## Task 24/67

In the series assembly of 1000 devices, which cost 100 MDN per device, 300 resistors are installed per device, and the inspection to ensure tolerance compliance costs 0.12 MDN per resistor.

The probability that a resistor does not meet the tolerance, according to statistical data, is 0.1%. The device functions properly only if all resistors are within tolerance. An end inspection is always required.

What is more economically advantageous - inspecting the resistors before assembly or foregoing the inspection?

The material costs for defective devices can be disregarded, as the material remains usable.
For the decision-making question, the costs $K_{1}$ for testing the resistors before assembly and the expected costs $K_{2}$ for assembling non-functional devices if control is waived are presented.

300 resistors per device $\cdot 1000$ devices $=300000$ resistors are needed, which must be flawless. Since only $99.9\%$ of the resistors are expected to be within specifications, $\frac{300000 \cdot 100}{99.9}=300300$ resistors need to be tested.

The costs for this amount to $K_{1}=36036$ MDN.

The probability $P(\bar{A})$ that only flawless resistors will be installed in a device if control is waived is $P(\bar{A})=0.999^{300}$; for the probability that a device will be defective due to faulty resistors, we have

$$
P(A)=1-P(\bar{A})=1-0.999^{300} \approx 0.256=25.6\%
$$

This means that $25.6\%$ of the devices are likely to be defective, so $\frac{1000 \cdot 100}{74.6} \approx 1340$ devices need to be produced, resulting in additional assembly costs for 340 devices. Thus, $K_{2}=34000$ MDN. It is evident: $K_{1}>K_{2}$. Waiving the prior control is economically more advantageous.