Question 17 of 23: Block Estimate from Samples A, B, C – Weight Sum and Likely Weights

Question 4.4: Block Estimate from Samples A, B, C – Weight Sum and Likely Weights (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

4.4.1 Sum of the weights. As in Question 4.3, this is still an ordinary-kriging block estimate, so $$w_A+w_B+w_C=\boxed{1.0}$$ regardless of the specific geometry — the unbiasedness constraint applies to any OK system.

4.4.2 Likely weight split. In the source sketch, samples B and C sit close together to one side of the block (roughly similar, moderate distance from the block, and close to EACH OTHER), while sample A sits alone on the opposite side at a broadly similar distance to the block as B and C individually. With the stated range at three times the longest sample-to-block/sample-to-sample distance, every pairwise separation (sample-to-block and sample-to-sample) falls well inside the variogram's rising portion, so distance alone would suggest roughly equal thirds (≈0.33 each). However, because B and C are also close TO EACH OTHER, they partially screen/duplicate one another's information (the same screen effect as Question 4.3), so the OK solution should shift weight modestly AWAY from the closer B/C pair and TOWARD the more independent, unclustered sample A — a reasonable qualitative split is roughly w_A≈0.40, w_B≈0.32, w_C≈0.28 (summing to 1.0), rather than a naive equal 0.33/0.33/0.33, with the exact numbers only obtainable by actually solving the 4×4 (3 samples + Lagrange multiplier) OK system against the measured coordinates and the fitted variogram.