**Evaluating maintenance policy** I need to analyze the problem from the image, which presents a Markov Decision Process (MDP) regarding maintenance and its states: Good (G), Degraded (D), and Failure (F). The failure state is absorbing, and it resets to G in the following week. There's an immediate cost associated with failure, and various transition probabilities to consider regarding maintenance actions. My goal is to determine the optimal policy for maintenance and compute the expected costs over time.**Calculating Q values for maintenance** I need to compute the thresholds and Q values for the states: Good (G), Degraded (D), and Failure (F). Since F is an absorbing event that resets to G next week, it doesn't last long. When failure occurs, it incurs a cost \( c_F = 30 \). While in D without maintenance, failure might happen with a probability of 0.2, leading to that cost and a reset to G. In G, if I don’t maintain, I face transition possibilities to D. When I choose to maintain in non-F states, it guarantees a reset to G for a cost of 5.**Setting up dynamic programming** Since F is an event rather than an active state, there's no option to maintain in F. However, I could model F as deterministically transitioning back to G. My focus now is on computing the value \( V \) under the optimal policy. In states G and D, I need to determine whether to maintain or do nothing, each with associated costs and transitions. I’ll define optimal cost-to-go for G and D. For D, the expected failure cost incurs with probability \( p \), which results in that $30 cost if I choose not to maintain. For "do nothing" in G, there’s no regular cost to consider.**Analyzing failure costs** I realize there's no failure cost when starting from G, as it degrades to D with a 0.1 probability but incurs no immediate cost. Here's what I’ve broken down: For maintaining in G, the Q value is \( Q_G^{maintain} = 5 + 0.95 V(G) \). If I choose to do nothing, then \( Q_G^{do} = 0 + 0.9 V(G) + 0.1 V(D) \). In D, maintaining gives \( Q_D^{maintain} = 5 + 0.95 V(G) \), while doing nothing includes the expected failure cost of \( 0.2 * 30 \) plus further expected values. Since it resets to G next week, I need to ensure I find the optimal decisions by comparing these values. I’ll compute \( V(G) \) and \( V(D) \) as the minimums of those Qs.**Computing optimal policies and costs** I'm analyzing the system's policies now. I’ll compute \( V \) under different policies to determine their optimality. For Policy π1, where I maintain in both G and D, I find \( V(G) \) and \( V(D) \) both equal 100, but maintenance in G is not optimal. Then I explore Policy π2, maintaining only in D and doing nothing in G. After setting up the equations and solving, I determine that \( V(G) \approx 8.676 \) and \( V(D) \approx 13.242 \). Next, I'll assess the long-run expected discounted weekly cost under the optimal policy, noting key values for both states.**Calculating weekly costs and policies** If they want the "weekly cost," I can compute it using the formula \( C_{\text{bar}} = (1-\gamma) V(G) \). This comes out to be approximately 0.43379, which seems small due to the discount factor. Starting from G, I find \( V^*(G) \approx 8.676 \) and \( V^*(D) \approx 13.242 \). The chosen policy is to do nothing in G and maintain in D. I should also detail the steps and include relevant equations while showing why the policy of maintaining in D is optimal.**Finalizing the optimal policy and costs** After substituting values, I found that \( V(D) \approx 51.95 \) and subsequently \( V(G) \approx 34.07 \). The maintenance cost for D is approximately \( 37.37 \), which is better than doing nothing. Thus, the optimal policy is to do nothing in G and maintain in D. The value functions are \( V^*(G) \approx 8.676 \) and \( V^*(D) \approx 13.242 \). The long-run expected discounted cost from G equals \( 8.676 \), and the equivalent average discounted weekly cost is approximately \( 0.4338 \). I’ll craft a message with explanations and headings, adhering to the instructions.