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Unformatted text preview: Definitions: Symbol MGMT 36100 Operations Management (OM) Lead time On Hand Inventory Inventory Level Inventory Position On Order hand size 1 50 50 50 50 45 75 75 75 2 LT OH IL IP Time between placing & receiving order. Physical stock. Units available to meet future demand. IL plus ordered quantity in transit Shortage Comments 0 25 25? 75? Result OH 5 0 0 50 IL 5 25 0 50 VIVI Inventory Management  I
Add Value Operations Input Output Let’s consider 4 different situations
Action Deliver 45 45 Deliver 50 Deliver 50 Deliver 0 Back order Lost sales Lost sales 3 4 Inventory / quality / monitoring
ChVI‐InventoryManagementI‐Sp10 Control Process Schematic
1 IL = OH – Quantity on backorder IP = IL + Quantity in transit We will not deal with lost sales case.
ChVI‐InventoryManagementI‐Sp10 2 MGMT 36100 MGMT 36100 Definitions:
Lead time LT Time between placing and receiving order. On Hand Inventory OH Physical stock Inventory Level IL IL = OH – Quantity on backorder Inventory Position IP IP = IL + Quantity in transit 1 The EOQ Model – Economic Order Quantity
Developed in 1915 by Harris. Very simple, robust. Used by thousands of companies around the world. Objective: To determine order quantity (lot size) to minimize annual inventory cost.
1 Demand is known, occurs at uniform rate. 2 3 4 5 6 Ordering cost S is fixed for every order. fi Holding cost H charged on average inventory Lot size Q fixed, delivered in one lot No shortages permitted. Leadtime (LT) is fixed (equal to zero) LeadC: Purchase price per unit [$] Holding cost: Carrying cost D: [ Q / T ]. S: [ $ ] H: [ $ / (Q*T) ] Q: [ Q ] LT: [ T ] Let’s consider different situations
Given OH = 50, LT = 0 IL = 20, LT = 0 Calculate IL, IP OH, IP IL = 50, IP = 50 OH = 0, IP = 20 IL = 100 IP = 100 + 75 OH = 0 IP =  40+80+50 OH = 100, LT = 3 days. 75 units ordered IL, IP yesterday IL =  40, LT = 5 days, 80 units ordered OH, IP yesterday, 50 units ordered 3 days ago. We will now consider many inventory control models involving a single product.
ChVI‐InventoryManagementI‐Sp10 3 ChVI‐InventoryManagementI‐Sp10 4 MGMT 36100 MGMT 36100 Suppose demand for a product is uniform  6000 units/year…. D 500 What is the monthly demand? 500 units. Suppose we order 250 units every time (Q), how often will we place order (this is called the order cycle)? Once a year / once a month / once every half month / ?? Order cycle = Q / D = 250/6000 = 1/24 [years] = 1/2 [month] Now let’s plot a graph of inventory vs. time 250 250
Inventory Inventory Example 1.1 AIC(Q) = (S . D / Q) + (H . Q / 2) D = 12,000 units / year, H = $3.6 per unit per year, S = $140 year, year, How does one determine H and S? C = $15 /unit and annual holding cost: 24% of C, H = 15 * 0.24 We will now calculate the Annual Inventory Cost (AIC) for different lot sizes. AIC = annual ordering cost + annual inventory holding cost. Q # of Ordering Average Holding orders: cost: inventory: cost: (D/Q) S * (D/Q) (D/Q) Q/2 H * Q/2 Q/2 30.0 15.0 10.0 7.5 4200.0 2100.0 1400.0 1050.0 200.0 400.0 600.0 800.0 720.0 1440.0 2160.0 2880.0 Annual Inventory cost (AIC): Ordering Holding Ordering + Holding 4920.0 3540.0 3560.0 3930.0 The average inventory for this model is Q/2 when the lot size is Q. Average inv = 250/2 = 125 units 400 400 800 1200 1600 Order cycle 1m Time
5 Larger lot size means lower ordering cost but higher inventory holding cost.
ChVI‐InventoryManagementI‐Sp10 6 ChVI‐InventoryManagementI‐Sp10 MGMT 36100 MGMT 36100 ChVIInventoryManagementISp10 1 Q1 Inventory Q2 Annual inventory cost = ordering cost + holding cost + H.Q1/ 2 AIC(Q1) = S.D/Q1 + H.Q2/ 2 AIC(Q2) = S.D/Q2 Let optimal quantity be Q0.
