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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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