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1 ANALYSIS OF INVENTORY MODEL Notes 1 of 2 By: Prof. Y.P. Chiu 2011 / 09 / 01.

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Presentation on theme: "1 ANALYSIS OF INVENTORY MODEL Notes 1 of 2 By: Prof. Y.P. Chiu 2011 / 09 / 01."— Presentation transcript:

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2 1 ANALYSIS OF INVENTORY MODEL Notes 1 of 2 By: Prof. Y.P. Chiu 2011 / 09 / 01

3 2 § I1 :Types of Inventories § I1 : Types of Inventories (A) Raw Materials (B) Components (C) Work in process (D) Finished goods ◇ § I2 : Inventory Relevant Costs (A) Holding Cost (B) Order Cost (C) Penalty Cost (D) Outdate Cost Analysis of Inventory Model

4 3 § I2(A) : Holding Cost § I2(A) : Holding Cost Opportunity cost of alternative investment Taxes and insurance Breakage, spoilage, deterioration, obsolescence Cost of physical space Eg: 28% = cost of capital 2% = Taxes and insurance 6% = cost of storage 1% = breakage and spoilage 37% = Total interest charge h = i . c ◇

5 4 § I2(B) : Order Cost § I2(B) : Order Cost Cost of procuring x items : (fixed cost plus proportional cost) ◇

6 5 § I2(C) : Penalty Cost § I2(C) : Penalty Cost Cost when demand exceeds supply ( per unit of excess demand ) § I2(D) : Outdate Cost Cost when inventories spoiled (or outdated) ( including cost of discarding the spoilage items ) ◇

7 6 § I3 : Motivation for Holding Inventories 1.Economies of Scale 2.Uncertainties . Excess demand 3.Speculation 4.Transportation 5.Smoothing 6.Logistics . safety stock . minimum purchasing quantity 7.Control Costs . record keeping & . management costs ◇

8 7 § I4 : Characteristics of Inventory System § I4 : Characteristics of Inventory System (A) Demand . Constant versus Variable . Known versus Random (B) Lead Time . Zero, Constant, Variable, Random (C) Review Time . Continuous, Periodic (D) Excess Demand . Backordered, Lost (E) Ordering Policy. (r,Q), (s,S), etc. (F) Issuing Policy. FIFO, LIFO, etc. (G) Changing Inventory . Shelf life (expired), obsolete ◇

9 8 § I5 : EOQ (Economic Order Quantity) Q T Slope = - Inventory( I( t ) ) Fig.1 Time t◇

10 9 § I5: EOQ(Economic Order Quantity) K : setup (ordering) cost (per cycle) c : unit cost (per item) h : holding cost ( per item per year ) λ: demand per unit time (eg. year ) Q T I(t) t where T = Q / λ (cycle length) ◇

11 10 G(Q) = Inventory Costs per unit time (per year) G(Q) = Gc(Q) / T = (K+cQ) / T + h Q/2 Inventory Costs per cycle T = Q / λ ……[Eq.5.1] § I5.1 : Inventory Costs G(Q) in EOQ Model ◇

12 11 Set up cost / unit time Purchase cost / unit time Holding cost / unit time ….[Eq.5.2] § I5.1 : Inventory Costs G(Q) in EOQ Model ◇

13 12 § I5.2 : Minimizing G(Q) n To find Q that Minimizes G(Q) … n If G”(Q) > 0 then G(Q) is a convex function with a minimum. n Let G’(Q*) = 0, we can solve Q* for minimum of G(Q). ◇

14 13 § I5.2: To find Q* Let G’(Q) = 0, solve Q* ? ……[Eq.5.3]◇

15 14 § I5.3 : EOQ ~ Discussion EOQ model : Balances order cost and holding cost The basic model: 1. The demand rate is known and is constant λ items per unit time 2. Shortages are not permitted 3. No order lead time 4. Costs include ◆ setup (ordering) cost K per order ◆ holding cost h per item held per unit time ◆ unit cost c per item ordered.

16 15 §. I 5. Problems & Discussion ( # N4.1, N4.9 ) ( # S5.1, S5.2 ) Preparation Time : 20 ~ 30 minutes Discussion : 15 ~ 25 minutes

17 16 § I6 : Sensitivity n How cost increases if not using Q* recall [Eq.5.2] Drop λC for now if we use Q rather than Q*, then

18 17 § I6 : Sensitivity

19 18 §. I 6. Problems & Discussion ( # C.1, # C.2 ) Preparation Time : 20 ~ 30 minutes Discussion : 15 ~ 25 minutes

20 19 § I7 : Order Lead Time : τ To order “τ” time in advance. or to consider a reorder point (I.e. level of inventory = “R”) (A) For τ < T 4 month R =1040 Q * =3870 t I(t) T = 1.24 year ◇

