Lot size/eorder level, Models Esma Gel, Pınar Keskinocak, 0 ISYE 0 Fall 0 ecap: Basic EO Inventory It -d =d T d T T T T Lead time time Place an order when the inventory level is. The order arrives after time periods was the only decision variable could be computed easily because D was deterministic
Uncertain demand Both and are decision variables Cycle time is no longer constant! Inventory s s s time T T T, Decisions We choose to meet the demand during lead time Service levels: Protect against uncertainties in demand or lead time Balance the costs: stock-outs and inventory Tradeoff in : Fixed cost versus holding cost Objective: Minimize fixed cost + holding cost + stockout backorder cost
Demand during lead time Inventory eorder level What will happen if demand follows one of these patterns? Excess inventory Stockout Time Demand during lead time Inventory Often the probability distribution of demand during lead time follows a Normal pattern PD>= Probability of stockout Time Expected demand during lead time
, Model Assumptions Continuous review Demand is random and stationary. Expected demand is d per unit time. Lead time is Costs K: Setup cost per order h: Holding cost per unit per unit time c: Purchase price cost per unit p: Stockout backorder cost per unit Demand during lead time is a continuous random variable D with pdf density function fx and cdf distribution function Fx Mean= and standard deviation=, Model Expected total cost per unit time Holding cost Shortage cost Fixed cost K n C h s p ecap : T T T d s Average inventory level before an order arrives n eorder level Expected expected demand during leadtime - shortage per cycle D shortage D- D shortage 0 n 0 0 f x dx x f x dx Standard loss function x f x dx L z
, Model Expected total cost per unit time Holding cost Shortage cost Fixed cost K n C h s p ecap : T T T d s Average inventory level before an order arrives n eorder level Expected expected demand during leadtime - shortage per cycle D shortage D- D shortage 0 n 0 0 f x dx x f x dx Standard loss function x f x dx L z, Model Expected total cost per unit time Holding cost Shortage cost Fixed cost K n C h s p ecap : T T T d Same expression as s Average inventory level before an order arrives the expected eorder level expected number demand of during leadtime - n Expected shortage per cycle stockouts in the newsvendor model replaced by D shortage D- D shortage 0 n 0 0 f x dx x f x dx Standard loss function x f x dx L z 5
6, Model Expected total cost per unit time pd h F F pd h G h pn K d pn K d h pdn Kd h G d n p d K d h G ] [ ] [ 0 Shortage cost Fixed cost Holding cost, Model Expected total cost per unit time pd h F h p n K d d n p d K d h C ] [ Optimalsolution: Shortagecost Fixedcost Holdingcost How do we pull and from these equations? Solve iteratively!!
Solving for optimal and Start with a 0 value and iterate until the values converge 0 =EO emember: To find, you need n = Lz Lookup for z in the Normal tables Example ainbow Colors ainbow Colors paint store uses a, inventory system to control its stock levels. For a popular eggshell latex paint, historical data show that the distribution of monthly demand is approximately Normal, with mean 8 and standard deviation 8. eplenishment lead time for this paint is about weeks. Each can of paint costs the store $6. Although excess demands are backordered, each unit of stockout costs about $0 due to bookkeeping and loss-of-goodwill. Fixed cost of replenishment is $5 per order and holding costs are based on a 0% annual interest rate. What is the optimal lot size order quantity and reorder level? What is the expected inventory level safety stock just before an order arrives? 7
Example ainbow Colors Input Monthly demand Normal mean=8 std.dev.=8 = weeks c=$6, p=$0, K=$5 h=ic=0.6=$.8/unit/year Computed input d=? Expected annual demand Expected demand during lead time? Variance of demand during lead time? Example ainbow Colors Input Monthly demand Normal mean=8 std.dev.=8 = weeks c=$6, p=$0, K=$5 h=ic=0.6=$.8/unit/year Computed input d=8=6 units/year Expected annual demand Expected demand during lead time 6 units/year weeks 90 units 5 weeks/year Variance of demand during lead time Annual variance 8 768 Variance of lead time demand 768 06.77.8 5 8
