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. By: Marco Antonio Guimarães Dias Technical Consultant by Petrobras Doctoral Research by PUC-Rio 5 th 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

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Undelineated Field: Option to Appraise Appraisal Investment Revised Volume = 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) Investment Oil/Gas Success Probability = p Expected Volume of Reserves = B Delineated Undeveloped Reserves: Options to Invest in Additional Information and to Develop Development Investment

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Alternatives of Investment in Additional Information into a Dynamic Framework u Motivation: Answer the following questions related to a discovered and delineated oilfield, but with remaining technical uncertainties about the reserve size and quality l Is better to invest in information, to develop, or to wait? l What is the best alternative to invest in information? u The presented model considers two information costs: l the cost of investment in information itself; l the cost to delay the development (time to learn) u Benefit: adequation of development cost to the true reserve size and quality, and eluding to develop revealed non-economic oilfield. Optimization under uncertainty l Each costly learning alternative has different revelation potential, modeled as distributions of probabilities

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The Oilfield Development Option and NPV Equation u By incurring the cost of information with certainty, we have the option to develop the oilfield, which depends: l 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 è q = economic quality of the reserve, which has technical uncertainty, revelation modeled with a triangular distribution –depends of rocks perm-porosity quality, oil quality, etc. è 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); è B = reserve size (million barrels), which has technical uncertainty, revelation modeled with a triangular distribution; è D = oilfield development cost, function of the reserve size B: ó D = Fixed Cost + Variable Cost x B D = FC + VC. B

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Alternatives of Investment in Information (0, 1 and 2) u Alternative 0: Not invest in information (wait or develop, until the expiration) u Alternative 1: long-term well testing l Cost = 25 days of rig x 120 M$/day = US$ 3 million l Time to Learn (development delay) = 30 days l Information revelation (reserve size B and quality q): è B ~ Triangular (B 0 x 0.8 ; B 0 ; B 0 x 1.2) è q ~ Triangular (q 0 x 0.9 ; q 0 ; q 0 x 1.1) u Alternative 2: an additional appraisal well (slim) l Cost = US$ 4 million l Time to Learn (development delay) = 30 days l Information revelation (investigate more the size B): è B ~ Triang (B 0 x 0.5 ; B 0 ; B 0 x 1.5) è q ~ Triang (q 0 x 0.9 ; q 0 ; q 0 x 1.1)

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Alternatives of Investment in Information (3 and 4) u Alternative 3: development drilling plan focused in information instead drilling cost, but without tests l Cost = 4 mob/demob = 40 days x 120 M$/day = US$ 4.8 million l Time to Learn (development delay) = 5 months l Information revelation: è B ~ Triang (B 0 x 0.4 ; B 0 ; B 0 x 1.6) è q ~ Triang (q 0 x 0.8 ; q 0 ; q 0 x 1.2) u Alternative 4: Idem altern. 3 but with well testing l Cost = 4 mob/demob + tests = 4 x (10 + 10) days x 120 M$/day = US$ 9.6 million l Time to Learn (development delay) = 6 months l Information revelation: è B ~ Triang (B 0 x 0.3 ; B 0 ; B 0 x 1.7) è q ~ Triang (q 0 x 0.6 ; q 0 ; q 0 x 1.4)

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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 = t revelation ) (in million of barrels) Minimum = 60 Most Likely = 120 Maximum = 180 Oil Price (P) ($/bbl) (from t = 0 until t = T) Mean = 20 US$/bbl Standard-Deviation: changes with the time ParameterDistributionValues (example) Economic Quality of the Developed Reserve (q) (only at t = t revelation ) Minimum = 15% Most Likely = 20% Maximum = 25% l 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

