Chapter 16 When demands are unknown, expected values are the keys for deciding how much to order and how often. Inventory Decisions with Uncertain Factors 1 Inventory Decisions with Uncertain Factors Two basic inventory decisions are evaluated: Single-period inventory—e.g., newspapers. Probability distribution is for period’s demand. Multi-stage inventory—e.g., birthday cards. Probability distribution is for lead-time demand. There are two demand probability distributions: Deterministic (tabular). Continuous (normal curve). There are two analytical approaches: Tabular: maximizing expected payoff Model: marginal analysis or EOQ. Two cases are modeled: 2 Backordering. Lost sales. Making an Inventory Decision: Maximizing Expected Payoff Problem: A drugstore stocks Fortunes.They sell for $3 and cost $2.10. Unsold copies are returned for $.70 credit. There are four levels of demand possible. Using profit as payoff, the following applies. 3 ACTS Demand ProbaEvent bility Q = 20 Q = 21 Q = 22 Q = 23 D = 20 .2 $18.00 $16.60 $15.20 $13.80 D = 21 .4 18.00 18.90 17.50 16.10 D = 22 .3 18.00 18.90 19.80 18.40 D = 23 .1 18.00 18.90 19.80 20.70 Making an Inventory Decision: Maximizing Expected Payoff Solution: The owner does not consider stocking less than the minimum demand or more than the maximum. (Why?) The expected payoffs are computed for each possible order quantity: Q = 20 Q = 21 Q = 22 Q = 23 $18.00 $18.44 $17.90 $16.79 maximum According to the Bayes decision rule, stocking 21 magazines is optimal. 4 If the probabilities were long-run frequencies, then doing so would maximize long-run profit. Maximizing expected payoff is assumed proper. The Single-Period Model: The Newsvendor Problem The payoff table approach can be cumbersome with many levels of demand. The same result is achieved with a marginal analysis model. The decision variable is Q = Order Quantity The model minimizes total expected cost for the period, using parameters: c = Unit procurement cost hE = Additional cost of each item held at end of inventory cycle pS = Penalty for each item short pR = Selling price The event variable is uncertain demand D. 5 The Single-Period Model: The Newsvendor Problem The shortage penalty here applies regardless of duration of stockout. Sales will equal D if demand falls at or below Q and Q if sales are greater. If D < Q, there are Q - D leftovers, each costing: hE + c If D > Q, there are D - Q shortages, each costing: pS + p R - c The objective is to minimize total expected cost: TEC(Q) = Q cm hE c Q - d Pr[ D d ] pS pR - c d - Q Pr[ D d ] d 0 6 d Q where m is the expected demand. The Single-Period Model: The Newsvendor Problem This is the expression for optimal order quantity: Q* is the smallest possible demand such that pS p R - c Pr[ D Q* ] pS pR - c hE c Problem: A newsvendor sells Wall Street Journals. She loses pS = $.02 in future profits each time a customer wants to buy a paper when out of stock. They sell for pR = $.23 and cost c = $.20. Unsold copies cost hE = $.01 to dispose. Demands between 21 and 30 are equally likely. How many should she stock? Solution: The expected demand is m = 25 copies. 7 The Single-Period Model: The Newsvendor Problem The following ratio is computed: pS p R - c .02 .23 - .20 .192 pS pR - c hE c .02 .23 - .20 .01 .20 Each demand level has probability .1. The smallest cumulative probability exceeding this is .20, corresponding to 22 papers. Thus, Q* = 22. The above is sensitive to the parameter levels. Raising pS to $.04 will increase Q* to 23. Raising pS to $.10 will increase Q* to 24. 8 Continuous Demand Distribution: Christmas Tree Problem When demand is continuous the marginal analysis involves areas under normal curve. Problem: Demand for noble firs is approximately normally distributed with m = 2,000 and s = 500. Trees sell for pR = $9 and cost c = $3. Loss of goodwill is pS = $1 per tree out of stock. Disposal cost is hE = $.50 per tree. How many trees should be stocked? Solution: The following applies: pS p R - c 1 9 - 3 .6667 pS pR - c hE c 1 9 - 3 .50 3 This normal curve area corresponds to z = .43, and the demand at or beyond this determines Q*. 9 Q* = m + zs = 2,000 + .43(500) = 2,215 trees Continuous Demand Distribution: Christmas Tree Problem The following is used in computing the total expected cost: TEC Q * cm hE c [Q - m BQ ] pS pR - c BQ The above uses the expected shortage: Q - m Qm s L s BQ m -Q m - Q s L Qm s where L(x) is the tabled loss function. 