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Question 1 Correct Mark 1.00 out of 1.00 Question 2 Correct Mark 1.00 out of 1.00 Started on Tuesday, 15 August 2023, 6:10 PM State Finished Completed on Tuesday, 15 August 2023, 6:32 PM Time taken 21 mins 58 secs Select one or more: Kriging allows the estimation of mean and variance of a regionalised variable at locations where the variable is not known utilizinig the variogram and data at known points. Tick all statements which are true: a. Kriging estimators are sometimes called according to the principle RED: real expectation data. b. Kriging provide smoothed estimates at unknown locations and do not preserve the histogram.  c. In ordinary Kriging the variance depends both on the structural model and the data. d. Kriging estimators are sometimes called BLUE: best linear unbiased estimator.  e. Simple Kriging estimators are neither BLUE nor RED.  Your answer is correct. The correct answers are: Kriging estimators are sometimes called BLUE: best linear unbiased estimator., Kriging provide smoothed estimates at unknown locations and do not preserve the histogram., Simple Kriging estimators are neither BLUE nor RED. Consider the case of a histogram for one variable - a bin width/spacing is set and the data lying within the range of each bin counted. A succint presentation of the data is a boxplot  . Expanding to two variables a scatter plot can show the partial  relationship between variables with some remaining scatter. Projections onto the individual axes result in marginal histograms  . Concentrating on a particular bin in such a projection results in a conditional distribution  . Two distributions may be compared using a Q-Q plot  . In higher dimensions (with more variables) transforming the set of variables into a new set of principal components  will result in variables with descending contributions of variance of the data. Distributions may be separated using cluster analysis  . 

Question 3 Partially correct Mark 0.50 out of 1.00 Question 4 Correct Mark 1.00 out of 1.00 Your answer is correct. The correct answer is: Consider the case of a histogram for one variable - a bin width/spacing is set and the data lying within the range of each bin counted. A succint presentation of the data is a [boxplot]. Expanding to two variables a scatter plot can show the [partial] relationship between variables with some remaining scatter. Projections onto the individual axes result in [marginal histograms]. Concentrating on a particular bin in such a projection results in a [conditional distribution]. Two distributions may be compared using [a Q-Q plot]. In higher dimensions (with more variables) transforming the set of variables into a new set of [principal components] will result in variables with descending contributions of variance of the data. Distributions may be separated using [cluster analysis]. Select one or more: Tick all true statements about boxplots. a. Upper extremes are values which are at least 3x the interquartile range larger than the upper wisker.  b. Lower outliers are values which are at least 3x the interquartile range lower than the lower wisker.  c. Extremes should always be removed from the data set during quality control. d. The range between lower and upper hinge represents 50% of the data. Your answer is partially correct. You have correctly selected 1. The correct answers are: The range between lower and upper hinge represents 50% of the data., Upper extremes are values which are at least 3x the interquartile range larger than the upper wisker. Select one: Consider the case of a typical geological setting where local correlations exist. Spatial correlations a. have no effect on the information content of data. b. reduce the information content of data.  c. increase the information content of data.

Question 8 Correct Mark 1.00 out of 1.00 Question 9 Correct Mark 1.00 out of 1.00 Your answer is correct. The correct answers are: Determine reservoir uncertainty, Estimate recovery factors, Simulate auid aow through the reservoir Select one or more: Consider the task of upscaling permeability. Which of the following statements are true accordingly (tick all that apply). a. For vertical communication calculations non-pay is ignored.  b. For a layered system horizontal permeability can be calculated using an arithmetic average.  c. For an isotropic heterogeneous reservoir layer internal averaging may be achieved using the geometric average.  d. For a layered system vertical permeability can be calculated using an arithmetic average. Your answer is correct. The correct answers are: For an isotropic heterogeneous reservoir layer internal averaging may be achieved using the geometric average., For a layered system horizontal permeability can be calculated using an arithmetic average., For vertical communication calculations non-pay is ignored. Select one or more: Upscaling of scalar quantities in reservoir characterisation is a common task and may follow specidc rules to reduce resulting errors. Consider below and tick all correct answers. a. Porosity upscaling is least error prone if using pore volume preservation.  b. Net-to-gross ratio is best upscaled assuming that bulk rock volume is preserved in upscaling. c. Depth values affect the gravity term and are best upscaled on bulk rock volume basis. d. Saturation is best upscaled using a pore volume weighted average to preserve dne scale saturation and thus relative permeability.  Your answer is correct. The correct answers are: Porosity upscaling is least error prone if using pore volume preservation., Saturation is best upscaled using a pore volume weighted average to