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Stochastic Processes for Oil Prices: the GBM u The oil price in a future time t is uncertain and follows a stochastic process u The paper use two stochastic processes. l The first is the popular Geometric Brownian Motion (GBM) l The second one is a special mean-reversion process (next slide) l In development the case combining mean-reversion and jumps u Used a risk-neutral simulation, in order to use the risk free discount rate (r) as the correct discount rate for the option u For the GBM, the price P t in a future time t is simulated: P t = P 0 exp{ (r ) t + is the convenience yield of P and is the volatility of P l By sampling the standard Normal N(0, 1) we get values for P t l Writing this format, detaching N(0, 1), is better for modular design

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u Sampling the N(0, 1), we get the risk-neutral paths for x(t). l Writing this format, detaching N(0, 1), is better for modular design u The variable x(t) reverts to a long run mean u Prices reverts to a long run equilibrium level. I set = 20 $/bbl u 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)])} l This is necessary due the log-normal properties u 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:

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Example of Computational Modular Design u C++ is suitable for modular software design. The module of random generation for Monte Carlo simulation can be: l (a) performed online as usual; (b) offline until perfection and saved in files; and (c) even by quasi-random numbers files. è Offline simulation with two separed table files is very flexible File with the revealed technical scenarios are different for each alternative of investment in information File with the N(0, 1) is used for most stochastic processes and for any alternative of investment in information

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Real Option Evaluation for a Given Decision Rule u The simulated sample paths are checked with the threshold A Option F(t = 1) = V D F(t = 0) = = F(t=1) * exp ( r*t) Present Value (t = 0) B F(t = 2) = 0 Expired Worthless

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Simulation and Technical Scenarios u 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.

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The Practical Power of Normalization u The combination of technical and economic uncertainties had a practical problem: l 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. u 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,, l 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 D è Thanks for McDonald & Siegel (1986, p.713 & appendix). è This was proved for the geometric Brownian motion (GBM) l 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, )

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The population of threshold curves are compared with the simulated project value/cost. Evaluate the rules Evolutionary Real Options Flowchart Genetic Algorithms (GA) Genetic operators generate n decision rules (n threshold curves) Monte Carlo Simulation Project value V and cost D are simulated along the time combining uncertainties Go to the next generation? The End The Best Decision Rule No u Stopping rule can be by computational time or by the user Yes New generation: GA reproduction GA operators: crossover, mutation, and variations Risk-neutral stochastic processes + technical uncertainties Exercise the option if the V/D reaches the threshold

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Why Evolutionary Real Options? u Why genetic algorithms (GA) as optimization tool? l Use with Monte Carlo simulation, which works better with the curse of dimensionality and curse of modeling è Alternative in Monte Carlo for American options problems è Monte Carlo is forward looking but dynamic programming is a backward looking: GA is not backward optimization l Complex real options: only near optimal is possible è Schwartz & Zozaya-Gorostiza (May, 2000): not known optimal exercise for path dependant and random expiration l Genetic algorithm has a high implicit parallelism è Schema theorem proved that processing each generation a population of only n chromosomes, GA processes usefully something like n 3 schemata! (see Goldberg, 1989, pp.40-45) u Genetic algorithms drawback: computational time l First paper version: 29.38 min (P-III, 800 MHz, population = 40, generations = 31) to get the optimal with 2% from the theoretical. è New version: this time is reduced to less than 2 min.! Forthcoming!

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Threshold Approximation with Function and Free Points u 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)

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

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Results for the Geometric Brownian Case u 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 u The solution was close to theoretical for both option value and threshold

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Real Options Results for Mean Reversion Case u The table shows the evaluation of the alternatives of investment in information and some output details u Even with different absolute values, again the best alternative was the number 4 (the alternatives ranking was the same)

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Evolution of the Best Threshold (Chromosome) u The chart shows the progress along generations of best chromosome (development threshold) for mean-reversion. u The points only appear if a new real options maximum occurs. Sometimes there is a jump as in generation 6.

