2 The Production Order Quantity Model USED WHENEVER THE VENDORCANNOT DELIVER THE ORDER( Q* ) ALL IN ONE DAYORUSED WHENEVER THE FACTORYCANNOT PRODUCE THE ORDER ( Q* ) ALL IN ONE DAY
3 Variable Interpretations Service SectorManufacturingP or pThe delivery rateofpurchased itemsP or pThe production rateofmanufactured items
4 Variable Interpretations Service SectorManufacturingQp*Qp*The optimal EOQwhenpurchased itemsarereceived in partialshipmentsThe optimal EOQwhenmanufactured itemscannotall be producedin a single day
5 Production Order Quantity Model Cycle Chart Graphically depicts the relationship betweenMaximum Inventory Level ( IMAX )Replenishment Rate ( P )Consumption Rate ( D )TimeCycle Chartsenhanceunderstandingof basicinventoryconcepts
6 Production Order Quantity Model Cycle Chart SAW TOOTH VERSIONREPLENISHMENTRATEMAXIMUM INVENTORY LEVELCONSUMPTIONRATE ONLYUNITSPDPDCONSUMPTION OCCURSEVEN ASREPLENISHMENTIS TAKING PLACEAVERAGE INVENTORY LEVELELAPSED TIME
7 Production Order Quantity Model Cycle Chart SAW TOOTH VERSIONREPLENISHMENTRATEMAXIMUM INVENTORY LEVELCONSUMPTIONRATE ONLYINVENTORYLEVEL PEAKSAT THIS POINTUNITSPDPDAn EOQ of 100 unitsis delivered at the rate of20 units per week overfive weeksAVERAGE INVENTORY LEVELINVENTORYFALLS TOZERO ORREORDER POINTELAPSED TIME
8 CYCLE CHART DISCUSSION The replenishment rate ( P or p ) is diminishedby the consumption rate ( D or d ).Average inventory will always be less than[ Q* / 2 ] .Average inventory is essentially [ IMAX / 2 ] .
9 Production Order Quantity Formula THE FINITE CORRECTION FACTOR PRODUCES A LARGERVALUE OF Q* IN ORDER TO COMPENSATE FOR PIECEMEALREPLENISHMENT AND CONSUMER DEMAND2DSH [ 1 – d / p ]√Qp* =THE CONSUMPTION ORUSAGE RATE ( D )THE REPLENISHMENT ORPRODUCTION RATE ( P )
10 Production Order Quantity Model EXAMPLE DA = 1000 units ( annual demand )S = $10.00 ( cost per order )H = $.50 ( annual unit carry cost )P = 8 units ( daily supply rate )D = 6 units ( daily usage rate )How manyunits shouldbe orderedat a time?How longto receivethe entireorder?
11 Production Order Quantity Model SOLUTION √2(1000)(10.00).50 [ 1 – 6 / 8 ]Qp* =√20,000.50 [ .25 ]√==160,000 = 400 units
12 The Basic EOQ Model SOLUTION √2(1000)(10.00).50Q* =√√20,000.5040,000 = 200 units==Q* IS ONLY HALF OF WHAT IT WAS UNDER THEPRODUCTION ORDERQUANTITY MODEL
13 Production Order Quantity Model EXAMPLE ORDER RECEIPT TIME PERIOD ( t )Qp*Pt === 50 days
14 Production Order Quantity Model EXAMPLE TOTAL VARIABLE COSTS ( TVC )Q*pDD1 -TVC =X H+X SX2PQ*p40061,000X1 -X .50+X=82400= [ 200 x (.25)(.50) ] + [ 2.5 x ]= [ ] + [ ] = $50.00
15 Post Solution Comments The Qp* must be larger than Q* since it is beingdrawn down even as it arrives on a piecemeal basis.The larger Qp* produces no increase in carry costs.Annual carry costs are actually less than they areunder the basic EOQ model ( Q* ) .“P” and “D” can be expressed asdaily, weekly, monthly, and annualfigures without changing the valueof the finite correction factor [1-d/p]
31 The Backorder Inventory Model USED WHENEVER THEFIRM IS APPEALINGTO ITS CUSTOMERSTO BUY AND THENWAIT FOR THEIRPURCHASES UNTILA NEW SHIPMENT( Q* ) IS ORDEREDAND RECEIVED