21 20 Let τ = 4 months = 0.3333 year R = λτ = 3120 ( 0.3333) = 1040 (B) When τ > T (1) From the ratio τ / T (2) Consider only the fractional remainder (f - r) of the ratio. Convert this (f - r) back to year. (3) use R = λ * τ (f-r) § I7 : Order Lead Time : τ ◇

22 21 if τ = 6 weeks Q* = 25 T = 2.6 weeks λ = 500 per year ( a ) τ / T = 6 / 2.6 = 2.31 periods ( b ) 0.31: fractional remainder 0.31 (2.6) / 52 = 0.0155 years ( c ) R = 500 (0.0155) = 7.75 ≒ 8 [Eg.7.1] ◇

23 22 (C) Summary : (1) EOQ FORMULA (2) REORDER LEVEL For τ< T, R =λτ For τ> T, R =λ where (3) Rules for Computing If 2T > τ > T, then = τ - T If 3T > τ > 2T, then = τ - 2T etc. § I7 : Order Lead Time : τ § I7 : Order Lead Time : τ◇

24 23 §.I 7. Problems& Discussion ( # N4.12, N4.14 N4.14-2 ) N4.14-2 ) Preparation Time : 25 ~ 30 minutes Discussion : 20 ~ 25 minutes

25 24 § I8 : Model for Shortages Permitted S Q S-λt S/λ Q/λ Demand = λ t T = Q/ λ (a) Cycle length T = Q/ λ ; hp h : holding costs; p : shortage costs/item/unit time (Q-S)/λ (b) Shortage occurs for a time : (Q-S)/λ [0+(Q-S)]/2 (c) Average amount of shortages : [0+(Q-S)]/2 p(Q-S) / 2 (d) Shortage cost : p(Q-S) / 2

26 25 § I8 : Model for Shortages Permitted Total costs per cycle. [Eq.8.2] Total costs per unit time [Eq.8.1]

27 26 [Eq.8.3a] [Eq.8.3c] [Eq.8.3b] § I8: Model for Shortages Permitted [Eq.8.2]

28 27 §. I8. Problems & Discussion ( # C.3 ) ( # C.3 ) Preparation Time : 20 ~ 30 minutes Discussion : 15 ~ 25 minutes

29 28 [Eq.8.3d] ■ The fraction of time that no shortage exists. ■ Maximum shortage Q*-S* [Eq.8.3e] § I8: Model for Shortages Permitted

30 29 n A television manufacturing company produces its own speakers, which are used in production of its television sets. The television sets are assembled on a continuous line at the rate of 8,000 per month. The speakers are produced in batches because they do not warrant setting up a continuous production line, and relatively large quantities can be produce in a short time. The company is interested in determining when and how many to produce. Several costs must be considered: [Eg. 8.1] 1.Each time a batch is produced, a setup cost of $12,000 is incurred. This cost includes the cost of “tooling up,” administrative costs, record-keeping, and so forth. Note that the existence of this cost argues for producing speakers in large batches.

31 30 2.The production of speakers in large batches leads to a large inventory. The estimated cost of keeping a speaker in stock is 30 cents / month. This cost includes the cost of capital tied up, storage space, insurance, taxes, protection, and so on. The existence of a storage or holding cost argues for producing small batches. [Eg. 8.1] 3.The production cost of a single speaker (excluding the setup cost) is $10 and can be assumed to be a unit cost independent of the batch size produced. (In general, however, the unit production cost need not be constant and may decrease with batch size.)

32 31 4. Company policy prohibits deliberately planning for shortages of any of its components. However, a shortage of speakers occasionally crops up, and it has been estimated that each speaker that is not available when required costs 1.10 / month. This cost includes the cost of installing speakers after the television set is fully assembled, storage space, delayed revenue, record keeping, and so forth. [Eg. 8.1] n K = $12,000 / order λ = 8000 / month h = $0.3 / item / month c = $10 / item p = $1.10 / item / month unit of time = a month Solution to [Eg.8.1]

33 32 Solution to [Eg.8.1] (A) E.O.Q

34 33 per unit of time → ie ”month” in this case Solution to [Eg.8.1]

35 34 n (B) When shortage permitted p = $1.10 per speaker k = $12,000 h = $0.3 λ=8000 Use [Eq.8.3 a] Solution to [Eg.8.1] Use [Eq.8.3b] Use [Eq.8.3c] T* = Q* / λ = 28540 / 8000 = 3.57 (months)

36 35 Use [Eq.8.3d] n Maximum shortage Q* - S* = 6116 n Time shortage occurs ( Q* - S*) / λ = 6116 / 8000 = 0.76 months Time no shortage occurs S* / λ = 22424 / 8000 = 2.8 months Solution to [Eg.8.1]

