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Yaomin Jin01-03-2002
Design of Experiments------Morris Method
Outline of the presentation Introduction of screening
technique Morris method Examples Conclusions
Screening technique large-scale models requirement of considerable computer time for each run depend on a large number of input variables
Input factors
x1
x2
x3
…xk
Model Output y=f(x1,x2,…,xk)
Which factor is important?
Morris method(1991) OAT(one factor at a time) the baseline changes at each step
wanders in the input factors space Estimate the main effect of a factor by computing
r number of local measures at x1,x2,…xr in the
input space then take average.
Elementary effects Reduce the dependence on the specific
point that a local experiment has. Determine which factor have: negligible effects linear and additive effects non-linear and interaction effects
Elementary effects k dimensional factors vector x for the
simulation model has components xi that have p-values in the set {0, 1/(p-1),…,1}
The region of the experiment is a k dimensional p level grid. In practical applications, the values sampled in are subsequently rescaled to generate the actual values of the simulation factors.
Δ=1/(p-1).
Elementary effects of i-th factor at given point x
1 2 1 1[ ( , ,..., , , ,..., ) ( )]( ) i i i k
i
y x x x x x x y xd x
where x is any value in selected such that the perturb point is still in . A finite distribution Fi of elementary effects for the i-th input factor is obtained by sampling x from . The number of elements of each Fi is
x
1[ ( 1)]kp p p
Economy of Morris method In the simplest form, the total computational effect
required for a random sample of r values from each distribution Fi is n=2rk runs. Each elementary effect requires the evaluation of y twice.
The simplest form of Morris design has an economy rk/2rk=1/2.
# of elementary effects estimated by the design
# of runsEconomy of design
Based on the construction of a matrix B* with rows that
represent input vectors x, for which the corresponding experiment provides k elementary effects (one for each input factor) from k+1 runs. Economy of the design is k/(k+1). assume that p is even and , each of the elementary effects for the i-th input factor has equal probability of being selected. The key idea is: Base value x* is randomly chosen from the vector x, each component xi being sampled from the set
One or more of the k components of x* are increased by such that vector x(1) still in
/[2( 1)]p p 1[ ( 1)] 2
kk pp p p
10, ,...,1-(p-1)
Economical design
The estimated elementary effect of the i-th component of x(1) (if the i-th component of x(1) has been changed by ) if x(1) increased by Δ; if x(1) decreased by Δ. Let x(2) be the new vector , select a third vector x(3) such that differs from x(2) only one component j:
(1) (1) (1) (1) (1) (1) (1)(1) 1 2 1 1[ ( , ,..., , , ,..., ) ( )]
( ) i i i ki
y x x x x x x y xd x
(1) (1) (1) (1) (1) (1) (1)(1) 1 2 1 1[ ( ) ( , ,..., , , ,..., )]
( ) i i i ki
y x y x x x x x xd x
(1) (1) (1) (1)1 1( ,..., , ,..., )i i kx x x x
(3) (2) ,j jx x j i
Economical design (continue)
Economical design (continue)
Repeat the upper step get the k+1 input vectors x(1),x(2),…,x(k+1) , any component i of x* is selected at least once to be increased by . To estimate one elementary effect for each factor.
(2) (3)(2) [ ( ) ( )]
( )j
y x y xd x
(3) (2)(2) [ ( ) ( )]
( )j
y x y xd x
else
Economical design (continue)
The rows of orientation matrix B* are the vectors
describe above. This provides a single elementary effect per factor.
To build B*, Restrict attention: a. p is even; b. Firstly, selection of sampling matrix B with
elements that are 0 or 1, such that every column there are two rows of that differ in only one element.
(1) (2) ( 1), ,..., kx x x
2 1p
p
In particular, B may be chosen to be a strictly lower triangular matrix of 1, consider B’ given by
( 1)
0 0 0 ... 0
1 0 0 ... 0
1 1 0 ... 0
... ... ... ... ...
1 1 1 ... 1k k
B
Economical design (continue)
Jk+1,1 is a matrix of 1. as a design matrix(i.e. each row a value for
x) x* is randomly chosen base value of x. B’ could be used as a design matrix. Each element is randomly assigned a value from with equal probability. Since it would provide k elementary effects; one effect each input factor, with a computation cost of k+1 runs. However the problem is that the k elementary effects B’ produces would not be randomly selected.