Question 10 10 Partially correct Mark 0.50 out of 1.00 Question 11 11 Correct Mark 1.00 out of 1.00 Question 12 12 Correct Mark 1.00 out of 1.00 preservation., Saturation is best upscaled using a pore volume weighted average to preserve dne scale saturation and thus relative permeability. Select one or more: Consider statements below - tick all which are true. a. A symmetric distribution is leptokurtic when IQR equals standard deviation. b. A symmetric distribution is platykurtic when IQR equals standard deviation.  c. For a Gaussian distribution mode, median, and mean are equal.  d. A lognormal distribution is negatively skewed. Your answer is partially correct. You have correctly selected 1. The correct answers are: For a Gaussian distribution mode, median, and mean are equal., A symmetric distribution is leptokurtic when IQR equals standard deviation. Drag the words from the list below into the correct spaces: Reservoir characterisation aims at quantifying the behaviour  of the system. It uses geological data  (sedimentology, diagenesis, grain size…) to describe, quantify, and model the heterogeneous elements in a reservoir  (facies, aow barriers,…). It provides a spatial continuous assignment of physical properties  relevant for reservoir modelling. Your answer is correct. The correct answer is: Drag the words from the list below into the correct spaces: Reservoir characterisation aims at quantifying the [behaviour ] of the system. It uses [geological data] (sedimentology, diagenesis, grain size…) to describe, quantify, and model the [heterogeneous elements in a reservoir] (facies, aow barriers,…). It provides a spatial continuous assignment of [physical properties] relevant for reservoir modelling. Consider the computational costs of stochastic simulation based on Kriging. How can you speed-up the calculation of a realisation for a particular scenario? Assume for simplicity that the variogram has a sill equal to the theoretical sill.

Question 15 15 Correct Mark 1.00 out of 1.00 Question 16 16 Correct Mark 1.00 out of 1.00 great for faults and fractures  network models consist of a set of points with links in between  pure object models treat reservoir as collection of entities with different shapes  Your answer is correct. The correct answer is: data point sets → cost for calculation of distances between points is high, data point sets → are suitable for dense data, pure object methods → great for faults and fractures, network models → consist of a set of points with links in between, pure object models → treat reservoir as collection of entities with different shapes Select one or more: Resampling of grids may be applied for various reasons and may cause certain issues impacting on the accuracy of transfering/rescaling grid values from one grid to another. Mark all that apply. a. Resampling to centre cells may be used to assign some value to a dne grid from a coarse grid, followed by stochastic simulation to generate multiple dne-scale realisations.  b. Aliasing is no problem under grid rotation, since there is a 1-to-1 relationship between grid cells. c. Resampling to all cells after upscaling and differentiating element-wise with the original dne-scale grid values may reveal coarse cells where upscaling is more error prone.  d. Resampling to centre cells may be used to assign some value to a dne grid from a coarse grid, setting a coarse scale trend.  Your answer is correct. The correct answers are: Resampling to centre cells may be used to assign some value to a dne grid from a coarse grid, setting a coarse scale trend., Resampling to centre cells may be used to assign some value to a dne grid from a coarse grid, followed by stochastic simulation to generate multiple dne-scale realisations., Resampling to all cells after upscaling and differentiating element-wise with the original dne-scale grid values may reveal coarse cells where upscaling is more error prone. Select one or more: Consider the case of a sequential indicator simulation (SIS). The procedure is somewhat analogue to SGS, with some pecularities as one deals with categorical variables. Lets assume we have three rock types corresponding to three indicator variables. Which of the following statements are true? a. A layered system can be constructed by projecting the indicator expectation values on a single Gaussian distribution. 