32 Backorder Model Variables Backorder cost – BOptimal number of backorders – S*Optimal order quantity under a backorderingscenario – Qb*Number of units going into stock after allbackorders have been filled – b or [ Qb* - S* ]
33 Backorder Model Cycle Chart Optimal Order Quantity,Optimal Number of Backorders,Remaining Units Going IntoStock after BackordersHave Been FilledGraphically depicts the relationship betweenCycle chartsenhanceunderstandingof basic inventoryconcepts
34 Backorder Model Cycle Chart PICKET FENCE VERSION Q* or EOQb = Q*- S*POSITIVEINVENTORYINITIALCYCLECYCLE2CYCLE3CYCLE4NEGATIVEINVENTORYS*
35 Backorder Model Cycle Chart PICKET FENCE VERSION 88 UNITS GOINTO INVENTORY AFTER THE FOUR STOCKOUTSHAVE BEENFILLEDQ* or EOQ = 92 unitsb = Q*- S* = 88 unitsASSUME ANEOQ OF92 UNITSPOSITIVEINVENTORYTHIS MODELASSUMESTHAT THENEW EOQ IS ORDERED WHENBACKORDERSEQUAL FOURUNITSINITIALCYCLECYCLE2CYCLE3CYCLE4THE FIRM ALLOWS BACKORDERSTO REACH4 UNITSNEGATIVEINVENTORYS* = 4 UNITS
36 Backorder Model Formula OPTIMAL ORDER QUANTITYIN BACKORDER-TOLERATEDSITUATIONSANNUAL CARRY COSTPER UNITHUNIT BACKORDERCOST
37 Backorder Model Formula THEOPTIMALNUMBEROF BACKORDERSOPTIMAL ORDER QUANTITY UNDERBACKORDERINGUNIT BACKORDERCOST
38 The Backorder Model EXAMPLE DA = 500 units ( annual demand )S = $4.00 ( order cost )H = $ .50 ( annual unit carry cost )B = $10.00 ( unit backorder cost )How manyunits shouldbe ordered?What are thenumber ofbackorders?
39 The Backorder Model SOLUTION √2(500)(4.00) ( )Qb* =X√= x 1.05=≈ 92 units√
40 The Backorder Example SOLUTION .50S* = x= ( x )≈ 4 units
41 Relationship Between ROP and S* If leadtime ( L ) = 0 , then ROP = S*IF LEADTIME IS ZERO, THE REORDER POINT OCCURS WHENTHE OPTIMAL NUMBER OF BACKORDERS IS REACHED
42 Relationship Between ROP and S* If leadtime ( L ) > 0, then ROP > S*IF LEADTIME IS NOT ZERO, THE REORDER POINT OCCURSBEFORE THE OPTIMAL NUMBER OFBACKORDERS IS REACHED
43 Backorder Reorder Point EXAMPLE Given: L = 0 days Qb* = 92 units S* = 4 unitsOrder 92 units whenthe number of backordersaccumulate to 4
44 Backorder Reorder Point EXAMPLE Given: d = 2 units, L = 6 days, Qb* = 92 units, S* = 4 unitsOrder 92 units when there are8 units still left in theaccount balance.THE INITIAL ROP = d x L = [ 2 units x 6 days ] = 12 unitsTHE FINAL ROP = Initial ROP – S* = [ 12 units – 4 units ] = 8 units
45 ROP under Backordering EXAMPLE WHERE L IS POSITIVEQb*Qb*POSITIVEINVENTORYREGIONFINAL ROPFINAL ROP12 – 4 = +812 – 4 = +8INITIAL ROP = 12 UNITSINITIAL ROP = 12 UNITSNEGATIVEINVENTORYREGIONS* = 4 units ( - 4 )
46 MAY OR MAY NOT BE AVAILABLE ON YOUR PARTICULAR SOFTWARE THE BACKORDER MODELMAY OR MAY NOT BE AVAILABLE ON YOUR PARTICULAR SOFTWARE
50 ABC Classification System RATIONALEPurchasing personnel are relatively few in numberin the firm.There are thousands of components, materials,and finished good inventory accounts in mediumand large firms.There needs to be a priority system for establishingand updating inventory control doctrines ( Q* / R ).