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Conclusion and Extensions u Genetic algorithm optimization is a good option for Monte Carlo simulation of complex American options l 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. l The evolutionary real options approach performed better than expected in terms of computational time (now 2 min.) u The threshold normalization (V/D)* eases a lot the combination of technical with market uncertainties u In the case study, the costlier alternative of investment in information reached the maximum real options value due the larger information revelation potential u There are several extensions in execution or consideration: l Stochastic process: jumps, random equilibrium level, random D l Optimal timing search also for investment in information

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Anexos APPENDIX SUPPORT SLIDES

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Paper New Contributions and Limitations u The main new contributions of this paper are: l Use of evolutionary approach (genetic algorithms + data structures) for optimization under uncertainty in E&P è Flexibility: non-backward optimization and choice of stochastic process è Paper used a non-conventional mean-reversion process l Combination of market uncertainty (oil price) with technical uncertainties made simple using a normalized threshold curve è Revealed technical uncertainty affects both the underlying asset V and the development cost D. Normalization V/D uses the same threshold! l Modular computational design, including offline simulation u The main limitations (for future improvement): l 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 l Considered only alternatives with single learning stage, not sequential investment in information è Focus is additional information decision, not the whole appraisal phase

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Case Study and the Model Equations u Consider an undeveloped oilfield l The expected reserve size before revelation is estimated in: E[B] = 120 million barrels l 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 u 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 l q = economic quality of the developed reserve, with expected value of E[q] = 20% l P is the current oil price (P 0 = 20 $/boe) and is the current source of market uncertainty. Hence, the static NPV is zero (= 0.2 x 20 x 120 480)

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Alternatives of Investment in Information under Uncertainties u The alternatives considered in this study are: l Alternative 0: Not invest in information (wait or develop) l Alternative 1: long-term well testing l Alternative 2: an additional appraisal well (slim) l Alternative 3: development drilling plan focused in information instead drilling cost, but without tests l Alternative 4: Idem alternative 3 but with well testing u Previous job must eliminate alternatives of investment in information which are dominated by the others: è For more expensive alternatives, is useful to analyze only if they reveal more information than the cheaper one or present other economic benefit.

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Alternative 5: Future Case Study u Not included in the paper, an extension. u Alternative 5: Pilot Production System l Pilot of NPV = zero Cost = Benefit of early revenue + salvage value l Impact over the development phase: è Pilot production decrease the reserve in 20 million barrels è Investment cost D is reduced in US$ 40 million l Time to Learn (development delay) = 4 years l Information revelation: B ~ Triang [(B 0 20) x 0.25 ; B 0 20 ; (B 0 20) x 1.75] è q ~ Triang [q 0 x 0.4 ; q 0 ; q 0 x 1.6] Investment Cost (in million) D = CF 40 + CV x B

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Technical Uncertainty u Technical uncertainty decreases as the investment in information is performed (step-by-step approach). u Geological uncertainty is reduced by the investment of the whole industry in a basin l 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 additional information (t = T) Lower Risk Expected Value Current project evaluation (t=0) Higher Risk Expected Value confidence interval Lack of Knowledge Trunk of Cone

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Technical Uncertainty u But in addition to the risk reduction process, there is another important issue: revelation of the true value l The expected value after the investment in information can be very different of the initial estimative è Investment in information reveals good or bad news Value with good revelation Value with bad revelation Current project evaluation (t=0) Investment in Information Project value after investment t = T Value with neutral revelation

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Technical Uncertainty u The number of possible scenarios to be revealed is proportional to the cumulative investment in information l Information can be costly (our investment) and/or free, from the other firms investment (free-rider). Investment in information (wildcat drilling, etc.) Investment in information (costly and free-rider) Today technical and economic valuation t = 0 t = 1 Possible scenarios after the information arrived during the first year of option term t = T Possible scenarios after the information arrived during the option lease term u The arrival of information process leverage the option value of a tract

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Investment in Information and Revealed Scenarios u Suppose the following stylized delineation of an oilfiled (appraisal phase of a discovered field). MM = million l The drilled well in the area A proved 100 MM bbl AB D C Area A: proved B A = 100 MM bbl Area B: possible 50% chances of B B = 100 MM bbl & 50% of nothing Area D: possible 50% chances of B D = 100 MM bbl & 50% of nothing Area C: possible 50% chances of B C = 100 MM bbl & 50% of nothing u 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)