Question 17 17 Correct Mark 1.00 out of 1.00 Question 18 18 Correct Mark 1.00 out of 1.00 values on a single Gaussian distribution. b. Expectation values for local probabilities are formulated with regard to the categorical values, not the indicator variables. c. Each indicator variable must be modeled spatially e.g. by using an indicator variogram.  d. We may consider more than three categorical variables, where each categorical variable may be represented by multiple indicators.  Your answer is correct. The correct answers are: We may consider more than three categorical variables, where each categorical variable may be represented by multiple indicators., Each indicator variable must be modeled spatially e.g. by using an indicator variogram., A layered system can be constructed by projecting the indicator expectation values on a single Gaussian distribution. Select one or more: The pressure solver method in the context of upscaling allows testing of upscaling rules and the determination of the best upscaling strategy. However, certain limits apply. Mark all true statements below. a. The method can only be used with conforming grids.  b. Upscaled values using the pressure solver may be pessimistic.  c. The method has diiculties with pinch outs in dne grid cells. Your answer is correct. The correct answers are: Upscaled values using the pressure solver may be pessimistic., The method can only be used with conforming grids. Select one: Many averaging formula utilized in geostatistical interpolation or estimation techniques need as one input the distance between points. A common differentiation is the choice of stratigraphic or cartesian distances. Which one is larger for a dipping horizon - choose the right answer? a. The cartesian distance is larger.  b. The stratigraphic distance follows the horizon and is larger. c. Both distances are the same. Your answer is correct.

Question 21 21 Correct Mark 1.00 out of 1.00 Question 22 22 Correct Mark 1.00 out of 1.00 function] with some degree of auto-correlation. A further component of is some [random noise]. Often we will model the regionalised variable as the sum of a [polynomial] trend and a correlated higher-frequency [random variable]. Select one or more: Consider the case of stochastic Gaussian simulation. A set of known points are used to constraint possible realisations of a random function in space. Tick one or more true statements below. a. The chosen path (or order of estimation of values at each point) determines the expected mean at each location of the random function. b. The Kriging variance for the estimation of a value at a particular location decreases if points around it have already been estimated via SGS.  c. In a simple implementation the Kriging system continually grows with the number of known points.  d. Taking a sample from a map generated by SGS will be expected to show a histogram similar to the original one; there is no narrowing of the distribution function.  e. The Kriging variance represents the total uncertainty in the data. Your answer is correct. The correct answers are: In a simple implementation the Kriging system continually grows with the number of known points., The Kriging variance for the estimation of a value at a particular location decreases if points around it have already been estimated via SGS., Taking a sample from a map generated by SGS will be expected to show a histogram similar to the original one; there is no narrowing of the distribution function. Conductive  fractures may provide pathways for extraneous water production. Furthermore, at fault locations different lithologies can become discontinous or be enhanced in continuity due to false juxtaposition  . Fracture density is often a factor in horizontal  well performance, which depends on kv/kh. When intersected by high angle wells, vertical  fractures are of particular importance. Water aoods may alter the conductivity  of fractures due to a change in stress regime. Examples where fractures are important include: a) fractures might provide signidcantly higher permeability than surrounding rock matrix. b) naturally fractured reservoirs. c) conductivity of fractures can change with stress regimes, e.g. during water-aood. d) vertical fractures might be intersected in particular by high angle or horizontal wells. e) conductive fractures may provide pathways for extraneous water production. The juxtaposition of different lithologies across faults can either cause interruption or enhancement in continuity.

Question 23 23 Correct Mark 1.00 out of 1.00 Question 24 24 enhancement in continuity. Your answer is correct. The correct answer is: [Conductive] fractures may provide pathways for extraneous water production. Furthermore, at fault locations different lithologies can become discontinous or be enhanced in continuity due to false [juxtaposition]. Fracture density is often a factor in [horizontal] well performance, which depends on kv/kh. When intersected by high angle wells, [vertical] fractures are of particular importance. Water aoods may alter the [conductivity ] of fractures due to a change in stress regime. Examples where fractures are important include: a) fractures might provide signidcantly higher permeability than surrounding rock matrix. b) naturally fractured reservoirs. c) conductivity of fractures can change with stress regimes, e.g. during water-aood. d) vertical fractures might be intersected in particular by high angle or horizontal wells. e) conductive fractures may provide pathways for extraneous water production. The juxtaposition of different lithologies across faults can either cause interruption or enhancement in continuity. Select one or more: Consider the case of structural modeling using variograms. Assume that a theoretical sill is known as well as the range of the variogram. A small non-zero nugget is also predetermined. Given these conditions, which of the following statements are true (tick all that apply) when comparing the mentioned variogram models? a. A nested variogram is constituted of the sum of two simpler variogram models.  b. The exponential variogram has a sill and rises fastest from h=0.  c. The spherical variogram has a sill and has the steepest rise at h=0. d. A nested variogram has two component functions where the drst one is applied over range 1 and the second one over range 2. e. The Gaussian variogram has a sill and exhibits the strongest correlations at h -> 0.  Your answer is correct. The correct answers are: The exponential variogram has a sill and rises fastest from h=0., The Gaussian variogram has a sill and exhibits the strongest correlations at h -> 0., A nested variogram is constituted of the sum of two simpler variogram models. Consider the case of estimating a mean and error variance at a location where actual data

