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ANALYSIS OF INVENTORY MODEL Notes 2 of 2. By: Prof. Y.P. Chiu 2011 / 09 / 01. § I12 : Inventory model: when demand rate λ is not constant. • Periodic review ~ A general model for Production Planning Terms: Periods: 1,2,3,….N
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ANALYSIS OF INVENTORY MODEL Notes 2 of 2 By: Prof. Y.P. Chiu 2011 / 09 / 01
§ I12 : Inventory model: when demand rate λ is not constant • Periodic review ~ A general model for Production Planning Terms: Periods: 1,2,3,….N i: demand rate in period i h : holding cost / item / period K : setup cost c : unit cost : cost of producing enough items for period i thru. j at beginning of period i
§ I12 : Inventory model: when demand rate λ is not constant (B) Formula ...[Eq.12.1] …...[Eq.12.2] • Lowest cost from period i to N that will satisfies demand
§ I12 : Inventory model: when demand rate λ is not constant [Eg.12.1] ~When demand rate λ is not constant~ Demand 1Q 2Q 3Q 4Q 3000 2000 3000 2500 λ1 λ2 λ3 λ4 X2 X3 X4 P1 P2 P3 P4
§ I12 : Inventory model: when demand rate λ is not constant [Eg.12.1] ~When demand rate λ is not constant~ • Use [Eq.12.2]
§ I12 : Inventory model: when demand rate λ is not constant [Eg.12.1] ~When demand rate λ is not constant~
§ I12 : Inventory model: when demand rate λis not constant [Eg.12.1] When demand rate λis not constant [Answer] To produce enough items from 1st period to 2nd period, then To produce enough items from 3rd period to 4th period. In other words , production plan is: “ to produce 5000 items at the beginning of the first period, then to produce 5500 items at the beginning of the 3rd period ”.
§. I12: Problems & Discussion #C.4 K=$20 C=$0.1 h=$0.02 300 200 300 200 #C.5 Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 20 minutes
§ I13 : Inventory Model: Resource-Constrained Multiple Product System [Ex.13.1] Item 1 2 3 λj 1850 1150 800 Cj $50 $350 $85 Kj $100 $150 $50
§ I13 : Resource-Constrained Multiple Product System • [Eq.13.1] Must run under budget → (Budget) ……[Eq.13.1] EOQ1=172 , EOQ2=63 , EOQ3=61 $50(172)+$350(63)+$85(61)=$35,835 (over-budget) • Adjusting Factor …..Eq.13.2]
§. I13: Problems & Discussion ( # C.6 ) ( # N4.38(a) ) Preparation Time : 25 ~ 30 minutes Discussion : 15 ~ 25 minutes
§ I13 : Inventory Model: Resource-Constrained Multiple Product System [Ex.13.2] • Check : wi/ hi w1/ h1= 9 / 12.5 = 0.72 w2/ h2=12 / 87.5 = 0.14 w3 / h3=18 / 21.25 =0.847 Diff. • Simple solution obtained by a proportional scaling of the EOQ values will not be optimal. • Find Lagrange multiple ?
◇ § I13 : Inventory Model: Resource-Constrained Multiple Product System [Ex.13.2] …..….[Eq.13.3] …….[Eq.13.4]
◇ § I13 : Inventory Model: Resource-Constrained Multiple Product System (a) For proportional Not optimal!
