Using the least squares method, determine the correction, in millimeters, to the field difference in elevation along route 2.

• A. -60
• B. -50
• C. -55
• D. -65

Answer: (C)

In a least-squares adjustment for a route, you’re finding the correction to apply to the observed field differences so that the overall set of differences best fits by minimizing the sum of squared residuals. If you assume a single, uniform correction applies to all the differences, the optimal value is the one that makes the total adjusted difference zero. Denoting the observed differences as d1, d2, ..., dN, applying the same correction x gives adjusted values di + x. Setting the sum of these to zero yields x = - sum(di) / N, which is the negative average of the observed differences. Therefore, the least-squares correction is the negative of the average misfit.

So, the result being -55 millimeters means the average observed misfit across route 2 was 55 mm too large, and the correction should be applied as a reduction of 55 mm to those field differences. The other numerical options would correspond to different averages and would produce larger overall residuals, not the minimum.