Presentation on theme: "Inventory Control Models"— Presentation transcript:
1Inventory Control Models Ch 4 (Known Demands)R. R. LindekeIE 3265, Production And Operations Management
2Reasons for Holding Inventories Economies of ScaleUncertainty in delivery lead-timesSpeculation. Changing Costs Over TimeSmoothing: to account for seasonality and/or BottlenecksDemand UncertaintyCosts of Maintaining Control System
3Characteristics of Inventory Systems DemandMay Be Known or UncertainMay be Changing or Unchanging in TimeLead Times - time that elapses from placement of order until it’s arrival. Can assume known or unknown.Review Time. Is system reviewed periodically or is system state known at all times?
4Characteristics of Inventory Systems Treatment of Excess Demand.Backorder all Excess DemandLose all excess demandBackorder some and lose someInventory who’s quality changes over timeperishabilityobsolescence
5Real Inventory Systems: ABC ideas This was the true basis of Pareto’s Economic Analysis!In a typical Inventory System most companies find that their inventory items can be generally classified as:A Items (the % of sku’s) that represent up to 80% of the inventory valueB Items (the 20 – 30%) of the inventory items that represent nearly all the remaining worthC Items the remaining 50 – 70% of the inventory items sku’s) stored in small quantities and/or worth very little
6Real Inventory Systems: ABC ideas and Control A Items must be well studied and controlled to minimize expenseC Items tend to be overstocked to ensure no runouts but require only occasional reviewSee mhia.org – there is an “e-lesson” on the principles of ABC Inventory management – check it out! – do the on-line lesson!
7Relevant Inventory Costs Holding Costs - Costs proportional to the quantity of inventory held.Includes:Physical Cost of Space (3%)Taxes and Insurance (2 %)Breakage Spoilage and Deterioration (1%)Opportunity Cost of alternative investment. (18%)Here these holding issues total: 24%Therefore, in inventory systems, the holding cost would be taken as:h .24*Cost of product
8Lets Try one: Problem 4, page 193 – cost of inventory Find h first (yearly and monthly)Total holding cost for the given period:THC = $Average Annual Holding Costassumes an average monthly inventory of trucks based on on hand data$3333
9Relevant Costs (continued) Ordering Cost (or Production Cost).Includes both fixed and variable componentsslope = cK C(x) = K + cx for x > 0; 0 for x = 0.
10Relevant Costs (continued) Penalty or Shortage Costs. All costs that accrue when insufficient stock is available to meet demand. These include:Loss of revenue due to lost demandCosts of book-keeping for backordered demandsLoss of goodwill for being unable to satisfy demands when they occur.
11Relevant Costs (continued) When computing Penalty or Shortage Costs inventory managers generally assume cost is proportional to number of units of excess demand that will go unfulfilled.
12The Simple EOQ Model – the most fundamental of all! Assumptions:1. Demand is fixed at l units per unit time – typically assumed at an annual rate (use care).2. Shortages are not allowed.3. Orders are received instantaneously. (this will be relaxed later).
13Simple EOQ Model (cont.) Assumptions (cont.):4. Order quantity is fixed at a value “Q” per cycle. (we will find this as an optimal value)5. Cost structure:a) includes fixed and marginal order costs (K + cx)b) includes holding cost at h per unit held per unit time.
18Finding an Optimal Level of ‘Q’ – the so-called EOQ Requires us to take derivative of the G(Q) equation with respect to QThen, Set derivative equal to Zero:Now, Solve for Q
19Properties of the EOQ (optimal) Solution Q is increasing with both K and and decreasing with hQ changes as the square root of these quantitiesQ is independent of the proportional order cost, c. (except as it relates to the value of h = I*c)
20Try ONE!A company sells 145 boxes of BlueMountain BobBons/week (a candy)Over the past several months, the demand has been steadyThe store uses 25% as a ‘holding factor’Candy costs $8/bx and sells for $12.50/bxCost of making an order is $35Determine EOQ (Q*) and how often an order should be placed
21Plugging and chugging: = 145*52 = 7540In your teams: Compute Pr. 10, pg 201
22But, Orders usually take time to arrive! This is a realistic relaxation of the EOQ ideas – but it doesn’t change the modelThis requires the user to know the order “Lead Time”And then they trigger an order at a point before the delivery is needed to assure no stock outsIn our example, what if lead time is 1 week?We should place an order when we have 145 boxes in stock (the one week draw down)Note make sure lead time units match units in T!
23But, Orders usually take time to arrive! What happens when order lead times exceed T?We proceed just as before (but we compute /T) is the lead time is units that match THere, lets assume = 6 weeks then:/T = 6/3.545 = 1.69 – in the Blue Mount Bon-Bon casePlace order: 1.69 cycles before we need product!Trip Point is then 0.69*Q* = .69*514 = 356 boxesThis trip point is not for the next stock out but the one after that (1.69 T from now!) – be very careful!!!
