Stochastic (Random) Demand Inventory Models

Transcription

1 Stochastic (Random) Demand Inventory Models George Liberopoulos 1

2 The Newsvendor model Assumptions/notation Single-period horizon Uncertain demand in the period: D (parts) assume continuous random variable Density function and cumulative distribution function of D: f(x) and F(x) a df( x) F( a) = P( D a) = f ( x) dx f ( a) = dx x= 0 x= a Infinite production/replenishment rate (instantaneous replenishment) Zero lead time Overage cost rate: cost per unit of positive inventory remaining at the end of the period: c o ( per left-over part) Underage cost rate: cost per unit of unsatisfied demand (negative ending inventory) : c u ( per missing part or unsatisfied demand) No fixed setup production/order cost Decision Order quantity at the beginning of the period: (parts) 2

3 Definitions The Newsvendor model (Positive) inventory remaining at the end of the period: I + Unsatisfied demand (negative inventory) at the at the end of the period: I + + I = ( D) max( D,0) + I = ( D ) max( D,0) Total overage and underage cost at the end of the period: G(, D) Expected cost: G() GD (, ) = ci + ci = c( D) + c( D ) o u o u G ( ) = EGD [ (, )] = Gx (, ) f( xdx ) D x= 0 + o u x= 0 x= 0 + = c ( x) f ( x) dx + c ( x ) f ( x) dx = c ( x) f ( x) dx + c ( x ) f ( x) dx o x= 0 u x= 3

4 The Newsvendor model + + I = ( D) max( D,0) f( x) + I = ( D ) max( D,0) c o x= 0 ( x) f( x) dx μ = Ε[D] c u x= ( x ) f( x) dx D 4

5 Problem The Newsvendor model First derivative of cost function dg( ) d d = co ( x) f ( x) dx cu ( x ) f ( x) dx d + x= 0 x= d = c ( x) f ( x) dx + 1( x) f( x) 0( x) f( x) + x= x= 0 d o x= 0 d cu ( x ) f ( x) dx + 0( x ) f( x) 1( x ) f( x) x d x= = c 1 f ( x) dx + c ( 1) f ( x) dx o u x= 0 x= = cf ( ) c [ 1 F ( )] o u Minimize G ( ) = x= 5

6 The Newsvendor model In deriving the previous formula, we used Leibnitz s rule for taking the derivative of an integral whole limits are functions of the variable with respect to which the derivative is taken. Leibnitz s rule is: b( y) b( y) d df ( x, y) db( y) da( y) f ( x, y) dx dx f ( b( y), y) f ( a( y), y) dy = + dy dy dy a( y) a( y) 6

7 Second derivative The Newsvendor model 2 dg ( ) d 2 [ cf o c u F ] c o c u f d = ( ) [1 ( )] = ( + ) ( ) 0 d First-order condition for minimization dg( ) d = 0 = 0 cf ( ) c[1 F ( )] = 0 ( c+ c) F ( ) = c o u o u u c * * u * 1 u : F ( ) = = F co + cu co + cu Note: F() is the fill rate, i.e. the probability that a demand will be satisfied! c Recall: EO model with backorders: F = h + b * b 7

8 The Newsvendor model Special case: D Normal(μ, σ) D µ µ µ F( ) = P( D ) = P σ σ = Φ σ Normal(0,1) µ F ( ) =Φ ( z), z= σ Φ ( z) standardized Normal cumulative distribution function and hence F ( ) can be evaluated from standardized Normal tables z Φ( z) Φ( z)

9 The Newsvendor model Special case: D Normal(μ, σ) cont d * * * * cu µ cu µ 1 cu : F ( ) = Φ = =Φ co + cu σ co + cu σ co + cu * zc u ( co+ cu ) = µ + σ z c ( c + c ) u o u Example D Normal(120, 45) buying price c = 30 selling price S = 110 salvage price s = 10 cu = S c= = 80 co = c s = = 20 cu 80 = = 0.80 z0.80 = 0.85 c + c o u * = µ + σz0.80 = + = + =

