 Solution

The first binned empirical semivariogram rises strongly through the displayed distances, approximately quadratically, with no visible sill. The spatial data plot shows higher depth at larger $y$. A stationary residual covariance should not be asked to explain that deterministic drift.

For a model $Z(s)=m(s)+W(s)$ with stationary zero-mean residual field, each raw squared difference has expectationWith linear drift $m(x,y)={\beta }_0+{\beta }_{1}y$, the added term is $\frac{1}{2}{\beta }_1^2(y_i-y_j{)}^2$. Its average within distance bins can grow with separation, producing the shape seen in the first plot. The empirical semivariogram uses half the average squared difference over the pairs in each bin, so removing only a common intercept does not remove this drift contribution.

The formula depth ~ y estimates and removes a linear drift before computing a residual empirical semivariogram. The second graph levels off near $0.47$ with a fitted zero nugget and a finite scale, making the Gaussian residual covariance a plausible working model. Estimate the drift and fit spatial dependence to residuals rather than the raw depth differences. Estimated residuals are not the true errors, so their semivariogram has estimation effects; sparse long-distance bins are also noisy. These graphs support the model but do not establish stationarity or isotropy.

The subpart refers to svgm and svgm2, while the code calls the objects smvg and smvg2. Also the printed code omits the line that fits gauss.model2 before plotting or using it. The supplied output must be interpreted as the result of that missing residual-variogram fit, rather than as an executable complete script.