10 Multiperiod Inventory Policies When demand is uncertain, multiperiod inventory might look like this over time. 11 Multiperiod Inventory Policies The multiperiod decisions involve two variables: Order quantity Q Reorder point r The following parameters apply: 12 A = mean annual demand rate k = ordering cost c = unit procurement cost pS = cost of short item (no matter how long) h = annual holding cost per dollar value m = mean lead-time demand Multiperiod Inventory Policies: Discrete Lead-Time Demand The following is used to compute the expected shortage per inventory cycle: Br d - r PrL D d d r The following is used to compute the total annual expected cost: (With Backordering) Q A A TEC r ,Q k hc r - m pS Br 2 Q Q 13 Multiperiod Inventory Policies: Discrete Lead-Time Demand Solution Algorithm. Calculate the starting order quantity: 2 Ak Q1 hc Determine the reorder point r*: r* is smallest level such that hcQ PrD r * 1 (with backordering) pS A Determine optimal order quantity: 14 2 Ak pS Br Q* hc Multiperiod Inventory Policies: Discrete Lead-Time Demand Recompute r* after getting Q*, and vice versa, until one of them stops changing. Problem: Annual demand for printer cartridges costing c = $1.50 is A = 1,500. Ordering cost is k = $5 and holding cost is $.12 per dollar per year. Shortage cost is pS = $.12, no matter how long. Lead-time demand has the following distribution. 15 Find the optimal inventory policy. Multiperiod Inventory Policies: Discrete Lead-Time Demand 16 Solution: The starting order quantity is: 21,5005 Q1 289 .121.50 Using the above, we compute: hcQ .12.15289 1 1 .93 pS A .51,500 The smallest cumulative lead-time demand probability > .93 is .95, corresponding to 7 cartridges. Thus, r* = 7 cartridges. We compute: B(7) = (8–7)(.03) + (9–7)(.01) + (10–7)(.01)=.08 and the optimal order quantity is: 21,5005 .50.08 Q* 290 .121.50 Multiperiod Inventory Policies: Discrete Lead-Time Demand Solution (continued): Substituting the above into the expression used for finding r* the same value as before is found. r* does not change, and the optimal inventory policy is: r* = 7 Q* = 290 The Lost Sales Case: There is a new parameter: pR = selling price The condition for reorder point changes to: r* is smallest level such that pS p R - c A PrD r * hcQ pS pR - c A 17 (lost sales) Multiperiod Inventory Policies: Discrete Lead-Time Demand The Lost Sales Case (continued): The optimal order quantity expression is: 2 Ak pS pR - c Br Q* hc (lost sales) 18 Multiperiod Inventory Policies: Continuous Lead-Time Demand The formulas and algorithms for the continuous case are the same, except for the expected shortage: r - m rm s L s B r m -r m - r s L rm s where m and s are the parameters of the normal lead-time demand distribution and L(x) is the tabled losss function. 19 Inventory Spreadsheet Templates Payoff Table Newsvendor Christmas Tree Multiperiod Discrete Backordering Multiperiod Discrete Lost Sales Multiperiod Normal Backordering Multiperiod Normal Lost Sales 20 Payoff Table (Figure 16-1) 1. Enter problem name in B3. A B C D 2. Enter data in B9:F12 and labels in A9:A12 and C8:F8. E PAYOFF TABLE EVALUATION 1 2 3 PROBLEM: Fortune Magazine 4 5 Problem Data C 6 18 =SUMPRODUCT($B$9:$B$12,C9:C12) 7 Act 1 Act 2 Act 3 8 Events Probability Q = 20 Q = 21 Q = 22 9 1 D = 20 0.2 $18.00 $16.60 $15.20 10 2 D = 21 0.4 $18.00 $18.90 $17.50 11 3 D = 22 0.3 $18.00 $18.90 $19.80 12 4 D = 23 0.1 $18.00 $18.90 $19.80 13 4. Expected payoffs 14 Act Summary 15 16 Act 1 Act 2 Act 3 17 Q = 20 Q = 21 Q = 22 18 Expected Payoff $18.00 $18.44 $17.96 F Copy cell C18 over to D18:F18. 21 Act 4 Q = 23 $13.80 $16.10 $18.40 $20.70 Act 4 Q = 23 $16.79 3. If more events or acts are required, expand the table by inserting additional rows and/or columns. Make sure the formulas in the Act Summary table include all the rows of the expanded table. 1. Enter the problem name in C3. 6. Optimal values: Q*, mu, TEC(Q*), B(Q*), PD>Q*. 3. Enter the demands and probabilities in C21:D40. 