24Sensitivity AnalysisLet G(Q) be the average annual holding and set-up cost function given byand let G* be the optimal average annual cost. Then it can be shown that:
25SensitivityWe find that this model is quite robust to Q errors if holding costs are relatively lowWe find, given a Q – error in ordering quantitythat Q* + Q has smaller error than Q* - Q (Error means extra inventory maintenance costs)That is, we tend to have a greater penalty cost if we order too little than too much
26Inventory Levels for Finite Production Rate Model
27EOQ With Finite Production Rate Suppose that items are produced internally at a rate P (> λ, the consumption rate). Then the optimal production quantity to minimize average annual holding and set up costs has the same form as the EOQ, namely:Except that h’ is defined as h’= h(1- λ/P)
29Lets Try one: We work for Sam’s Active Suspensions They sell after market kits for car “Tooners”They have an annual demand of 650 unitsProduction rate is 4/day (working at 250 d/y)Setup takes 2 technicians working 45 and requires an expendable tool costing $25
30Continuing: Each kit costs $275 Sam’s uses MARR of 18%, tax at 3%, insurance at 2% and space cost of 1%Determine h, Q*, H, T and break T down to:T1 = production time in a cycle (Q*/P)T2 = non producing time in a cycle (T – T1)Engineering Teams: (You can) Do It
31Quantity Discount Models All Units Discounts: the discount is applied to ALL of the units in the order. Gives rise to an order cost function such as that pictured in Figure 4-9
35Properties of the Optimal Solutions For all units discounts, the optimal will occur at the bottom of one of the cost curves or at a breakpoint. (It is generally at a breakpoint.). One compares the cost at the largest realizable EOQ and all of the breakpoints beyond it. (See Figure 4-11).For incremental discounts, the optimal will always occur at a realizable EOQ value. Compare costs at all realizable EOQ’s. (See Figure 4-12).
37To Find EOQ in ‘All Units’ discount case: Compute Q* for each cost levelCheck for Feasibility (the Q computed is applicable to the range) – we say it is “Realizable”Compute G(Q*) for each of the realizable Q*’s and the break points.Chose Q* as the one that has lowest G(Q)
38Lets Try one:Product cost is $6.50 in orders <600, $3.50 above 600.Organizational I is 34%K is $300 and annual demand is 900
39Lets Try one: Order 674 at a time! Both of these are Realizable (the value is ‘in range’)Compute G(Q) for both and breakpoint (600)G(Q) = c + (*K)/Q + (h*Q)/2Order 674 at a time!
40Average Annual Cost Function for Incremental Discount Schedule
41In an Incremental Case: Cost is a strictly varying function of Q -- It varies by interval!Calculate a C(Q) for the applied scheduleDivide by Q to convert it to a “unit cost” functionBuild G(Q) equations for each intervalFind Q* from each G(Q) EquationCheck if “Realizable”Compute G(Q*) for realizable Q*’s
42Trying the previous problem (but as Incremental Case): Cost Function: Basically states that we pay 6.50 for each unit up to 600 then 3.50 for each unit ordered beyond 601:C(Q) = 6.5(Q), Q < 600C(Q) = 3.5(Q – 600) + 6.5*600, Q 600C(Q)/Q = 6.5, Q < 600 (order up to 600)C(Q)/Q = ((3900 – 2100)/Q), Q 600 orC(Q)/Q = (1800)/Q (orders beyond 601)
43Trying the previous but as Incremental Case: For the First Interval:Q* = [(2*300*900)/(.34*6.50)] = 495 (realizable)For order > 600, find Q* by writing a G(Q) equation and then optimizing:[G(Q) = c + (*K)/Q + (h*Q)/2]
45Now Compute G(Q) for both and “cusp” Lowest cost – purchase 1783 about every 2 years!
46Properties of the Optimal Solutions Lets jump back into our teams and do some!TRY 21b and 22b on Pg 211
47Resource Constrained Multi-Product Systems Consider an inventory system of n items in which the total amount available to spend is C and items cost respectively c1, c2, . . ., cn. Then this imposes the following constraint on the system:
48Resource Constrained Multi-Product Systems When the condition that:is met, the solution procedure is straightforward. If the condition is not met, one must use an iterative procedure involving Lagrange Multipliers.
49EOQ Models for Production Planning Consider n items with known demand rates, production rates, holding costs, and set-up costs. The objective is to produce each item once in a production cycle. For the problem to be feasible the following equation must be true:
50Issues:We are interested in controlling Family MAKESPAN (we wish to produce all products within our chosen cycle time)Underlying Assumptions:Setup Cost (times) are not Sequence Dependent (this assumption is not always accurate as we will later see)Plants uses a “Rotation” Policy that produces a single ‘batch’ of each product each cycle – a mixed line balance assumption
51EOQ Models for Production Planning The method of solution is to express the average annual cost function in terms of the cycle time, T. The optimal cycle time has the following mathematical form:We must assure that this time allows for all setups and of production times.
52Working forward: This last statement means: (sj+(Qj/Pj)) TOf course: Qj = j*TSo – with substitution: (sj+((j*T )/Pj) TOr: T((sj/(1- j/Pj)) = TminFinally, we must Choose T(actual cycle time) = MAX(T*,Tmin)
53Given: 20 days/month and 12 month/year; $85/hr for setup Lets Try Problem 30ITEMMon Req’rDaily Req’rh = .2*c/Ph’Setup TimeSetup CostUnit CostDaily Pr. RateMon. Pr. RateJ55R1256.254.0453.821.2$102$201402800H2237.0326.780.8$68$352204400K-18R452.250.6.0230.5862.2$187$121002000Z-344240129.0738.343.1$263.5$451653300Given: 20 days/month and 12 month/year; $85/hr for setup