B
10, ,...,1-(p-1)
Economical design (continue) *
1,1' kB J X B
Economical design (continue)
A randomised version of the design matrix is given by
where
* * * *1,1 1, , 1, ,( 22k k k k k k k k kB J x B J D J P
•D* is diagonal matrix in which each diagonal element is either +1 or -1•P* is random permutation matrix, in which each column contains one element equal to 1 and all the others equal to 0, and no two columns have 1’s in the same position•B* provides one elementary effect per factor that is randomly selected.
Suppose that p=4, k=4 and , that is, four factors that may have values in the set {0,1/3,2/3, 1}. Then B5*4 is given by
Example2
3
0 0 0 0
1 0 0 0
1 1 0 0
1 1 1 0
1 1 1 1
B
and the randomly generated x*, D* and P* happen to be
1 1* (0, ,0, )
3 31 0 0 0
0 1 0 0*
0 0 1 0
0 0 0 1
x
D
0 0 1 0
0 0 0 1*
0 1 0 0
1 0 0 0
P
Example (continue)
*1, 1,
2 20 0
3 320 0 0 0 030 0 0
2 22 2 0 0 0 0
3 30 0 0
20 0 00 0 3
2 20 0
3 3
k k k kB J D J
Example (continue)
(1) (2) (3) (4) (5)1 1 1 1 1 1 1 1 1 1 1 1( ,1,1, ), ( ,1, , ), ( ,1, ,1), ( , , ,1), (1, , ,1).3 3 3 3 3 3 3 3 3 3 3 3
x x x x x
1 11 1
3 31 1 1
13 3 31 1
* 1 13 31 1 1
13 3 3
1 11 1
3 3
B
Example (continue)
To estimate the mean and variance of the distribution Fi(i=1,…,k), take a random sample of r elements;
that is sample r mutually independent orientation matrices. Since each orientation matrix provides one elementary effect for every factor, the r matrices together provide r×k dimensional samples, one for each Fi(i=1,…,k). We use the classic estimate for
every factor’s mean and standard deviation.
Elementary effects
Elementary effects The characterization of the distribution Fi through its
mean and standard deviation gives useful information about the influence on the output;
a high mean indicates a factor with an important overall influence on the output,
a high standard deviation indicates either a factor interacting with other factors or a factor whose effect is non-linear.
The lines constituting a wedge, are described by ; where is the standard deviation of the mean elementary effect. If the parameter has coordinates below the wedge, i.e. , this is a strong indication that the mean elementary effect of the parameter is non-zero. A location of the parameter coordinates above the wedge indicates that interaction effects with other parameters or non-linear effects are dominant.
Standard of importance
2 ii
Sdr
/iS r
2| | Sdr
1 2 3 410 25 40 75f x x x x
Example 1
Result of example 2(p=4, Δ=2/3 and r=4)
Example 2
10 10 10 10
0 , , , , , ,i i i j i j i j l i j l i j l s i j l si i j i j l i j l s
y w w w w w w w w w w
where wi=2(xi-0.5) except for i=3,
wi=2(1.1xi/(xi+0.1)-0.5), otherwise.
Coefficients of relatively large value were assigned as
20,i 1,...,5,i
, 15,i j , 1, 2,3i j
, , 10,i j l , , 1, 2,3i j l
, , , 5,i j l s , , 1, 2,3, 4i j l , , , , ,, 0,i j l i j l s others
,, ~ (0,1)i i j N }
}
•Input 1-5 are clearly separated from the cluster of remaining outputs, which have means and standard deviations close to 0. •In particular, inputs 4,5 have mean elementary effects that are substantially different from 0 while having small deviations. •Consider both means and standard deviations together, we conclude that the first 5 inputs are important, and that of these the first three appear to have effects that involve either curvature or interactions. •This is coincide with the model.
Result of example 2(from graph)
Conclusions Economical for models with a large
numbers of parameters Does not dependent on any
assumptions about the relationship between parameters and outputs
Results are easily explained in a graph Drawback is not consider the
dependencies between parameters(such as interactions)