# Warranty Cost Analysis for Multi-State Products Protected by Lemon Laws
## 2 Assumptions and Cost Analysis of \(\{\tau,N\}\) Model
### Probability Analysis of \(\{\tau,N\}\) Model
#### 2.3.1 Renewable Warranty Service Analysis of \(\{\tau,N\}\) Model
\begin{array}{c}\varphi^{l}(s)=\quad{((\mathbf{\mu}_{0}s\mathbf{I}-\mathbf{Q}_{UU })^{-1}\mathbf{Q}_{UD}\mathbf{P}_{SDU}e_{U})(\mathbf{\mu}_{1}((s\mathbf{I}- \mathbf{Q}_{UU})^{-1}\mathbf{Q}_{UD}\mathbf{P}_{SDU})^{l-2}\mathbf{e}_{U})\times }\\ \qquad\qquad\qquad\qquad\qquad\mathbf{\mu}_{1}(s\mathbf{I}-\mathbf{Q}_{UU})^{-1} \mathbf{Q}_{UD}\mathbf{P}_{LDU}e_{U},\qquad l=2,\cdots,N,\end{array}  \tag{3}\]  
\[\varphi^{1}(s)=\mathbf{\mu}_{0}(s\mathbf{I}-\mathbf{Q}_{UU})^{-1}\mathbf{Q}_{UD} \mathbf{P}_{LDU}e_{U}. \tag{4}\]  
_where \(\mathbf{P}_{LDU}\) is the state transition probability matrix after a long repair._  
Proposition 2 can be proved in a similar manner as Proposition 1. Let \(T^{l}(t)(l=1,\cdots,N)\) be the inverse Laplace transform of \(\varphi^{l}(s)\). Given the number of short repairs \(l\), the probability that the warrantty term is renewed is:  
\[p_{l}=\int_{0}^{W_{1}}T^{l}(t)\mathrm{d}tG(\tau)^{l-1}(1-G(\tau)),\,l=1,\cdots,N \tag{5}\]  
Let \(p_{\tau N}\) be the warrantty renewal probability under the \(\{\tau,N\}\) model. Then:  
\[p_{\tau N}=\sum_{l=0}^{N}p_{l}\]  
The expected length of a renewable warrantty for Case 1-1 is:  
\[L_{0}=\frac{\int_{0}^{W_{1}}tT^{0}(t)\mathrm{d}t}{p_{\tau N}}. \tag{6}\]  
The expected length of a renewable warrantty under Case 1-2 is:  
\[L_{I}=\frac{\int_{0}^{W_{1}}tT^{l}(t)\mathrm{d}t}{p_{\tau N}},\,l=1,\cdots,N. \tag{7}\]  
Note that after the product has gone through \(N\) short-duration repairs and transits to the failure states set \(D\) from the operating states set \(U\) within \(W_{1}\), the warrantty term will be renewed immediately under the obligation. Therefore, it is impossible that the warranttyterm is renewed due to exceeding the number of visits to the failure states set and the repair time. Nonrenewable Warranty Service Analysis of \(\{\tau,N\}\) Model  
According to the modeling assumptions, if the warranty term is not renewed, the product can visit the working state set \(l(l=1,\cdots,N+1)\) times and \(l(l=0,\cdots,N)\) short repairs can happen over the warranty period. The Laplace transform for the probability density functions of \(l(l=1,\cdots,N)\) visits to the working state set and \(l\) short repairs is  
\[\varphi^{l}(s)=\left(\mu_{0}(s\mathbf{I}-\mathbf{Q}_{UU})^{-1}\mathbf{Q}_{UD}\mathbf{P}_ {SDU}\mathbf{e}_{D}\right)(\mathbf{\mu}_{1}(s\mathbf{I}-\mathbf{Q}_{UU})^{-1}\mathbf{Q}_{UD}\bm {P}_{SDU}\mathbf{e}_{D})^{l-1}. \tag{8}\]  
The associated probability density functions are denoted by \(f_{l}(t)\). The probability that the product has gone through \(l(l=0,\cdots,N)\) short repairs over the warranty period, no repair action is performed, and warranty terms are not renewed is:  
\[\widetilde{p}_{l}=\left(\int_{0}^{W_{1}}\int_{W_{1}-t_{1}}^{+\infty}f_{l}(t_{ 1})f(t_{2})dt_{1}dt_{2}\right)G^{l}(\tau),l=1,\cdots,N, \tag{9}\]  
\[\widetilde{p}_{0}=\int_{W_{1}}^{+\infty}f_{1}(t_{1})dt_{1}. \tag{10}\]  
where \(f(t_{2})\) is the inverse Laplace transform of \(\mathbf{\mu}_{1}(s\mathbf{I}-\mathbf{Q}_{UU})^{-1}\mathbf{Q}_{UD}\mathbf{e}_{D}\). If the warranty term is not renewed and no repair action is performed, the duration in \(U\) will be longer than \(W_{1}\), which gives Equation (9). If warranty term is not renewed and \(l(l=0,\cdots,N)\) short repairs are performed during the warranty period, then the time taken by the \(l(l=0,\cdots,N)\) cycles will be shorter than \(W_{1}\) and the total time of the \(l\) cycles and the following visits to the working state set is longer than \(W_{1}\). According to the property of convolution, Equation (10) can be obtained. \(\square\)