22 Newsvendor Problem (Figure 16-3) A B C D E F G 1 SINGLE PERIOD INVENTORY MODEL -- NEWSVENDOR PROBLEM 2 3 PROBLEM: Wall Street Journal 4 5 Parameter Values: 6 Cost per Item Procured: c = 0.20 Additional Cost for Each Leftover Item Held: hE = 7 0.01 Penalty for Each Item Short: pS = 8 0.02 Selling Price per Unit: pR = 9 0.23 10 Number of demands for probability distribution = 11 11 12 Optimal Values: 13 Optimal Order Quantity: Q* = 47 14 Expected Demand: mu = 49.5 15 Total Expected Cost: TEC(Q*) = $10.07 16 Expected Shortages: B(Q*) = 2.66 17 Probability of Shortage: P[D>Q*] = 0.80 18 19 Cumulative Number of 20 Demand Probability Probability shortages 21 45 0.05 0.05 0.0 22 46 0.06 0.11 0.0 23 47 0.09 0.20 0.0 24 48 0.12 0.32 1.0 25 49 0.17 0.49 2.0 26 50 0.20 0.69 3.0 27 51 0.12 0.81 4.0 28 52 0.08 0.89 5.0 29 53 0.06 0.95 6.0 30 54 0.04 0.99 7.0 31 55 0.01 1.00 8.0 2. Enter the problem parameters in G6:G10. 4. If the number of demands for probability distribution is greater than 20 add the appropriate number of rows and copy the formulas in columns E and F down for these rows. 5. To calculate the expected profit, enter =SUMPRODUCT( C21:C40,D21:D4 0)*G9-G15 in cell G18. Newsvendor Formulas G 13 14 15 16 17 E F =IF(ROW(C21)=INDEX(C21:C40,MATCH(F13LOOKUP((G8+G9A B C D E F G =IF(ROW(C21)20<=$G$10,IF($G$13>C2 G6)/(G8+G9+G7),E21:E40,C21:C40),C21:C40)+1) 1 SINGLE PERIOD INVENTORY -- NEWSVENDOR PROBLEM 20<=$G$10,D21,"") 1,0,C21-$G$13),"") 21 MODEL =SUMPRODUCT(C21:C40,D21:D40) 2 =G6*G14+(G7+G6)*(G13-G14+G16)+(G8+G9-G6)*G16 =IF(ROW(C22)3 PROBLEM: Wall Street Journal =SUMPRODUCT(D21:D40,F21:F40) =IF(ROW(C22)20<=$G$10,IF($G$13>C2 4 2,0,C22-$G$13),"") 22 20<=$G$10,D22+E21,"") =1-VLOOKUP(G13,C21:F40,3) 5 Parameter Values: 6 Cost per Item Procured: c = 0.20 Additional Cost for Each Leftover Item Held: hE = 7 0.01 Penalty for Each Item Short: pS = 8 0.02 Selling Price per Unit: pR = 9 0.23 10 Number of demands for probability distribution = 11 11 12 Optimal Values: 13 Optimal Order Quantity: Q* = 47 14 Expected Demand: mu = 49.5 15 Total Expected Cost: TEC(Q*) = $10.07 16 Expected Shortages: B(Q*) = 2.66 17 Probability of Shortage: P[D>Q*] = 0.80 18 19 Cumulative Number of 20 Demand Probability Probability shortages 21 45 0.05 0.05 0.0 22 46 0.06 0.11 0.0 23 23 47 0.09 0.20 0.0 1. Enter the problem name in C3. Christmas Tree Problem (Figure 16-6) 2. Enter the problem parameters in G6:G11. A B C D E F G SINGLE PERIOD INVENTORY MODEL - CHRISTMAS TREE PROBLEM 3. Optimal values: Q*, mu, TEC(Q*), B(Q*), PD>Q*. 24 1 2 3 PROBLEM: Noble Fir 4 5 Parameter Values: 6 Mean of Demand Distribution: mu = 7 Stand. Deviation of Demand Distribution: sigma = 8 Cost per Item Procured: c = Additional Cost for Each Leftover Item Held: hE = 9 Penalty for Each Item Short: pS = 10 Selling Price per Unit: pR = 11 12 13 Optimal Values: 14 Optimal Order Quantity: Q* = 2215 15 Expected Demand: mu = 2000 16 Total Expected Cost: TEC(Q*) = $7,910.35 17 Expected Shortages: B(Q*) = 110.15 18 Probability of Shortage: P[D>Q*] = 0.33 The Normal Loss Table L(D) is on the next worksheet. It is used in the spreadsheet calculations. 2000 500 3.00 0.50 1.00 9.00 The Normal Loss Table L(D) The Normal Loss Table L(D) is used the calculations in the Christmas Tree template. 25 1 2 3 4 5 6 102 103 104 105 106 202 203 204 205 206 497 498 499 500 501 A D 0.00 0.01 0.02 0.03 0.04 1.00 1.01 1.02 1.03 1.04 2.00 2.01 2.02 2.03 2.04 4.95 4.96 4.97 4.98 4.99 B L(D) 0.3989 0.3940 0.3890 0.3841 0.3793 0.08332 0.08174 0.08019 0.07866 0.07716 8.491E-03 8.266E-03 8.046E-03 7.832E-03 7.623E-03 6.982E-08 6.620E-08 6.276E-08 5.950E-08 5.640E-08 Note that many rows have been hidden because the entire table is too big to show on one page. Christmas Tree Formulas A B C D E F G SINGLE PERIOD INVENTORY MODEL - CHRISTMAS TREE PROBLEM ‘L(D)’!A2:B501 refers to the normal loss table L(D) table located on the L(D) worksheet 26 1 2 3 PROBLEM: Noble Fir 4 5 Parameter Values: 6 Mean of Demand Distribution: mu = 7 Stand. Deviation of Demand Distribution: sigma = 8 Cost per Item Procured: c = Additional Cost for Each Leftover Item Held: hE = 9 Penalty for Each Item Short: pS = 10 Selling Price per Unit: pR = 11 12 13 Optimal Values: 14 Optimal Order Quantity: Q* = 2215 15 Expected Demand: mu = 2000 16 Total Expected Cost: TEC(Q*) = $7,910.35 17 Expected Shortages: B(Q*) = 110.15 18 Probability of Shortage: P[D>Q*] = 0.33 F 14 =NORMINV((G10+G11-G8)/(G10+G11+G9),G6,G7) 15 =G6 16 =G8*F15+(G9+G8)*(F14-F15+F17)+(G10+G11-G8)*F17 =IF(F14<G6,G6-F14+G7*VLOOKUP((G6F14)/G7,'L(D)'!A2:B501,2),G7*VLOOKUP((F1417 G6)/G7,'L(D)'!A2:B501,2)) 18 =1-NORMDIST(F14,G6,G7,TRUE) 2000 500 3.00 0.50 1.00 9.00 Multiperiod Discrete Backordering The solution to multiperiod models with discrete lead-time demand and backordering is based on the newsvendor spreadsheet. It varies in two respects: some formulas are a little different (described in Appendix 16-1) it contains many worksheets because of the iterative nature of the solution process. 