ChVI-InventoryManagementII_B-Sp10

ChVI-InventoryManagementII_B-Sp10 - MGMT 36100 Operations...

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Unformatted text preview: MGMT 36100 Operations Management (OM) 6 P system: Periodic Review System. system: Periodic • Inventory is reviewed at fixed interval P (every 2 days, 7 days, etc). • On the review day, we place an order so that the inventory position is raised to a specified value (called order-up-to level: OUL). order-upOUL We will assume that • Lead time is fixed. • Shortages are made up by backordering. • Order (OUL – IP) units at the review time. • 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.. VIVI- Inventory Management - II ChVI‐InventoryManagementII‐Sp10 1 ChVI‐InventoryManagementII‐Sp10 11 MGMT 36100 MGMT 36100 Example 6.1: Day 1 2 3 4 5 6 7 8 9 Total P system. OUL = 600, OI = 4 days Receipt 0 0 390 0 0 0 410 0 0 800 410 Order Withdraw 390 120 80 130 80 90 110 135 80 415 1215 120 945 Closing IL 90 10 270 190 100 -10 265 185 65 IP 480 400 270 190 510 400 265 185 480 6 Periodic Review System (P system) OI is fixed. On the review day, order Q = OUL – IP. OUL Notice that if we want to avoid shortage after period 2 We need to raise OUL and increase the value of Q1 OUL OUL OUL 0rder-up0rder-up-to level LT = 2 days, Opening stock = 210, no receipts on day 1, 2 Opening IL 210 90 10 270 190 100 -10 265 185 IP 210 480 400 270 190 510 400 265 185 Q1 Q2 Q3 Order Order interval is fixed, order quantity varies. tit Time 3 LT LT So, we need to consider demand over OI + LT units of time. This OI units is called the protection period. protection 1 OI 2 Receipt Verify formulas Closing IL(Y) = Opening IL(X) ++ΣΣOrder ––ΣΣWithdrawal ChVI‐InventoryManagementIIClosing IP(Y) = Opening IP(X) ‐Sp10 Withdrawal MGMT 36100 12 ChVI‐InventoryManagementII‐Sp10 13 MGMT MGMT 36100 6 P system Demand / time: N [ μ , σ]. Order Interval = OI and Lead time = LT We need the distribution of demand during protection period PP = (OI + LT) PP This is also distributed normally. Example 6.2 P system . Demand per week ∼ N(μ = 90, σ = 15), 15), 50 weeks per year. LT = 3 weeks. μPP = (OI + LT) * μ and σPP = (√(OI+LT)) * σ Order up to level (OUL) = μPP + ZSL * σPP = (OI + LT)* μ + ZSL * {√(OI + LT)} * σ = avg. demand during (OI + LT) + safety stock (SS) What should be the value of OI? Based on experience or EOQ approximation. ChVI‐InventoryManagementII‐Sp10 14 Part (a): Find a suitable order interval using EOQ approximation. Use S: $80, C: $12.00 per unit, holding cost: 27% per year. D = 50 * 90 = 4500, H = 0.27 * 12.0 = 3.24 Q* QO = √[ 2 (4500) (80) / (3.24)] = 471. Q* ≈ 450. OI = 450/90 = 5 weeks. Part (b): Use information in part (a) and find OUL for SL = 80% From the standard normal tables, Z0.8 = 0.85 OUL = (OI + LT) * μ + ZSL {√(OI + LT)} * σ LT) LT)} = (5+3) * 90 + 0.85 * (√8) * 15 = 720 + 36.1 = 756.1 ≈ 760 Part (c): Find OUL for SL = 98% From the standard normal tables, Z0.98 = 2.06 OUL = (OI + LT) * μ + ZSL {√(OI + LT)}* σ LT) = (5+3) * 90 + 2.06 * (√8) * 15 = 720 + 87.4 = 807.4 ≈ 810 ChVI‐InventoryManagementII‐Sp10 15 MGMT 36100 MGMT 36100 InventoryManagementII-Sp10 1 Example 6.3 Product: Oil, LT = 1 Day Demand / day ∼ N(μ = 2000, σ = 500) in gallons. Part (a): Suppose we use Q system and do not want stock-outs stock(shortages) to occur in more than 1 out of 10 cycles. Reorder point? Shortages in 1 