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Alternatives: Revealed Scenarios and Uncertainty u Alternative Zero: Not invest in information l This case there is only a single scenario, the current situation l 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 u But this value can be as low as 100 and as higher as 400 MM bbl. Hence, there is large uncertainty l Without learn, after the development you find one of the values: è 100 MM bbl with 12.5 % chances è 200 MM bbl with 37,5 % chances è 300 MM bbl with 37,5 % chances è 400 MM bbl with 12,5 % chances l The standard deviation (uncertainty) is 86.6 MM bbl

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Alternative 1: Invest in Information with Only One Well u Suppose that we drill the well in the area B. l This case generated 2 scenarios, because the well B result can be dry (50% chances) or success proving more 100 MM bbl l In case of positive revelation the expected value is: E 1 (B) = 100 + 100 + (0.5 x 100) + (0.5 x 100) = 300 MM bbl l In case of negative revelation the expected value is: E 2 (B) = 100 + 0 + (0.5 x 100) + (0.5 x 100) = 200 MM bbl u The uncertainty remains. In case of positive revelation, can occur: è 200 MM bbl with 25 % chances è 300 MM bbl with 50 % chances è 400 MM bbl with 25 % chances u The negative revelation case is analog: can occur 100 MM bbl (25% chances); 200 MM bbl (50%); and 300 MM bbl (25%) u The remaining standard-deviation in both cases of revelation is 70.7 MM bbl (decrease when compared with alternative zero)

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Alternative 2: Invest in Information with Two Wells u Suppose that we drill the wells in the areas B and C. l In this case 3 scenarios are generated because the results after the information (drilling) are: è Positive revelation: both wells proving oil (25%); è Neutral revelation: one dry hole + one with oil (50%); and è Negative revelation: two dry holes (25% chance). l The new expected values given the information are: è Positive revelation: E 1 (B) = 350 MM bbl; è Neutral revelation: E 2 (B) = 250 MM bbl; e è Negative revelation: E 3 (B) = 150 MM bbl. u Uncertainty still remains (a little bit). In any of revealed scenarios, the reserve can vary +/- 50 MM bbl è Eg.: a new positive revelation, B can reaches 400 MM bbl u 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)

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The Full Revelation Alternative 3: Invest in Three Wells u Suppose that we drill the wells in the areas B, C and D. l In this case 4 scenarios (the a priori scenarios) are revealed because the results after the information (drilling) are: è Positive revelation: all three wells proving oil (12.5%); è Neutral-positive revelation: two wells with oil and one dry hole (37.5 % chance for this scenario); è Neutral-negative revelation: one wells with oil and two dry holes (37.5 %); and è Negative revelation: all 3 wells are dry hole (12.5% chance). l The expected values given the information are, respectively: è E 1 (B) = 400 MM bbl; E 2 (B) = 300 MM bbl; E 3 (B) = 200 MM bbl; e E 4 (B) = 100 MM bbl. u In this case all the uncertainty is revealed. In any of the revealed scenarios, the reserve (100, 200, 300 or 400) become deterministic u The standard-deviation in any of the four revealed scenarios is zero (total reduction of the uncertainty on the reserve size)

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Alternatives: Revealed Scenarios and Uncertainty u The simple and stylized example shows: l As higher is the (non-redundant) investment in information, as higher is the number of revealed scenarios è Idea is to associate the investment in information with the distribution of probability of a revelation è For any scenario, the development investment (if happen) will take place in a more adjusted way (benefit) l As higher is the (non-redundant) investment in information, as lower is the remaining uncertainty in each revealed scenario è Inverted cone of uncertainty u In this example, all the information were totally non-redundant and independent (no correlation between the wells) l 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.

