Question 7.10: In a series of laboratory experiments to test the applicabil......
In a series of laboratory experiments to test the applicability of Hudson’s formula to a new armour unit, the relationship
\frac{H_{s} }{D_{n50}\Delta }=A\left(K_{D}\cot \left(\alpha \right) \right) ^{1/3}is determined where the parameter A is used to account for spread and uncertainty in the results. The reliability function in this case can be written as
G=AD_{n50}\Delta\left(K_{D}\cot \left(\alpha \right) \right) ^{1/3}-H_{s}The experimental values are such that K_{D} = 4, cot(α) = 2, Δ = 1.6 and D_{n50} = 1.5.
The parameter A is N(1, 0.18) and H_{s} is N(4.4, 0.7). In fitting their data to the formula the experimentalists note a positive correlation between the values of wave height and the parameter A of 0.2.
a. What is the probability of failure?
b. If, subsequently, new experiments are performed that weaken the correlation
between A and H_{s} to −0.2, what effect does this have on the probability of failure?
c. What is the result of ignoring the correlation?
a. The reliability function may be written as G = X – Y. We can determine directly:
\mu _{G} =\mu _{X}-\mu _{Y}=1×1.5×1.6×\left(8\right) ^{1/3} -4.4=0.4
\sigma_G=\left(\sigma_X^2-2 \rho_{X Y} \sigma_X \sigma_Y+\sigma_Y^2\right)^{1 / 2}=(0.746-2 \times(0.2) \times 0.864 \times 0.7+0.49)^{1 / 2}=0.997
[The following result has been used: if X is a random variable then the variable
Z = AX, where A is a constant, has mean = A x the mean of X and variance = A² x the variance of X].
P_f=\Phi\left\lgroup-\frac{0.4}{0.997}\right\rgroup=1-\Phi(0.401) \approx 1-0.655=0.34
So the probability of failure is 0.345.
b. The mean remains the same, but the standard deviation of G alters:
\begin{aligned} & \sigma_G=\left(\sigma_X^2-2 \rho_{X Y} \sigma_X \sigma_Y+\sigma_Y^2\right)^{1 / 2}=(0.746-2 \times(-0.2) \times 0.864 \times 0.7+0.49)^{1 / 2}=1.216 \\ & P_f=\Phi\left\lgroup-\frac{0.4}{1.216}\right\rgroup =1-\Phi(0.329) \approx 1-0.629=0.371 \end{aligned}
Thus, the inclusion of new data leads to a slight increase in the probability of failure.
c. Neglecting the correlation again leaves the mean unchanged but alters the standard deviation of G as
\begin{aligned} & \sigma_G=\left(\sigma_X^2+\sigma_Y^2\right)^{1 / 2}=(0.746+0.49)^{1 / 2}=1.112 \\ & P_f=\Phi\left\lgroup-\frac{0.4}{1.112}\right\rgroup=1-\Phi(0.360) \approx 1-0.641=0.359 \end{aligned}
Thus, in this case ignoring the correlation leads to an overestimate of the reliability of the armour (or underestimate the probability of failure) relative to the results with the additional measurements.