27 Ten iterations are done in this spreadsheet. This is sufficient for all problems in the book and will solve most other multiperiod, discrete, backordering models. However, addition iterations can be added whenever necessary. Multiperiod Discrete Backordering Each of the ten worksheets appear as tabs in the spreadsheet, numbered 1, 2, 3, . . . , 10. The problem data is entered in worksheet 1 (tab 1). Intermediate solution results for iteration 1 are on tab 1, the results for iteration 2 are on tab 2, and so forth up to the results for iteration 10 which appear on tab 10. An optimal solution is obtained when the results converge and do not vary with increasing iterations. Normally, an optimal solution is obtained after 2 or 3 iterations. A summary worksheet is provided after the iterations. It summarizes the intermediate results of all the iterations. 28 Multiperiod Discrete Backordering Iteration 1 1. Start with worksheet 1 (tab 1). It gives the results of the first iteration. 2. Enter the problem name in B3. 3. Enter the problem parameters in G6:G11. 6. Iteration 1 results are here 4. Enter the demands and probabilities in C23:D42. A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Backordering) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Printer Cartridges 4 5 Parameter Values 6 Fixed Cost per Order: k = 5 7 Annual Demand Rate: A = 1500 8 Unit cost of Procuring an Item: c = 1.5 9 Annual Holding Cost per Dollar Value: h = 0.12 Shortage Cost per Unit: pS = 10 0.5 11 Number of demands for probability distribution = 11 12 13 Optimal Values: 14 Optimal Order Quantity: Q* = 288.68 15 Optimal Reorder Point: r* = 7 16 Expected Lead-Time Demand: mu = 4 17 Total Expected Cost: TEC(Q*) = $ 52.7094 18 Expected Shortage: B(r*) = 0.08 19 Probability of Shortage: P[D>r*] = 0.05 20 21 Cumulative Number of 22 Demand Probability Probability Shortages 23 0 0.01 0.01 0 24 1 0.07 0.08 0 25 2 0.16 0.24 0 26 3 0.20 0.44 0 27 4 0.19 0.63 0 28 5 0.16 0.79 0 29 6 0.10 0.89 0 30 7 0.06 0.95 0 31 8 0.03 0.98 1 32 9 0.01 0.99 2 33 10 0.01 1.00 3 5. If the number of demands for probability distribution is greater than 20 add the 29 appropriate number of rows and copy the formulas in columns E and F down for these rows. Multiperiod Discrete Backordering (Figure 16-8) 1. Tab 2 gives the results of the second iteration, tab 3 the results of the 3rd iteration, etc. 2. The optimal solution occurs when the answers do not change from iteration to iteration. 3. To quickly find the optimal solution skip to the last iteration by clicking on tab 10 (shown here). 4. Optimal values: Q*, r*, mu, TEC(Q*), B(Q*), PD>Q*. 30 A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Backordering) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Printer Cartridges 4 5 Parameter Values 6 Fixed Cost per Order: k = 5 7 Annual Demand Rate: A = 1500 8 Unit cost of Procuring an Item: c = 1.5 9 Annual Holding Cost per Dollar Value: h = 0.12 Shortage Cost per Unit: pS = 10 0.5 11 Number of demands for probability distribution = 11 12 13 Optimal Values: 14 =SQRT((2 14 Optimal Order Quantity: Q* = 290 =INDEX(C 15 Optimal Reorder Point: r* = 7 ((G9*G8*G 16 Expected Lead-Time Demand: mu = 4 15 ),C23:C42, 17 Total Expected Cost: TEC(Q*) = $ 52.71 16 =SUMPRO 18 Expected Shortage: B(r*) = 0.08 =(G7/G14) 19 Probability of Shortage: P[D>r*] = 0.05 G16)+G10 17 20 18 =SUMPRO 21 Cumulative Number of 19 =1-VLOOK 22 Demand Probability Probability Shortages 23 0 0.01 0.01 0 24 1 0.07 0.08 0 25 2 0.16 0.24 0 26 3 0.20 0.44 0 27 4 0.19 0.63 0 28 5 0.16 0.79 0 29 6 0.10 0.89 0 30 7 0.06 0.95 0 31 8 0.03 0.98 1 32 9 0.01 0.99 2 33 10 0.01 1.00 3 Multiperiod Discrete Backordering Summary A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Backordering) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Printer Cartridges 4 5 Parameter Values To quickly find the 6 Fixed Cost per Order: k = 5 optimal solution click on7 Annual Demand Rate: A = 1500 8 Unit cost of Procuring an Item: c = 1.5 the Summary tab. It 9 Annual Holding Cost per Dollar Value: h = 0.12 provides a summary of all Shortage Cost per Unit: pS = 10 0.5 the 10 iterations. 