Consider a system that has one component in operation and $n-1$ components in stand by condition. To improve the reliability of this system, we use the preventive maintenance ($PM$) with block replacement policy ${\bf a}=(a_1 , \ldots a_n)$ such that $\sum_{i=1}^{n}a_{i}=a$. That is, we replace the $i^,$th component by $i+1^,$th component either it fails or at time $a_i$. Now, let $X_i, ~ i=1,\ldots , n$ be the life time of the $i^,$th component and every component is successfully switched with probability one. Then, the life time of this system is $\sum_{i=1}^{n} (X_i\wedge a_i)$ where $(X_i\wedge a_i)=\min (X_i, a_i)$. In this paper, we compare different allocation of $(a_1 , \ldots a_n)$ to find the best or worst allocation by the sense of expectation or usual stochastic ordering and prove the following results.\\ % \begin{itemize} % \item{1.} 1. If $X_1,\ldots,X_n$ are exchangeable, then $${\bf a}\geq_m {\bf a}^* \Longrightarrow \sum_{i=1}^{n} (X_i\wedge a_i) \leq_{icv} \sum_{i=1}^{n} (X_i\wedge a_i^*).$$ 2. If $X_1,\ldots,X_n$ are exchangeable and have log-concave joint density function, then $${\bf a}\geq_m {\bf a}^* \Longrightarrow \sum_{i=1}^{n} (X_i\wedge a_i) \leq_{st} \sum_{i=1}^{n} (X_i\wedge a_i^*).$$ As intuitively expected, under the conditions $(1)$ and $(2)$ the best allocation is $(\bar{a}, \ldots, \bar{a})$ by respect to expectation and survival function of life time of the system, respectively.\\ 3. If $X_1,\ldots,X_n$ have arrangement increasing joint density function and $(a_1^*, \ldots, a_n^*)$ is the best allocation by the sense of usual stochastic ordering, then $a_1^*\leq a_2^*, \ldots \leq a_n^*$.\\ 4. If $X_1,\ldots,X_n$ are have arrangement increasing and log-concave joint density function, then $${\bf a}\geq_m {\bf a}^* \Longrightarrow \sum_{i=1}^{n} (X_i\wedge a_{(n-i+1)}) \leq_{st} \sum_{i=1}^{n} (X_i\wedge a_i^*).$$