Question 27 27 Correct Mark 1.00 out of 1.00 Question 28 28 Correct Mark 1.00 out of 1.00 Question 29 29 Correct b. Use distance weighted averaging. c. Use an octant search to reduce clustering.  Your answer is correct. The correct answers are: Use a quadrant search to reduce clustering bias., Use an octant search to reduce clustering. Select one or more: Choose all answers that are true. a. The geometric mean is always equal or larger than the harmonic mean.  b. The power mean is a general averaging technique reducing to geometric mean for p=0.  c. The harmonic mean is always larger than the arithmetic mean. Your answer is correct. The correct answers are: The power mean is a general averaging technique reducing to geometric mean for p=0., The geometric mean is always equal or larger than the harmonic mean. Select one or more: Which of the following statements is true? Tick all which are true. a. Reservoir characterisation uses descriptive statistics  Indeed this is part of reservoir characterisation. However, reservoir characterisation is much more general while reservoir description is limited to descriptive statistics. b. Reservoir description targets system behaviour for the purpose of reservoir modelling. c. It places importance on the spatial relationship between different variables.  Your answer is correct. The correct answers are: Reservoir characterisation uses descriptive statistics, It places importance on the spatial relationship between different variables. Consider the case of a variogram describing the spatial arrangement of a reservoir. The variogram shows some undulations and at large lags the variance is increasing

Mark 1.00 out of 1.00 Question 30 30 Correct Mark 1.00 out of 1.00 Question 31 31 Correct Mark 1.00 out of 1.00 Select one or more: monotonically and is signidcantly larger than the sill. Which structural features cause this behaviour (mark all that apply)? a. There is an underlying trend causing the steady increase of variance with distance.  b. Anisotropy causes the undulations in the variogram. c. The undulations may originate from a too small sample.  d. Cyclic features cause undulations in the variogram.  Your answer is correct. The correct answers are: There is an underlying trend causing the steady increase of variance with distance., Cyclic features cause undulations in the variogram., The undulations may originate from a too small sample. Select one or more: Tick all statements which are true. a. PEBI grids due to their irregularity have diiculties in representing fractures b. PEBI grids are limited in representing wells c. Block-centered grids have variable cell dimensions in horizontal directions  d. Regular in plan, stretched in depth grids are excellent to represent proportional layering  e. Regular in plan, shifted in depth grids have constant cell dimensions  f. Regular in plan, shifted in depth grids store the top of the formation with dimension nk Your answer is correct. The correct answers are: Regular in plan, stretched in depth grids are excellent to represent proportional layering, Regular in plan, shifted in depth grids have constant cell dimensions, Block-centered grids have variable cell dimensions in horizontal directions Select one or more: Consider the case of conforming and unconforming grids, posing different upscaling challenges. Which of the following statements are true? a. For layered correspondence vertical correspondence is spatial and horizontal correspondence is logical. b. For conforming grids the cell correspondence is dedned by cell shape.

Question 34 34 Correct Mark 1.00 out of 1.00 The correct answers are: Mechanical discontinuities showing evidence of relative movements are fractures., Mechanical discontinuities showing evidence of relative movements are faults., Mechanical discontinuities showing no evidence of relative movement are fractures. Select one or more: Consider the case of interpolating values for porosity at a desired location. Five values around the point to estimate are known. Then the estimation problem can be written as a linear average of the values at the known points by applying a weighted average. Tick the statements below which are true. a. For a moving average with a window size larger than the largest distance to the known points calculated from the estimation location all weights are equal.  b. Inverse distance weighting results in a continuous differentiable estimation surface.  c. Triangulation cannot be used in cross-validation techniques. d. Applying a moving average method may change the mean of the data.  Your answer is correct. The correct answers are: For a moving average with a window size larger than the largest distance to the known points calculated from the estimation location all weights are equal., Applying a moving average method may change the mean of the data., Inverse distance weighting results in a continuous differentiable estimation surface. Previous Activity Jump to... Next Activity