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Dynamic programming for delayed product differentiation

Dynamic programming for delayed product differentiation.      課程老師 :陳茂生 教授           阮約翰 教授      報告學生 : 鄭漢中 937802. Reference. Karl Ulrich (1995), ” The role of product architecture in the manufacturing firm,” Research Policy , Vol. 24, pp. 419-440.

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Dynamic programming for delayed product differentiation

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  1. Dynamic programming for delayed product differentiation      課程老師:陳茂生 教授           阮約翰 教授      報告學生:鄭漢中 937802

  2. Reference • Karl Ulrich (1995), ” The role of product architecture in the manufacturing firm,” Research Policy, Vol. 24, pp. 419-440. • Hsi-Mei Hsu and Wen-Pai Wang (2004), “Dynamic programming for delayed product differentiation,” European Journal of Operational Research, Vol. 156, pp. 183–193. • David He, Andrew Kusiak* and Tzu-liang(Bill) Tseng (1998), “Delayed product differentiation:a design and manufacturing perspective,” Computer-Aided Design, Vol. 30, No. 2, pp. 105-113.

  3. Agenda • 簡介 • 動態規劃介紹 • Product differentiation dynamic programming(PDDP) model • 範例 • 結論

  4. 簡介 • 大量客製化的趨勢 • 必須重新設計產品以增加模組結構及零件共用性 • Multiple end-productsmay share some common processes and/orparts in the initial stages of the production process • At some point in the process, to customize the work-in-process into the different end-products • This research explore the differentiation point decisions in themanufacturing process

  5. Agenda • 簡介 • 動態規劃介紹 • Product differentiation dynamic programming(PDDP) model • 範例 • 結論

  6. 動態規劃簡介 • 動態規劃是進行一序列相關的決策,而它沒有線性問題的標準數學模式,是一種解題的通用方式 • 其特性為: • 問題可分成階段(stage),在每個階段要作策略決策 • 各階段有許多狀態(state),配合該階段的開始 • 已知目前狀態,剩下階段的最優策略與前階段的策略無關 • 找整個問題的最優策略,就是定出各階段的最優決策 • 解題方法是從找最後一階段的最優策略開始 • 已知階段n+1的策略時,用推返關係(recursive relationship)找出階段n的最優策略

  7. 動態規劃問題範例 • 馬車問題:起點A到終點J • 最安全的路徑就是總保費最便宜的路線 • 從地點i到另地點j的保費(c)如下:

  8. 動態規劃模式 • 建立決策變數xn (n=1,2,3,4),選擇的路徑為A→ x1 → x2 → x3 → x4 • fn (s, xn)為階段n地點s往前走,選xn為要去的地點,這樣策略的總成本 • xn*表示能使fn (s, xn)最小化,並以fn*(s)為此的最小值 fn (s, xn) = 階段n的成本+階段(n+1)之後的最少成本 =cij+ fn*(n+1)

  9. 動態規劃求解 n=4: n=3: n=1: n=2: 最佳決策 A→C→E→H→J A→D→E→H→J A→D→F→I→J 總成本=11

  10. Agenda • 簡介 • 動態規劃介紹 • Product differentiation dynamic programming(PDDP) model • 範例 • 結論

  11. 問題定義 • At each design stage, a decision should be made whether to continue with the common part assemblies or to proceed with specific component differentiation

  12. AND/OR graph

  13. Assumption • Each product requires processes performed in N stages • A buffer is held to store the WIP inventory right after each stage • Each stockpoint controls its inventory level using a periodic review policy • Different stockpoints have the same review period (e.g. 1 week), and a replenishment order is issued on the same day (e.g. Monday) • The state s(n) denotes a part, which is the WIP for some products • The stage means that designers ought to determine whether to continue with common processing or with some specific processing

  14. Notation N:total stages in system n :index of stage J :total product types j :index of the product Dj:demands for product j per time period is denoted by an i.i.d. random variable, where it is normally distributed with (μj ; σj) β:service level required for satisfying theneeds of back-end stage of the process lt(s(n)):processing lead-time required for executingoperations of the common part representedby state s(n) IC(s(n)): extra investment costs for executing operations of the common part represented by state s(n)

  15. Notation HC (s(n)):unit inventory holding cost of the specific components for executing operations of the common part represented by state s(n) I (s(n)):The buffer average inventory level of the specific part for states(n) where σ (s(n))denotes the aggregation standard deviation of products represented bys(n) z(s(n))that is the safety factor satisfiesβ

  16. Notation f (s(n) , x(n)): the cost function for stage n as s(n) and x(n) are its current state and the corresponding design decision, respectively c (s(n+1) , x(n)):the total operation cost when design decision x(n) is adopted and the differentiation process is moving to some state s(n+1) f*n+1 (s(n+1)): indicates the corresponding minimum costsof states(n+1)at next stage, the recursive relationship is

  17. Agenda • 簡介 • 動態規劃介紹 • Product differentiation dynamic programming(PDDP) model • 範例 • 結論

  18. 範例資料 • 四個產品(j = 1, 2, 3, 4) • The service level required for each stockpoint is 98% (β) • the lead-time for each stage takes 1 week (lt(s(n))) • Demands for each product per week have an identical normal distribution with mean μj =10,000 and standard deviation σj =1000 Product patterns and relative demand data z (s(n)) = 2.438

  19. Relative cost data EX: d(4)1的Cost為c(s(5)2,d(4)1)+c(s(5)3,d(4)1)+c(s(4)4, d(4)1) c(s(5)2,d(4)1)=IC(s(5)2)+PC(s(5)2)*D2+HC(s(5)2)*I(s(5)2) =1000+8*10000+12*(8449) =182318 c(s(5)3,d(4)1)=182318 c(s(5)4,d(4)1)=182318 d(4)1的Cost為547164

  20. 階段4的決策

  21. 階段3的決策

  22. 階段2的決策

  23. 階段1的決策

  24. Graphical display of the dynamic programming solution 第一階段:d(1)1 第二階段:d(2)2 第三階段:d(3)6, d(3)7 第四階段:d(4)4, d(4)5

  25. 結論 • Dynamic programmingapproach following an AND/OR graph to concretelyanalyzing the decisive design problem fordifferentiation deferment • 疑問: • 不是所有的產品都可以分成有相同的階段,這樣意謂這些產品的性質非常相近 • 在此研究中並沒有說明每個階段的決策方案是如何產生 • 產品重新的設計大多要改變其產品結構才行,而這方法覺得是評估產品重新設計後對於製程的影響

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