Annual Costs Holding Cost 18000 16000 14000 12000 10000 8000 6000 4000 2000 Ordering Cost Inventory Cost Example 1.2 S = $140, $140 ordering cost = holding cost Q0 = √{2 * D * S / H} S . (D/Q0) = H . (Q0 / 2) (D/Q (Q D = 12,000 units / year, P = $15 /unit. H = $3.6 per unit per year to within to Q0 = √{2 * 12000 * 140 / 3.6} = 966.1 Round off Q0 to within ± 5% to Q* {2 12000 966 Round off Selected order lot size Q* = 1000 Lot size # of orders Ord. cost Avg. inv. Holding cost Q0 …966.1 Q*… 1000
ChVI‐InventoryManagementI‐Sp10 AIC 0 0 200 400 600 800 1000 1200 1400 Lot Size (Q) 12.421 12 1738.94 1680 483.05 500 1738.98 3477.96 1800 3480.00
7 AIC(1000) = 140 * 12000 / 1000 + 3.6 * 1000 / 2 = 1680 + 1800 = $3480
ChVI‐InventoryManagementI‐Sp10 8 MGMT 36100 MGMT 36100 500 500
Inventory Inventory Annual demand: D Ordering cost: S Annual holding cost per unit: H Order quantity (lot size): Q Q* Q* Optimal lot size: Q0
Order cycle Time Ex 1.3: Variations of the basic EOQ model Q0 = 966.1 Policy [0,1000]
10000 9000 8000 7000 From Ex. 1.2: D = 12000 units / year, S = $140, H = 3.6$ / unit / year 1. At least 1500 per lot? Policy [0,1500] 2. At least 750 per lot? Policy [0,1000]
Annual Costs 6000 5000 4000 3000 2000 1000 0 150 300 450 600 750 900 105 120 135 150 165 180 0 0 0 0 0 0 Lot Size (Q) We want to answer two questions: when and how much to order. This is called inventory (ordering) policy. So, when we write (X, Y), for when and how much. (X X is the inventory position (IP) and Y is the order quantity (Q). For the EOQ model, lead time is zero. Therefore, IP = IL. Also, shortages are not permitted. Therefore IL = OH. So, the policy for basic EOQ model will be (0, Q*).
ChVI‐InventoryManagementI‐Sp10 9 3. Sold in boxes (150 per box). …, 600, 750, 900, 1050, 1200,… 900, 966.1, 1050, 900, 966.1, 1050, AIC(900) AIC(900) = 3486.67 Policy [0, 900]
ChVI‐InventoryManagementI‐Sp10 AIC(1050) = 3490 10 MGMT 36100 MGMT 36100 Example 1.4
From Ex. 1.2 D = 12000 units / year Policy: [0, 1000] We used lead time LT = 0 LT Symbols: Place order Receive order Q* Example 1.4
D = 12000 units / year Policy: [0, 1000] Time LT = 1½ months Policy: [1500, 1000] Inventory Q* 1 month Order when IP = demand during lead time = D * LT LT
Part 1: Suppose leadtime is ½ month leadQ* D * LT = LT 12000 [units / year] * 1½ [months] = 1500 units 1½ 1500 12 [months /year] Inventory We will still order 1000 units but ½ month earlier. Policy: [500, 1000] LT = 3 months ? D * LT = 3000 units Ordering Policy: [ IP = 3000, Q = 1000 ] ChVI‐InventoryManagementI‐Sp10 11 ChVI‐InventoryManagementI‐Sp10 Inventory 12 MGMT 36100 MGMT 36100 ChVIInventoryManagementISp10 2 2 The EPQ Model – Economic Production Quantity
Consider EOQ model first. D = 36000 [units/year] Suppose Q = 2000 [units] We define usage rate u = 100 units / day Now let’s consider the EPQ model.. u = 100 units/day. Q = 2000 units. Production rate p = 250 [units/day] It will take 8 days to produce the lot. In 8 days, we will also use 800 units. So max inventory will be 1200. For 12 more days, we will shut days, down production and use up inventory.
2000 2 EPQ model
Objective: Minimize annual inventory cost.