11 Number of demands for probability distribution = 11 12 Qi ri B(ri) TEC(Qi,ri) 13 Iteration, i 31 14 15 16 17 18 19 20 21 22 23 1 2 3 4 5 6 7 8 9 10 289 290 290 290 290 290 290 290 290 290 7 7 7 7 7 7 7 7 7 7 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 0.08 $ $ $ $ $ $ $ $ $ $ 52.71 52.71 52.71 52.71 52.71 52.71 52.71 52.71 52.71 52.71 Notice the answers do not change after the second iteration. Multiperiod Discrete Backordering Iteration 1 Formulas G 14 =SQRT((2*G7*G6)/(G9*G8)) =INDEX(C23:C42,MATCH(LOOKUP(1((G9*G8*G14)/(G10*G7)),E23:E42,C23: 15 C42),C23:C42,0)+1) 16 =SUMPRODUCT(C23:C42,D23:D42) =(G7/G14)*G6+G9*G8*(G14/2+G1517 G16)+G10*(G7/G14)*G18 18 =SUMPRODUCT(D23:D42,F23:F42) 19 =1-VLOOKUP(G15,C23:E42,3) 32 A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Backordering) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Printer Cartridges 4 5 Parameter Values 6 Fixed Cost per Order: k = 5 7 Annual Demand Rate: A = 1500 8 Unit cost of Procuring an Item: c = 1.5 9 Annual Holding Cost per Dollar Value: h = 0.12 Shortage Cost per Unit: pS = 10 0.5 11 Number of demands for probability distribution = 11 12 13 Optimal Values: 14 Optimal Order Quantity: Q* = 288.68 15 Optimal Reorder Point: r* = 7 16 Expected Lead-Time Demand: mu = 4 17 Total Expected Cost: TEC(Q*) = $ 52.7094 18 Expected Shortage: B(r*) = 0.08 19 Probability of Shortage: P[D>r*] = 0.05 20 21 Cumulative Number of 22 Demand Probability Probability Shortages 23 0 0.01 0.01 0 24 1 0.07 0.08 0 25 2 0.16 0.24 0 26 3 0.20 0.44 0 27 4 0.19 0.63 0 28 5 0.16 0.79 0 29 6 0.10 0.89 0 30 7 0.06 0.95 0 31 8 0.03 0.98 1 32 9 0.01 0.99 2 33 10 0.01 1.00 3 Multiperiod Discrete Backordering Iteration 10 Formulas Only one formula changes on the iteration 2 - 10 worksheets, in cell G14. The formula in this cell always refers back the the previous iteration. For example, the worksheet shown here is for iteration 10 so the formula in cell G14 refers back to iteration 9. A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Backordering) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Printer Cartridges 4 5 Parameter Values 6 Fixed Cost per Order: k = 5 7 Annual Demand Rate: A = 1500 8 Unit cost of Procuring an Item: c = 1.5 9 Annual Holding Cost per Dollar Value: h = 0.12 Shortage Cost per Unit: pS = 10 0.5 11 Number of demands for probability distribution = 11 12 13 Optimal Values: 14 =SQ 14 Optimal Order Quantity: Q* = 290 =IN 15 Optimal Reorder Point: r* = 7 ((G9 16 Expected Lead-Time Demand: mu = 4 15 ),C2 17 Total Expected Cost: TEC(Q*) = $ 52.71 16 =SU 18 Expected Shortage: B(r*) = 0.08 =(G 19 Probability of Shortage: P[D>r*] = 0.05 G16 17 20 18 =SU 21 Cumulative Number of 19 =1-V 22 Demand Probability Probability Shortages 23 0 0.01 0.01 0 24 1 0.07 0.08 0 25 2 0.16 0.24 0 26 3 0.20 0.44 0 27 4 0.19 0.63 0 28 5 0.16 0.79 0 29 6 0.10 0.89 0 30 7 0.06 0.95 0 31 8 0.03 0.98 1 32 9 0.01 0.99 2 33 10 0.01 1.00 3 G 14 =SQRT((2*G7*(G6+'9'!G18*G10))/(G9*G8)) The term ‘9’!G18 means the value of G18 (expected number of shortages) from iteration 9. 33 Multiperiod Discrete Lost Sales The solution to multiperiod models with discrete lead-time demand and lost sales is based on the backordering case just described. It varies only in that some formulas are different (described in Appendix 16-1). 34 Multiperiod Discrete Lost Sales (Figure 16-9) 1. Start with worksheet 1 (tab 1) and enter the problem name in B3, the problem parameters in G6:G12, and the demands and probabilities in C24:D43. 2. To quickly find the optimal solution skip to the last iteration by clicking on tab 10 (shown here). 35 A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Lost Sales) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Compact Disks 4 5 Parameter Values 6 Fixed Cost per Order: k = 20 7 Annual Demand Rate: A = 5000 8 Unit cost of Procuring an Item: c = 0.45 9 Annual Holding Cost per Dollar Value: h = 0.12 Shortage Cost per Unit: pS = 10 0.1 Shortage Cost per Unit: pR = 11 0.9 12 Number of demands for probability distribution = 10 13 14 Optimal Values: 15 Optimal Order Quantity: Q* = 1935 16 Optimal Reorder Point: r* = 160 17 Expected Lead-Time Demand: mu = 123 18 Total Expected Cost: TEC(Q*) = $ 106.48 19 Expected Shortage: B(r*) = 0.40 20 Probability of Shortage: P[D>r*] = 0.03 21 22 Cumulative Number of 23 Demand Probability Probability Shortages 24 90 0.05 0.05 0 25 100 0.12 0.17 0 3. Optimal 26 110 0.17 0.34 0 27 120 0.22 0.56 0 values: Q*, r*, 28 130 0.19 0.75 0 mu, TEC(Q*), 29 140 0.14 0.89 0 30 150 0.05 0.94 0 B(Q*), PD>Q*. 