Question 2.1: A semiconductor fabrication plant has an average output of 1......
A semiconductor fabrication plant has an average output of 10 million devices per week. It has been found that over the past year 100,000 devices were rejected in the final test.
(a) What is the unreliability of the semiconductor devices according to the conducted test?
(b) If the tests reject 99% of all defective devices, what is the chance that any device a customer receives will be defective?
The total number of devices produced in a year is:
(a) n_{0} = 52 \times 10 \times 10^{6} = 520 \times 10^{6}
The number of rejects (failures), n_{f} , over the same period is:
n_{f} = 1 \times 10^{5}.
Therefore, from Equation 2.2, an estimate for device unreliability is:
\hat{Q}(t)=\frac{n_f(t)}{n_0}, ( 2.2)
\hat{Q}(t)=\frac{n_f(t)}{n_0}=\frac{1 \times 10^5}{520 \times 10^6} \approx 1.92 \times 10^{-4},
or 1 chance in 5200.
(b) If the rejected devices represent 99% of all the defective devices produced, then the number of defectives that passed testing is:
x_d=\left[\frac{1 \times 10^5}{0.99}-\left(1 \times 10^5\right)\right] \approx 1010Therefore, the probability of a customer getting a defective device, or the unreliability of the supplied devices on first use, is:
\hat{Q}(t)=\frac{1010}{\left(520 \times 10^6\right)-\left(1 \times 10^5\right)} \approx 1.94 \times 10^{-6},or 1 chance in 515,000.