51 Class “A” Inventory Items Comprise only 15% of thetotal items in stock yetrepresent 70% - 80% ofthe total dollar volume.LARGE APPLIANCESAUTOMOBILESFURNITUREDIAMONDS
52 Class “B” Inventory Items Comprise 30% of thetotal items in stock andrepresent 15% - 25% ofthe total dollar volumeMID-SIZED APPLIANCESLAWN MOWERSMOST SUITS & COATS
53 Class “C” Inventory Items Comprise 55% of thetotal items in stock yetrepresent only 5% ofthe total dollar volumeTOASTERS / BLENDERSNUTS, BOLTS, SCREWSSTATIONERY SUPPLIESMOST ACCESSORIES
54 Policies Based On ABC“A” items would be inventoried in a more secure area
55 Policies Based On ABC“A” items would be inventoried in a more secure area“A” items would warrant more care in forecasting
56 Policies Based On ABC“A” items would be inventoried in a more secure area“A” items would warrant more care in forecasting“A” item records would be verified more frequently
57 Policies Based On ABC“A” items would be inventoried in a more secure area“A” items would warrant more care in forecasting“A” item records would be verified more frequently“A” items would justify closer attention to customerservice levels
58 Policies Based On ABC“A” items would be inventoried in a more secure area“A” items would warrant more care in forecasting“A” item records would be verified more frequently“A” items would justify closer attention to customerservice levels“A” items would qualify for real-time inventory trackingsystems and more sophisticated ordering rules
59 Additional Criteria for ABC Anticipated engineering changesDelivery problemsQuality problemsHigh unit production costsWE FOCUS ON THE RELATIVELY FEW ITEMS WITH MAJOR PROBLEMS
78 Reorder Point Models The original reorder point formula ROP = d x L computes the mean demandduring lead time, that is, theaverage demand during thewaiting period for the item.Where d = daily demandL = lead time( in days )
79 Reorder Point Models However, actual demand during lead time can be much higherthan the mean (average) demand.For this reason, the reorder pointshould contain a built-in safetystock ( SS or B ) that will meetunexpectedly higher demandsand consequently reducestockout costs.
80 Reorder Point Models The reorder point formula now becomes ROP = d x L + SSBut larger safety stocksinvolve higher carry costs.We must find a safetystock that minimizesboth carry costs andexpected stockoutcosts annually
81 Reorder Point Cost Tradeoff KNOWN STOCKOUT COST MODELREORDERPOINT TOTALCOSTSANNUALSAFETY( BUFFER )STOCKCARRYCOSTSCostTHESELECTEDSAFETY( BUFFER ) STOCKLEVEL( SS or B )ANNUALEXPECTEDSTOCKOUT COSTS∞Safety ( Buffer ) Stock in Units
82 Known Stockout Cost Model ASSUMPTIONSStockout cost per unit is known.Lead time is known and constant.Lead time demand is variable.IT IS ALSO ASSUMED THAT THE ANNUAL NUMBEROF ORDERING PERIODS IS KNOWN ( n )
83 Demand During Lead Time ( DL ) EXAMPLEProbability15% % % % %DEMAND IN UNITSLEAD TIME DEMAND IS APPROXIMATELY NORMALLY DISTRIBUTEDAND RANGES BETWEEN THIRTY AND SEVENTY UNITS
84 Demand During Lead Time ( DL ) THE HIGHESTPROBABILITYProbability15% % % % %DEMAND IN UNITSSINCE THERE IS A 30% CHANCE THAT LEAD TIME DEMAND WILL BE FIFTY ( 50 ) UNITS WE WOULD ATLEAST SET THE REORDER POINT AT 50 UNITS. OTHERWISE, WE ARE EXTREMELY VULNERABLE TOLARGE AND RECURRING STOCKOUTS.