1 Demand is known, occurs at uniform rate. 2 3 4 5 6 Setup cost S is fixed, incurred for each setup. Holding cost H charged on average inventory Production lot size Q is fixed No shortages permitted. Production rate (p) is uniform. D: [ Q / T ] S: [ $ ] H: [ $ / (Q*T) ] Q: [ Q ] p: [ Q / T ] 2000 1200
Inventory Inventory Inventory Inventory Time 20 20 days
ChVI‐InventoryManagementI‐Sp10 8 “D” usually denotes annual demand. Usage Rate : u is the daily demand and Production rate: p is also production per day. Time 20 20 days
13 ChVI‐InventoryManagementI‐Sp10 14 MGMT 36100 MGMT 36100 EPQ
Q
Inventory Annual Demand: D, Usage Rate : u, Production rate: p Only production Production & consumption Only consumption EPQ EPQ
Q
Inventory Annual Demand: D, Usage Rate : u, Production rate: p Only production Production & consumption Only consumption Annual number of runs = D / Q Average inventory = ½ Q” = ½ Q . (p – u) / p Q” t Q” t 0 t’’ Time 0 t’’ Time Run Run time = Q / p. Run time is the interval of time when machine is running.
Cycle Cycle Time = Q / u. Cycle time is time between u. beginning of runs. AIC(Q) = S . D + H. Q. (p  u) Q 2 p QO = 2*S*D H p (p  u) D, H: same units p, u: same units What if p is very large compared to u?
16 ChVI‐InventoryManagementI‐Sp10 15 ChVI‐InventoryManagementI‐Sp10 MGMT 36100 MGMT 36100 Inventory D = 18000 units / year, S = $175. 1 year = 360 days. H = $4.2 / unit / year. p = 90 units / day. Run time = Q / p (a) Find optimal lot size Q Cycle time = Q / u u = 18000 / 360 = 50 [units / day] Q” = Q . (p – u)/ p QO = √[2 * D * S / H] * √[ p /(p – u)] Q” = √[2 * 18000 * 175 / 4.2] *√ 90 /(9050) /(90 Ex. 2.1 Old final D = 45000 units / year, S = $120. p = 250 units / day. Assume 25 days/month. H = $1 / unit / month. How much higher AIC if run time = 4 days instead of optimal? H = $12 / unit / year There are 25 x 12 = 300 days per year. u = 45000 / 300 = 150 [units / day] QO = √[2 * D * S / H] * √[ p /(p – u)] /(250= √[2 * 45000 * 120 / 12] *√ 250 /(250150) = 1500 AIC(1500) = S.D/Q + ½H.Q.(pu)/p = 3600 + 3600 = $7200 per year. (1500) p3600 3600 $7200 Run time = Q / p Optimal Optimal run time = 1500 / 250 = 6 days. For run time of 4 days, Q = 4 x 250 = 1000 AIC(1000) = S.D/Q + ½H.Q.(pu)/p H.Q.(pu)/p = 5400 + 2400 = $7800 per year Extra cost = 7800 – 7200 = $600 per year. = 1837 Q* = 1800 (b) Calculate run time, cycle time and AIC.
Run time = 1800 [units] / 90 [units/day] = 20 days, 20 Cycle Cycle time = 1800 [units] / 50 [units/day] = 36 days, 36 Q” Q” = maximum inventory = 800 AIC(1800) = S.D/Q + ½H.Q.(pu)/p H.Q.(pu)/p = 1750 + 1680 = $3430 $3430 (c) What if machine cannot be run for more than 15 days?
Now run time = 15 days,. Then Q* = 1350 , 15 Then Q* Cycle time = 27 days and AIC(1350) $3593.33 CycleInventoryManagementI‐Sp10 and AIC(1350) = $3593.33 27 ChVI‐
MGMT 36100
17 ChVI‐InventoryManagementI‐Sp10 18 MGMT 36100 ChVIInventoryManagementISp10 3 EPQ: Two products with the same cycle time
Product A: Run time 15 days, Cycle time 40 days t Product B: Run time 23 days, Cycle time 40 days 3. EOQ with quantity discounts All unit discount
Price $5 per unit if Q ≤ 100 and $4.50 per unit for all units if Q > 100. For Q = 100, purchase price for the lot = 100 x 5.00 = 500 dollars. For Q = 200, purchase price for the lot = 200 x 4.50 = 900 dollars This model is identical to the original EOQ model except the purchase price depends on the lot size. So we cannot ignore annual purchase price and the objective is to minimize annual total cost. Annual total cost = annual purchase price + annual inventory cost 38 40
ChVI‐InventoryManagementI‐Sp10 19 ChVI‐InventoryManagementI‐Sp10 20 MGMT 36100 MGMT 36100 Example 3.1: From an old exam QO = √{2 * D * S / H}
