The document discusses various inventory management concepts including types of inventory, inventory costs, inventory control systems, and ABC classification. It describes the economic order quantity (EOQ) model, which helps determine optimal order quantities to minimize total inventory costs given annual demand, ordering costs, and holding costs. The reorder point indicates when to place a new order based on daily demand, lead time, and a safety stock to protect against variability in demand.

1. Logistics management INVENTORY MANAGEMENT - CONTROL Joanna Oleśków-Szłapka

4. Water Tank Analogy for Inventory Supply Rate Inventory Level Demand Rate Inventory Level Buffers Demand Rate from Supply Rate

6. Inventory Hides Problems Poor Quality Unreliable Supplier Machine Breakdown Inefficient Layout Bad Design Lengthy Setups

15. ABC CLASSIFICATION

18. ABC Classification: Example 1 $ 60 90 2 350 40 3 30 130 4 80 60 5 30 100 6 20 180 7 10 170 8 320 50 9 510 60 10 20 120 PART UNIT COST ANNUAL USAGE

19. ABC Classification: Example (cont.) Example 10.1 1 $ 60 90 2 350 40 3 30 130 4 80 60 5 30 100 6 20 180 7 10 170 8 320 50 9 510 60 10 20 120 PART UNIT COST ANNUAL USAGE TOTAL % OF TOTAL % OF TOTAL PART VALUE VALUE QUANTITY % CUMMULATIVE 9 $30,600 35.9 6.0 6.0 8 16,000 18.7 5.0 11.0 2 14,000 16.4 4.0 15.0 1 5,400 6.3 9.0 24.0 4 4,800 5.6 6.0 30.0 3 3,900 4.6 10.0 40.0 6 3,600 4.2 18.0 58.0 5 3,000 3.5 13.0 71.0 10 2,400 2.8 12.0 83.0 7 1,700 2.0 17.0 100.0 $85,400 A B C % OF TOTAL % OF TOTAL CLASS ITEMS VALUE QUANTITY A 9, 8, 2 71.0 15.0 B 1, 4, 3 16.5 25.0 C 6, 5, 10, 7 12.5 60.0

24. Economic Order Quantity Model

28. EOQ Inventory Order Cycle Demand rate 0 Time Lead time Lead time Order Placed Order Placed Order Received Order Received Inventory Level Reorder point, R Order qty, Q As Q increases, average inventory level increases, but number of orders placed decreases ave = Q/2

29. Total Cost of Inventory – EOQ Model

31. The EOQ Model Q = Number of pieces per order Q* = Optimal number of pieces per order (EOQ) D = Annual demand in units for the Inventory item S = Setup or ordering cost for each order H = Holding or carrying cost per unit per year

32. An EOQ Example Determine optimal number of needles to order D = 1,000 units S = $10 per order H = $.50 per unit per year Q* = 2 DS H Q* = 2(1,000)(10) 0.50 = 40,000 = 200 units

33. An EOQ Example Determine optimal number of needles to order D = 1,000 units Q* = 200 units S = $10 per order H = $.50 per unit per year = N = = Expected number of orders Demand Order quantity

34. An EOQ Example Determine optimal number of needles to order D = 1,000 units Q* = 200 units S = $10 per order N = 5 orders per year H = $.50 per unit per year = T = Expected time between orders Number of working days per year N

35. An EOQ Example Determine optimal number of needles to order D = 1,000 units Q* = 200 units S = $10 per order N = 5 orders per year H = $.50 per unit per year T = 50 days Total annual cost = Setup cost + Holding cost TC = S + H D Q * Q * 2

37. Reorder Point: Example Demand = 10,000 kg /year Store open 311 days/year Daily demand = 10,000 / 311 = 32.154 kg /day Lead time = L = 10 days R = dL = (32.154)(10) = 321.54 kg = 322 kg

39. Variable Demand with a Reorder Point Reorder point, R Q LT Time LT Inventory level 0

40. Reorder Point with a Safety Stock Reorder point, R Q LT Time LT Inventory level 0 Safety Stock

41. Reorder Point With Variable Demand R = dL + z  d L where d = average daily demand L = lead time  d = the standard deviation of daily demand z = number of standard deviations corresponding to the service level probability (service factor) z  d L = safety stock

42. Reorder Point for a Service Level Probability of meeting demand during lead time = service level Probability of a stockout R Safety stock d L Demand z  d L

43. Safety factor values for CSL

44. Reorder Point for Variable Demand The carpet store wants a reorder point with a 95% service level and a 5% stockout probability For a 95% service level, z = 1.6 4 d = 30 m per day L = 10 days  d = 5 m per day R = dL + z  d L = 30(10) + (1.6 4 )(5)( 10) = 32 5.9 m Safety stock = z  d L = (1.6 4 )(5)( 10) = 2 5 . 9 m