10 The Newsvendor model Extension: Discrete demand Uncertain demand in the period: D (parts) assume discrete random variable Probability mass function and cumulative distribution function of D: p(x) and F(x) Fa ( ) = PD ( a) = px ( ) pa ( ) = Fa ( ) Fa ( 1) x a Expected cost: G() Problem: Minimize G ( ) First-order difference G ( ) = EGD [ (, )] = Gx (, ) px ( ) = c ( x) px ( ) + c ( x ) px ( ) D First-order condition for minimization 1 o u x x= 0 x= G ( + 1) G ( ) = c px ( ) c px ( ) = cf ( ) c[1 F ( )] o u o u x= 0 x= + 1 * G G cf o cu F :smallest such that ( + 1) ( ) 0 ( ) [1 ( )] 0 * * :smallest such that F ( ) c o cu + c u 10

11 The Newsvendor model Extension: Staring inventory y > 0 Still want to be at * after ordering, because * is the minimizer of G() Order quantity: U Optimal policy now depends on starting inventory: U * y, if y< ( y) = 0, if y * * Note: U * optimal order quantity * optimal order-up-to point inventory target level base stock level * 11

12 The Newsvendor model Interpretation of c o and c u for the single-period model S = selling price ( per part) c = variable cost ( per part) h = holding cost ( per part per period) p = loss-of-goodwill cost ( per part short per period) G(, D) = c + h( D) + + p( D ) + S min( D, ) order cost leftover inventory cost shortage cost sales sales revenue G( ) = E[ G(, D)] = c + h ( x) f ( x) dx + p ( x ) f ( x) dx S[ xf ( x) dx + f ( x) dx] D 0 0 = c + h ( x) f ( x) dx + ( p + S) ( x ) f ( x) dx Sµ µ xf ( x) dx 0 dg( ) + = 0 c+ hf ( ) ( p+ S)(1 F ( )) = 0 µ ( x ) f ( x) dx= µ E[( D ) ] d p+ S c = = + = + p+ S + h * : F ( ) cu p S c, co h c 12

13 The Newsvendor model Extension: infinite periods (infinite horizon) with backorders Same assumptions as single-period model except that: D t = demand in period t; D 1, D 2, D 3, are i.i.d. with distribution f(x), F(x) U t = amount ordered in period t Optimal policy in each period is order-up-to In the long-run, the inventory can never be higher than In steady-state (long run): U t = D t 1 Inventory/ backorders Steady state Transient U t = D t 1 time t 13

14 The Newsvendor model Extension: infinite periods (infinite horizon) with backorders (cont d) Total cost in a period with demand D Expected average cost per period First-order condition for minimizing G() c and S play no role in determining *, because in the long run, all demands are satisfied regardless of ; therefore, the expected average ordering cost minus revenue per period is (c S)μ regardless of. GD (, ) = ( c SD ) + h ( D) + + pd ( ) + dg( ) d order cost sales revenue inventory holding cost backorder cost + + G( ) = E[ G(, D)] = ( c S) µ + he[( D) ] + pe[( D ) ] D = 0 hf ( ) p(1 F ( )) = 0 p = p+ h = = Note: * * : F ( ) Newsventor formula: cu p, co h 14

15 The Newsvendor model Extension: infinite periods (infinite horizon) with lost sales Same assumptions as infinite horizon with backorders except that: Unmet demand is not backordered but is lost In steady-state (long run): U t = min(, D t 1 ) Inventory/ lost sales Transient U t = min(, D t 1 ) time t 15

16 The Newsvendor model Extension: infinite periods (infinite horizon) with lost sales (cont d) Total cost in a period with demand D GD (, ) = ( c S) min( D, ) + h ( D) + pd ( ) Expected average cost per period First-order condition for minimizing G() dg( ) d G( ) = E[ G(, D)] = ( c S) µ E[( D ) ] + he[( D) ] + pe[( D ) ] D = 0 hf ( ) ( p+ S c)(1 F ( )) = 0 p+ S c = p+ S c+ h = + = Note: * * : F ( ) Newsvendor formula: cu p S c, co h c and S now play a role in determining *, because in the long run, the demand satisfied and the orders are min(, D), so they depend on ; therefore, the expected average ordering cost minus revenue per period is (c S)E[min(, D)]. 16

