Inventory Control Models Ch 5 (Uncertainty of Demand) R. R. Lindeke IE 3265, Production And Operations Management
Lets do a ‘QUICK’ Exploration of Stochastic Inventory Control (Ch 5) • We will examine underlying ideas – • We base our approaches on Probability Density Functions (means & std. deviations) • We are concerned with two competing ideas: Q and R • Q (as earlier) an order quantity and R a stochastic estimate of reordering time and level • Finally we are concerned with Servicing ideas – how often can we supply vs. not supply a demand (adds stockout costs to simple EOQ models)
The Nature of Uncertainty • Suppose that we represent demand as: • D = Ddeterministic + Drandom • If the random component is small compared to the deterministic component, the models of chapter 4 will be accurate. If not, randomness must be explicitly accounted for in the model. • In this chapter, assume that demand is a random variable with cumulative probability distribution F(t) and probability density function f(t).
Single Period Stochastic Inventory Models • These models have the objective of properly balancing the cost of Underage – having not ordered enough products vs. Overage – having ordered more than we can sell • These models apply to problems like: • Planning initial shipments of ‘High-Fashion’ items • Amount of perishable food products • Item with short shelf life (like the daily newspaper) • Because of this last problem type, this class of problems is typically called the “Newsboy” problem
The Newsboy Model • At the start of each day, a newsboy must decide on the number of papers to purchase. Daily sales cannot be predicted exactly, and are represented by the random variable, D. • The newsboy must carefully consider these costs: co = unit cost of overage cu = unit cost of underage • It can be shown that the optimal number of papers to purchase is the fractile of the demand distribution given by F(Q*) = cu / (cu + co).
Computing the Critical Fractile: • We wish to minimize competing costs (Co & Cu): • G(Q,D) = Co*MAX(0, Q-D) + Cu*MAX(0, D-Q) • D is actual (potential) Demand • G(Q) = E(G(Q,D)) (an expected value) • Therefore:
Applying Leibniz’s Rule: • d(G(Q))/dQ = CoF(Q) – Cu(1 – F(Q)) • F(Q) is a cumulative Prob. Density Function (as earlier – of the quantity ordered) • Thus: G’(Q*) = (Cu)/(Co + Cu) • This is the critical fractile for the order variable as stated earlier
Lets see about this: Prob 5 pg 241 • Observed sales given as a number purchased during a week (grouped) • Lets assume some data was supplied: • Make Cost: $1.25 • Selling Price: $3.50 • Salvageable Parts: $0.80 • Co = overage cost = $1.25 - $0.80 = $0.45 • Cu = underage cost = $3.50 - $1.25 = $2.25
Continuing: • Compute Critical Ratio: • CR = Cu/(Co + Cu) = 2.25/(.45 + 2.25) = .8333 • If we assume a continuous Probability Density Function (lets choose a normal distribution): • Z(CR) 0.967 when F(Z) = .8333 (from Std. Normal Tables!) • Z = (Q* - )/) • From the problem data set, we compute • Mean = 9856 • St.Dev. = 4813.5
Continuing: • Q* = Z + = 4813.5*.967 + 9856 = 14511 • Our best guess economic order quantity is 14511 • (We really should have done it as a Discrete problem -- Taking this approach we would find that Q* is only 12898)
Newsboy’s Extensions • Assuming we have a certain number of parts on hand, u > 0 • This extends the problem compared to our initial u = 0 assumption for the single period case • This is true only if the product under study has a shelf life that extends beyond one period • Here we still compute Q* will order only Q* - u (or 0 if u > Q*)
Try one (in your Engineering Teams) : • Do Problem 11a & 11b (pg 249)
Lot Size Reorder Point Systems • Earlier we considered reorder points (number of parts on hand when we placed an order) they were dependent on lead times as a dependent variable on Q, now we will consider R as an independent variable just like Q • Assumptions: • Inventory levels are reviewed continuously (the level of on-hand inventory is known at all times) • Demand is random but the mean and variance of demand are constant. (stationary demand)
Lot Size Reorder Point Systems Additional Assumptions: • There is a positive leadtime, τ. This is the time that elapses from the time an order is placed until it arrives. • The costs are: • Set-up cost each time an order is placed at $K per order • Unit order cost at $C for each unit ordered • Holding at $H per unit held per unit time (i.e., per year) • Penalty cost of $P per unit of unsatisfied demand
Describing Demand • The response time of the system (in this case) is the time that elapses from the point an order is placed until it arrives. Hence, • The uncertainty that must be protected against is the uncertainty of demand during the lead time. • We assume that D represents the demand during the lead time and has probability distribution F(t). Although the theory applies to any form of F(t), we assume that it follows a normal distribution for calculation purposes.