85 Demand During Lead Time ( DL ) Probability15% % % % %SAFETYSTOCKISZEROHEREDEMAND IN UNITSTHE SAFETY OR BUFFER STOCK LEVEL AT THE MINIMUM REORDER POINT IS ZERO BY DEFINITION
86 Reorder Point Carry Cost ANNUAL UNITCARRY COSTSAFETY OR BUFFERSTOCK in unitsXANNUAL SAFETY STOCKCARRY COSTS=
87 Reorder Point of 50 Units ‘0’ Safety Stock x $5.00 per unit = $0.00 ANNUAL SAFETY STOCK CARRY COSTS‘0’ Safety Stock x $5.00 per unit = $0.00per yearcarry cost( BUFFER STOCK )
88 CREATE A STOCKOUT OF 10 UNITS CREATE A STOCKOUT OF 20 UNITS Reorder Point of 50 UnitsEXPECTED STOCKOUTS PER ORDER PERIODEXPECTED STOCKOUTSProbability15% % % % %10Stockouts20StockoutsReorderPointActualDemandA DEMAND OF 60 UNITS WOULDCREATE A STOCKOUT OF 10 UNITSWITH A 20% PROBABILITYA DEMAND OF 70 UNITS WOULDCREATE A STOCKOUT OF 20 UNITSWITH A 10% PROBABILITY
89 Reorder Point Stockout Cost Lead Time ExpectedStockouts in unitsXStockout Cost per unitAnnual ExpectedStockout Costs=XAnnual Number ofLead Time Periods( ordering periods )
90 NUMBER OF ORDER PERIODS Reorder Point of 50 UnitsANNUAL EXPECTED STOCKOUT COSTSSTOCKOUTCOST( per unit )NUMBER OF ORDER PERIODS( per year )NUMBEROFSTOCKOUTSEXPECTEDSTOCKOUTSCOSTLead TimeDemandof 60102$40.006$480.00Lead TimeDemandof 70202$40.006$480.00Σ = $960.00
91 Reorder Point Total Cost ANNUAL SAFETY STOCKCARRY COSTS+ANNUAL EXPECTEDSTOCKOUT COSTSTHE REORDER POINT THAT HAS THE LOWEST TOTAL COST IS SELECTED
93 RAISING THE REORDER POINT TO 60 UNITS AUTOMATICALLY Reorder Point of 60 UnitsANNUAL SAFETY STOCK CARRY COSTS‘10’ Safety Stock x $5.00 per unit = $50.00per yearcarry cost( BUFFER STOCK )RAISING THE REORDER POINTTO 60 UNITS AUTOMATICALLYCREATES A SAFETY STOCKOF TEN UNITS
94 CREATE A STOCKOUT OF 10 UNITS Reorder Point of 60 UnitsEXPECTED STOCKOUTS PER ORDER PERIODEXPECTEDSTOCKOUT1Probability15% % % % %10StockoutsReorderPointActualDemandA DEMAND OF 70 UNITS WOULDCREATE A STOCKOUT OF 10 UNITSWITH A 10% PROBABILITY
95 NUMBER OF ORDER PERIODS Reorder Point of 60 UnitsANNUAL EXPECTED STOCKOUT COSTSSTOCKOUTCOST( per unit )NUMBER OF ORDER PERIODS( per year )NUMBEROFSTOCKOUTSEXPECTEDSTOCKOUTSCOST101$40.006$240.00LEAD TIME DEMANDof 70Σ = $240.00
97 RAISING THE REORDER POINT TO 70 UNITS AUTOMATICALLY Reorder Point of 70 UnitsANNUAL CARRY COSTS‘20’ Safety Stock x $5.00 per unit = $100.00per yearcarry cost( BUFFER STOCK )RAISING THE REORDER POINTTO 70 UNITS AUTOMATICALLYCREATES A SAFETY STOCKOF 20 UNITS
98 Reorder Point of 70 Units 15% 25% 30% 20% 10% 30 40 50 60 70 EXPECTED EXPECTED STOCKOUTS PER ORDER PERIODEXPECTEDSTOCKOUTSZEROProbability15% % % % %StockoutsReorderPointBASED ON HISTORICAL DEMAND DATA,NO DEMAND SHOULD OCCUR THAT CREATES A STOCKOUT
99 NUMBER OF ORDER PERIODS Reorder Point of 70 UnitsANNUAL EXPECTED STOCKOUT COSTSSTOCKOUTCOST( per unit )NUMBEROFSTOCKOUTSNUMBER OF ORDER PERIODS( per year )EXPECTEDSTOCKOUTSCOST$40.006$0.00LEAD TIME DEMANDof 70Σ = $0.00
114 Template and Sample Data Insert SelectedReorder Point, etc.This is the conditional payoff matrix for this problem. The strategiesare the safety stocks, the events are the five possible demands, andthe conditional payoffs are the cost consequences (expectedstockouts + carry costs) of selecting a particular safety stockand a certain demand materializing.
