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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

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
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
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
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
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
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
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
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
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
continuing11
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
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
Try one (in your Engineering Teams) :
  • Do Problem 11a & 11b (pg 249)
lot size reorder point systems
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 systems15
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
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
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 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
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
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
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
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
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
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
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
Example: Type 2 Service Pr 5-17
  • Require Iterative Solution:
s s policies
(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.)