out of 10 cycles means service level SL = 9/ 10 = 0.9 Z0.9 = 1.29 ROP = μLT+ ZSL σLT = (1) * 2000 + 1.29 * {√(1)} *500 = 2,645 g. 1.29 Part (b): Suppose we switch to P system and order sufficient quantity to fill 10,750-gallon tank. 10,750(i) What is order up to level OUL = ? What is order up to level OUL OUL 10 OUL = 10,750 gal. gal (ii) If we order every 4 days (OI = 4), how much is the service level? OUL = (OI + LT)* μ + ZSL {√(OI + LT)}* σ LT)* 10,750 = (5) * 2000 + ZSL{√(5)}* 500 = 10,000 + ZSL* 1118 ZSL = (10,750 – 10,000)/1118 = 0.6708, SL = 74.86% (iii) What if we ORDER every 3 days? ZSL = 2.75 SL = 99.7% 10,750 = (4)*2000 + ZSL{(4)}* 500 ChVI‐InventoryManagementII‐Sp10 16 Example (From an old test) test) Part A: Use Q system. Demand per week: N (250,20) LT = 4 weeks. LT SL SL = 90%. Calculate the reorder point (round it up to the nearest multiple of 10) and the safety stock. ROP = μLT + ZSL * σLT = 4 * 250 + 1.29 * √(4) * 20 = 1000 + 51.6 = 1051.6 1,060 1,060 units. SS = 1060 – 1000 = 60 units. Part B: Suppose you switch to P system with OI = 4 weeks. Use OI the safety stock value from above. What is the service level and service what is the value of the order-up-to level (OUL)? order-upFrom above SS = 60. SS SS SS = ZSL * σOI+LT ZSL= 60 / [√(8) * 20] =1.06 SL SL = 0.8554 = 85.5% Order-up-toOrder-up-to-level OUL = μOI+LT + SS = 8 * 250 + 60 = 2,060 units OUL ChVI‐InventoryManagementII‐Sp10 17 MGMT 36100 MGMT 36100 5 Q system: Continuous Review System •Inventory reviewed continuously but fixed quantity ordered. •When inventory position falls below a specified level (called rereorder point), an order is placed. •Reorder Point: ROP ROP •Order Q units when IP ≤ ROP IP •Q is selected based on approx. EOQ formula or experience. 6 P system: Periodic Review System •Inventory ordered at fixed at fixed interval interval. (OI : order interval) •Order sufficient quantity to raise inventory inventory position to a specified level (called order-up-to level). order-up•Order up to level: OUL OUL •Order (OUL – IP) units. units. •OI is selected based on approx. OI EOQ formula or experience. Q and P systems: Demand/unit time: N[μ, σ], Lead Time: T, Order Interval: OI, Service Level: SL, Protection period PP = (OI + LT) OI SL, Protection PP Q system Demand during lead-time leadReorder point: ROP ROP = μLT + ZSL * σLT ROP = LT * μ + ZSL* (√LT)* σ SS = ROP – LT * μ Order Q units when IP ≤ ROP IP P system Demand during protection period Order-upOrder-up-to level: OUL OUL = μPP + ZSL * σPP OUL = (OI+LT)* μ + ZSL {√(OI+LT)}* σ (OI+LT)* SS = OUL – (OI+LT) * μ Order OUL – IP units on review date. OUL Fixed lead time. Backordering permitted when shortages occur. Q1 OUL Q2 Q3 ROP Q ROP LT Q Q LT LT OI + LT MGMT 36100 19 ChVI‐InventoryManagementII‐Sp10 LT MGMT 36100 18 ChVI‐InventoryManagementII‐Sp10 7 Newsvendor Model: This is a single period model with probabilistic demand. Buy Q units Time Selling Selling period 7 Newsvendor Model: Objective: How many units to buy to minimize expected cost (or maximize maximize expected profit)? •To find optimal lot size and corresponding profit. 1 Single period. 