31 160 0.03 0.97 0 32 170 0.02 0.99 10 33 180 0.01 1.00 20 Multiperiod Discrete Lost Sales Summary A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Backordering) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Compact Discs 4 5 Parameter Values 6 Fixed Cost per Order: k = 20 7 Annual Demand Rate: A = 5000 8 Unit cost of Procuring an Item: c = 0.45 on 9 Annual Holding Cost per Dollar Value: h = 0.12 Shortage Cost per Unit: pS = 10 0.1 To quickly find the optimal solution click the Summary tab. It provides a summary of all 11 the 10 iterations. 12 36 Shortage Cost per Unit: pR = Number of demands for probability distribution = 13 14 15 Iteration, i Qi ri B(ri) TEC(Qi,ri) 16 17 18 19 20 21 22 23 24 25 1 2 3 4 5 6 7 8 9 10 1925 1935 1935 1935 1935 1935 1935 1935 1935 1935 160 160 160 160 160 160 160 160 160 160 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 0.40 $ $ $ $ $ $ $ $ $ $ 106.48 106.48 106.48 106.48 106.48 106.48 106.48 106.48 106.48 106.48 0.9 10 Notice the answers do not change after the second iteration. Multiperiod Discrete Lost Sales Iteration 1 Formulas G 15 =SQRT((2*G7*G6)/(G9*G8)) =INDEX(C24:C43,MATCH(LOOKUP((G10+G1 1-G8)*G7/(G9*G8*G15+(G10+G1116 G8)*G7),E24:E43,C24:C43),C24:C43,0)+1) 17 =SUMPRODUCT(C24:C43,D24:D43) =((G7*(1G19/G15))/G15)*G6+G9*G8*(G15/2+G16G17+G19)+(G10+G11-G8)*(G7*(118 G19/G15)/G15)*G19 19 =SUMPRODUCT(D24:D43,F24:F43) 20 =1-VLOOKUP(G16,C24:E43,3) 37 A B C D E F 1 MULTI-PERIOD EOQ MODEL (Lost Sales) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Compact Discs 4 5 Parameter Values 6 Fixed Cost per Order: k = 7 Annual Demand Rate: A = 8 Unit cost of Procuring an Item: c = 9 Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = 10 Shortage Cost per Unit: pR = 11 12 Number of demands for probability distribution = 13 14 Optimal Values: 15 Optimal Order Quantity: Q* = 16 Optimal Reorder Point: r* = 17 Expected Lead-Time Demand: mu = 18 Total Expected Cost: TEC(Q*) = $ 19 Expected Shortage: B(r*) = 20 Probability of Shortage: P[D>r*] = 21 22 Cumulative Number of 23 Demand Probability Probability Shortages 24 90 0.05 0.05 0 25 100 0.12 0.17 0 26 110 0.17 0.34 0 27 120 0.22 0.56 0 28 130 0.19 0.75 0 29 140 0.14 0.89 0 30 150 0.05 0.94 0 31 160 0.03 0.97 0 32 170 0.02 0.99 10 33 180 0.01 1.00 20 G 20 5000 0.45 0.12 0.1 0.9 10 1925 160 123 106.48 0.40 0.03 Multiperiod Discrete Lost Sales Iteration 10 Formulas Only one formula changes on the iteration 2 - 10 worksheets, in cell G15. The formula in this cell always refers back the the previous iteration. For example, the worksheet shown here is for iteration 10 so the formula in cell G15 refers back to iteration 9. A B C D E F 1 MULTI-PERIOD EOQ MODEL (Lost Sales) - DISCRETE LEAD-TIME DEMAND 2 3 PROBLEM: Compact Disks 4 5 Parameter Values 6 Fixed Cost per Order: k = 7 Annual Demand Rate: A = 8 Unit cost of Procuring an Item: c = 9 Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = 10 Shortage Cost per Unit: pR = 11 12 Number of demands for probability distribution = 13 14 Optimal Values: 15 Optimal Order Quantity: Q* = 16 Optimal Reorder Point: r* = G 17 Expected Lead-Time Demand: mu = 15 =SQRT((2*G7*(G6+'9'!G19*(G10+G11-G8))/(G9*G8))) 18 Total Expected Cost: TEC(Q*) = $ 19 Expected Shortage: B(r*) = 20 Probability of Shortage: P[D>r*] = The term ‘9’!G19 21 22 Cumulative Number of means the value 23 Demand Probability Probability Shortages of G19 (expected 24 90 0.05 0.05 0 25 100 0.12 0.17 0 number of 26 110 0.17 0.34 0 shortages) from 27 120 0.22 0.56 0 28 130 0.19 0.75 0 iteration 9. 29 140 0.14 0.89 0 30 150 0.05 0.94 0 31 160 0.03 0.97 0 38 32 170 0.02 0.99 10 33 180 0.01 1.00 20 G 20 5000 0.45 0.12 0.1 0.9 10 1935 160 123 106.48 0.40 0.03 Multiperiod Normal Backordering The solution to multiperiod models with normal lead-time demand and backordering is a variation of the Christmas Tree template and it incorporates features from the multiperiod, discrete leadtime template. The formulas are described in Appendix 16-1. 