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Visualization of Revealed Scenarios: Revelation Distribution. t = 0 Information (1 well) E(B) = 250 (MM boe) E 1 (B) = 300 E 2 (B) = 200 Alternative 1. t = 0 Information (2 wells) E(B) = 250 (MM boe) E 1 (B) = 350 E 2 (B) = 250 E 3 (B) = 150 Alternative 2. t = 0 Information (3 wells) E(B) = 250 (MM boe) E 1 (B) = 400 E 2 (B) = 300 E 3 (B) = 200 E 4 (B) = 100 Alternative 3

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Mean Reversion Process u Consider the arithmetic mean reversion process u 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: u We want a mean reversion process for the oil prices P with lognormal distribution with mean E[P(T)] = exp{E[x(T)]}

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u Sampling the N(0, 1), we get the risk-neutral paths for x(t). l Writing this format, detaching N(0, 1), is better for modular design u It is similar but not equal the known Schwartzs model 1 (1997). l Schwartz used x = ln P, but with a more complex relation between and l here is the long-run equilibrium (mean) level, which the oil price reverts u 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 u 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)])} u 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:

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Mean Reversion Forecasting

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Genetic Algorithm Chromosome u Adopted a simple chromosome: heuristic studying American options thresholds. u The chromosome has two free points (a, b) with linear interpolation + logarithmic function with 2 coefficient (c, d). è Vector (a, b, c, d) T (V/D) t through a generation function of (V/D) t è 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 : l b a c + d * ln(0.2 t) b c d Threshold a at = 0.1 year Function = c + d ln( ) Free Points a, b Threshold b at = 0.2 year

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Genetic Algorithm Flowchart The End Best Chromosome No Go to the new Generation? 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 created Generation change: n = N Initialize the Population P(g = 0) Using N chromosomes (eg. N = 40) Random with seed

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Genetic Operators u The program uses 5 genetic operators, being two for mutation and 3 for crossover (see Michalewicz, p.127) l 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: è Boundary mutation was eliminated because here is known that values at the limit of constrains are sub-optimal u Operator 1: Uniform Mutation l All genes have the same probability of mutation l Mutation for gene k: uniform distribution in the selection of the mutant gene inside the valid domain for k. 5.26.37.48.5 k Mutation 5.2 6.3 2.1 8.5 k

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Genetic Operators u Operator 2: Non-Uniform Mutation l Operator responsible for the fine tuning capabilities Mutation of gene k: mutant gene value x k has a variation of, in function of the relation (1 g/ G) b, where: è g = actual generation, G = maximum generation (to be set) and b is a parameter that gives the non-uniformity degree The value decreases along the generation (g getting close of G). For g G 0 è In words, the operator search uniformly the search space initially, and very locally at later generations Mutation 5.2 6.37.48.5 6.3 2.6 8.5 Initially: Mutation 5.2 6.37.48.5 6.3 7.3 8.5 Later:

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Genetic Operators u Operator 3: Arithmetical Crossover l Given two selected chromosomes (parents) J 1 e J 2 the resultant offspring J 1 e J 2 have genes values that are linear combination of parents genes: J 1 = a. J 1 + (1 a). J 2 J 2 = a. J 2 + (1 a). J 1 l a is random and chosen on the interval [0, 1] u Ex.: for a = 0.5 (average crossover), twin offsprings: J1J1 11.0 1.02.03.04.0 12.0 13.0 14.0 J2J2 + 6.07.08.09.0 J1J1 J2J2 6.07.08.09.0

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Genetic Operators u Operator 4: Simple Crossover l The cut position k is selected for the selected parents J 1 e J 2 resulting in the offspring J 1 e J 2 l Example: in case of k = 3 (after the third gene) J1J1 11.0 1.02.03.04.0 12.0 13.0 14.0 J2J2 + 1.02.03.014.0 J1J1 J2J2 11.0 12.0 13.0 4.0 u Michalewicz uses a linear combination for the new genes in order to guarantee the viability in terms of constrains. u The case above is for a = 1 in the linear combination. If it is not viable, the algorithm reduces this value until reach 0

