Appendix Methods for Calculating the Mean of a Lognormal Distribution All of the following methods are based upon a dis- NOTE: The statistical mean is theoretically the pre- tribution in which n = 9, and the constituent values ferred expression. However, I recommend against its are: general use, for the following reasons: NATURAL LOGS 1. For field-size distributions, we do not ordinarily P900/0 = 0.6MM KEY VALUES (In) anticipate that a given trend or basin will have an P800/0 = 1.25MM P99% = O.lMM infinite number of fields; instead we generally P700/0 = 2.1MM P900/0 = 0.6MM find tens to perhaps a few hundreds of fields. P600/0 = 3.3MM P84% = 0.95MM Accordingly, the continuous distribution, when n P500/0 = 5.0MM P500/0 = 5.0MM 1.6094 = 00, seems inappropriate, and leads to an inflated P400/0 = 7.6MM PI6% = 26.5MM mean field size. P300/0 = 12.1MM P100/0 = 43.0MM 2. For prospect parameters such as Area, Average P200/0 = 20.0MM PI% = 250.0MM Net Pay, Gross Rock Volume, HC-recovery in P100/0 = 43.0MM barrels per acre-foot (bbl/af) or thousand cubic Total = 95.95MM feet per acre-foot (mcf/ af), Prospect Reserves, Initial Production Rates, etc., the recommended 1. Arithmetic Mean (x) probabilistic estimating connotations of P99%, x = 95.95 + 9 = 10.66 MM P900/0, P500/0, P100/0, and PI% treat those very 2. Swanson's Mean (Msw): 0 0 0 large outcomes greater than PI% as practically 0.3 (P90 /0) + 0.4 (P50 /0) + 0.3 (P10 /0) and geologically impossible. Because the statisti- Msw 0.3 (0.6) + 0.4 (5.0) + 0.3 (43.0) = cal mean includes contributing values larger than I5.08MM PI%, such untruncated means are unrealistically 3. Statistical Mean (Mst): assumes a continuous dis- large. The P99% and PI% estimates then become tribution, (i.e., n = 00), with contribution from val- very useful as lower and upper plausibility or ues greater than PI% credibility checks that encourage iterations lead- =eli + (52/2 ing to greatly improved estimates. 2 =e1n(s.O) + In(26.5/S.0) /2 3. When such distributions are truncated at PI% =e1n(S.O) + In(S.3)2/2 and the truncated mean is calculated by incre- = 5.0e(1.6677)2/2 mental summing and averaging, such truncated =5.0e(2.7812)/2 mean values approach values obtained by using =5.0e(1.3906) Swanson's Mean. Truncation at the small end of = 5.0 x 4.0173 reserves distributions, reflecting commercial or =20.01MM economic thresholds, reinforces the practical util- ity of Swanson's Mean. Where Jl = In median (P500/0) value % 0 (J = In standard deviation (P16 + P50 /0) 4. Statistical Mean, truncated above Pl% (Mst [t > 1%]): Mst(t> 1%) = 15.77 MM [Compare with item 2, Swanson's Mean (= 15.08 MM)]. NOTE: Upper truncation is also discussed on pages 14 and 20. 125 Downloaded from http://pubs.geoscienceworld.org/books/book/chapter-pdf/3834110/9781629810652_backmatter.pdf by guest on 25 September 2021 Downloaded from http://pubs.geoscienceworld.org/books/book/chapter-pdf/3834110/9781629810652_backmatter.pdf by guest on 25 September 2021 Appendix Graphical Method for Combining Probabilistic Distributions by Multiplication In conjunction with their long-running AAPG b. P50-the median-proportional pat- School, "Managing and Evaluating Petroleum Risk," terns in lognormal distributions Capen, Megill, and Rose developed a graphical proce- (P10/P50 = P50/P90) . dure to carry out the analytical method by which sev- c. Don't start the estimating process eral probabilistic distributions may be combined by with a P500/0 estimate, to resist multiplication. This procedure gives the "ideal" "anchoring." results that would occur through multi-trial Monte C. Iteration and Reiteration Carlo or Latin Hypercube simulation, without any 1. Project the sloping line downward to the minor variations caused by random sampling. The P99% intersection; what is the resulting method assumes that all distributions to be combined P99% value? Is it consistent with a mini- are lognormal, and that differences in their variances mum value barely sufficient to support are not excessive. However, even normal distributions flow into the borehole? Is it consistent - may be combined with satisfactory results, as long as with the structural and stratigraphic their distributions have low variance (i.e., P10/P90 geometry of the prospect? Is it consistent ranges are less than about 5). with existing geotechnical data? 2. Project the sloping line upward to the P1% I. Develop ranges for reserves parameters (Produc- intersection; what is the resulting P1% tive Area, Average Net Pay, HC-recovery Factor value? Is it consistent with available struc- in bbl/af or mcf/af-see Figure B-1): tural, stratigraphic, or reservoir