本論文探討整合製造商和零售商的存貨問題，研究中假設製造商和零售商有長期商業夥伴的關係，並且製造商提供信用交易優惠且零售商進貨檢驗時會發生誤判的情況。分別在第二章和第三章建立不同交易條件的單一製造商與單一零售商的整合存貨模式，其目的在尋求使得供應鏈整體總利潤有最大值的最適運送及補貨策略。為符合不同商業情境，第二章考慮零售商每次訂購量必須達到某一水準或以上時，製造商才會提供延遲付款的優惠；否則，於收到貨品時，必須馬上付清貨款。第三章則假設製造商採用二選一信用交易策略，即若零售商在M1時點付清貨款，則可享受δ(0<δ<1)比例的價格優惠；否則，必須在M2時點付清貨款，其中0≦M1<M2。 對所建立的整合存貨模式利用嚴謹的數量方法找出最適解。進一步，分別發展出演算法以利求解。最後，以數值範例說明求解程序，並探討信用交易策略、產品品質及檢驗能力等參數的敏感度分析，得到製造商應慎選信用交易策略並降低產品不良率，而零售商則應提升產品檢驗能力的結論，以達到供應鏈整體總利潤有最大值的目標。 This study discusses the integrated manufacturer-retailer inventory problems. In this thesis we assume that the manufacturer and the retailer are long-term business partners, and we also assume that the manufacturer offers trade credits and the retailer’s screening errors may occur. In order to make the supply chain model has the maximum profit, we try to find the optimal delivery and replenishment strategies. To meet the different business scenarios, in the second and third chapters we consider respectively different credit trades. In chapter 2, we assume that the manufacturer provides a permissible delay to the retailer if the order quantity is greater than or equal to a predetermined quantity. In chapter 3, we assume that the manufacturer adopts two-part trade credit. The retailer can obtain the δ(0<δ<1) proportion discount off the price of the merchandise if the payment is made within M1 days; otherwise the full price of the merchandise is due within M2 days, with 0≦M1<M2. For each integrated inventory model, we use a rigorous quantitative method to find optimal solution. Further, this thesis develops algorithms to determine the optimal solutions. Finally, numerical examples are provided to show the solution procedure, the sensitivity analysis on the credit trading strategies, product quality and testing capabilities are given to analyse the theoretical results. We make the proposals that the manufacturer should carefully choose the credit trade policy and reduce product defect proportion, the retailer should promote product screening capabilities, to meet the supply chain profit has maximum goal.
