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- 1. Inventory Control Models Ch 5 (Uncertainty of Demand) R. R. Lindeke IE 3265, Production And Operations Management
- 2. Lets do a ‘QUICK’ Exploration of Stochastic Inventory Control (Ch 5) <ul><li>We will examine underlying ideas – </li></ul><ul><li>We base our approaches on Probability Density Functions (means & std. deviations) </li></ul><ul><li>We are concerned with two competing ideas: Q and R </li></ul><ul><li>Q (as earlier) an order quantity and R a stochastic estimate of reordering time and level </li></ul><ul><li>Finally we are concerned with Servicing ideas – how often can we supply vs. not supply a demand (adds stockout costs to simple EOQ models) </li></ul>
- 3. The Nature of Uncertainty <ul><li>Suppose that we represent demand as: </li></ul><ul><ul><ul><li>D = D deterministic + D random </li></ul></ul></ul><ul><li>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. </li></ul><ul><li>In this chapter, assume that demand is a random variable with cumulative probability distribution F(t) and probability density function f(t). </li></ul>
- 4. Single Period Stochastic Inventory Models <ul><li>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 </li></ul><ul><li>These models apply to problems like: </li></ul><ul><ul><li>Planning initial shipments of ‘High-Fashion’ items </li></ul></ul><ul><ul><li>Amount of perishable food products </li></ul></ul><ul><ul><li>Item with short shelf life (like the daily newspaper) </li></ul></ul><ul><li>Because of this last problem type, this class of problems is typically called the “Newsboy” problem </li></ul>
- 5. The Newsboy Model <ul><li>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. </li></ul><ul><li>The newsboy must carefully consider these costs: c o = unit cost of overage </li></ul><ul><li>c u = unit cost of underage </li></ul><ul><li>It can be shown that the optimal number of papers to purchase is the fractile of the demand distribution given by F(Q*) = c u / (c u + c o ). </li></ul>
- 6. Determination of the Optimal Order Quantity for Newsboy Example
- 7. Computing the Critical Fractile: <ul><li>We wish to minimize competing costs (Co & Cu): </li></ul><ul><ul><li>G(Q,D) = Co*MAX(0, Q-D) + Cu*MAX(0, D-Q) </li></ul></ul><ul><ul><ul><li>D is actual (potential) Demand </li></ul></ul></ul><ul><ul><li>G(Q) = E(G(Q,D)) (an expected value) </li></ul></ul><ul><ul><li>Therefore: </li></ul></ul>
- 8. Applying Leibniz’s Rule: <ul><li>d(G(Q))/dQ = C o F(Q) – C u (1 – F(Q)) </li></ul><ul><li>F(Q) is a cumulative Prob. Density Function (as earlier – of the quantity ordered) </li></ul><ul><li>Thus: G’(Q*) = (Cu)/(Co + Cu) </li></ul><ul><li>This is the critical fractile for the order variable as stated earlier </li></ul>
- 9. Lets see about this: Prob 5 pg 241 <ul><li>Observed sales given as a number purchased during a week (grouped) </li></ul><ul><li>Lets assume some data was supplied: </li></ul><ul><ul><li>Make Cost: $1.25 </li></ul></ul><ul><ul><li>Selling Price: $3.50 </li></ul></ul><ul><ul><li>Salvageable Parts: $0.80 </li></ul></ul><ul><li>Co = overage cost = $1.25 - $0.80 = $0.45 </li></ul><ul><li>Cu = underage cost = $3.50 - $1.25 = $2.25 </li></ul>
- 10. Continuing: <ul><li>Compute Critical Ratio: </li></ul><ul><ul><li>CR = Cu/(Co + Cu) = 2.25/(.45 + 2.25) = .8333 </li></ul></ul><ul><li>If we assume a continuous Probability Density Function (lets choose a normal distribution): </li></ul><ul><ul><li>Z(CR) 0.967 when F(Z) = .8333 (from Std. Normal Tables!) </li></ul></ul><ul><ul><li>Z = (Q* - )/ ) </li></ul></ul><ul><ul><li>From the problem data set, we compute </li></ul></ul><ul><ul><ul><li>Mean = 9856 </li></ul></ul></ul><ul><ul><ul><li>St.Dev. = 4813.5 </li></ul></ul></ul>
- 11. Continuing: <ul><li>Q* = Z + = 4813.5*.967 + 9856 = 14511 </li></ul><ul><li>Our best guess economic order quantity is 14511 </li></ul><ul><li>(We really should have done it as a Discrete problem -- Taking this approach we would find that Q* is only 12898) </li></ul>
- 12. Newsboy’s Extensions <ul><li>Assuming we have a certain number of parts on hand, u > 0 </li></ul><ul><ul><ul><li>This extends the problem compared to our initial u = 0 assumption for the single period case </li></ul></ul></ul><ul><li>This is true only if the product under study has a shelf life that extends beyond one period </li></ul><ul><li>Here we still compute Q * will order only Q * - u (or 0 if u > Q * ) </li></ul>
- 13. Try one (in your Engineering Teams) : <ul><li>Do Problem 11a & 11b (pg 249) </li></ul>