115 5 discrete demand possibilities during any given lead time ( with probabilities )Selected Reorder PointEMV Criterion Maxi-Min CriterionConditional PayoffsFor ROP = 50 , total cost = $ , H = $For ROP = 60 , total cost = $ , H = $For ROP = 70 , total cost = $ , H = $100.00Safety stocks associatedwith all 5 discrete demands
116 Service Level Model When the stockout cost per unit ( Cs ) is REORDER POINT DETERMINATIONWhen the stockout cost per unit ( Cs ) isdifficult or impossible to determine, thefirm may elect to establish a policy ofkeeping enough safety stock on hand tosatisfy a prescribed level of customerservice.FOR EXAMPLE, THE FIRM MAY DESIRE A SERVICE LEVELTHAT MEETS 95% OF THE DEMAND, OR CONVERSELY,RESULTS IN STOCKOUTS ONLY 5% OF THE TIME.
117 Service Level Model Assuming that the demand during lead time REORDER POINT DETERMINATIONAssuming that the demand during lead time( DL ) follows a normal distribution, only themean ( µ ) and standard deviation ( σ ) arerequired to define the reorder point as wellas the safety stock ( SS or B ) for any givenservice level.SALES DATA ARE USUALLY ADEQUATE FOR COMPUTINGTHE MEAN AND STANDARD DEVIATION
118 Service Level ModelREORDER POINT DETERMINATION EXAMPLEA firm stocks an item that has a normally-distributeddemand during the lead time period. The average ormean demand during the lead time period is 400 unitsand the standard deviation is 15 units. The firm wantsa reordering policy that limits stockouts to only 5% ofthe time.Requirement:1. How much safety stock should the firm maintain?2. What should be the reorder point?
119 Service Level Model Z AREAS UNDER THE STANDARD NORMAL CURVE .03 .04 .05.06.071.4.92364.92507.92647.92785.929221.5.93699.93822.93943.94062.941791.6.94845.94950.95053.95154.952541.7.95818.95907.95994.96080.96164Z95% SERVICE LEVEL IS REPRESENTED BY z = 1.65 standard normal deviates
121 Service Level Model μ ROP Z = + 1.65 0 units REORDER POINT AND SAFETY STOCK FOR 95% SERVICE LEVELZ =5%STOCKOUTPROBABILITY95% STOCKAGE PROBABILITYSAFETY STOCK0 unitsμROPLEAD TIME MEAN DEMAND = 400 UNITS
122 Service Level Model Reorder Point ( R ) = μ + ( z )( σ ) FORMULAEReorder Point ( R ) = μ + ( z )( σ )Safety Stock ( SS ) = ( z )( σ )μ = MEAN DEMAND DURING LEAD TIMEz = NUMBER OF STANDARD NORMAL DEVIATESσ = STANDARD DEVIATION OF LEAD TIME DEMAND
123 Service Level Model Reorder Point ( R ) = 400 + ( 1.65 )( 15 ) EXAMPLE – 95% SERVICE LEVELReorder Point ( R ) = ( 1.65 )( 15 )= 425 unitsSafety Stock ( SS ) = ( 1.65 )( 15 )= 25 unitsμ = 400 unitsz = 1.65 ( 95% service level )σ = 15 units
124 Service Level Model Reorder Point ( R ) = 400 + ( 2.33 )( 15 ) EXAMPLE – 99% SERVICE LEVELReorder Point ( R ) = ( 2.33 )( 15 )= 435 unitsSafety Stock ( SS ) = ( 2.33 )( 15 )= 35 unitsμ = 400 unitsz = 2.33 ( 99% service level )σ = 15 unitsZ.03
130 IF THE DAILY DEMAND AND/OR THEN THEIR STANDARD DEVIATION(S) = 0 THE DATA INPUT TABLE REQUIRESDAILY DEMAND DURING LEAD TIMETHE STANDARD DEVIATION OF DAILYDEMAND DURING LEAD TIMETHE SERVICE LEVEL DESIRED ( i.e. 95% )LEAD TIME IN DAYSTHE STANDARD DEVIATION OF LEAD TIMEIF THE DAILY DEMAND AND/ORLEADTIME ARE CONSTANT,THEN THEIR STANDARD DEVIATION(S) = 0