2 Demand: discrete / continuous distribution [$ /Q] 3 Shortage cost Cs (same as per unit “profit”) (same 4 Excess cost Ce (same as per unit loss on unsold units) [$ /Q] (same 5 Units leftover can be sold at salvage value. leftover can be sold at salvage value All quantities are per unit. Selling price: Revenue: R, Purchase price: Cost: C , Salvage value: V Shortage cost Cs = R – C … this is per unit unrealized “profit”. Excess cost Ce = C – V … this is per unit loss. We buy Q units in advance, in anticipation of demand. Suppose demand is expected to be between 100 to 500. We buy 350 units. buy 350 units If actual demand is 420, we sell 350 and stop (lose potential profit for 70 units). If actual demand is 300, we sell 300 and assume that remaining 50 units are sold at discount (called the salvage value). If this situation keeps on recurring, what should be the lot size Q to maximize expected (i.e. average) profit? ChVI‐InventoryManagementII‐Sp10 20 ChVI‐InventoryManagementII‐Sp10 21 MGMT 36100 MGMT 36100 InventoryManagementII-Sp10 2 Revenue: R, Purchase price: Cost: C , Salvage value: V Shortage cost Cs = R – C … this is per unit unrealized “profit”. Excess cost Ce = C – V … this is per unit loss. Suppose we buy Q units and sell X out of these Q. Then number of unsold units = (Q – X). Then the profit is = X * (R – C) – (Q – X) (C – V) Since we want to maximize profit, we would like to buy Q0 units. This is done by setting service level to a specific value and equating th that to the cumulative probability of the demand distribution. th th di We will consider the following three distributions: Uniform, normal and discrete (empirical). Revenue: R, Purchase price: Cost: C , Salvage value: V Suppose we buy Q units and sell X out of these Q. Number of unsold units = (Q – X) & profit = X * (R – C) – (Q – X) (C – V) We will consider three different distributions of demand. Uniform Normal Discrete X A Average demand 10 20 30 40 Pr(X) .05 .35 .39 .21 X→ (A+B)/2 B μ μ (.05*10+.35*20+.39*30+.21*40) = 27.6 Set Set service level to: SL = SL Cs Cs + Ce R-C = R-V For uniform distribution: Avg. number of units sold = Q - (Q-A)2 / [2 * (B-A)] (Q(BFor normal distribution: Number of units sold = ??? For discrete distribution: In the example above, if we stock up Q = 30. Avg. number of units sold = (.05*10+.35*20+.39*30+.21*30) = 25.5 (.05*10+.35*20+.39*30+.21*30) 22 ChVI‐InventoryManagementII‐Sp10 23 ChVI‐InventoryManagementII‐Sp10 MGMT 36100 MGMT 36100 Ex. 7.1: Demand distribution - Uniform 7.1: Revenue = $10 Purchase cost = $5.5, Savage value = $5.0 Demand X ~ U[A = 800, B = 1200] Ex. 7.2: Ex. 7.2: Demand distribution – normal. Use data from 7.1 except demand distribution is normal. Demand X ∼ N[ μ = 1000 , σ = 100 ]. Solution method: Step 1: Calculate SL = (R – C) / (R – V) SL SL = (10 – 5.5) /(10 – 5.0) = 0.9 800 X→ 1200 Solution method: SL Step 1: Calculate SL = (R – C) / (R – V) SL = (10 – 5.5) / (10 – 5.0) = 0.9 Step 2: Set cum. probability F(Q0) = SL. Solve this equation. Solve this equation. Q0 = A + SL * (B – A) SL (B A) Q0 = 800 + 0. 9 * (1200 – 800) = 1160 Q0 Step Set Step 2: Set F(Q0) = SL. SL Solve this equation Solve this equation. Q0 = μ + ZSL σ Q0 Step 3: Expected profit (BP(Q) = (R – C)*Q – (R - V)*(Q -A)2 / [2 * (B-A)] P(1160) = (10 – 5.5)*1160 – (10 – 5.0)*(1160-800)2 / [2 * (1200 – 800)] 5.0)*(1160[2 = 4410 Z0.9 = 1.29 Q0 = 1000 + 1. 