An Example of Augmenting a Latin Hypercube
knitr::opts_chunk$set(
collapse = TRUE,
comment = "#>"
)
source("VignetteCommonCode.R")
require(lhs)
graph2DaugmentLHS1 <- function(sims, extras)
{
A <- randomLHS(sims, 2)
B <- augmentLHS(A, extras)
plot.default(A[,1], A[,2], type = "n", ylim = c(0,1),
xlim = c(0,1), xlab = "x1", ylab = "x2", xaxs = "i", yaxs = "i", main = "")
for (i in 1:length(A[,1]))
{
rect(floor(A[i,1]*sims)/sims, floor(A[i,2]*sims)/sims,
ceiling(A[i,1]*sims)/sims, ceiling(A[i,2]*sims)/sims, col = "grey")
}
points(A[,1], A[,2], pch = 19, col = "red")
abline(v = (0:sims)/sims, h = (0:sims)/sims)
return(list(A = A, B = B, sims = sims, extras = extras))
}
graph2DaugmentLHS2 <- function(X)
{
A <- X$A
B <- X$B
sims <- X$sims
extras <- X$extras
plot.default(A[,1], A[,2], type = "n", ylim = c(0,1),
xlim = c(0,1), xlab = "x1", ylab = "x2", xaxs = "i", yaxs = "i", main = "")
N <- sims + extras
for (i in 1:length(B[,1]))
{
rect(floor(B[i,1]*N)/N, floor(B[i,2]*N)/N,
ceiling(B[i,1]*N)/N, ceiling(B[i,2]*N)/N, col = "grey")
}
points(A[,1], A[,2], pch = 19, col = "red")
points(B[((sims + 1):(sims + extras)), 1], B[((sims + 1):(sims + extras)), 2],
pch = 19, col = "blue")
abline(v = (0:N)/N, h = (0:N)/N)
}
# X <- graph2DaugmentLHS1(5,5)
# graph2DaugmentLHS2(X)
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 lhs
package.
First create a random Latin hypercube using the randomLHS(n, k)
command:
A <- randomLHS(5,2)
An example of this hypercube is shown in r registerFigure("X")
. 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.
r addFigureCaption("X", "A randomly produced Latin Hypercube with uniform marginal distributions for 2 parameters with 5 simulations", register=FALSE)
set.seed(10)
X <- graph2DaugmentLHS1(5, 5)
Next, in order to augment the design with more points use augmentLHS(lhs, m)
. The following will add 5 more points to the design:
B <- augmentLHS(A, 5)
The 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 r registerFigure("Y")
.
r addFigureCaption("Y", "A randomly produced Latin Hypercube of 5 points (red) with 5 augmented points (blue). Each parameter has a uniform marginal distribution.", register=FALSE)
graph2DaugmentLHS2(X)
The augmentLHS
function uses the following algorithm (see the documentation for augmentLHS
):
- Create a new
(n+m)
byk
matrix to hold the candidate points after the design has been re-partitioned into(n+m)^2
cells, wheren
is number of points in the originallhs
matrix. - Then randomly sweep through each
column (1...
k
) in the repartitioned design to find the missing cells. - For each column (variable), randomly search for an empty row, generate a
random value that fits in that row, record the value in the new matrix.
The new matrix can contain more than
m
points unlessm = 2n
, in which case the new matrix will contain exactlym
filled rows. - Finally, keep only the first
m
rows of the new matrix. It is guaranteed that there will bem
full 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 r registerFigure("Z")
and r registerFigure("W")
.
The commands were:
A <- randomLHS(7, 2)
B <- augmentLHS(A, 3)
r addFigureCaption("Z", "Original design with 7 points", register=FALSE)
set.seed(12)
X <- graph2DaugmentLHS1(7, 3)
r addFigureCaption("W", "Augmented design with 3 additional points. Note that row 9 has 2 points and row 3 has none.", register=FALSE)
graph2DaugmentLHS2(X)