Suppose that a computer simulation study is being designed that
requires expensive runs. A Latin hypercube design is desired for this
simulation so that the expectation of the simulation output can be
estimated efficiently given the distributions of the input variables.
Latin hypercubes are most often used in highly dimensional problems, but
the example shown is of small dimension. Suppose further that the total
extent of funding is uncertain. Enough money is available for 5 runs,
and there is a chance that there will be enough for 5 more. However, if
the money for the additional 5 runs does not materialize, then the first
5 runs must be a Latin hypercube alone. A design for this situation can
be created using the
First create a random Latin hypercube using the
randomLHS(n, k) command:
An example of this hypercube is shown in Figure 1. Note that the Latin property of the hypercube requires that each of the 5 equal probability intervals be filled (i.e. each row and each column is filled with one point). Also notice that the exact location of the design point is randomly sampled from within that cell using a uniform distribution for each marginal variable.
Next, in order to augment the design with more points use
augmentLHS(lhs, m). The following will add 5 more points to
<- augmentLHS(A, 5)B
augmentLHS function works by re-dividing the
original design into
n+m intervals (e.g. 5+5=10) keeping
the original design points exactly in the same position. It then
randomly fills the empty row-column sets. The results are shown in
augmentLHS function uses the following algorithm
(see the documentation for
kmatrix to hold the candidate points after the design has been re-partitioned into
nis number of points in the original
k) in the repartitioned design to find the missing cells.
m = 2n, in which case the new matrix will contain exactly
mrows of the new matrix. It is guaranteed that there will be
mfull rows (points) in the new matrix. The deleted rows are partially full. The additional candidate points are selected randomly because of the random search used to find empty cells.
Also notice that because the original points are randomly placed within the cells, depending on how you bin the marginal distributions, a histogram (of x1 for example) will not necessarily be exactly uniform.
Now, the augmenting points do not necessarily form a Latin Hypercube themselves. The original design and augmenting points may form a Latin Hypercube, or there may be more than one point per row in the augmented design. If the augmented points are equal to the number of original points, then a strictly uniform Latin hypercube is guaranteed. An example of an augmented design which is not uniform in the marginal distributions is given in Figure 3 and Figure 4. The commands were:
<- randomLHS(7, 2) A <- augmentLHS(A, 3)B