data (i.e., A. Remember that all estimates express subjec- is it a "high-side value so large as to be tive probability judgments. barely possible," honoring the data? B. Getting started 3. Taking steps C1 and C2 into account, 1. Assume lognormality adjust the position and slope of the distri- 2. Make reasonable low-side and high-side bution. Work back and forth between estimates of the parameter, remembering maps, cross-sections, production data, the prevalent conservative bias to set pre- etc., and log probability graphs, adjusting, dictive ranges too narrow. Plot the low- examining implications of different cases, side estimate at P90% and the"high-side iterating, and reiterating until the distribu- estimate at P10o/0 on a cumulative log tion represents a best-fit to P1%, P100/0, 0 0 % probability graph. Draw a straight sloping P50 /0, P90 /o, and P99 requirements. line connecting them on the graph. 3. What is the resulting P50o/0 value? Is that a II. Steps in combining distributions with log proba- plausible "best guess" value? bility paper (three variables). See Figure B-2: 4. Estimating a middle value A. Remember that multiplying the three 10th a. What's wrong with "most likely"? percentile values for Area, Pay, and HC- • Geologists don't intuitively iden- recovery will not give a 10% reserves prod- tify it-confuse with median, uct-as a matter of fact, it gives a product of mean, mode, etc.; 1.3%! Similarly, multiplying the three 90th • Not identifiable on cumulative percentile values gives a 98.7% reserves prod- curves plotted on log probability uct, not 90%. However, multiplying the three paper; 50th percentile values does indeed yield a • Useless in calculating the mean. 50% reserves product! 127 Downloaded from http://pubs.geoscienceworld.org/books/book/chapter-pdf/3834110/9781629810652_backmatter.pdf by guest on 25 September 2021 128 Graphical Method for Combining Probabilistic Distributions by Multiplication 0 B. WE CAN USE THIS: Multiply the three PI0 /0 A Proof values for Area, Net Pay, and HC-recovery to yield a reserves product that should be plotted How do we know that multiplying the 77% points for at 1.3%. Similarly, multiply the three P900/0 area, pay, and recovery will lead to the 90% point for values for Area, Net Pay, and HC-recovery reserves? and plot the resulting reserves product at 98.7%. Also multiply the three P500/0 values to recovery x net pay x area =reserves 0 give reserves P50 /0 and plot that. Draw a best- YI xY2 xY3 =R fit line through these three points: this line is Xi =In Yi the graphical reserves distribution. Pick off the P900/0 and PI0O/o reserves values (and the Where Xi is normally distributed with mean Ili and 0 P50 /0 reserves value) and employ them using variance (J2. All the variances are equal. Swanson's Rule, to find mean reserves. (This assumption causes no trouble. Our tests c. ALTERNATIVE APPROACH: What are the showed that we would have errors of no worse than appropriate percentile values that, when mul- 4% using this approximation. In oil and gas explo- tiplied, will generate 10% and 90% reserves ration, a 4% error amounts to a direct hit.) values? It turns out that multiplying the three 23% values gives a 10% reserves product, and Multiplying Ys is equivalent to adding Xs. multiplying the three 77% values gives a 90% reserves product (see Ed Capen's A Proof- In R =In (YIX Y2 X Y3) following). We can put this to work: =Xl + X2 + X3 1. Find the 23% values for Area, Pay, and HC-recovery from the plots you made on What is the variance of the sum of the Xs? log probability paper. Multiply them to Assuming independence, the variance of a sum is produce the 10% value for reserves; plot the sum of the variances. this value on log probability paper. 2 2. Similarly, find the 77% values for Area, (JL = variance of the sum = 3 (J2 Pay, and He-recovery from the same And (JL =rrV3 plots; multiply them to produce the 90% (J =(JL/-Y3 value for reserves; plot this value also. 3. Plot a "reserves" line on your log proba- This says that the st. dev. of area, for example, equals bility paper, using the derived and plotted the st. dev. of reserves divided by the square root of 3. 10% and 90% values plus the 50% value We know that the std. dev., (J, is just the scale factor you determined earlier. for the standard normal, Z. "Standard" means zero 4. Calculate the mean of this reserves distri- mean and variance =1. bution using Swanson's Rule. So a Zarea would equal Zreserves/-Y3. D. For combining two variables: multiplying two P90O/o's gives P96.5%; multiplying two PI00/0's Example: Say we want Zreserves to be at the 90% gives P3.5%.