29 * 100 = 1129 Calculation of expected sales and profit for normal distribution is somewhat involved and we will omit it. ChVI‐InventoryManagementII‐Sp10 24 ChVI‐InventoryManagementII‐Sp10 25 MGMT 36100 MGMT 36100 Ex. 7.3: discrete distribution. Revenue = $100/ unit, Cost = $30/unit and salvage value = $20/unit. Demand distribution Demand X Pr(X) 100 0.1 200 0.15 300 0.275 400 0.225 500 600 0.185 0.065 Example 7.3: discrete distribution. R = $100, C = $30 and V = $20 Part (d): What is the optimal lot size? Expected profit for this ? Now we apply the three step procedure. R-C Step 1: SL = Step Step 2: R-V = (100 – 30) (100 – 20) = 0.875 Part (a): What is the average demand? 10.0 30.0 82.5 90.0 92.5 39.0 X * Pr(X) Pr(X) Part (b): If we buy 400 units w hat w ill be average sale? Part (b): If w e buy 400 units, what will be average sale? Sale (Y) Pr(Y) 100 100 0.10 200 0.15 300 0.275 400 0.225 400 400 344.0 Demand distribution 100 0.10 0.10 200 0.15 0.25 300 0.275 0.525 400 0.225 0.75 500 600 0.185 0.065 0.935 1.00 Demand X Pr(X) F(X) 0.185 0.065 Y * Pr(Y) 10.0 30.0 82.5 90.0 74.0 26.0 312.5 Part (c): If we buy 400 units, what will be average profit? Average sale = 312.5 Average unsold (leftover) = 400 – 312.5 = 87.5 Average profit = 312.5 * (100 - 30) – 87.5 * (30 - 20) = $21,000 $21,000 ChVI‐InventoryManagementII‐Sp10 26 Find smallest X so that F(X) ≥ SL Set Optimal Q = X Optimal Q = 500 Step 3: Average sales = 337.5, Avg. unsold = 500 – 337.5 = 162.5 We use the procedure from part (b) to get sales. Average profit = 337.5 * (100 - 30) – 162.5 * (30 - 20) = $22,000 $22,000 ChVI‐InventoryManagementII‐Sp10 27 MGMT 36100 MGMT 36100 InventoryManagementII-Sp10 3 Example 7.4 - From an old test. R = $35, C =$20 and V = $15 Demand distribution Demand X Pr(X) F(X) 100 0.05 200 0.12 300 0.15 0.32 400 0.23 0.55 500 600 700 800 0.03 1.00 Ex 7.4 continued. Fall 2002 test. R = $35, C =$20 and V = $15 Demand distribution Demand X Pr(X) F(X) 100 0.05 200 0.12 300 0.15 0.32 400 0.23 0.55 500 600 700 800 0.03 1.00 0.17 0.15 0.10 0.72 0.87 0.97 0.17 0.15 0.10 0.72 0.87 0.97 0.05 0.17 0.05 0.17 Part (a): What is the optimal lot size? SL = (35 - 20) / (35 – 15) = 0.75 Buy 600 units Part (c): What is the Expected profit? Units bought = 600, average sold = 419, average unsold = 181. Average profit = 419 * (35 – 20) – 181 * (20 – 15) = $5380 Part (d): What is Q0 if there is no salvage value? Salvage value = 0, SL = (35-20) / (35 – 0) = 0.42856 0, SL (35Can salvage value be negative? What if Salvage value ≥ Unit cost? ChVI‐InventoryManagementII‐Sp10 29 Part (b): How many units will be sold if we buy the optimal lot size? Y (units sold) 100 200 300 400 500 600 600 600 Q0 = 400 units Average units sold = .05*100 + .12*200 + .15*300 + .23*400 + .17*500 + .28 * 600 = 419 ChVI‐InventoryManagementII‐Sp10 28 MGMT 36100 MGMT 36100 Inventory: Stock or store of goods. Stock Manufacturing: raw material, purchased parts, partially finished and finished goods. Machine spares, tools, other supplies. Department Stores: Clothing, furniture, carpeting, white goods, shoes, gift items, toys, cards, etc. Hospitals: drugs, surgical supplies, life-monitoring lifeequipment, sheets, etc. Supermarkets: fresh and canned food, frozen food, dairy products, produce, magazines, etc. Inventory Inventory Classification 1. Raw material / purchased parts 2. WIP 3. Finished goods / merchandise 4. Replacement parts / tools / supplies 5. Goods in transit (pipeline) Functions of inventory 1. Meet anticipated customer demand. 2. Smooth production requirements. 3. Decouple operations (buffer) aga 4. Protect against shortages. 5. Take advantage of order cycle. 