17 Lot size Reorder point (, R) model Assumptions Infinite horizon Continuous review (as opposed to periodic review) D t : random stationary demand per unit time (e.g., daily demand) mean λ Ε[D t ], variance σ t2 = E[(D t λ) 2 ] Unmet demand is either backordered or lost τ: Fixed replenishment order lead time Costs: Variable unit production/order cost: c ( per part) Fixed setup production/order cost: K ( per production run/order) Inventory holding cost rate: h ( per part per unit time) Stock-out (shortage/penalty) cost rate (2 cases / 4 situations: see next) Order policy (, R) policy: order when inventory position falls below R Decision variables : lot size (reorder quantity) R: reorder point 17

18 (, R) model Assumptions on stock-outs and stock-out cost rate Case 1: Backordered demand p 1 ( per stock-out occasion) p 2 ( per part short ) p 3 ( per part short per unit time) Case 2: Lost sales p L ( per lost sale) 18

19 (, R) model: Backordered demand R+ Inventory position R On-hand inventory time τ Backorders 19

20 (, R) model: Backordered demand Analysis D: demand during lead time τ Density function and cumulative distribution function of D: f(x) and F(x) D = D 1 + D D τ Mean: μ E[D] = Ε[D 1 + D D τ ] = τ Ε[D t ] = τ λ Variance: σ 2 Var[D] = Ε[(D μ) 2 ] = Var[D 1 + D D τ ] = τ Var[D t ] = τ σ t 2 Inventory / backorders R τ μ f(x) x time 20

21 (, R) model: Backordered demand Inventory holding cost Safety stock ss R μ = R λ τ Expected average inventory approximation Ī ss + /2 = R λ τ + /2 (underestimates true value) Expected average inventory hold cost = h Ī = h(r λ τ + /2) Inventory / Backorders ss + = R λ τ + R ss = R λ τ τ λ μ = λ τ time 21

22 (, R) model: Backordered demand To see why expected average inventory approximation underestimates true expected average inventory: I R E[ D] + 2 = R xf ( x) dx + 2 appr R 0 + I = E[( R D) ] + 2 = ( R x) f ( x) dx + 2 = RF( R) + xf ( x) dx + 2 true 0 0 R I I = R(1 F( R)) xf ( x) dx appr true = R f ( x) dx xf ( x) dx R = Rf ( x) dx xf ( x) dx R = ( R x) f ( x) dx R + = E[( D R) ] 0 I appr I true R R R 22

23 (, R) model: Backordered demand Setup cost Expected average order frequency = λ/ Expected average setup cost per unit time = K λ/ Stock-out (penalty) cost Assumption: τ << /λ stock-out per cycle depends only on R B(R) : Expected stock-out cost per cycle (depends on definition of stockout cost rate) Expected average stock-out cost per unit time = B(R) λ/ Total expected average cost per unit time λ λ G(, R) = h + R λτ + K + B( R) 2 23

24 (, R) model: Backordered demand Optimization problem λ λ Minimize G(, R) = h + R λτ + K + B( R), R 2 Optimality conditions G(, R) h Kλ λbr ( ) 2 2 λ[ K+ BR ( )] = = 0 = h = 2 λ[ K+ BR ( )] (1) h G(, R) λ db( R) = h + = R dr db( R) h = (2) dr λ 0 24

25 (, R) model: Backordered demand Optimality condition (2): 3 cases Case 1: Stock-out cost p 1 per stock-out occasion db( R) BR ( ) = p1 [1 FR ( )] = pf 1 ( R) dr probabilty of stock-out per cycle h Condition (2) : p1 f( R) = f( R) = λ Case 2: Stock-out cost p 2 per part short h p λ + db( R) dr n( R) expected number x= R of stock-outs per cycle n( R) BR ( ) = p E[( D R) ] = p ( x R) f ( x) dx = p [1 F( R)] h Condition (2) : p2[1 F( R)] = F( R) = 1 λ 1 h p λ 2 25