Do calculations for lowest price/unit Is Q0 feasible? No. Calculate for next higher price. price. Is Q0 feasible? Yes! Stop. Pick lowest cost solution. Policy [0, 700] Cost: $18204.57 / yr Ex. 3.2 D = 6000 units / year, S = $200, Holding cost per year: 20%
A B $2.45 $0.49 C $2.40 $0.48 D = 10,000 units / year, S = $5.5, H: 20% Lot size 1 to 399 400 to 699 700 and up price /unit H $2.20 $0.44 $2.00 $0.40 $1.80 $0.36 Lot size Unit price H Q0 Lot Lot size: Q* Purchase $ Ordering $ Holding $ Total $ 1 to 1000 1001 to 5999 6000 & up $2.50 $0.50 Do calculations for lowest cost column “C” Is Q0 feasible? No. Calculate for next vendor. “B” Is Q0 feasible? Yes! Stop. Pick lowest cost column. Q0 552.80 524.40 700.00 500.00 Lot Lot size Q* Annual Cost 20000.00 18000.00 Purchase $ 110.00 78.57 Ordering $ 100.00 126.00 Holding $ 20210.00 18204.57 Total $ This This Q0 value is without lot size constraint 2213.33 2250.00 533.33 551.25 2236.00 6000.00 200.00 1440.00 14700.00 14400.00 15784.58 16040.00 Part (b): What if LT = 2 weeks ( assume 50 weeks/per year )? LT Demand during lead time = 400. Policy [400, 700]
ChVI‐InventoryManagementI‐Sp10 21 This This Q0 value is without lot size constraint
ChVI‐InventoryManagementI‐Sp10 Policy (0, 2250), annual total cost: $15,784.58
22 MGMT 36100 MGMT 36100 Graphs of total annual cost vs. lot size
“Optimal” for entire range feasible range EOQ EOQ discount EPQ 18000 17500 17000 16500 16000 15500 15000 0 1000 2000 3000 4000 5000 6000 7000 8000 H
A A B C B C May be proportional to purchase price within ± 5%, within range
D.C + S.D/Q + H.Q / 2 Q0 (optimal) √{2 * D * S / H} √{2 * D * S / H} √[2*D*S /H] * √[p / (p – u)] (optimal) Q* Q* (lot size) within ± 5% Round off Annual cost S.D/Q + H.Q / 2 Lead time Constraint Yes Yes within ± 5%
S.D/Q + H. (Q/2) (p  u) / p Yes Yes Yes ChVI‐InventoryManagementI‐Sp10 23 ChVI‐InventoryManagementI‐Sp10 24 MGMT 36100 MGMT 36100 ChVIInventoryManagementISp10 4 4 Lot sizing models
1 2 3 4 5 There are n periods. Demand for each period is known. No shortages are permitted. Leadtime is zero (we will modify this later.). LeadWe maintain specified inventory (called Safety Stock) Safety Method 1 Lot for lot 2 Fixed order quantity. 3 Period order quantity. L4L FOQ POQ Text Class Comments Order Order every period Order Order fixed quantity Order Order every x periods These These models require cost parameters. Example 4.1
Period Demand Initial Initial inventory = 30, SS = 10 0 1 2 3 4 5 6 20 7 0 0 0 8 0 9 10 75 70 10 35 40 30 15 40 30 50 50  0 80 20 0 80  100 0 0 75 70 4.14.1a: L4L 4.14.1b: FOQ FOQ 50 4.14.1c: POQ 3 Q
End End inventory 30 10 10 10 10 10 10 10 10 10 Q
Demand  100 50 0 75 145 70  End inventory End inventory 30 45 55 25 25 45 25 25 25 50 30 80 20  Skip100  35 40 30 85 Q 4 Economic order quantity EOQ 5 Least total cost LTC 6 Least unit cost LUC End End inventory 30 80 40 10 10 30 10 10 10 80 10 ChVI‐InventoryManagementI‐Sp10 25 4.14.1a: What would be Q for P1 if the initial inventory was 60? Notice that the inventory cannot fall below SS. 4.14.1c: Notice that the inventory goes down to the SS value at the end ofInventoryManagementI‐Sp10 3 periods. ChVI‐ 26
MGMT 36100 MGMT 36100 The (4) models we considered so far were deterministic models (demand pattern known). No. Model
1 2 3 4 EOQ EPQ Lot Sizing Objective
Min. Annual Inventory Cost ?? Used for
Purchase Production Purchase Purchase / production EOQ with discounts Min. Total cost In the next phase we will study 3 probabilistic models.
No Model Objective Used for 5 Reorder point Achieve desired service level 6 Fixed order interval Purchase 7 Newsvendor Min. expected cost
ChVI‐InventoryManagementI‐Sp10 27 MGMT 36100 ChVIInventoryManagementISp10 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 UniversityWest Lafayette.
 Spring '10
 panwalker
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