# Decision Problem on Imperfect Inspections Combined under Two-Stage Inspection Policy
## 5 Discussion
 \tag{21}\]  
Because the failure and preventive replacement probabilities are complementary, the expected number of preventive replacements may be estimated by the formula:  
\[\Delta ND(T,n\_I\_1,n\_II\_1,n\_II\_2)=ND_{II}(T,n\_II\_1,n\_II\_2)-ND_{I}(T,n \_I\_1)=-\Delta NF(T,n\_I\_1,n\_II\_1,n\_II\_2) \tag{22}\]  
The change in the expected number of the first-stage inspections during a renewal cycle if the two-stage policy is applied in the system may be described by the formula:  
\[\begin{array}{l}\Delta NI\_1 st(T,n\_I\_1,n\_II\_1,n\_II\_2)=NI\_1 st_{II}(T,n \_II\_1,n\_II\_2)-NI\_1 st_{I}(T,n\_I\_1)=\\ \sum_{i=1}^{\infty}(1-F_{U}(t_{i}))\cdot\left[(1-\alpha_{n\_II\_1}\cdot\alpha_ {n\_II\_2})^{z-1}-(1-\alpha_{n\_I  
The total increase in costs if the two-stage inspection policy is applied in comparison with the one-level policy is the sum of cost components:  
\[\Delta C(T,n\_I\_1,n\_II\_1,n\_II\_2)\] \[=c_{F}\cdot\Delta NF(T,n\_I\_1,n\_II\_1,n\_II\_2)+c_{D}\cdot \Delta ND(T,n\_I\_1,n\_II\_1,n\_II\_2)+c_{In\_1} \tag{24}\] \[\cdot\Delta NI\_1 st(T,n\_I\_1,n\_II\_2)+c_{In\_2}\cdot NI\_2 st_ {II}(T,n\_I\_1,n\_II\_1,n\_II\_2)\]  
The expected number of the second-stage inspections (included as the last component of Formula (24)), if the two-stage policy is chosen, can be calculated according to Equation (9) and it constitutes a unique component of the two-stage policy. Formula (24) depends, among others, on the reliability features of the inspected system. However, if we treat reliability features as constant in a given system (independent of any inspection policy), the cost change in Equation (24) depends only on the relation between false positive mistakes of both policies for a given inspection interval \(T\) and does not depend on the false negative mistake of the inspection method used in the second stage. Most of the components of Formula (24) have positive values under the condition \(\alpha_{In\_II,1}\cdot\alpha_{In\_2}\leq\alpha_{n\_I,1}\) and the only decrease that is present in Function 24 represents the expected number of preventive replacements. In all cases where \(c_{D}<c_{F}\), the function \(\Delta C(T,n\_I\_1,n\_II\_1,n\_II\_2) is positive. However, one should remember that the expected renewal period must satisfy the following constraint for every \(T\), if \(\alpha_{n\_II\_1}\cdot\alpha_{n\_II\_2}\leq\alpha_{n\_I,1}\):  
\[\tau_{II}(T,n\_II\_1,n\_II\_2)\geq\tau_{I}(T,n\_I\_1) \tag{25}\]  
The above constraint results from the fact that the objective of the two-stage policy is to neglect unnecessary preventive replacements. Formula (24) cannot directly imply the condition for the two-stage policy to be more economical than the one-stage policy. Nonetheless, based on the results of the case study presented in the previous section of the paper, one can notice that the two-stage policy is always more profitable than the policy of a one-stage inspection for every inspection period \(T\) shorter than the expected length of the system delay time (Figure 3). In this range of \(T\), for which the probability of system failure is low (close to 0) if the two-stage policy is applied, the increase in Formula (21) is also low (close to 0). At the same time, the expected lifetime of a diagnosed system is much longer for the two-stage policy than for the one-stage policy in this range. It implies the fact that within this area of \(T\), the two-stage policy should be the cheaper policy than the one-stage inspection variant. The conclusions from the practical case and the analysis of Equations (20)-(25) allow one to define the main dependencies between the one- and two-stage policies which can support the research process for a profitable version of the two-stage inspection policy. They are valid if the available imperfect inspection methods are significantly cheaper than the available perfect methods.