26 (, R) model: Backordered demand Case 3: Stock-out cost p 3 per part short per unit time Average waiting time per backorder when the demand is D: + ( D R) 2λ Average total waiting time of all backorders when the demand is D: ( D R) [( D R) ] ( D R) = 2λ 2λ 2 ( x R) p3 2 B( R) = p3 f ( x) dx = ( x R) f ( x) dx 2λ 2λ x= R x= R db( R) dr expected total waiting time of all backorders per cycle p3 p3 + p3 = ( x R) f ( x) dx E[( D R) ] n( R) λ = = λ λ x= R p h h nr = nr = λ λ p 3 Condition (2) : ( ) ( )

27 (, R) model: Backordered demand Simultaneous solution of conditions (1) and (2) Solve by fixed-point iteration: 1. Given R, solve (1) to find 2. Given, solve (2) to find R 3. Repeat until convergence 27

28 (, R) model: Backordered demand Illustration for case 2: Stock-out cost p 2 per part short Optimality conditions for case 2 2 λ[ K + p2nr ( )] h = (1 ) F( R) = 1 (2) h p λ Assumption: D~Normal(μ, σ) Use standardized cumulative distribution function (cdf) Φ(z) to compute F(R) D µ R µ R µ F( R) = P( D R) = P σ σ =Φ σ Normal(0,1) ( ) F( R) =Φ z, z = R µ σ Φ(z) and hence F(R) can be evaluated from standardized Normal cdf tables 2 28

29 (, R) model: Backordered demand Use standardized loss function L(z) to compute n(r) + Y ~ Normal(0,1) L( z) E[( Y z) ] = ( y z) ϕ( y) dy L(z) and hence n(r) can be evaluated from standardized loss function tables It can be shown that Lz ( ) = ϕ( z) z[1 Φ( z)] y= z + nr ( ) = σ Lz ( ) = σϕ ( z) + ( µ R)[1 Φ ( z)], z= Normal(0,1) density function + D µ R µ R µ nr ( ) = E[( D R) ] = E σ = σl σ σ σ Normal(0,1) nr ( ) = σ Lz ( ), z= R µ σ R µ σ 29

30 (, R) model: Backordered demand Under the assumption D~Normal(μ, σ), the optimality conditions become 2 λ[ K+ p2σlz ( )] = h h Φ ( z) = 1 (2) p λ 2 (1) z = R µ (3) σ 30

31 (, R) model: Backordered demand Fixed point iteration algorithm for case 2 under the assumption D~Normal(μ, σ) 2λK 1 h 0 0 =, z0 =Φ 1, R0 µ σz0, n 1 h = + = pλ Step 1: Step 2 Step 3: Step n 2 λ[ K+ p2σlz ( n 1)] = h 1 : zn 1 R n h n =Φ pλ = µ + σz n 4: ε OR R R ε n n+ 1, GOTO Step 1 n n 1 n n 1 31

32 (, R) model: Lost sales R+ Inventory position R On-hand inventory Lost sales τ time 32

33 = (, R) model: Lost sales Modify total expected average cost per unit time function λ λ GR (, ) = h + R λτ + n( R) + K + pl nr ( ) 2 expected lost sales per cycle Optimality conditions L expected lost sales per unit time 2 λ[ K+ pnr L ( )] (1) (same as case 2 of backordered demands) h G(, R) dn( R) plλ dn( R) plλ dn( R) = h 1+ + = = 0 R dr dr h dr plλ 1 1+ [1 F( R)] = 0 h h plλ F( R) = 1 = (2) (Newsvendor formula: c h + p λ h + p λ L ( h << p λ same as case 2 of backordered demands) L o = h, c = p λ) u 33 L

34 Note (, R) model: Lost sales G(, R) is an approximation, because it overestimates the order frequency In the lost sales model, not all demand is met; on average, demands are met and n(r) demands are not met Average number of demands per cycle = + n(r) parts per cycle Number of demands per unit time = λ parts per unit time Order frequency = λ/[ + n(r)] instead of λ/ A more accurate expression for G(, R) is: λ λ G( R, ) = h + R λτ + nr ( ) + K + pl nr ( ) 2 + n( R) + nr ( ) Usually, n(r) <<, so the order frequency can still be approximated by λ/ 34