131 The mean demand during lead time was given as 400 units. Since this model requires both daily demand during thelead time, and the lead time ( in days ), we will assumehere that daily demand = 50 units, and lead time = 8 days.
132 THE 95% “SERVICE LEVEL” REORDER POINT = 423 UNITS THE 95% “SERVICE LEVEL” SAFETY STOCK = 23 UNITS
138 Stochastic Reorder Point Models WHEN DEMAND AND / OR LEAD TIME ARE VARIABLEThese models generally assume that any variability ineither the demand rate or lead time can be adequatelydescribed by a normal distribution. This, however, isnot a strict requirement.These models will provide approximate reorder pointseven in cases where the actual probability distributionsdepart substantially from normal.In all models shown, stockout costs are assumed to beunknown.
139 Stochastic Reorder Point Models MODELS CONSIDERED IN THIS PRESENTATIONvariable demand rate / constant lead timeconstant demand rate / variable lead timevariable demand rate / variable lead time
140 Stochastic Reorder Point Models THE VARIABLESd = constant demand rated = average demand rateL = constant lead timeL = average lead timeσD = standard deviation of demand rateσL = standard deviation of lead time
141 Stochastic Reorder Point Models LEAD TIME IS CONSTANTDEMAND RATE IS VARIABLEJack’s Pizza Parlor uses 1,000 cans of tomatoes per month at anaverage rate of 40 per day for each of 25 days per month. Usagecan be approximated by a normal distribution with a standard de-viation of 3 cans per day. Lead time is constant at 4 days. Jackdesires a service level of 99% , that is, a stockout risk of only 1%Requirement:1. Determine the reorder point ( ROP )2. Determine the safety or buffer stock ( SS or B )
142 Reorder Point Solution PIZZA PARLOR_ROP = ( d x L ) + ( z ) ( L ) ( σD )= ( 40 x 4 ) ( 4 ) ( 3 )= ( 6 )== 174 cansGiven:d = 40 cans dailyσD = 3 cans dailyL = 4 days_
143 Safety Stock Solution SS or B = ( z ) ( L ) ( σD ) = 2.33 ( 4 ) ( 3 ) PIZZA PARLORSS or B = ( z ) ( L ) ( σD )= 2.33 ( 4 ) ( 3 )= 2.33 ( 6 )≈ 14 cans
149 Stochastic Reorder Point Models LEAD TIME IS VARIABLEDEMAND RATE IS CONSTANTAn oil-driven generator uses 2.1 gallons per day. Lead time isnormally distributed with a mean of 6 days.The standard deviation of lead time is 2 days. The service levelis 98%, that is, the stockout risk is 2%Requirement:1. Determine the reorder point ( ROP )2. Determine the safety or buffer stock ( SS or B )
150 Reorder Point Solution THE GENERATOR_ROP = ( d x L ) + ( z ) ( d ) ( σL )= ( 2.1 x 6 ) ( 2.1 ) ( 2 )= ( 4.2 )== gallonsGiven:d = 2.1 gallons dailyσL = 2 daysL = 6 days_
151 Safety Stock Solution SS or B = ( z ) ( d ) ( σL ) THE GENERATORSS or B = ( z ) ( d ) ( σL )= ( 2.1 ) ( 2 )= ( 4.2 )= 8.63 gallons
156 Stochastic Reorder Point Models LEAD TIME IS VARIABLEDEMAND RATE IS VARIABLEBeer consumption at a local tavern is known to be normallydistributed with a mean of 150 bottles daily and a standarddeviation of 10 bottles daily. Delivery time is also normallydistributed with a mean of 6 days and a standard deviationof 1 day. The service level is 90%Requirement:1. Determine the reorder point ( ROP )2. Determine the safety or buffer stock ( SS or B )
172 Quantity Discount Model Solved ProblemQuantity Discount ModelIvonne CallenComputer-BasedManual
