Inventory Control Models

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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 –

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### 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
• 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

• 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.
• 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.)