一般而言，(製造)系統會因運轉而逐漸退化或蒙受外在的致命衝擊而導致提前完全失效，因而增加運轉成本(如製損)。為維持在較佳的狀況下運轉並降低上述成本，必須採取(預防性之)維修動作(如維修和置換)因而另增加維修成本。故適當的維修策略應以能降低單位時間的總成本之多少為考量。本研究以一連續時間多狀態的馬可夫退化過程來描述此未採任何維修策略下之(製造)系統，並分別建構連續檢測下和定期檢測下的預防性維修策略。根據檢測確認系統狀態後即決定應採取略過、維修或置換動作，而維修動作之種類(或不確定性)以不同(或相同)之機率表示。而其中所考慮的決策變數前者有維修管制門檻、置換管制門檻及使用期限，後者有維修管制門檻、置換管制門檻及定期檢測週期。兩種預防性維修策略都以長期單位時間總成本(含閒置、檢測、製損、維修及置換成本等等)的最小化來決定最佳維修策略，透過演算法以解得最佳的決策變數。此兩種維修策略涵蓋範圍廣泛，許多見諸文獻中的維修策略實為特例。本論文並以一(製造)系統作各種維修策略數值上優缺點的比較，除此之外，並針對維修機率(表示維修種類、程度或要求)及維修次數作敏感度分析。

Generally, a (manufacturing) system either deteriorates continuously due to aging or fails suddenly due to fatal shocks, and so it can’t remain in good operating states without taking maintenance. Maintenance (such as repair, replacement taken preventively) may keep manufacturing systems within good operating states to reduce the scrapping cost, but on the expense of maintenance cost. Therefore, proper maintenance policy should be able to reduce the expected total cost per unit time. For a continuous-time multi-state Markovian deteriorating (manufacturing) system subject to aging and fatal shocks, two control limit maintenance policies under continuous inspection and periodic inspection respectively are proposed. Under continuous inspection, the system is inspected continuously to identify the state upon which an action from {do-nothing, repair, replacement} is to be taken. Under periodic inspection, the system is inspected periodically to identify the state upon which an action from {do-nothing, repair, replacement} is to be taken. Such two policies are generalization of many maintenance policies in the literature. The first policy considers repair threshold, replacement threshold and time limit as decision variables and the second policy considers repair threshold, replacement threshold and inspection interval as decision variables. Both are evaluated in terms of expected long-run cost per unit time. For each policy, an iterated algorithm is developed to obtain the optimal policy(or optimal decision variables) and an illustration of it in terms of a numerical example is also presented.

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