Example ainbow Colors Input Monthly demand Normal mean=8 std.dev.=8 = weeks c=$6, p=$0, K=$5 h=ic=0.6=$.8/unit/year Computed input d=8=6 units/year Expected annual demand Expected demand during lead time 8 units/year weeks 90 units 5 weeks/year Variance of demand during lead time Annual variance 8 768 Variance of lead time demand 768 06.77.8 5 Example ainbow Colors Input Monthly demand Normal mean=8 std.dev.=8 = weeks c=$6, p=$0, K=$5 As the lead time increases, so does the mean and variance h=ic=0.6=$.8/unit/year of demand during lead time Computed input d=8=6 units/year Expected annual demand Shorter lead times Less variability of demand during lead time Expected demand during lead time 8 units/year weeks 90 units 5 weeks/year Variance of demand during lead time Annual variance 8 768 Variance of lead time demand 768 06.77.8 5 9
Example ainbow Colors Iteration 0: Compute EO Kd h 56.8 0 75 Example ainbow Colors Iteration : Compute given 0 and then compute given 0h F pd 75.8 06 0.96 z z.75 D ecap: F P D P P Z z z Z z Standard Normal z z.8.75 90 5 Safety Stock Expected Demand during Lead time 0 =75 From standard Normal table =5 0
Example ainbow Colors Iteration continued: Compute given d[ K p n ] h n L z.80.06 0. 6[5 00.] 80.8 0 =75 =80 0 and not close, continue iterations =5 Example ainbow Colors Iteration : Compute given and then compute given h 80.8 F 0.957 z z.7 pd 06 z.8.7 90 5 0 =75 =80 =5 STOP! values converged, optimal,=80,5 =5
Example ainbow Colors,=80,5 eorder level is larger than expected demand during lead time. Why? Optimal order quantity is larger than EO. Why? Safety stock s=-=5-90=5 units Weekly demand~5.6 Avg cycle time T=/d=80/6.6=.8 weeks. Lead time= weeks. Cycle time shorter than lead time Impact of on the costs and inventory, therefore Total cost As Holding cost Fixed cost Shortage cost 5 5
Optimal as a function of F h pd As the order quantity increases, the reorder level decreases holding cost and setup cost, therefore so that we can bring the holding cost, although a lower means shortage cost The Impact of Holding Cost on the Optimal, As h goes up, both and go down, but in this example drops at a faster rate! i 0. 97 6 0. 8 5 0. 7 0.5 6 0.6 59 0.7 55
The Impact of Stockout Cost on the Optimal, As p goes up, goes??? and goes??? The Impact of Stockout Cost on the Optimal, As p goes up, goes down and goes up! p 8 0 6 8 0 8 5 8 7 8 80 8 80 0
Summary:, Models Balance between holding cost, setup/fixed cost, and shortage cost To save on the shortage cost, we want large To save on the holding cost, we want small and small To save on the fixed cost, we want large Choose and to strike a good balance among these three costs!!! Service Levels in, Models Esma Gel, Pınar Keskinocak, 0 ISYE 0 Fall 0 5
Service objectives Type I service level The proportion of cycles in which no stockouts occur Example: 90% Type I service level There are no stockouts in 9 out of 0 cycles on average Type II service level fill rate, Fraction of demand satisfied on time Service objectives - Example Order cycle Demand Stock-outs 80 0 75 0 5 5 0 0 5 80 0 6 00 0 7 50 0 8 90 0 9 60 0 0 0 0 TOTAL: 50 55 Fraction of periods with no stock-outs = 8/0 Type I service = 80% = 0.8 Fraction of demand satisfied on time = 50-55/50=0.96 Type II service = 96% = 0.96 In general, is it easier to achieve an x% Type I service or Type II service level? 6
7 Type I service level, : Long-run average proportion of cycles with no stock-outs : Probability of having no stock-outs in a cycle : Probability of having no stock-outs during lead time : Probability that demand during lead time is less than!!! ecap : z z Z P D P D P D P Set =EO Find z that satisfies z= Set =z+ safety stock + expected demand during lead time Type I service level, : Long-run average proportion of cycles with no stock-outs : Probability of having no stock-outs in a cycle : Probability of having no stock-outs during lead time : Probability that demand during lead time is less than!!! ecap : z z Z P D P D P D P Set =EO Find z that satisfies z= Set =z+ safety stock + expected demand during lead time