173 Ivonne Callen ProblemIvonne Callen sells beauty supplies. Her annual demandfor a particular skin lotion is 1,000 units.The cost of placing an order is $20.00, while the holdingcost per unit per year is 10 percent of the cost.The item currently costs $10.00 if the order quantity isless than 300. For orders of 300 units or more, the costfalls to $9.80.To minimize total cost, how many units should Ivonneorder each time she places an order?What is the minimum total cost?
174 Ivonne Callen Problem D = 1,000 units S = $20.00 I = 10% P = $10.00 if Q < 300 unitsP = $ if Q > 300 units
175 Ivonne Callen Problem To minimize total costs, how many units should Ivonne order each time she placesan order ?What is the minimum total cost ?
176 √ √ √ EOQ at $10.00 Item Cost 2 (1,000 ) ( 20 ) 2 x D x S ( .10 ) ( )2 x D x SI x P√=Q*1 =√40,0001.00= 200 units=
177 Total Cost at $10.00 Item Cost TC = [ Q* / 2 ] (I)(P) + [ D / Q* ] (S) + [ D x P ]= [ 200 / 2 ] (.10)(10.00) + [ 1,000 / 200 ] (20.00) + [1,000 x 10.00]= $ $ $10,000.00= $10,200.00
178 Total Cost at $9.80 Item Cost TC = [ Q* / 2 ] (I)(P) + [ D / Q* ] (S) + [ D x P ]= [ 300 / 2 ] (.10)(9.80) + [ 1,000 / 300 ] (20.00) + [1,000 x 9.80]= $ $ $9,800.00= $10,013.67
179 Ivonne Callen Problem Therefore, she should order 300 units. CONCLUSIONTherefore, she shouldorder 300 units.
180 Serial Rate Production Model The Handy Manufacturing Company Solved ProblemSerial Rate Production ModelThe Handy Manufacturing CompanyComputer-BasedManual
181 The Handy Mfg. CompanyThe Handy Manufacturing Company manufactures smallair conditioner compressors. The estimated demand forthe year is 12,000 units. The setup cost for the productionprocess is $ per run, and the carrying cost is $10.00per unit per year. The daily production rate is 100 units perday, and demand has been 50 units per day.Determine the number of units to produce in each batch,the number of batches that should be run each year, andthe time interval, in days, between each batch.( Assume 240 operating days. )
182 The Data D = 12,000 units S = $200.00 H = $10.00 / unit / year p = 100 units / dayd = 50 units / day240 working days / yearUseProduction OrderQuantity Model
183 √ √ Production Run Model H ( 1 - d / p ) Qp* = 2 (12,000)(200.00) = 2 (D)(S)H ( 1 - d / p )Qp* =√2 (12,000)(200.00)(10.00)( / 100 )=
184 √ √ √ Production Run Model (10.00)(.5) Qp* = 4,800,000 5 = = 960,000 = 979.8units
185 Batches Run Annually n = D / Q*p n = 12,000 / 980 n ≈ 12 production runs
186 Time Between Runst = 240 days / n= 240 days / 12= every 20 days
188 Solved ProblemReorder Point ModelMr. BeautifulComputer-BasedManual
189 Mr. Beautiful ProblemMr. Beautiful, an organization that sells weight training sets, hasan ordering cost of $40.00 for the BB-1 set. The carry cost for theBB-1 set is $5.00 per set per year.To meet demand, Mr. Beautiful orders large quantities of BB-1 (7)seven times a year. The stockout cost for BB-1 is estimated to be$50.00 per set. Over the past several years, Mr. Beautiful has ob-served the following demand during lead time for BB-1:Lead Time Demand ProbabilityStarting with a ROP of 60 units, what level of safety stockshould be maintained for BB-1?
