Problem for Kriging Estimation500 mN There are 4 drillholes of x1, x2, x3, and x4 which located randomly in the middle of regular grid where the grid spacing is 100100 m (see figure). Those drillholes were derived from exploration of primary Au deposit. The Au grades for each known points are: z(x1) = 10 ppm, z(x2) = 20 ppm, z(x3) = 5 ppm, and z(x4) = 15 ppm. If the variogram model for Au grades distribution is Spherical Model with range = 200 m, nugget variance (C0) = 5 ppm2, and sill (C) = 25 ppm2, then: 400 x4 300 x3 V ? 200 x2 100 x1 0 100 200 300 400 500 mE 1. Please estimate the Au grade for Block V or z(V)* by using Ordinary Block Kriging, by assuming the searching radius for estimation is 250 m from the central of Block V! The average variogram between data points and estimated block are: x1 ,V 25 ppm2 x3 ,V 20 ppm2 x2 ,V 30 ppm2 x4 ,V 30 ppm2 2. Please calculate kriging variance from the estimation result! The average variogram within block is: V ,V 15 ppm2 . Also calculate the relative standard deviation of estimation (error) on Block V! Note: relative to the estimated grade in Block V. 3. Please compare the estimated grade of Au in Block V by using conventional estimation methods such as IDS and NNP with searching radius of estimation is 250 m from the central of Block V! 4. Please compare the result of block kriging estimation by using Ordinary Point Kriging to estimate the point located in the central of Block V (named x0)! Please also calculate its kriging variance and relative standard deviation! x . and . x 3 x 1 . x 1 2 x 1 . x 2 3 x 1 . x x .61 m or h > a. V (1) 1 x 2 .Solution for Kriging Estimation Variogram model for Spherical with nested structure of: nugget effect (C0) = 5 ppm2. x 3 x 2 . To solve the equation we need to calculate the variogram between point to point using the formula and condition in item (a) above. sill (C) = 25 ppm2. so (xi. The distance x1 to x2 = x2 to x1 = 316. x 2 3 x 2 . the formula is: 3 h h 3 γ(h ) C 0 C 3 for h a 2a 2a C 0 C for h a 0 for h 0 Then the data used for estimation are only z(x1). x 1 2 x 3 . range (a) = 200 m. consequently the z(x4) is not used for estimation. V n i i1 i j i n 1 i1 i We could expand the formula above as follow: 1 x 1 . xj) = C0 + C The distance x1 to x3 = x3 to x1 = 223. Ordinary Block Kriging System: . x 3 x 3 . V (3) 1 (4) 2 3 0 1 Our task is to find 1. xj) = C0 + C . xj) = C0 + C The distance x2 to x3 = x3 to x2 = 223. z(x2).23 m or h > a.61 m or h > a. so (xi. and z(x3) due to the constraint on searching neighborhood 250 m. Z(V)*. so (xi. x 2 3 x 3 . 2. 3. x 1 2 x 2 . n n i1 i1 Estimator formula: z(V)* i zx i for block kriging or z(x 0 )* i zx i for point kriging 1. V (2) 1 x 3 . 333 + 2 = 1 32 = 0.175 ppm Au 1. Kriging Variance is given as: Block kriging variance: n V.499 5) = 9.5 2 = 0.499 Finally: z(V)* = 1.334 10) + (0. x i .167 20) + (0.z(x3) = (0.z(x2) + 3.167 1 = 0.2 = 0.302 = 5 1 = 0.334 3 = 1 .167 + 0.Then the solution is: 1 0 2 30 3 30 25 (1) 130 2 0 3 30 30 (2) 130 2 30 3 0 20 (3) 1 (4) 2 3 0 1 Subtracting (2) from (1) then: 301 . V i .z(x1) + 2.167 + 2 (5) Subtracting (2) from (3) then: -302 + 303 = 10 3 = 0.1 .167 + 2 + 2 + 0.333 + 2 (6) Substituting (5) and (6) to (4) then: 0. V 2 K i1 .167 = 0. Ordinary Point Kriging System: . x n i i1 i j i 0 n 1 i1 i We could expand the formula above as follow: 1 x 1 .42)2) + (20 / (200)2) + (5 / (100)2)] / [(1 / (141.To solve the equation. x 0 (1) .02 + (0. Lagrange multiplier from the kriging system in item (b): (1): 302 303 25 then = 5. we need to calculate . x 1 2 x 1 . x 2 3 x 1 . x .84 % 9.175 2.36 ppm2 To calculate relative error or relative kriging standard deviation in block V: (V )* 13.334 25) + (0.42)2) + (1 / (200)2) + (1 / (100)2) = 8.57 ppm Au 3.167 30) + (0.02 2 = -15 + 5.65 100% 39.36 3.65 ppm (V ) *relative (V ) * z (V ) * 100% 3.499 20) = 13. x x . Estimation in V or x0 using NNP (Nearest Neighborhood Point) and IDS (Inverse Distance Square): NNP z(V)* = z(x0)* = z(x3) = 5 ppm Au IDS z(V)* = z(x0)* = [(z(x1) / (hx1-x0)2) + (z(x2) / (hx2-x0)2) + (z(x3) / (hx3-x0)2)] / [1 / (hx1-x0)2) + (1 / (hx2-x0)2) + (1 / (hx3-x0)2)] = [(10 / (141. x 3 x 1 . 097 + 2 (5) Subtracting (2) from (3) then: -302 + 303 = 7. x 1 2 x 3 .z(x1) + 2.1 x 2 . x 1 2 x 2 .2 = 1 – 0.z(x3) = (0.097 + 2 + 2 + 0.214 = 0. x 0 (2) 1 x 3 . To solve the equation we need to calculate the variogram between point to point using the formula and condition in item (a) above.311 – 0.214 1 = 0.311 10) + (0.10 (1) 130 2 0 3 30 30 (2) 130 2 30 3 0 22. x 3 x 3 . x 3 x 2 .z(x2) + 3. Z(x0)*.19 (3) 1 (4) 2 3 0 1 Subtracting (2) from (1) then: 301 .260 + 2 (6) Substituting (5) and (6) to (4) then: 0.1 . So the solution is: 1 0 2 30 3 30 27.643 2 = 0. x 2 3 x 3 .214 20) + (0. 2. x 2 3 x 2 .260 + 2 = 1 32 = 0.097 + 0. 3.475 Finally: z(x0)* = 1. and .214 = 0.302 = 2.81 3 = 0.475 5) .311 3 = 1 .9 1 = 0. x 0 (3) 1 (4) 2 3 0 1 Our task is to find 1. Estimated value of point kriging > block kriging > IDS > NNP 2.43 2(x0)* = 6.10 then = 6. x 0 i1 To solve the equation.82 5.10) + (0. x i . Estimated value of block kriging IDS 3.64 ppm ( x0 ) *relative ( x0 ) * z ( x0 ) * 100% 5.82 ppm2 To calculate relative error or relative kriging standard deviation in point x0: ( x0 )* 31.475 22. Standard deviation of error of block kriging < point kriging .77 % 9.765 ppm Au Point kriging variance: n K2 i .43 + (0.64 100% 57.= 9. we need to calculate .214 30) + (0. Lagrange multiplier from the kriging system in item (b): (1): 302 303 27.311 27.19) = 31.765 Conclusion: 1.