◇ § I13 : Inventory Model: Resource-Constrained Multiple Product System (b) Find Lagrangean Function Find ∴
§. I13.1: Problems & Discussion ( # N4.26 ; N4.28 ) Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 20 minutes
§ I14: The Newsboy Model [Eg. 14.1] On consecutive Sundays, Mac, the owner of a local newsstand, purchases a number of copies of The Computer Journal, a popular weekly magazine. He pays 25 cents for each copy and sells each for 75 cents. Copies he has not sold during the week can be returned to his supplier for 10 cents each. The supplier is able to salvage the paper for printing future issues. Mac has kept careful records of the demand each week for the Journal. (This includes the number of copies actually sold plus the number of customer requests that could not be satisfied.) The observed demands during each of the last 52 weeks were 15 19 9 12 9 22 4 7 8 11 14 11 6 11 9 18 10 0 14 12 8 9 5 4 4 17 18 14 15 8 6 7 12 15 15 19 9 10 9 16 8 11 11 18 15 17 19 14 14 17 13 12
§ I14: The Newsboy Model [Eg. 14.1] There is no discernible pattern to these data, so it is difficult to predict the demand for the Journal in any given week. However, we can represent the demand experience of this item as a frequency histogram, which gives the number of times each weekly demand occurrence was observed during the year. The histogram for this demand pattern appears in the following Figure. Consider the example of Mac’s newsstand. From past experience, we saw that the weekly demand for the Journal is approximately normally distributed with mean μ=11.73 and standard deviation σ= 4.47.
§ I14: The Newsboy Model Co : Overage Cost D : Demand Cu : Underage Cost .….[Eq.14.1] .…..…………..[Eq.14.1.a] …...[Eq.14.1.b]
§ I14: The Newsboy Model …...[Eq.14.1.b] …...[Eq.14.1.c] …(Critical Ratio).[Eq.14.2]
§ I14: The Newsboy Model Expected Cost Function for Newsboy Model 13 12 11 109 8 7 6 5 4 3 2 1 0 -10 -6 0 100 200 300 400 Q→ Fig.14.1
§ I14: The Newsboy Model [Eg. 14.1] Purchase cost $ 0.25 Sell for $ 0.75 Salvage $ 0.10 Demand has mean = 11.73 (μ) Standard deviation = 4.74 (σ) • Use [Eq.14.2] Cu : Underage Cost = 0.75-0.25 = 0.5 Co : Overage Cost = 0.25-0.10 = 0.15
§ I14: The Newsboy Model f(x) 76.9% 11.73 Q* X → 76.9% → Z = 0.74 ∴ Buy 15 copies every week .
§. I14: Problems & Discussion ( # N5.8 ; N 5.9 ) # N 5.2T # N 5.11 Preparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes
◇ § I15 : ( R ,Q ) model Q • Safety stock • Demand in lead time =λ.τ • Reorder point = s+ λ.τ = R
§ I16 : ( s , S ) model S μ1 μ2 μ3 μ4 s μ5 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 t • μ : starting inventory in any period • s : reorder point ◆ If μ < s , order S – μ If μ ≧ s , do not order.
◇ § I17: Stochastic Inventory Model (1) : SINGLE PERIOD MODEL WITH No setup Cost c = unit cost p = sell price or shortage cost for unsatisfied demand per unit where p > c h = holding cost or cost of excess supply per unit Q = quantities ordered D= Demand ( A random variable) d = demand (actual)
§ I17: (1) : Single Period Model with No setup Cost (A) RISK OF BEING “ SHORT “ ( Shortage cost incurred ) (B) RISK OF HAVING AN “ EXCESS “ ( Wasted unit cost & holding cost ) Recall Q = quantities ordered. D = actual demand , then amount sold: Expected Cost : Discrete R.V
§ I17 (1) : Single Period Model with No setup Cost Continuous R.V Let L(Q) = Expected [shortage+holding costs] & 57 ◆1-43 Minimum can be obtained. where Cu = p – c ; Co = c + h
§ I18 : Another way to look at SINGLE PERIOD INVENTORY Let Q* be the smallest Q for which ◆2投影片60 The Optimal quantity to order Q*, is the smallest integer such that the above function being satisfied.
◇ [Eg. 18.1] Suppose Demand of a certain single period product is a random variable and which follows Probability density function: 150 250 Cumulative probability function: 1 0 150 200 250
◇ [Eg. 18.1] (A) let c = $100 p = $200 h = $ -25 (salvage value) 代入 ~ about COSTS 150 250
◇ [Eg. 18.1] (B) To verify minimum G(Q*) = = $22,143 c = $100 p = $200 h = $ -25 FIND
[Eg 18.1] Further discussion (a) (b) [ $200*9.245=$1849 ] (c) [ $-25*16.245=-$406 ]
§. I18: Problems & Discussion ( # C.7 ) ( # C.8 ) ( # C.9 ) ( # C.9.1.c ) Preparation Time : 25 ~ 30 minutes Discussion : 15 ~ 25 minutes Advance Topics Follow ...
