ChVI-InventoryManagementII_A-Sp10

ChVI-InventoryManagementII_A-Sp10 - MGMT 36100 Operations...

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Unformatted text preview: MGMT 36100 Operations Management (OM) Models involving probability distributions. Earlier models (EOQ, EPQ, etc) had deterministic demand. In all of these models, no shortages were permitted. This resulted in 100% service level. Now we will consider probabilistic demand models. We will have occasional shortages and service level may be less than 100%. 100%. VIVI- Inventory Management - II No. Model type 5 6 7 ChVI‐InventoryManagementII‐Sp10 1 When to reorder: re-order point model (Q system) reHow much to order: Fixed order interval model (P system) Single period model (Newsvendor) ChVI‐InventoryManagementII‐Sp10 2 MGMT 36100 MGMT 36100 5 Q system: Continuous Review System. system: Continuous • Inventory is reviewed continuously (from practical point of view, we may do this every morning). • When inventory position (IP) falls below a specified value (called reorder point: ROP), we place an order. ROP We will assume that • We will order a fixed quantity (Q). • Lead time is fixed. • Shortages are made up by backordering. Order Q units when IP ≤ ROP IP •Demand may vary from day to day and shortages can occur occasionally while we wait for an order to arrive. •We will start with a manual example. ChVI‐InventoryManagementII‐Sp10 3 Example 5.1 ROP = 320, Q = 450, Order Q units when IP ≤ ROP IP LeadLead-time LT = 4 days, Opening stock = 190, Receipt on day 2 190, Date Opening Receipt Order Withdrawal Closing IL 1 2 3 4 5 6 7 8 Total 190 120 460 320 230 120 70 -10 IP 640 570 460 320 680 570 520 440 0 450 0 0 0 0 0 450 900 450 450 70 110 140 90 110 50 80 70 720 IL 120 460 320 230 120 70 -10 370 IP 570 460 320 680 570 520 440 370 Decision is based on IP, not IL. Always IL ≤ IP Closing IL (or IP) on Day X = Opening IL (or IP) on Day X+1 ChVI‐InventoryManagementII‐Sp10 4 MGMT 36100 MGMT 36100 Ex 5.1 Following formulas are always true for any X, Y Following Closing IL(Y) = Opening IL(X) + ΣReceipt – ΣWithdrawal Closing IP(Y) = Opening IP(X) + ΣOrder – ΣWithdrawal 5 Continuous Review System (Q system) Continuous When IP ≤ ROP, order “Q” units. Lead time: LT LT Notice shortage. Shortage occurs when the demand during lead time exceeds ROP. ROP Shortages affect service level. Order quantity is fixed, order interval varies. Inventory Let’s verify IL formula for X = 2, Y = 6 70 = 120 + 450 - 500 Verify IP formula for X = 1, Y = 8 370 640 450 720 370 = 640 + 450 - 720 Date 1 2 3 4 5 6 7 8 Total ChVI‐InventoryManagementII‐Sp10 Opening Recd. Order With- Closing Withdraw IL IP IL IP 190 120 460 320 230 120 70 -10 640 570 460 320 680 570 520 440 0 450 0 0 0 0 0 450 900 0 0 0 450 0 0 0 0 450 70 110 140 90 110 50 80 120 460 320 230 120 70 -10 570 460 320 680 570 520 440 Q Q Q ROP LT LT LT Time 70 370 370 720 5 What should be the value of Q? Q? Based on experience. EOQ approximation. ChVI‐InventoryManagementII‐Sp10 6 MGMT 36100 MGMT 36100 InventoryManagementII-Sp10 1 Let’s compare EOQ with the Q system EOQ: EOQ: Q system: daily daily demand: 100 Q = 375, LT = 1 day daily daily demand: N~[100, 10] Q = 375, LT = 1 day. If If we set ROP to 100, demand during lead time will be met only 50% of the time. This means we offer only 50% service level (SL). For 95% SL we need to change ROP to 100 + 1.65 * 10 = 116.5 When IP ≤ 116.5, order 375 If LT = 5 days, the policy will be [IP = 500, 375] When to order : ROP ChVI‐InventoryManagementII‐Sp10 5. Q system Demand Demand / unit time : N [ μ , σ]. We need the to know demand during lead time LT. LT μLT : mean , σLT : std. deviation of this new distribution. μLT + ZSL*(√LT)* σ Demand during lead time = 100. Hence, policy [IP =100, 375] μY= n*μx , σY = σx* (√n) 100 μLT = = LT * μ and σLT = (√LT) * σ LT LT *μ Reorder point “ROP” 116.5 μLT + ZSL * σLT σ = LT * 536.89 μ + ZSL * (√LT)* For LT = 5 and 95% SL, ROP = 500 + 1.65 * 10 * √5 = 536.89 When IP ≤ 536.89, order 375 = avg. demand during LT + safety stock (SS) safety EOQ: ROP = demand during LT For Q system: ROP = avg. demand during LT + safety stock 7 ChVI‐InventoryManagementII‐Sp10 8 MGMT 36100 MGMT 36100 Example 5.2 Demand per week N[μ = 500, σ = 20] , SL = 98% 20] SL LeadLead-time LT = 4 weeks ROP = μ + Z * σ LT LT SL Example 5.3 LT Demand per week N[μ = 200, σ = 25] LeadLead-time LT = 5 weeks, SL = 95% ROP. Part (a): Calculate ROP. Round up to the nearest multiple of 10. 2.06. For SL For SL = 98%, ZSL = Z0.98 = 2.06. LT = 4. ROP = LT * μ + ZSL* (√LT)* σ = 4 * 500 + 2.06 * (√4)*20 = 2082.4 ≈ 2090. Why round up? Increases SL slightly. Part (a): What should be reorder point (ROP)? For SL = 95%, ZSL = Z0.95 = 1.65. LT = 5. ROP = 5 * 200 + 1.65 * (√5)*25 = 1000 + 92.235 ≈ 1100. Part Part (b): safety stock (SS)? SS = 1100 – 1000 = 100 Ordering cost $60, Purchase price $12.50/unit, annual holding cost: 25% of purchase price. Assume 1 year = 50 weeks. Part (b): Calculate safety stock. stock. SS = ZSL* (√LT)* σ = 82.4. However, when we round up ROP, safety stock goes up. SS = ROP – LT * μ = ROP – LT * μ = 2090 – 2000 = 90 Part (c): What lot size based on EOQ approximation? approximation? D = 50 * 200 = 10000, H = 0.25 * 12.5 = 3.125 QO = √[ 2 (10000) (60) / (3.125)]. Q* ≈ 600 Q* Part (c): Calculate ROP and SS for LT = 2. Same SL and rounding . ROP LT ROP = 2 * 500 + 2.06 * (√2) * 20 = 1058.27 ≈ 1060. 500 1060. SS = 1060 – 1000 = 60 ChVI‐InventoryManagementII‐Sp10 9 Part (d): Average time between orders? 600 [units] / 200 [units / week] = 3 weeks ChVI‐InventoryManagementII‐Sp10 10 MGMT 36100 MGMT MGMT 36100 InventoryManagementII-Sp10 2 ...
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This note was uploaded on 12/14/2010 for the course MGMT 361 taught by Professor Panwalker during the Spring '10 term at Purdue University-West Lafayette.

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