Type I service level, : Long-run average proportion of cycles with no stock-outs : Probability of having no stock-outs in a cycle : Probability of having no stock-outs during lead time : Probability that demand during lead time is less than!!! P D ecap : Why is =EO optimal in this case? D P D P P Z z z Set =EO Find z that satisfies z= Set =z+ safety stock + expected demand during lead time Example ainbow Colors ainbow Colors paint store uses a, inventory system to control its stock levels. For a popular eggshell latex paint, historical data show that the distribution of monthly demand is approximately Normal, with mean 8 and standard deviation 8. eplenishment lead time for this paint is about weeks. Each can of paint costs the store $6. Although excess demands are backordered, each unit of stockout costs about $0 due to bookkeeping and loss-of-goodwill. Fixed cost of replenishment is $5 per order and holding costs are based on a 0% annual interest rate. What is the optimal lot size order quantity and reorder level? What is the expected inventory level safety stock just before an order arrives? 8
Example ainbow Colors ainbow Colors is not sure whether the $0 estimate for the shortage cost is accurate. Hence, they decided to use a service level approach. What are the optimal, values if they want to achieve no stockouts in 90% of the order cycles? satisfy 90% of the demand on time? Example ainbow Colors Input Monthly demand Normal mean=8 std.dev.=8 = weeks, c=$6, K=$5 h=ic=0.6=$.8/unit/year = 0.9 or = 0.9 Computed input d=8=6 units/year Expected annual demand Expected demand during lead time 8 units / year 5 weeks / year Variance of demand during lead time Annual variance 8 768 weeks 90 units Variance of lead time demand 768 06.77.8 5 9
ainbow Colors Type I service Find, to have 90% Type I service level =EO=75 z= = 0.9 z=.8 = z+ =.8.8+90=08 For 90% Type I service level,=75,08 emember: With unit penalty cost of $0, we found,=80,5. What is the Type I service level that corresponds to,=80,5? = z+ 5=.8z+90 z=.785.785=0.96 96% Type I service level when,=80,5 Type II service level : Fraction of demand met on time - : Fraction of demand not met on time stock-outs ecap: Expected # of stockouts n d n per unit time T since T d Expected # of stockouts per unit time Expected demand per unit time n n With this information, for a given,, we can compute. 0
ainbow Colors For 90% Type I service level we found,=75,08 What is the Type II service level which corresponds to this policy? The same policy results in 90% Type I service and 99% Type II service!! 08 90.5 z.8 n L z L.5.80.0506 0.776 n 0.776 75 0.0097 0.99 Finding the optimal, for a desired Type II service level emember : stock - out cost p : Optimal solution when we have d[ K p n ] h F h pd
Finding the optimal, for a desired Type II service level Optimal solution when we have stock - out cost From Substitute d[ K p n ] h : p into h p d F : h F pd p : 5 Imputed shortage cost n Kd n F h F To be solved simultaneously with n Impact of service level on For a given As n=- i.e., As the service level, the reorder level as well
Finding the optimal, for a desired Type II service level n Kd n n F h F 0 =EO Start with a 0 value and iterate until the values or the values converge Example ainbow Colors Iteration 0: Compute EO Kd h 56.8 0 75
Example ainbow Colors Iteration : Compute given 0 and then compute given n 0 0.975 7.5 L z L z 0.56 z 0. z.8 0. 90 86.8 87 To find weneed -F.Look at thenormaltable. F 0. F0. 0.587 89 0 =75 =89 =87 Example ainbow Colors Iteration : Compute given and then compute given
Example ainbow Colors Iteration : Compute given and then compute given n 0.989 8.9 L z L z 0.69 z 0.8 z.8 0.8 90 8.5 85 To find F 0.8 F0.8 0.68 90 weneed -F.Look at thenormaltable. 0 =75 =89 =90 =87 =85 Example ainbow Colors Iteration : Compute given and then compute given n 0.985 8.5 L z L z 0.59 z 0. z.8 0. 90 85. 85 0 =75 =89 =90 STOP! values converged, optimal,=90,85 =87 =85 =85 5