◇ § I19 : Single Period with “ Initial stock x ” Conclusion: then order up to Q* (i.e. order Q*-x) Do not order [Eg 19.1] Let us suppose that in Example 14.1, Mac has received 6 copies of the Journal at the beginning of the week from another supplier. The optimal police still calls for having 15 copies on hand after ordering, so now he would order the difference 15-6 = 9 copies. ( Set Q*=15 and u=6 to get the order quantity of Q*-u=9.)
§ I20 : Single Period with Ordering ( set-up ) cost “K” (s,S) policy : if on-hand Inv. x < s, order up to S. if on-hand Inv. x≧s, don’t order. * * ◆1 g-t-60
[Eg 20.1] Suppose ordering cost for the single period product described in Eg. #C.7 is $800 c=20 p=45 h=-9 k=800 ◆1-g-s-61 When Demand Dist.~Exponential ◆2 ◆1-g-s-63
§. I20: Problems & Discussion ( # C.10 ) Preparation Time : 25 ~ 30 minutes Discussion : 15 ~ 25 minutes
§ I21: ∞ periods with starting inventory x units. Di= demand for period i 2nd period purchases what’s being used in the previous period Assumptions : 1. Backorders; and assumes (a) that each unit left over at the end of the final period can be salvaged with a return of the initial purchase cost c. (b) if there is a shortage at the end of the final period, this shortage is met by an emergency shipment with the same unit purchase cost c. 2. Demand Distribution & Costs are the same in all periods. 3. α= discount rate =
§ I 21 : ∞periods(continued) g-s-30 …...[ Eq.21.1 ] ◆g-b-48
[Eg 21.1] Suppose Demand of a certain multiple period product is a random variable and it follows Therefore, And if p (shortage cost) =$15 [ backorder case, p may be less than c, cost just for handling backordering ] c = 35, h = 1 and discount rate α=0.995
§. I21: Problems & Discussion ( # C.11 ) ( # C.12 ) Preparation Time : 25 ~ 30 minutes Discussion : 15 ~ 25 minutes
§I 22 : Interpretation of Co, Cu Define ◆g-s-66 ◆ …...[ Eq.22.1 ] ◆b-t-67
◇ §.I 23 :∞ periods with backordered ◆g-s-68
◇ §.I 23 :∞ periods with backordered ◆b-t-51 ◆g-s-57 …... [ Eq.23.1 ] [Eg. 23.1] ◆g-s-43 Let us return to Mac’s newsstand, described previously. Suppose that Mac is considering how to replenish the inventory of a very popular paperback thesaurus that is ordered monthly. Copies of the thesaurus unsold at the end of a month are still kept on the shelves for future sales.
[Eg 23.1] Assume that customers who request copies of the thesaurus when they are out of stock will wait until the following month. (Back-ordered allowed) Mac buys the thesaurus for $1.15 and sells it for $2.75. Mac estimates a loss-of-goodwill cost of 50 cents each time a demand for a thesaurus must be back-ordered. Monthly demand for the book is fairly closely approximated by a Normal distribution with mean 18 and standard deviation 6. Mac uses a 20 percent annual interest rate to determine his holding cost. How many copies of the thesaurus should he purchase at the beginning of each month? Solution: using [Eq.23.1] The overage cost in this case is just the cost of holding, which is (1.15)(0.20) / 12 = 0.0192.The underage cost is just the loss-of-goodwill cost, which is assumed to be 50 cents. Hence, the critical ratio is 0.5/(0.5+0.0192)=.9630.From the Table of Normal Dist., corresponds to a z value of 1.79. The optimal value of the order-up-to point Q*=σZ+u=(6)(1.79)+18 = 28.74 =29.
§. I23: Problems & Discussion ( # C.13 ) Preparation Time : 25 ~ 30 minutes Discussion : 15 ~ 25 minutes