On a class of reserved devices

In this paper, we consider three models of redundancy:

• By use of the mean time between system failures on a finite interval;
• By use of the mean time between system failures on a infinite interval;
• By use of the system reliability on a finite interval.

For all three models, the redundancy criterion has the following form:
$$ T(k,r)=\sum_{i=0}^{k-m}C_k^i p^{k-i}q^i T(r-i). \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad{(1)} $$
Using the sigma-operator turns out to be an effective way for proving many properties of optimal strategies. Let $T(r) > 0$ and $T(r)$ increase.
The following properties are proved:

• If $p\geqslant\frac{k}{2k-m+1}$, then the function $T_m(k,r)$ for the system $S_m$ has at most two maximums for a fixed $r$, it is convex on the interval $m\leqslant k\leqslant k_0^m(r)+1$ and nonincreasing on $k_0^m(r)<k\leqslant r$.
• Since $T(r+1)/T(r)>1$ and strictly decreases, $\lim\limits_{r\to\infty}T(r+1)/T(r)$ exists by the Bolzano–Weierstrass theorem and this limit is equal to $1$.

From convexity of the function $T(r)$, it is easy to prove that

• $\frac{T(r+2)}{T(r+1)}<\frac{T(r+1)}{T(r)}$.

Under more restrictive conditions, this inequality was obtained in the thesis of L. V. Ushakova.

• The function $\ln T(r)$ is convex;
• $T(k+1,r)-T(k,r)$ increases with an increase in $r$.

To find the optimal strategy, a simplified algorithm is obtained using the properties. This algorithm is based on a modification of the Bellman dynamic programming method mentioned in V. V. Travkina's work. The essence of the algorithm is as follows.

• If there are $m$ elements, then we have $k_0(m)=m$. For each model, $T(m)$ is calculated.
• All further calculations for the three models are similar. Then it is necessary to calculate the values of the function $T(r)$ \emphby means of its previous values using the formula
  $$ T(k,r)=\frac{1}{1-p^k}\sum_{i=1}^{k-m}C_k^i p^{k-i}q^i T(r-i), \qquad\qquad\qquad\qquad\qquad\qquad{(2)} $$
  
• Suppose that $k_0(m)$, $k_0(m+1)$, …, $k_0(r-1)$, $T(m)$, $T(m+1)$, …, $T(r-1)$ have already calculated.

To find $k_0(r)$, we need to calculate $k_0(r-1)$ and $k_0(r-1)+1$. Then, using (2), we calculate $T(k_0(r-1),r)$ and $T(k_0(r-1),r)$ and compare them. If $T(k_0(r-1),r)\geqslant T(k_0(r-1)+1,r)$ then $k_0 (r) = k_0(r-1)$ and $T(r) = T(k_0 (r-1), r)$. If $T(k_0 (r-1)+1, r) \geqslant T(k_0 (r-1), r)$ then $k_0 (r) = k_0 (r-1)+1$, and $T(r) = T(k_0 (r-1)+1, r)$.