, 1) is possible to use in almost all stochastic processes of interest.

Alternative Evaluation (Previous Version)

END: Print log and best solutionNoYes

GA Create J Chromosomes : (V/D)j, t

Value Scenario (see detail); Increment k++

Initialize Technical Scenario: k = 1

Read Technical Scenarios File (q.B)k e DkOBS: Fixed by alternative

Print (log): (a, b, c, d)j , t [0, T] ; Fitnessj = 1,..J

Create morethresholds?

Calculate J Chromosomes Fitness: Fitnessj = cost_inf + k (i F /NI)/NIT

Yes

k > NIT?No

Real Options Results for Geometric Brownian The table shows the evaluation of the alternatives of

investment in information and some output details

The best alternative was the number 4 (the most expensive but revealing more information)

Vector X AlternativeCase

gen/pop_sizeBest Chromosome

ValueGeneration of the best

1.07775891 0 46/40 62.367 201.184335711.495958570.139035111.07506621 1 30/40 58.8850 281.254851461.419501070.115396621.10754013 2 30/40 60.372 301.171581981.500482680.16309881.12806344 3 30/40 62.041 291.204050421.46492350.154918641.13385618 4 30/40 63.133 291.175323251.45274830.09308907

Results for the Mean-Reversion Process The thresholds generated by the GA for alternatives 3 and 4

are very close

Alternative 3 Alternative 4

The Timing of Investment in Information Thresholds for both investment in information and in development could be something like the figure

The threshold curve for investment in information is a future research