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3 SINGLE-ECHELON SYSTEMS WITH INDEPENDENT ITEMS

3 SINGLE-ECHELON SYSTEMS WITH INDEPENDENT ITEMS. Different items can be controlled independently. The items are stocked at a single location, i.e., not in a multi-stage inventory system. 3.1 Costs. Holding costs  -  Opportunity cost for capital tied up in inventory

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3 SINGLE-ECHELON SYSTEMS WITH INDEPENDENT ITEMS

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  1. 3 SINGLE-ECHELON SYSTEMS WITH INDEPENDENT ITEMS • Different items can be controlled independently. • The items are stocked at a single location, i.e., not in a multi-stage inventory system.

  2. 3.1 Costs • Holding costs  - Opportunity cost for capital tied up in inventory    - Material handling costs -Costs for storage    - Costs for damage and obsolescence    - Insurance costs    - Taxes

  3. Ordering or Setup Costs - setup and learning - administrative costs associated with the handling of orders   - transportation and material handling.

  4. Shortage Costs or Service Constraints - extra costs for administration    - price discounts for late deliveries    - material handling and transportation. • If the sale is lost, the contribution of the sale is also lost. In any case it usually means a loss of goodwill. - component missing    - rescheduling, etc. • Because shortage costs are so difficult to estimate, it is very common to replace them by a suitable service constraint.

  5. 3.2 Different Ordering Systems • 3.2.1 Inventory position • Inventory position = stock on hand + outstanding orders - backorders. • Inventory level = stock on hand - backorders.

  6. 3.2.2 Continuous or Periodic Review • As soon as the inventory position is sufficiently low, an order is triggered. We denote this continuous review. • L = lead-time. • T = review period, i. e., the time interval between reviews.

  7. Inventory position R+Q R Inventory level L L 3.2.3 Different Prdering Policies • (R, Q) policy • Figure 3.1 (R, Q) policy with continuous review. Continuous demand.

  8. (R, Q) Policy • When the inventory position declines to or below the reorder point R, a batch quantity of size Q is ordered. (If the inventory position is sufficiently low it may be necessary to order more than one batch to get above R) .

  9. Inventory position Inventory level Time L L • (s, S) Policy Figure 3.2 (s, S) policy, periodic review. • When the inventory position declines to or below s, we order up to the maximum level S.

  10. 3.3.1 Classical Economic Order Quantity Model • Harris (1913), Wilson (1934), Erlenkotter (1989)  - Demand is constant and continuous.  - Ordering and holding costs are constant over time.   - The batch quantity does not need to be an integer.   - The whole batch quantity is delivered at the same time.   - No shortages are allowed.

  11. Notation: H = holding cost per unit and time unit A = ordering or setup cost D = demand per time unit Q = batch quantity C = costs per time unit

  12. Stock level Q Q/d Time Figure 3.3 Development of inventory level over time.

  13. (3.1) (3.2) (3.3) (3.4) • How important is it to use the optimal order quantity? From (3.1), (3.3), and (3.4) (3.5)

  14. If Q/Q*= 3/2 (or 2/3), then C/C* = 1.08 from (3.5) • The cost increase is only 8 percent. • Costs are even less sensitive to errors in the cost parameters. For example, if ordering cost A is 50 percent higher than correct ordering cost, from (3.3), batch quantity is Q/Q*= (3/2)1/2 =1.225 , relative cost increase  2 percent.

  15. Example 3.1 A = $200, d = 300, unit cost $100, holding cost is 20 percent of the value. Then, • h = 0.20  100 =$20 • Applying (3.3), Q* = (2Ad/h)1/2 = (2  200  300/20)1/2 = 77.5 units. • In practice Q often has to be an integer.

  16. Stock level Q Q(1-d/p) Q/p Time Q/d 3.3.2 Finite Production Rate • If there is a finite production rate, the whole batch is not delivered at the same time Figure 3.4 Development of the inventory level over time with finite production rate.