- 14. Lot Size Reorder Point Systems <ul><li>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 </li></ul><ul><li>Assumptions: </li></ul><ul><ul><li>Inventory levels are reviewed continuously (the level of on-hand inventory is known at all times) </li></ul></ul><ul><ul><li>Demand is random but the mean and variance of demand are constant. (stationary demand) </li></ul></ul>
- 15. Lot Size Reorder Point Systems <ul><ul><li>There is a positive leadtime, τ. This is the time that elapses from the time an order is placed until it arrives. </li></ul></ul><ul><ul><li>The costs are: </li></ul></ul><ul><ul><ul><li>Set-up cost each time an order is placed at $K per order </li></ul></ul></ul><ul><ul><ul><li>Unit order cost at $C for each unit ordered </li></ul></ul></ul><ul><ul><ul><li>Holding at $H per unit held per unit time (i.e., per year) </li></ul></ul></ul><ul><ul><ul><li>Penalty cost of $P per unit of unsatisfied demand </li></ul></ul></ul>Additional Assumptions:
- 16. Describing Demand <ul><li>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, </li></ul><ul><li>The uncertainty that must be protected against is the uncertainty of demand during the lead time. </li></ul><ul><li>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. </li></ul>
- 17. Decision Variables <ul><li>For the basic EOQ model discussed in Chapter 4, there was only the single decision variable Q . </li></ul><ul><li>The value of the reorder level, R , was determined by Q. </li></ul><ul><li>Now we treat Q and R as independent decision variables. </li></ul><ul><li>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) </li></ul>
- 18. Changes in Inventory Over Time for Continuous-Review (Q, R) System
- 19. The Cost Function <ul><li>The average annual cost is given by: </li></ul><ul><li>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 </li></ul><ul><li>The optimal values of (Q,R) that minimizes G(Q,R) can be shown to be: </li></ul>
- 20. Solution Procedure <ul><li>The optimal solution procedure requires iterating between the two equations for Q and R until convergence occurs (which is generally quite fast) </li></ul><ul><ul><ul><li>We consider that the problem has converged if 2 consecutive calculation of Q and R are within 1 unit </li></ul></ul></ul><ul><li>A cost effective approximation is to set Q=EOQ and find R from the second equation. </li></ul><ul><li>A slightly better approximation is to set Q = max(EOQ, σ) </li></ul><ul><ul><li>where σ is the standard deviation of lead time demand when demand variance is high. </li></ul></ul>
- 21. Ready to Try one? Lets! <ul><li>Try Problem 13a & 13b (pg 261) </li></ul><ul><li>Start by computing EOQ and then begin iterative solution for optimal Q and R values </li></ul>
- 22. Service Levels in (Q,R) Systems <ul><li>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: </li></ul><ul><ul><li>Type 1 service: Choose R so that the probability of not stocking out in the lead time is equal to a specified value. </li></ul></ul><ul><ul><li>Type 2 service. Choose both Q and R so that the proportion of demands satisfied from stock equals a specified value. </li></ul></ul>
- 23. Computations <ul><li>For type 1 service, if the desired service level is α then one finds R from F(R)= α and Q=EOQ. </li></ul><ul><li>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. </li></ul>
- 24. Comparison of Service Objectives <ul><li>Although the calculations are far easier for type 1 service, type 2 service is generally the accepted definition of service. </li></ul><ul><li>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. </li></ul><ul><li>Refer to the example in section 5-5 to see the difference between these objectives in practice (on the next slide). </li></ul>
- 25. Comparison (continued) <ul><li>Order Cycle Demand Stock-Outs </li></ul><ul><li>1 180 0 </li></ul><ul><li>2 75 0 </li></ul><ul><li>3 235 45 </li></ul><ul><li>4 140 0 </li></ul><ul><li>5 180 0 </li></ul><ul><li>6 200 10 </li></ul><ul><li>7 150 0 </li></ul><ul><li>8 90 0 </li></ul><ul><li>9 160 0 </li></ul><ul><li>10 40 0 </li></ul><ul><li>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. </li></ul>
- 26. Example: Type 1 Service Pr 5-16 <ul><li>Desire 95% Type I service Level </li></ul><ul><li>F(R) = .95 Z is 1.645 (Table A4) </li></ul><ul><li>From Problem 13: was found to be 172.8 and was 1400 </li></ul><ul><li>Therefore: R = Z + = 172.8*1.645 + 1400 R = 1684.256 1685 </li></ul><ul><li>Use Q = EOQ = 1265 </li></ul>
- 27. Example: Type 2 Service Pr 5-17 <ul><li>Require Iterative Solution: </li></ul>
- 28. Example: Type 2 Service Pr 5-17 (cont.)
- 29. (s, S) Policies <ul><li>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 </li></ul><ul><li>(In general, computing the optimal values of s and S is much more difficult than computing Q and R.) </li></ul>

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