How to Analyze Latin Hypercube Results After running an experiment generated by a Latin Hypercube design, what next? Just pick the best point? I guess there’s no ‘modeling’ or ‘curve fitting’ going on right? So you basically just choose the optima of the runs as your parameters?
Is there more to it? Thanks.
 A: The next step is dependent on the type of problem you are solving.  It sounds like the problem you are interested in is an optimization problem.  In this case, you can model your response against the Latin hypercube design points and then optimize the fitted curve or surface.  This can be a much more accurate optimum point (obviously depending on where your design was created).  An example is below using R.
require(lhs)

# true underlying function of the parameters you are measuring
true_function_of_parameters <- function(x1, x2)
{
  stopifnot(length(x1) == length(x2))
  return(5 - 2*x1^2 - 3*x2^2 + rnorm(length(x1), 0, 1))
}

# sample Latin hypercube with 20 samples of two input parameters
set.seed(1976)
X <- lhs::randomLHS(20, 2)

# transform the lhs margins to a distribution that matches your data
X[,1] <- qunif(X[,1], 0, 10)
X[,2] <- qnorm(X[,2], 5, 1)
X <- as.data.frame(X)
names(X) <- c("V1", "V2")

# conduct the experiment and get output
Y <- true_function_of_parameters(X$V1, X$V2)

# model to determine the functional form
lm1 <- lm(Y ~ V1 + I(V1^2) + V2 + I(V2^2), data = X)
# Result:  Y = -0.2275 + -0.4186*V1 + -1.9514*V1^2 + 1.9765*V2 + -3.1794*V2^2

# find the optimum
## using the best point in the lhs
ind <- which.max(Y)
X[ind,]
# V1 = 0.1255928, V2 = 2.991991

## find the optimum from the fitted linear function
### V1
-coef(lm1)[2]/2/coef(lm1)[3]
# -0.1072474
### V2
-coef(lm1)[4]/2/coef(lm1)[5]
# 0.3108302

## true optimum
### V1 = 0, V2 = 0