  17. P = production rate (p > d). • The average inventory level is now Q(1 - d/p)/2 . (3.6)  . (3.7)

  18. 3.3.4 Quantity Discounts • v = price per unit for Q < Q0, i.e., the normal price, • v´= price per unit for Q  Q0, where v´ < v. • Holding cost • h = h0 + rv for Q < Q0, • h´= h0 + rv´ for Q  Q0,

  19. 3.3.3 More General Models • Example 3.2 d: constant customer demand Two machine production rates: p1 > p2 > d, Two machine set up costs: A1and A2 The fixed cost of the transportation of a batch of goods from machine 2 to the warehouse: A3 The holding cost before machine 1 is h1 per unit and time unit. The holding cost is h2 and after machine 2 it is h3. Optimal Batch size?

  20. 1 M1 2 M2 3 4 Stock level before machine 1 (1) Q Time Q/d Q/p1 Stock level between machines 1 and 2 (2) Time Q/p1 Q/p2 Stock level after machine 2 (3) Time Q/p2 Stock level at warehouse (4) Time Transportation Q/d

  21. Holding Cost Calculation: …… (3.8) (3.9)

  22. C Q´´Q´ Q0 Q (3.10) (3.11) Figure 3.6 Costs for different values of Q.

  23. Two steps • 1. First we consider (3.11) without the constraint Q Q0. We obtain (3.12) and . (3.13) • If Q´  Q0, (3.12) and (3.13) give the optimal solution, i.e., Q* = Q´, and C* = C´.

  24. Example 3.3v = $100, v´ = $95 for Q  Q0 = 100. h0= $5 per unit and year, and r = 0.2, i.e., h = $25 and h´= $24 per unit and year. • d = 300 per year, A = $200. • From (3.12) and (3.13), Q´= 70.71 and C´= 30197. Since Q´< Q0, go to step 2. From (3.14) and (3.15), Q´´ = 69.28 and C´´ = 31732. Applying (3.16), C(100) = 30300. Q* = Q0 = 100.

  25. 2. If Q´ < Q0 we need to determine . (3.14) . (3.15) • Since v > v´ we know that Q´´ < Q´< Q0 . (3.16) • The optimal solution is the minimum of (3.15) and (3.16).

  26. v3=0.7 A3 v2=0.8 A2 v1=1 A1 Q1=1000 Q2=2500 Incremental Discounts d =36500 , r =0.3

  27. A1=A=15 A2=A1 +(c1-c2)Q1=15+(1-0.8)*1000=215 A3=A2 +(c2-c3)Q2=215+(0.8-0.7)*2500=465 • EOQi= Procedure • Candidate oi from each segment EOQi if feasible • Oi = Qi if EOQi > Qi Qi-1 if EOQi < Qi-1 Q*=best of all Oi

  28. Example • EOQ1=1910 > Q1  O1=Q1=1000 • EOQ2= =8087  O2=Q2=2500 • EOQ3= =12714 feasible

  29. TC(Qi) =vid+Qivir /2+dAi / Qi • TC(1000)=36500*1+1000*0.5*0.3*1+36500*15/1000 =36500+150+547.5=37197.5 • TC(2500)=(1000*1+1500*0.8)*36500/2500 +2500*0.5*0.3*(1000*1+1500*0.8)/2500 +36500*15/2500=32120+330+219=32669 • TC(12714)=[1000*1+1500*0.8 +(12714-2500)*0.7+15]*36500/12714 =26884.9+1404.7=28289.6 • Q*=12714

  30. Inventory level Q(1-x) Time -Qx Q/d 3.3.5 Backorders Allowed b1= shortage penalty cost per unit and time unit. x = fraction of demand that is backordered. Figure 3.7 Development of inventory level over time with backorders.