37 36 (C) Discussion: When “shortage permitted” S Q S/λ T=Q/λ Q-S t

38 37 Eg. from 8.1(b) When “shortage permitted” Q*=28540 T* = Q*/λ = 3.57 month S*=22424

39 38 When “shortage not-permitted” Eg. from 8.1(a) Class work ( #C.3.4 ;#C.3.5 ) Preparation Time : 20 ~ 30 minutes Discussion : 15 ~ 25 minutes

40 39 § I9 : Inventory Management for Finite Production Rate Finite Production Rate Inventory Levels for Finite Production Rate Model Fig.9.1 Slope=P- λ Slope= - λ

41 40 P : production rate (per unit time) λ: demand rate (per unit time) P > λ § I9 : Finite Production Rate

42 41 ………...[Eq.9.3]..[Eq.9.1] …….[Eq.9.2] § I9: Finite Production Rate

43 42 A local company produces a programmable EPROM for several industrial clients. It has experienced a relatively flat demand of 2,500 units per year for the product. The EPROM is produced at a rate of 10,000 units per year. The accounting department has estimated that it costs $50 to initiate a production run, each unit cost the company $2 to manufacture, and the cost of holding is based on a 30 % annual interest rate. Determine the optimal size of a production run, the length of each production run, and the average annual cost of holding and setup. What is the maximum level of the on-hand inventory of the EPROMs? [Eg. 9.1]

44 43 λ = 2500 / year P = 10,000 /year K = $50 / setup c = $ 2 / unit i = 30% Solution to [Eg.9.1] : ~ Finite Production Rate ~ Use [Eq.9.3]

45 44 Solution to [Eg.9.1] : ~ Finite Production Rate ~ Use [Eq.9.2] setup cost / year holding cost / year

46 45 § I9.1: Finite Production Rate with Backordering with Backordering

47 46 §. I 9. Problems & Discussion Preparation Time : 25 ~ 30 minutes Discussion : 20 ~ 25 minutes ( # N4.17 ; N4.20 ) # C.3.8 ; #C.3.9 # C.3.8 ; #C.3.9 ( # S5.3 ; S5.4 ; S5.7 )

48 47 § I10 : All-Units Discount Inventory Model Inventory Model Fig.10.1 All-units Discount order cost function All Units Discount 0.30Q for 0 ≦ Q < 500 C(Q) = 0.29Q for 500 ≦ Q < 1000 0.28Q for 1000 ≦ Q ﹛◇

49 48 Assume λ= 600 k = $ 8 h = 0.20( C j ) = (0.2)(0.3) = (0.2)(0.29) = (0.2)(0.28) (A)Use [Eq.5.3] to find optimal for each Q j -intervals. § I10 : All-Units Discount Inventory Model Inventory Model ◇

50 49 Q (1) ≒ 406 ∴ Q (1) * = 500 Q (2) ≒ 414 ∴ Q (2) * = 1000 100 200 300 400 500 600 700 800 900 1000 § I10 : All-Units Discount Inventory Model Inventory Model ◇

51 50 (B) Plug in Q* i to find min. cost ﹛ ∴ The optimal solution is to place a standing order for 500 units at annual cost of $198.1 § I10 : All-Units Discount Inventory Model Inventory Model ◇

52 51 §. I10. Problems & Discussion Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 20 minutes ( # N4.22 ; N4.24 ) ( # N4.22 ; N4.24 ) ( # S5.9 ) ( # S5.9 )

53 52 § I11: Incremental Discount Inventory Model Inventory Model Fig.11.1 Incremental Discount order cost function

54 53 ﹛ (A) (B) ﹛ § I11: Incremental Discount Inventory Model Inventory Model

55 54 (C) Find Q* for each Q j -intervals [C.1] Cj = 8(600)/400 + 180+ (0.06)(400)/2 ≒ 204 § I11: Incremental Discount Inventory Model Inventory Model

56 55 [C.2] § I11: Incremental Discount Inventory Model Inventory Model

57 56 [C.3] 100 200 300 400 500 700 1000 § I11: Incremental Discount Inventory Model Inventory Model

58 57 (D) Discussion : Incremental Discount vs. All Units Discount ◆ All units discount optimal at Q* = 500 units & cost of G(Q*) =$198.1 ◆ Incremental discount optimal at Q* = 400 units & cost of G(Q*) =$204 § I11: Incremental Discount Inventory Model Inventory Model

59 58 (E) Incremental solution technique: There are other discount schedules. There are other discount schedules. § I11: Incremental Discount Inventory Model Inventory Model

60 59 §. I11. Problems & Discussion Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 20 minutes ( # N4.23 ; N4.35 ) ( # N4.23 ; N4.35 ) ( # S5.14 ) ( # S5.14 ) The End of Class Notes 1 of 2


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