35 (, R) model: Summary Eq. (1): 2 λ [ K+ BR ( )] R = z = µ h σ Situation p 1 ( per stockout occasion) p 2 ( per part short) p 3 ( per part short per unit time) p L ( per lost sale) 1 BR ( ) p [1 F ( R )] BR ( ) D~Normal(μ,σ) Eq. (2) f( R) = h p λ 1 1 pnr ( ) 2 p3 2 ( ) ( ) 2λ x = R x R f x dx pnr ( ) L p [1 Φ ( z )] p2 σ Lz ( ) pσ Lz ( ) L F( R) = 1 h nr ( ) = p h p λ plλ F( R) = h + p λ 3 2 L Eq. (2) D~Normal(μ,σ) σ h ϕ( z) = p λ Φ ( z) = 1 1 h Lz ( ) = σ p h p λ plλ Φ ( z) = h + p λ 3 2 L 35

36 (, R) model: Service Levels Service levels in (, R) systems Type 1 Service (replaces stock-out cost p 1 per stock-out occasion) S 1 Probability of not stocking out during the lead time S 1 = P(D R) = F(R) Optimization problem λ Minimize G(, R) = h + R λτ + K, R 2 subject to F( R) α (i.e., subject to S α) Solution * 2Kλ = = EO h = minimum such that ( ) α * R R F R * 1 continuous r.v. ( ) D R = F 1 α 36

37 (, R) model: Service Levels Type 2 Service (replaces stock-out cost p 2 per part short) S 2 Proportion of demands met from stock S 2 = 1 n(r)/ Optimization problem λ Minimize G(, R) = h + R λτ + K, R 2 n( R) subject to 1 β (i.e., subject to S2 β) Note: Now the constraint depends on both R and 37

38 (, R) model: Service Levels Type 2 Service (cont d) Approximate solution * 2Kλ = EO h * * R= minimum Rsuch that nr ( ) (1 β ) D nr = β * * continuous r.v. ( ) (1 ) D µσ nr σlz = β * * * ~ Normal(, ) ( ) ( ) (1 ) (1 β ) σ * * * * 1 R = µ + σz, z = L 38

39 (, R) model: Service Levels Type 2 Service (cont d) More accurate solution Consider first-order conditions (1) and (2) for case 2 2 λ[ K + pn( R)] h = (1), F( R) = 1 (2) h pλ h (2) p = imputed stock-out cost [1 F( R)] λ 2 λ{ K + hn( R) [1 F( R)] λ} (1) = quandratic function in h nr ( ) 2 Kλ nr ( ) positive root: = + + (3) 1 F( R) h 1 F( R) 2 (1 β ) n( R) = (4) σ 39

40 (, R) model: Random Lead Time Extension: Random lead-time L: random lead time Mean: τ Ε[L], variance σ L2 Ε[(L τ) 2 ] D: demand during lead time L Density function and cumulative distribution function of D: f(x) and F(x) D = D 1 + D D L, where L is a random variable It can be shown (see next page) that: Mean: μ E[D] = τ λ Variance: σ 2 Var[D] = Ε[(D μ) 2 ] = τ σ t2 + λ 2 σ 2 L Everything else holds!! 40

41 (, R) model: Random Lead Time Derivation of μ and σ 2 Mean: µ ED [ ] = EED [ [ L]] = EL [ λ] = τλ L DL L Variance: [ ] [( ) ] [ 2 ] [ ] 2 [ ] [ ] where we used: [ 2 ] = [ DL D L σ VarD = E D µ = ED µ D+ µ = ED µ ED+ Eµ = EED [ [ L]] 2µ + µ = τσ + λ σ + λ τ µ = τσ + λ σ + λ τ τ λ L = τσ DL + λ σ t L t L t L E D L E Var[ D L] + E[ D L] ] = Lσ + Lλ t E[ E[ D L]] = E[ Lσ + L λ ] = τσ + λ ( Var[ L] + τ ) = τσ + λ ( σ + τ ) = τσ + λ σ + λ τ t t t L t L L DL L 41