Decision Variables • For the basic EOQ model discussed in Chapter 4, there was only the single decision variable Q. • The value of the reorder level, R, was determined by Q. • Now we treat Q and R as independent decision variables. • Essentially, R is chosen to protect against uncertainty of demand during the lead time, and Q is chosen to balance the holding and set-up costs. (Refer to Figure 5-5)
The Cost Function The average annual cost is given by: • Interpret n(R) as the expected number of stockouts per cycle given by the loss integral formula (see Table A-4 (std. values)). And note, the last term is this cost model is a shortage cost term • The optimal values of (Q,R) that minimizes G(Q,R) can be shown to be:
Solution Procedure • The optimal solution procedure requires iterating between the two equations for Q and R until convergence occurs (which is generally quite fast) • We consider that the problem has converged if 2 consecutive calculation of Q and R are within 1 unit • A cost effective approximation is to set Q=EOQ and find R from the second equation. • A slightly better approximation is to set Q = max(EOQ,σ) • where σ is the standard deviation of lead time demandwhen demand variance is high.
Ready to Try one? Lets! • Try Problem 13a & 13b (pg 261) • Start by computing EOQ and then begin iterative solution for optimal Q and R values
Service Levels in (Q,R) Systems • In many circumstances, the penalty cost, p, is difficult to estimate. For this reason, it is common business practice to set inventory levels to meet a specified service objective instead. The two most common service objectives are: • Type 1 service: Choose R so that the probability of not stocking out in the lead time is equal to a specified value. • Type 2 service. Choose both Q and R so that the proportion of demands satisfied from stock equals a specified value.
Computations • For type 1 service, if the desired service level is α then one finds R from F(R)= α and Q=EOQ. • Type 2 service requires a complex interative solution procedure to find the best Q and R. However, setting Q=EOQ and finding R to satisfy n(R) = (1-β)Q (which requires Table A-4) will generally give good results.
Comparison of Service Objectives • Although the calculations are far easier for type 1 service, type 2 service is generally the accepted definition of service. • Note that type 1 service might be referred to as lead time service, and type 2 service is generally referred to as the fill rate. • Refer to the example in section 5-5 to see the difference between these objectives in practice (on the next slide).
Comparison (continued) Order Cycle Demand Stock-Outs 1 180 0 2 75 0 3 235 45 4 140 0 5 180 0 6 200 10 7 150 0 8 90 0 9 160 0 10 40 0 • For a type 1 service objective there are two cycles out of ten in which a stockout occurs, so the type 1 service level is 80%. For type 2 service, there are a total of 1,450 units demand and 55 stockouts (which means that 1,395 demand are satisfied). This translates to a 96% fill rate.
Example: Type 1 Service Pr 5-16 • Desire 95% Type I service Level • F(R) = .95 Z is 1.645 (Table A4) • From Problem 13: was found to be 172.8 and was 1400 • Therefore: R = Z + = 172.8*1.645 + 1400R = 1684.256 1685 • Use Q = EOQ = 1265
Example: Type 2 Service Pr 5-17 • Require Iterative Solution:
(s, S) Policies • The (Q,R) policy is appropriate when inventory levels are reviewed continuously. In the case of periodic review, a slight alteration of this policy is required. Define two levels, s < S, and let u be the starting inventory at the beginning of a period. Then (In general, computing the optimal values of s and S is much more difficult than computing Q and R.)