  31. . (3.17) , (3.18) and inserting in (3.17) we get . (3.19) , (3.20) .(3.21)

  32. Example 3.4 demand = 1000 units per year, production rate = 3000 units per year, holding cost before the machine = $10 per unit and year, holding cost after the machine =$15 per unit and year, shortage cost = $75 per unit and year, ordering cost = $1000 per batch.

  33. Stock level before the machine Q Time Q/3000 Q/1000 Stock level after the machine Q Time Q/3000 Q/1000 Stock level at warehouse Q(1-x) Time -Qx Q/1000 The optimal solution is x* = 15/90 = 1/6, and Q* = 310.

  34. 3.3.6 Time-varying Demand • T = number of periods • Di= demand in period i, i = 1,2,...,T, (Assume that d1 > 0, since otherwise we can just disregard period 1) • A = ordering cost, • h = holding cost per unit and time unit.

  35. A replenishment must always cover the demand in an integer number of consecutive periods. • The holding costs for a period demand should never exceed the ordering cost. Case: when backorders are not allowed

  36. 3.3.7 The Wagner-Whitin Algorithm fk=minimum costs over periods 1, 2, ..., k, i.e., when we disregard periods k + 1, k + 2, ..., T, • fk,t=minimum costs over periods 1, 2, ..., k, given that the last delivery is in period t (1  t  k). , (3.22) . (3.23)

  37. Period t dt 1 50 2 60 3 90 4 70 5 30 6 100 7 60 8 40 9 80 10 20 k=t 300 600 660 840 1030 1090 1370 1450 1530 1770 k=t+1 360 690 730 870 1130 1150 1410 1530 1550 k=t+2 540 830 790 1070 1250 1230 1570 1570 k=t+3 750 920 1090 1250 1370 1470 1630 k=t+4 870 1330 1410 1550 k=t+5 1530 • Example 3.6 T =10, A = $300, h = $1 per unit and period. Table 3.1 Solution, fk,t , of Example 3.6.

  38. Period t 1 2 3 4 5 6 7 8 9 10 Solution 1 110 190 200 100 Solution 2 110 190 300 Table 3.2 Optimal batch sizes in Example 3.6.

  39. Rolling horizon • Whether it is possible to replace an infinite horizon by a sufficiently long finite horizon such that we still get the optimal solution in the first period.

  40. DYNAMIC DETERMINISTIC MODELS

  41. Q t t1 t2 t3 …

  42. STOPPING RULE A PROCEDURE WHICH CHECKS WHETHER OR NOT THE INITIALDECISION IS A 'GOOD ENOUGH‘ APPROXIMATION EVERY TIME T IS INCREASED BY ONE PERIOD. A) APPARENT DECISION HORIZON: Q1 HAS CHANGED LITTLE OR NONE FOR THE LAST FEW VALUES OF T. B) DECISION HORIZON: Q1 CAN BE GUARANTEED NEVER TO CHANGE IF T WERE FURTHER INCREMENTED (INDEPENDENTOF DEMAND IN PDS AFTER T).

  43. STOPPING RULE (CONTINUED) C) NEAR-COST DECISION HORIZON: Q1 CAN BE GUARANTEED TO BE WITHIN 5% OF THE OPTIMAL COST. D) NEAR-POLICY DECISION HORIZON: Q1 CAN BE GUARANTEED TO WITHIN 5% OF Q1*.

  44. FORECAST HORIZON, DECISION HORIZON SUPPOSE AFTER SOLVING A FORWARD ALGORITHM OUT TO T, WE CAN GUARANTEE THAT DECISIONS FOR THE FIRST J PERIODS ARE CORRECT FOR ANY (T +K) - PROBLEM, K≥ 1 (INDEPENDENT OF DEMAND IN T+1, T+2, ... PDS), THEN THE FIRST T PERIODS IS CALLED A FORECAST HORIZON WHILE THE FIRST J PERIODS ARE CALLED A DECISION HORIZON.

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