190 Mr. Beautiful Problem ROP 40 50 60 70 80 90 10 sets 20 sets 30 sets 10 sets20 sets30 setsExpected Stockouts[.20 x 10] + [.20 x 20] + [.10 x 30] = 9 sets
191 Mr. Beautiful Problem Carry Cost / Set / Year = $5.00 Stockout Cost / Set = $50.00Order Periods / Year = 7At ROP = 60 sets , SS = 0 setsStockout Cost:[.20 x 10 sets x 20 sets x 30 sets] x 7 orders x $50.00[ ] x 7 x $50.00 = $3,150.00Carrying Cost:SS = 0 sets x $5.00 / set / year = $0.00TOTAL COST = $3,150.00
192 Mr. Beautiful Problem ROP 40 50 60 70 80 90 10 sets 20 sets 10 sets20 setsExpected Stockouts[.20 x 10] + [.10 x 20] = 4 sets
193 Mr. Beautiful Problem Carry Cost / Set / Year = $5.00 Stockout Cost / Set = $50.00Order Periods / Year = 7At ROP = 70 sets , SS = 10 setsStockout Cost:[.20 x 10 sets x 20 sets] x 7 orders x $50.00[ ] x 7 x $50.00 = $1,400.00Carrying Cost:SS = 10 sets x $5.00 / set / year = $50.00TOTAL COST = $1,450.00
194 Mr. Beautiful Problem ROP 40 50 60 70 80 90 10 sets 10 setsExpected Stockouts[.10 x 10] = 1 set
195 Mr. Beautiful Problem Carry Cost / Set / Year = $5.00 Stockout Cost / Set = $50.00Order Periods / Year = 7At ROP = 80 sets , SS = 20 setsStockout Cost:[.10 x 10 sets] x 7 orders x $50.00[ 1 ] x 7 x $50.00 = $350.00Carrying Cost:SS = 20 sets x $5.00 / set / year = $100.00TOTAL COST = $450.00
196 Mr. Beautiful Problem ROP 40 50 60 70 80 90 .10 .20 .20 .20 .20 .10 Expected Stockouts[ .00 x 0 ] = 0 sets
197 Mr. Beautiful Problem Carry Cost / Set / Year = $5.00 Stockout Cost / Set = $50.00Order Periods / Year = 7At ROP = 90 sets , SS = 30 setsStockout Cost:[.0 x 0 sets] x 7 orders x $50.00[ 0 ] x 7 x $50.00 = $0.00Carrying Cost:SS = 30 sets x $5.00 / set / year = $150.00TOTAL COST = $150.00
198 Mr. Beautiful Problem Reorder Point Stockout Cost Carry Total 60 sets $3,150.00$0.0070 sets$1,400.00$50.00$1,450.0080 sets$350.00$100.00$450.0090 sets$150.00Mr. Beautiful should select a ROP = 90 sets with a SS = 30 sets
202 THERE ARE SIX ( 6 ) DEMANDS THAT CAN OCCUR DURING LEADTIME
203 PROBABILITY DISTRIBUTION OF THE DATA INPUT TABLE REQUIRES:STARTING REORDER POINT WHERETHE SAFETY STOCK IS ZERO ( 0 )CARRY COST PER UNIT PER YEARSTOCKOUT COST PER UNIT PER YRNUMBER OF ORDERS PER YEAR ( 7 )THE DISCRETEPROBABILITY DISTRIBUTION OFLEAD TIME DEMAND