6. Hedge against price increase. 7. To permit operations. 8. Take advantage of quantity discount. ChVI‐InventoryManagementII‐Sp10 30 ChVI‐InventoryManagementII‐Sp10 31 MGMT 36100 MGMT MGMT 36100 % cumulative wealth Pareto (1848 – 1923) Population vs. Wealth Population divided in three classes: rich, middle class, poor 120 100 80 60 40 20 0 0 20 40 60 80 100 120 Inventory related costs: Purchase price: [$ / unit].There may be quantity discount Ordering Cost [$] Order preparation and tracking. Transport, load unload. Inspection. Payment. Set up cost: [$] Remove old set up. Clean. Machine set up. Trials. ABC classification: Pareto Pareto analysis applied to inventory usage. % Of % Dollar Control items usage A 5-10 70-80 7010-15 105-10 Tight Medium Little Models Advanced Common % population Holding Cost (carrying cost): [$/ (Q*T)] B 10-20 10C 70-80 70- AB C Safety Safety stock Pipeline On On Hand ?? Cum. Number of items 32 Shortage cost: [$/ (Q*T)] or [$ / Q] ChVI‐InventoryManagementII‐Sp10 33 ChVI‐InventoryManagementII‐Sp10 MGMT 36100 MGMT 36100 InventoryManagementII-Sp10 4 Performance Measures Financial (Carrying Costs, InventoryInventory-Sales Ratio) Turnover (Inventory Turns) Fill Fill Rate or Service Level (SL) Inventory Accuracy Forecast Accuracy Benchmarking: involves comparing the performance with companies involved in similar business with respect to selected measures. ChVI‐InventoryManagementII‐Sp10 34 Importance of Inventory Management Assume 1 Yr = 360 days. A grocery store buys cartons of milk and sells those every 3 days. Inv. Turns = 120. A store buys and sells a batch of washers every batch 30 days. Inv. Turns = 12. JanJan-2007 Total assets inventory revenue COGS Inventory / assets Inventory/revenue Inv/COGS = WIP/TPR = FT Inv. Turns = 1/FT ChVI‐InventoryManagementII‐Sp10 Walmart 151.2 B$ 33.7 B$ 348.7 B$ 264.2 B$ 22.3 % 9.7% Target 37.3 B$ 6.2 B$ 59.5 B$ 39.4 B $ 16.6% 10.4% 0.128 years 0.157 years = 1.5 months = 1.9 months 7.8 times / year 6.4 times / year 35 MGMT MGMT 36100 MGMT 36100 Inventory turns are a common measure of inventory productivity but vary widely Retail Industry Segment1 Apparel And Accessory Stores Catalog, Mail-Order Houses Department Stores Drug & Proprietary Stores Food Stores St Hobby, Toy, And Game Shops Home Furniture & Equip Stores Jewelry Stores Radio, TV, Consumer Electronic Stores Variety Stores Inventory Turnover 4.57 8.60 3.87 5.26 10.78 2.99 5.44 1.68 4.10 4.45 Gross Margin 37% 39% 34% 28% 26% 35% 40% 42% 31% 29% Objective: Provide right material at the right time with Provide right quality at right price Typical Activities Materials Materials planning Purchasing & control Inventory Control Material Planning ABC Analysis Budgeting Material Research Codification Standardization Source Selection Purchase Systems Price Forecasting Seasonal Buying Capital Equipment International Buying Legal Aspects Stores management Systems & Procedures Incoming Material Material Handling Transportation Scrap Management Stock Checking Obsolete Items Mgmt. Materials Management 1: Information from S&P’s Compustat data base: 1985-2000 ChVI‐InventoryManagementII‐Sp10 36 ChVI‐InventoryManagementII‐Sp10 37 MGMT 36100 MGMT 36100 InventoryManagementII-Sp10 5 ...
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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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