5. Independent vs. Dependent Demand E(1) Independent Demand (not related to other items or final end-product) Dependent Demand (derived from component parts, sub-assemblies, raw materials, etc.)
6. Independent versus Dependent Demand Dependent demand Work in progress Components and raw materials Time Demand/usage Independent demand - finished goods - spare parts Time Demand/usage
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9. Material-Flows Process From Suppliers To Customer Production Processes Inventory in transit Stores warehouse Finished goods WIP WIP Work in process
10. Stock : Input (Flow in), Storage (Holding) and Flow out (Usage) Supply Rate Inventory Level Rate of Demand (Usage) Stock Level
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12. Order Quantities & Reorder Points R = Reorder point L = Lead time L L q R Time No. of units on hand safety or buffer level Average stock q/2 q
13. Simple inventory system Raise order for ROQ Orders MRP Check stock level Yes No Receive/inspect. Accept into stock Send back? Part-delivery No Next Check point Yes No Yes <=ROL? Outstanding Order? Due now?
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15. EOQ Aim = Cost Minimisation Cost Holding + ordering costs = total cost curve. Find Q eoq inventory order point to minimise total costs. Ordering Costs Holding Costs Q eoq Order Quantity (Q) Total Cost
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17. EOQ Solution Q = 2DS H = 2(1,000 )(10) 2.50 = 89.443 units or 90 units eoq d = 1,000 units p.a. 365 days p.a. = 2.74 units/day Reorder point D L = 2.74 units/day = 19.18 or 20 for 7 day lead time EOQ order = 90 units. When only 20 units left, place next order for 90 units.
18. EOQ and ROQ example 2 Annual Demand = 10,000 units Days per year considered in average daily demand = 365 Cost to place an order = £10 Holding cost per unit per year = 10% of cost per unit Lead time = 10 days Cost per unit = £15 365.148 (366 units) = 1.50 2(10,000)(10) = H 2DS = Q eoq If lead time = 10 days, ROL= 273.97 = 274 units Place order for 366 units. When 274 left, place next order for 366. D = 10,000 units/year 365 days = 27.397 units/day
19. Total variable cost Find point of minimum TVc Avg.stock Demand 2 x unit cost x Hc% + Oc 1200 2 x £3 x 25% = £450 + £10 Once per year = £460 1200/52 2 x £3 x 25% = £9 + £510 Once per week = £519 approx
20. EOQ Table – minimum TVc Avg.stock x item £ x hc % Oc + Hc
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24. Order Point with Safety Stock Units Days Safety Stock Actual lead time is 3 days! (at day 21) 2200 2000 Order Point 400 200 0 18 21 Dip into safety stock
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28. Normal Distribution of Demand over Lead Time m = mean demand r = reorder point s = safety stock frequency probability of stock out demand over lead time service level probability m s r
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33. Price-Break Model Assumptions similar to as EOQ model i = % of unit cost as carrying cost C = cost per unit “ C” varies for each price-break so apply the formula to each price-break cost value. Holding cost per annum 2(Demand p.a.)(Order or Setup-cost) = iC 2DS = Q OPT
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35. Solution = 1,826 units 0.02(1.20) = iC 2DS D = 10,000 units Order cost (S) = £4 Put data into formula for each price-break of “C”. =2,000 = =2,020 4) 2(10,000)( = Carrying cost % (i) = 2% Cost per unit (C) = £1.20, £1.00, £0.98 Q opt 0 - 2499 Feasible 2500-3999 and 4000+ Not feasible Are Q opt values feasible for the price breaks? 2(10,000)(4) 0.02(1.00) 0.02(0.98) 2(10,000)(4)
36. U-shaped function True Q opt values occur at the start of each price-break interval.The total annual cost function is a “u” shaped function 0 1826 2500 4000 Order Quantity Total annual costs Price-breaks
37. Price-Break Solution Now apply the Q opt values to total annual cost & identify the total cost for each price-break. TC(0-2499)= (10000x1.20)+(10000/1826)x4+(1826/2)(0.02x1.20) = £12,043.82 TC(2500 -3999) = £10,041 TC(4000+) = £9,949.20 Least cost Q opt = 4000
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