Chapter 5 Manage Class A items (high value – high volume) Model Assumption Formular Notation Slow – moving item (Q=1) Continuous – review, order point s (or R) Q=1 if D́1.5 then s=s p and S=s p+ Q p Step 2: if ^x R z= √ Otherwise, S0 =x R + L+k σ R +L Where k satisfies: r B3 +r Then: s=min { s p , S0 } ; S=min { s p +Q p , S } Pr ( U ≥ k ) = Note that B3 in $/$ short/ review interval; r in $/$/review interval; D in units per year, and R and L in years (page 332 textbook 4th ed) R : Review time (time unit), L: Lead time S: Upper ordering level (Q=S−s ¿ s: Reorder point (units) B3: Fraction of cost per unit short at the end of each period ^x R + L , σ R + L: mean and std deviation of demand during (R+L) interval r : carrying rate (% or $ per $ per time) D : Demand rate per time Chapter 6: Perishable items (seasonal, perishable) Model General newsvendor problem Formular The optimal solution satisfies: ¿ ¿ G ( Q )=Pr ( X ≤Q )= Notation cu =α c o +c u How to find Q ¿ : ¿ −ln (1−α ) λ Exponential distribution with mean λ ( units per time ): Q = Uniform distribution with (a,b): Q=α∗( b−a ) +a ¿ Normal distribution with N ( μ , σ ): Q =μ+ z α σ (tra bảng z dựa α ) Discrete distribution: Choose the smallest integer Q that satisfies Q c ∑ Pr ( X= x ) ≥ c +cu x=1 u o n n , σ ( X )=∑ ( x i−μ ) P(x i); μ=¿ i=1 Geometric distribution with prob of success ρ : ∑ x i P ( x i) i=1 co ) c u +c o ln (ρ) ln ( Q¿ = Expected profit: E [ P ( Q ¿ ) ]= ( p−g ) ⋅ ¯x −( v −g ) Q−( p−g+ B ) ES Where: c u: shortage cost per unit c o : overage cost per unit v = acquisition cost ($/unit) p = revenue per sale ($/unit) B = penalty (beyond the profit) for not satisfying demand ($/unit) ( B2 v ¿ g = salvage value ($/unit) σ : Standard deviation of demand X during the season Q¿ : Optimal order quantity c u= p−v + B ; c o=v−g ; ES=σ G u ( k )=σ ¿ ) (G u ( k ) can be looked up in the normal distribution table by k) Simplest case: Unconstrained, single item, newsvendor problem (page 383 textbook 4th ed) Choose Q* satisfies: Pr ( X W => return to step with larger M i=1 Expected profit is calculated the same formular in two model above, but for each item, and sum them Chapter 7: Coordinated ordering Model for Lot sizing with multiple product S1: Lots are ordered and delivered independently for each product Formular Notation Order quantity for each product: Index i : 1…N where N: number of products Other notations can be refered in EOQ problem Q i=EO Q i= √ Ai D i vi r Average flow rate = Cycle inventory = Average flow time = S2: Lots are ordered and delivered jointly for all products Di Qi ; Order frequency n= Qi Cycle inventory Demand Order frequency (the same for all products) ¿ n= √ ∑ Di r C i i S¿ Capacity consideration case: n¿ =max ( ∑ Di i Capacity ; n¿ : Optimal order frequency for all products (time unit) ∑ D i r Ci C i: Unit cost of product i ($ per S¿ item) S¿: Combined fixed order cost ($ √ ) i ¿ Where S =S+ ∑ si i Order quantity: Q i= Di n¿ Total annual cost: TC=∑ i S3: Lots are ordered and delivered jointly for selected subset of the products Di r C i ¿ ¿ +n S n¿ Step 1: identify the most frequently ordered product, assuming each product is ordered independently n´ =´n { √ i∗¿=max n´ i = } r C i Di ¿ 2( S +s i ) Step 2:For all products i ≠ i*, evaluate the ordering frequency: n´ i = √ r Ci D i si per order) si: Product-specific fixed order cost for product i ($ per order for each product) (example: cost for pickup at one more location to take product i ) S: Common fixed order cost ( $ per order) (example: administration cost for placing order) r : carrying rate per time ($/$/time or %) Step 3: For all i i*, evaluate the frequency of product i relative to the most frequently ordered product i* to be: mi=⌈ n´ / n´ i ⌉ where m i =1 (round up) Step 4:Having decided the ordering frequency of each product i, recalculate the ordering frequency of the most frequently ordered product i* to be: ¿ n= √ ∑ r C i mi Di i S+ ∑ si /mi ( i ) Step 5: evaluate an order frequency of ni = n/mi and the total cost of such an ordering policy TC=nS+ ∑ ni s i+ ∑ i Model for inventory risk pooling Product pooling/ Inventory pooling (applicable for products with similar component structure, different labelling only,…) i Di rCi ni ( ) Assumption Formular Notation Demand is normally distributed Situation: Pooling products (customize later) Specialized inventory Leadtime demand is normal ( μ , σ ) Basestock for each item: R=μ + zσ Total safety stock: SS=znσ Pooled inventory Leadtime is normal ( nμ , √ n σ ) μ: mean of lead time demand variable for every model σ : std dev of lead time demand variable for every model z : z score for standard normal distribution n: Number of models Basestock level (for the same service): R=nμ+ z √ n σ Ratio of safety stock to specialized safety stock ¿ 1/ √ n Assumption: use basestock model Location pooling (centralize the warehouses/ manufacture) Situation: Pooling components (assembly later) Specialized inventory (stock product) Basestock level for each product type : R+1=μ+ zσ (normal); or =R+1 where poisson dist ( R , μ ,TRUE )=fillrate (poisson) On-hand inventory: I ( R+1 ) =R+1−μ+ B( R) Where B ( R )=μg ( R ) +(μ−R)(1−G ( R ) ) Pooled inventory (Stock components) Leadtime mean demand = μ∗L∗n1 Service level for each component: S1=S1 / m Basestock level: G ( R )=S1 => R+1 On-hand inventory: I ( R+1 ) =R+1−μ+ B( R) Market 1: D1 ( μ , σ ); Market 2: D2 μ2 , σ 22 ¿ Centralize: D + D : ( μ ,σ ) where μ=μ 1+ μ ; σ 2=σ 21 +σ 22+ ρ σ σ Where ρ=CORREL ( array ,array ) or given If D1, D2 positively correlated (tỉ lệ thuận), > If D1, D2 are independent , = If D1, D2 negatively correlated (tỉ lệ nghịch), < Then, if we have n Di, centralizing will lead to D +…+ D n : ( μ , σ ) if Di is similarly distributed: μ=n μ i and σ 2=n σ 2i +2∗C2n∗σ 2i ∗ρ General formular: μ=μ 1+ …+ μn σ = ∑ σ 2i +2 ∑ ρij σ i σ j i=1 n i< j Safety stock SS=z σ i √ L Reorder point R=μ i L+ SS A Di where Di: average demand of market type i vr Order up-to-level S=R +Q Q Average inventory I =SS+ Order quantity Q= √ n1: Number of models if using pooling components m : Number of components for a model S: Given fill rate for a model S1: Fill rate for components = correlation coefficient of D1, D2 1 Coefficient of variance COV = (centralized benefit) Standard deviation => higher – greater Average demand Chapter 8: Multi-echelon inventory Model PO2 for single stage Formular Decision variable: T ¿: reorder interval l¿ Step 1: ≥ Notation T L: base planning ∗T E=¿ 2l¿ ≥ ¿ (1) √2 T L period (given or assumed) K : Fixed ordering cost λ : constant and known rate of demand h : holding cost of unit for year/month, … Step 2: l ¿: smallest value satisfies (1) Ex: 2l¿ ≥ 32=> l ¿ =5 ¿ l Step 3: T =2 ∗T L where T L is given ¿ where T E = PO2 for multiple stage: serial system √ K ; g= hλ g h'i =hi−h i+1 , gi = λ h'i K ( Gk ) = ∑ K i K i: Fixed ordering cost of stage i hi : holding cost of unit for year/month, h'i : echelon holding cost (value added due to activity at stage i ) λ : constant and known rate of demand i∈ N (G k ) g ( Gk ) = ∑ gi i∈ N ( Gk ) Process No h'i gi Firstly, Obtain optimal solution to Relax problem (RP) as follows: Then, Obtain k subgraphs N ( G ) , N ( G ) , , N ( G k ) =¿ T ( )= Finally Obtain the solution to the problem (P) by √ K ( G1 ) g ( G1 ) ; T ( )= √ K ( G 2) g ( G 2) ,… 2l ≥ ∗T ( k )=¿ l=¿ T ¿ =2l∗T L √ 2T L (all stages will follow the same plan in PO2 solution even though solutions for subgraph in RP are not the same)