39 Multiperiod Normal Backordering (Figure 16-10) A 1. Start with worksheet 1 (tab 1) and enter the problem name in B3 and the problem parameters in G6:G12. B C D E F G MULTI-PERIOD EOQ MODEL (Backordering) - NORMAL LEAD-TIME DEMAND 1 2 3 PROBLEM: Unleaded Gas at Oil Refinery 4 5 Parameter Values: 6 Mean of Demand Distribution: mu = 7 Stand. Deviation of Demand Distribution: sigma = 8 Fixed Cost per Order: k = 9 Annual Demand Rate: A = 10 Unit Cost of Procuring an Item: c = 11 Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = 12 13 14 Optimal Values: 2. To quickly find15the Optimal Order Quantity: Q* = optimal solution 16 skip Optimal Reorder Point: r* = 17 Expected Demand: mu = to the last iteration 1810 Total Expected Cost: TEC(Q*) = by clicking on tab Expected Shortages: B(r*) = (shown here). 19 20 Probability of Shortage: P[D>r*] = 40 200,000 20,000 1,000 40,000,000 0.40 0.20 0.05 1,008,256 234,937 200,000 83,455.46 331.60 0.04 3. Optimal values: Q*, r*, mu, TEC(Q*), B(Q*), P D>Q*. Multiperiod Normal Backordering Summary A B C D E F 1 MULTI-PERIOD EOQ MODEL (Backordering) - NORMAL LEAD-TIME DEMAND 2 3 PROBLEM: Unleaded Gas at Oil Refinery 4 5 Parameter Values: Mean of Demand Distribution: mu = To quickly find the6 7 Stand. Deviation of Demand Distribution: sigma = optimal solution click on 8 Fixed Cost per Order: k = the Summary tab. 9It Annual Demand Rate: A = provides a summary 10 of all Unit Cost of Procuring an Item: c = Annual Holding Cost per Dollar Value: h = the 10 iterations. 11 Shortage Cost per Unit: pS = 12 13 14 Qi ri B(ri) TEC(Qi,ri) 15 Iteration, i 41 16 17 18 19 20 21 22 23 24 25 1 2 3 4 5 6 7 8 9 10 1,000,000 1,008,053 1,008,256 1,008,256 1,008,256 1,008,256 1,008,256 1,008,256 1,008,256 1,008,256 235,014 234,939 234,937 234,937 234,937 234,937 234,937 234,937 234,937 234,937 323 332 332 332 332 332 332 332 332 332 $ $ $ $ $ $ $ $ $ $ 83,447.90 83,455.61 83,455.46 83,455.46 83,455.46 83,455.46 83,455.46 83,455.46 83,455.46 83,455.46 G 200,000 20,000 1,000 40,000,000 0.40 0.20 0.05 Notice the answers do not change after the third iteration. Multiperiod Normal Backordering Iteration 1 Formulas 15 16 17 18 A B C D E F 1 MULTI-PERIOD EOQ MODEL (Backordering) - NORMAL LEAD-TIME DEMAND 2 3 PROBLEM: Unleaded Gas at Oil Refinery 4 5 Parameter Values: 6 Mean of Demand Distribution: mu = 7 Stand. Deviation of Demand Distribution: sigma = 8 Fixed Cost per Order: k = 9 Annual Demand Rate: A = 10 Unit Cost of Procuring an Item: c = 11 Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = 12 13 14 Optimal Values: 15 F Optimal Order Quantity: Q* = 1,000,000 16 Optimal Reorder Point: r* = 235,014 =SQRT((2*G9*G8)/(G11*G10)) =NORMINV(1-((G11*G10*F15)/(G12*G9)),G6,G7) 17 Expected Demand: mu = 200,000 =G6 18 Total Expected Cost: TEC(Q*) = $ 83,447.90 =(G9/F15)*G8+G11*G10*(F15/2+F1619 Expected Shortages: B(r*) = 323.40 G6)+G12*(G9/F15)*F19 20 Probability of Shortage: P[D>r*] = 0.04 =IF(F16<F17,F17-F16+G7*VLOOKUP((F17F16)/G7,'L(D)'!A2:B501,2),G7*VLOOKUP((F1619 F17)/G7,'L(D)'!A2:B501,2)) 20 =1-NORMDIST(F16,G6,G7,TRUE) 42 G 200,000 20,000 1,000 40,000,000 0.40 0.20 0.05 Multiperiod Normal Backordering Iteration 10 Formulas Only one formula changes on the iteration 2 - 10 worksheets, in cell F15. AThe B C D E F G formula in this cell 1 alwaysMULTI-PERIOD EOQ MODEL (Backordering) - NORMAL LEAD-TIME DEMAND 2 previous refers back the the 3 PROBLEM: iteration. For example, the Unleaded Gas at Oil Refinery 4 worksheet shown5 here is for Parameter Values: iteration 10 so the6 formula in Mean of Demand Distribution: mu = 200,000 cell F15 refers back 7 to Stand. Deviation of Demand Distribution: sigma = 20,000 8 Fixed Cost per Order: k = 1,000 iteration 9. The term ‘9’!F19 means the value of F19 (expected number of shortages) from iteration 9. 43 9 10 11 12 13 14 15 16 17 18 19 20 Annual Demand Rate: A = Unit Cost of Procuring an Item: c = Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = Optimal Values: Optimal Order Quantity: Q* = Optimal Reorder Point: r* = Expected Demand: mu = Total Expected Cost: TEC(Q*) = Expected Shortages: B(r*) = Probability of Shortage: P[D>r*] = F 15 =SQRT((2*G9*(G8+G12*'9'!F19))/(G11*G10)) 40,000,000 0.40 0.20 0.05 1,008,256 234,937 200,000 83,455.46 331.60 0.04 Multiperiod Normal Lost Sales The solution to multiperiod models with normal lead-time demand and lost sales is based on the backordering case just described. It varies only in that some formulas are different (described in Appendix 16-1). 