Probability of Machine Failing with 5 Engines

Question

The other day I posted this math question Coin Toss Games with Changing Coin Probabilities but I think it was too long. I am now trying to make a shortened and simpler version.

• Suppose there is a machine with 5 engines.
• At turn=0, the probability of each engine failing is 0.5
• Each turn, the probability of a given engine failing increases by p_current+0.01
• Once an engine has failed, it can never be repaired.
• The machine stops working when all 5 engines have failed

Question: Given the current state of the system (e.g. turn = $k$, number_of_working_engines = $j$ , $P_1 = p_1, P_2 = p_2, ... P_j = p_j$), is it possible to derive a probability distribution for the expected number of turns before the entire machine fails?

I tried to represent this problem mathematically in 3 parts:

Part 1: Probability of each engine ($E_i$) working (1 = working) at time $t+1$ given the state at $t$

$$ P(E_{i,t+1} = 1 | E_{i,t} = 1) = 1 - (p_t - 0.01) \quad \text{for} \quad i = 1, 2, 3, 4, 5 $$

$$ P(E_{i,t+1} = 0 | E_{i,t} = 1) = p_t - 0.01 \quad \text{for} \quad i = 1, 2, 3, 4, 5 $$

$$ P(E_{i,t+1} = 0 | E_{i,t} = 0) = 1 \quad \text{for} \quad i = 1, 2, 3, 4, 5 $$

Part 2: Probability of having $k$ working engines at time $t+1$ given the state at $t$

$$ P(N_{t+1} = k | N_t = k) = 1 - k \cdot (p_t - 0.01) \quad \text{for} \quad k = 1, 2, 3, 4, 5 $$

$$ P(N_{t+1} = k-1 | N_t = k) = k \cdot (p_t - 0.01) \quad \text{for} \quad k = 1, 2, 3, 4, 5 $$

Part 3: Probability of the machine ($M$) working in general at time $t+1$ given the state at $t$

$$ P(M_{t+1} = 1 | N_t = k) = 1 - (k \cdot (p_t - 0.01))^k \quad \text{for} \quad k = 1, 2, 3, 4, 5 $$

But from here, I am unsure how to solve this problem apart from numerical simulation. Perhaps a Birth-Death process can be used to model this problem?

Can someone please show me how to start working on this problem from a theoretical perspective?

Answer 1

For a single engine, the random variable $X$ for the number of turns until failure has probability distribution $$\Pr[X = x] = (p_0 + ix) \prod_{k=0}^{x-1} (1 - p_0 - ik), \quad x \in \{0,1,2,\ldots,m\}$$ where $p_0$ is the initial probability of failure at turn $0$; $i$ is the incremental amount by which the failure probability increases each turn; and $m = \lfloor (1-p_0)/i \rfloor$ is the maximum value for $X$, after which point the probability of failure is guaranteed. In your case, $p_0 = 0.5$ and $i = 0.01$, so $m = 50$.

Then the failure time for the entire machine is given by the maximum order statistic $X_{(5)} = \max_i X_i$ for iid engine failure times $X_1, \ldots, X_5$. Hence $$\Pr[X_{(5)} \le x] = \prod_{n=1}^5 \Pr[X_n \le x] = \Pr[X \le x]^5$$ and $$\Pr[X_{(5)} = x] = \Pr[X_{(5)} \le x] - \Pr[X_{(5)} \le x-1] = \Pr[X \le x]^5 - \Pr[X \le x-1]^5.$$ At this point, it is better to use a computer to compute this probability distribution. We obtain $$\begin{array}{c|c} x & \Pr[X_{(5)} = x] \\ \hline 0 & 0.03125000000 \\ 1 & 0.2140710180 \\ 2 & 0.2896466031 \\ 3 & 0.2175799115 \\ 4 & 0.1266289561 \\ 5 & 0.06491113406 \\ 6 & 0.03099366486 \\ 7 & 0.01414199647 \\ 8 & 0.006239115394 \\ 9 & 0.002675187318 \\ 10 & 0.001117114593 \\ 11 & 0.0004545792505 \\ 12 & 0.0001802380138 \\ 13 & 0.00006960208635 \\ 14 & 0.00002616254832 \\ 15 & 9.56581844148743 \times 10^{-6} \\ 16 & 3.39956233840373 \times 10^{-6} \\ 17 & 1.17336623824378 \times 10^{-6} \\ 18 & 3.92990371392391 \times 10^{-7} \\ 19 & 1.27606311871750 \times 10^{-7} \\ 20 & 4.01312630928546 \times 10^{-8} \\ \vdots & \vdots \end{array}$$

This method can be generalized to compute the conditional probability distribution for the number of additional turns $Y_{(N)} \mid N, T$ until machine failure, given that there are $N = n$ engines exposed to failure at turn $T = t$. Then one can compute the conditional expectation of $Y_{(N)} \mid N, T$. For a single operational engine, the random number of additional turns $Y \mid T = t$ until failure has probability distribution $$\Pr[Y = y \mid T = t] = (p_0 + i(t + y)) \prod_{k=0}^{y-1} (1 - p_0 - i(t+k)), \quad y \in \{0, 1, \ldots, m^*\}$$ where now $m^* = m - t$ (since $t$ turns have already been taken), hence $$\Pr[Y_{(N)} = y \mid N = n, T = t] = \Pr[Y \le y \mid T = t]^n - \Pr[Y \le y-1 \mid T = t]^n.$$