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Genetic Operators u Operator 5: Heuristic Crossover l 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. l Fine tuning operator and to search the + promising direction Maximization: if f(J 2 ) > f(J 1 ) J = J 2 + a. (J 2 J 1 ) l a is a random number on the interval (0, 1) l Only one child is generate, but no child is possible (constrains) u Example: f(J 2 ) > f(J 1 ) and a = 0.5: J1J1 11.0 1.02.03.04.0 12.0 13.0 14.0 J2J2 + J 16.0 17.0 18.0 19.0

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Genetic Algorithm under Uncertainty u The GA application must consider some specific issues for the case of uncertainties l The most expensive and complex item is the chromosomes evaluation, performed under uncertainty (MC simulation) l The evaluation (fitness) function is a noisy type è Davis (Handbook of GAs): steady-state reproduction does not work well for noisy evaluation function l 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) è GAP = number of chromosomes replaced in each generation u Adopted the language C++, adapting the source of the software Genocop, described in Michalewiczs book. l GAP is controled by the frequency of operators utilization

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Adaptation of Genocop Program u After the decision to build the software in C++, was adopted the Genocop I (version 3.0), because: l C source available, but needing to adapt to this application l It works with floating point (real) not binary as other l It works with linear constrains in reliable and efficient form è Creator of Genocop (Michalewicz) is the greatest expert in terms of genetic algorithms under constrains (linear and non-linear) u Why not to use the latest Genocop version, Genocop III? l Genocop III is more complex because works with non-linear constrains, but my application demands only linear constrains è Costlier adaptation for version III, without advantage here u Problems in Genocop adaptation l Insufficient documentation: read me file and book gives an idea l Integration with the complex evaluation program is non-trivial l Input file replaced by a minimum interface mínima, fixing immutable

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Adaptation of Genocop and C++ Software u Difficulty of integration with the complex evaluation system, reading two files, using pointers, and performing 2 million calculation by chromosome l Problems when passing from C to C++ è Eg.: new and try are reservated words for C++ but not for C and Genocop was using as variable Genocop code adaptation u The Genocops chromosome evaluation method was changed. Genocop values individually. The software evaluate in parallel all the population (thanks to Prof. Pacheco) u Performance Pentium III, 800 Mhz: l Total time for a basic experiment (31 generations and population of 40 chromosomes): ~ 29 min. u Current program improvement: less than 2 min. for GBM l 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

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Hybrid Approch for Simulation u Simulate scenarios and oil prices in C++ or only read a file with a previously simulated distributions (eg. @Risk)? l Simulation in C++: without a good variance reduction technique requires several iterations (high computation cost) l Simulation in C++ only with good variance reduction (what?) l Using @Risk for offline simulation: it has a good method of variance reduction (Latin Hypercubic) l Problem with @Risk to save all the simulations in the same file: giant file, poor C++ performance, loose design flexibility u Hybrid solution is more feasible and more flexible: l @Risk only supply the very basic input distributions, that is, the triangular distributions and the Normal(0, 1) è Independentes distributions can be divided into several files l C++ program uses module (function) to generate the output of the simulation, especially oil price simulation and the V/D path l Flexibility: with N(0, 1) is possible to use in almost all stochastic processes of interest.

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Alternative Evaluation (Previous Version) END: Print log and best solution No Yes 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 D k OBS: Fixed by alternative Print (log): (a, b, c, d) j, t [0, T] ; Fitness j = 1,..J Create more thresholds? Calculate J Chromosomes Fitness: Fitness j = cost_inf + k ( i F /NI)/NIT Yes k > NIT? No

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Real Options Results for Geometric Brownian u The table shows the evaluation of the alternatives of investment in information and some output details u The best alternative was the number 4 (the most expensive but revealing more information)

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Results for the Mean-Reversion Process u The thresholds generated by the GA for alternatives 3 and 4 are very close Alternative 3 Alternative 4

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The Timing of Investment in Information u Thresholds for both investment in information and in development could be something like the figure l The threshold curve for investment in information is a future research

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