44 Multiperiod Normal Lost Sales (Figure 16-11) A B C D E F MULTI-PERIOD EOQ MODEL (Lost Sales) - NORMAL LEAD-TIME DEMAND 1 2 1. Start with 3 PROBLEM: Roger's Sentinel Station worksheet 1 4 (tab 1) and 5 Parameter Values: 6 Mean of Demand Distribution: mu = enter the 7 Stand. Deviation of Demand Distribution: sigma = problem 8 Fixed Cost per Order: k = name in B3 9 Annual Demand Rate: A = and the 10 Unit Cost of Procuring an Item: c = problem 11 Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = 12 parameters in Selling Price per Unit: pR = 13 G6:G13. 14 15 Optimal Values: 2. To quickly find 16 Optimal Order Quantity: Q* = the optimal solution 17 Optimal Reorder Point: r* = Expected Demand: mu = skip to the last18 19 Total Expected Cost: TEC(Q*) = iteration by clicking 20 Expected Shortages: B(r*) = on tab 10 (shown 21 Probability of Shortage: P[D>r*] = here). 45 G 1,000 50 100 500,000 1.48 0.19 0.25 1.75 18,876 1,103 1,000 5,336.87 0.37 0.02 3. Optimal values: Q*, r*, mu, TEC(Q*), B(Q*), P D>Q*. Multiperiod Normal Lost Sales Summary A B C D E F G 1 MULTI-PERIOD EOQ MODEL (Lost Sales) - NORMAL LEAD-TIME DEMAND 2 3 PROBLEM: Roger's Sentinel Station 4 5 Parameter Values: 6 Mean of Demand Distribution: mu = 1,000 To quickly find the 7 Stand. Deviation of Demand Distribution: sigma = 50 8 on Fixed Cost per Order: k = 100 optimal solution click 9 Annual Demand Rate: A = 500,000 the Summary tab. It 10 Unit Cost of Procuring an Item: c = 1.48 provides a summary of all 11 Annual Holding Cost per Dollar Value: h = 0.19 Shortage Cost per Unit: p = 12 0.25 S the 10 iterations. 13 14 15 Iteration, i Qi ri B(ri) TEC(Qi,ri) 16 1 18,858 1,103 0.37 $ 5,336.87 17 2 18,876 1,103 0.37 $ 5,336.87 18 3 18,876 1,103 0.37 $ 5,336.87 Notice the 19 4 18,876 1,103 0.37 $ 5,336.87 answers do not 20 5 18,876 1,103 0.37 $ 5,336.87 change after the 21 6 18,876 1,103 0.37 $ 5,336.87 22 7 18,876 1,103 0.37 $ 5,336.87 second iteration. 23 8 18,876 1,103 0.37 $ 5,336.87 24 9 18,876 1,103 0.37 $ 5,336.87 25 10 18,876 1,103 0.37 $ 5,336.87 46 Multiperiod Normal Lost Sales Iteration 1 Formulas 16 17 18 19 A B C D E F 1 MULTI-PERIOD EOQ MODEL (Lost Sales) - NORMAL LEAD-TIME DEMAND 2 3 PROBLEM: Roger's Sentinel Station 4 5 Parameter Values: 6 Mean of Demand Distribution: mu = 7 Stand. Deviation of Demand Distribution: sigma = 8 Fixed Cost per Order: k = 9 Annual Demand Rate: A = 10 Unit Cost of Procuring an Item: c = 11 Annual Holding Cost per Dollar Value: h = F Shortage Cost per Unit: pS = 12 =SQRT((2*G9*G8)/(G11*G10)) Selling Price per Unit: pR = 13 =NORMINV((G12+G1314 G10)*G9/(G11*G10*F16+(G12+G13Optimal Values: G10)*G9),G6,G7) 15 16 Optimal Order Quantity: Q* = 18,858 =G6 17 Optimal Reorder Point: r* = 1,103 =(G9*(118 Expected Demand: mu = 1,000 F20/F16))/F16*G8+G11*G10*(F16/2+F1719 Total Expected Cost: TEC(Q*) = 5,336.87 F18+F20)+(G12+G13-G10)*(G9*(1Expected Shortages: B(r*) = 0.37 F20/F16)/F16)*F20 20 21 Probability of Shortage: P[D>r*] = 0.02 =IF(F17<F18,F18-F17+G7*VLOOKUP((F18- F17)/G7,'L(D)'!A2:B501,2),G7*VLOOKUP((F1720 F18)/G7,'L(D)'!A2:B501,2)) 21 =1-NORMDIST(F17,G6,G7,TRUE) 47 G 1,000 50 100 500,000 1.48 0.19 0.25 1.75 Multiperiod Normal Lost Sales Iteration 10 Formulas Only one formula changes on A B C D E F G the iteration 2 - 10 1 F16. The MULTI-PERIOD EOQ MODEL (Lost Sales) - NORMAL LEAD-TIME DEMAND worksheets, in cell 2 formula in this cell always 3 PROBLEM: Roger's Sentinel Station refers back the the 4 previous iteration. For example, the 5 Parameter Values: Mean of Demand Distribution: mu = 1,000 worksheet shown6here is for Stand. Deviation of Demand Distribution: sigma = 50 iteration 10 so the7 formula in 8 Fixed Cost per Order: k = 100 cell F16 refers back to 9 Annual Demand Rate: A = 500,000 iteration 9. 10 Unit Cost of Procuring an Item: c = 1.48 The term ‘9’!F20 means the value of F20 (expected number of shortages) from iteration 9. 48 11 12 13 14 15 16 17 18 19 20 21 Annual Holding Cost per Dollar Value: h = Shortage Cost per Unit: pS = Selling Price per Unit: pR = Optimal Values: Optimal Order Quantity: Q* = Optimal Reorder Point: r* = Expected Demand: mu = Total Expected Cost: TEC(Q*) = Expected Shortages: B(r*) = Probability of Shortage: P[D>r*] = 0.19 0.25 1.75 18,876 1,103 1,000 5,336.87 0.37 0.02 F 16 =SQRT((2*G9*(G8+'9'!F20*(G12+G13-G10))/(G11*G10)))