Illustrative numerical examples

Example 1. Polynomial with two random variables. Consider a simple example with Z defined by,

z=1X4+ZZX; Xj 2

where 5 = 2, X and X2 are independent normally distributed random variables with a mean of zero and standard deviation of one. It can be shown that the exact mean equals 7.

Table 1. Relation between the locations, xi, and probability concentrations, p, in the point estimate method and the abscissas, xt’, and weights, wi, of well known integration formulas.

Guassian Integration formulas

n

1w(t)h(t)dt« ^ wth(tt)

П, "=1

Point estimate methods

n

1 f (x)h(x)dx « ^p, h(xt)

Ox i=1

Jacobi integration w(t) = (1 -1 )“(1 +t/ Qt є (-1,1)

Beta distribution

f (x) = (b – x)a (a + xf /((b – a)a+/i+lB(a, ft)j, QX є (a, b) pi = wt /((b – a)a+^+1 B(a, fl)), xt = ((b – a)tt + (b + a))/2

Legender integration w(t) = 1, Qt є (-1,1)

Uniform distribution

f (x) = 1/(b – a), QX є (a, b), pt = wt /2, xi = ((b – a)tt + (b + a))/2

Generalized Laguerre integration

w(t) = tae^, Qt є (0, да)

Gamma distribution

f (x) = Za+lxae-Ax / T(a +1), aX є (0, да) pt = wi /T(a +1), xt = ti IX

Laguerre integration

w(t) = e~l, Qt є (0, да),

Exponential distribution

f (x) = Xe~**, QX є (0, да), pt = wt, xt = tt / X

Hermite integration

w(t) = e, Qt e (-да, да)

Normal distribution

f (x) = ,— exp ^ ( ‘) , Qx є ( да, да) <2лс7 2 а ) J

pt = wt/^, xt = m + t^2a

If we use the LHS technique, samples for a typical run with 100 cycles are shown in Figure 2. These samples are obtained by partitioning the domain of each random variable into 100 intervals each with the same probability. Now if we replace the original probability distribution function of each of the random variables by 5 concentrations obtained using the point estimate method, the samples for a typical run with 100 cycles obtained by using the HPCS technique are also shown in Figure 2. Note that by using the HPCS, one need to carry out only 13 evaluations for this particular run because many samples fall into the same point.

By repeating the above analysis many times, the average value and the standard deviation of the mean of Z are shown in Figure 3. The results shown in the figure indicate that the accuracy of the proposed HPCS technique is comparable to that of LHS the technique.

Example 2. Polynomial with many random variables. Consider Eq. (13) but with s equal to 5. Xi, i -1,2, • • •, s, are independent normally distributed variates with a mean of zero and standard deviation of one.

We replace the original probability distribution function of each of the random variables by k concentrations based on the point estimate method. For k equal to 3, 4 and 5, the joint probability distribution of the random variables is represented, respectively, by 243, 1024 and 3125 probability concentrations. Using n partitions with equal probability of 1/n for each of the random variables and carrying out simulation analysis using the HPCS, the obtained results are shown in Table 2a. Also shown in the table is the number of evaluations of h(X), ne, needed by the HPCS for the simulation runs carried out. Clearly, the results show that ne is less than n
which represents the number of functional evaluations if the LHS technique is employed. The reduction in the computational effort is most significant if к is small and/or n large. Note that in all cases the obtained mean values are close to the exact value which is equal to 25.

4

4

-4 -2 0

Xi

Number of runs (each with 100 cycles)

Figure 3. Mean and standard deviation of the mean of the performance function.

Table 2b. Results for Example 2 with s = 10.

Repeat this analysis but with 5 = 10. The obtained results are shown in Table 2b. Comparison of the results shown in Tables 2a and 2b suggests that ne for 5 = 5 is smaller than that for 5 = 10. This is expected since in the latter for к equal to 3, 4 and 5, the joint probability distribution function of the random variables is represented, respectively, by 59049, 1048576, and 9765625 concentrations.