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Inventory Control Models

Inventory Control Models. Ch 4 (Known Demands) R. R. Lindeke IE 3265, Production And Operations Management. Reasons for Holding Inventories. Economies of Scale Uncertainty in delivery lead-times Speculation. Changing Costs Over Time Smoothing: to account for seasonality and/or Bottlenecks

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Inventory Control Models

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  1. Inventory Control Models Ch 4 (Known Demands) R. R. Lindeke IE 3265, Production And Operations Management

  2. Reasons for Holding Inventories • Economies of Scale • Uncertainty in delivery lead-times • Speculation. Changing Costs Over Time • Smoothing: to account for seasonality and/or Bottlenecks • Demand Uncertainty • Costs of Maintaining Control System

  3. Characteristics of Inventory Systems • Demand • May Be Known or Uncertain • May be Changing or Unchanging in Time • Lead 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?

  4. Characteristics of Inventory Systems • Treatment of Excess Demand. • Backorder all Excess Demand • Lose all excess demand • Backorder some and lose some • Inventory who’s quality changes over time • perishability • obsolescence

  5. Real 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 10 - 20% of sku’s) that represent up to 80% of the inventory value • B Items (the 20 – 30%) of the inventory items that represent nearly all the remaining worth • C Items the remaining 50 – 70% of the inventory items sku’s) stored in small quantities and/or worth very little

  6. Real Inventory Systems: ABC ideas and Control • A Items must be well studied and controlled to minimize expense • C Items tend to be overstocked to ensure no runouts but require only occasional review • See mhia.org – there is an “e-lesson” on the principles of ABC Inventory management – check it out! – do the on-line lesson!

  7. Relevant 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

  8. Lets Try one: • Problem 4, page 193 – cost of inventory • Find h first (yearly and monthly) • Total holding cost for the given period: • THC = $26666.67 • Average Annual Holding Cost • assumes an average monthly inventory of trucks based on on hand data • $3333

  9. Relevant Costs (continued) • Ordering Cost (or Production Cost). Includes both fixed and variable components slope = c KC(x) = K + cx for x > 0; 0 for x = 0.

  10. Relevant 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 demand • Costs of book-keeping for backordered demands • Loss of goodwill for being unable to satisfy demands when they occur.

  11. Relevant 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.

  12. The 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).

  13. Simple 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.

  14. Inventory Levels for the EOQ Model

  15. The Average Annual Cost Function G(Q)

  16. Modeling Inventory:

  17. Subbing Q/ for T

  18. Finding an Optimal Level of ‘Q’ – the so-called EOQ • Requires us to take derivative of the G(Q) equation with respect to Q • Then, Set derivative equal to Zero: • Now, Solve for Q

  19. Properties of the EOQ (optimal) Solution • Q is increasing with both K and  and decreasing with h • Q changes as the square root of these quantities • Q is independent of the proportional order cost, c. (except as it relates to the value of h = I*c)

  20. Try ONE! • A company sells 145 boxes of BlueMountain BobBons/week (a candy) • Over the past several months, the demand has been steady • The store uses 25% as a ‘holding factor’ • Candy costs $8/bx and sells for $12.50/bx • Cost of making an order is $35 • Determine EOQ (Q*) and how often an order should be placed

  21. Plugging and chugging: • h = $8*.25 = $2 •  = 145*52 = 7540 Then, In your teams: Compute Pr. 10, pg 201

  22. But, Orders usually take time to arrive! • This is a realistic relaxation of the EOQ ideas – but it doesn’t change the model • This 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 outs • In 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!

  23. But, 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 T • Here, lets assume  = 6 weeks then: • /T = 6/3.545 = 1.69 – in the Blue Mount Bon-Bon case • Place order: 1.69 cycles before we need product! • Trip Point is then 0.69*Q* = .69*514 = 356 boxes • This trip point is not for the next stock out but the one after that (1.69 T or 6 weeks from now!)– be very careful!!!

  24. Sensitivity Analysis Let G(Q) be the average annual holding and set-up cost function given by and let G* be the optimal average annual cost. Then it can be shown that:

  25. Sensitivity • We find that this model is quite robust to Q errors if holding costs are relatively low • We find, for a given Q – a mistake in the ordering quantity • that Q* +Q has smaller error than Q* - Q (Error means we incur extra inventory maintenance costs – a penalty cost) • That is, we tend to have a smaller penalty cost if we order too much compared to too little

  26. Inventory Levels for Finite Production Rate Model

  27. EOQ 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)

  28. This is based on solving:

  29. Lets Try one: • We work for Sam’s Active Suspensions • They sell after market kits for car “Tooners” • They have an annual demand of 650 units • Production rate is 4/day (working at 250 d/y) • Setup takes 2 technicians working 45 minutes @$21/hour and requires an expendable tool costing $25

  30. Continuing: • 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

  31. Quantity 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

  32. All-Units Discount Order Cost Function

  33. Quantity Discount Models • Incremental Discounts: the discount is applied only to the number of units above the breakpoint. Gives rise to an order cost function such as that pictured in Figure 4-10.

  34. Incremental Discount Order Cost Function

  35. Properties 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).

  36. All-Units Discount Average Annual Cost Function

  37. To Find EOQ in ‘All Units’ discount case: • Compute Q* for each cost level • Check 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)

  38. Lets 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

  39. Lets Try one: • 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)/2 Order 674 at a time!

  40. Average Annual Cost Function for Incremental Discount Schedule

  41. In an Incremental Case: • Cost is a strictly varying function of Q -- It varies by interval! • Calculate a C(Q) for the applied schedule • Divide by Q to convert it to a “unit cost” function • Build G(Q) equations for each interval • Find Q* from each G(Q) Equation (derivative!) • Check if “Realizable” • Compute G(Q*) for realizable Q*’s

  42. Trying 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 < 600 • C(Q) = 3.5(Q – 600) + 6.5*600, Q  600 • C(Q)/Q = 6.5, Q < 600(order up to 600) • C(Q)/Q = 3.5 + ((3900 – 2100)/Q), Q  600 or • C(Q)/Q = 3.5 + (1800)/Q(orders beyond 601)

  43. Trying 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]

  44. Differentiating G2(Q) Realizable!

  45. Now Compute G(Q) for both and “cusp” • G(495) = 900*6.5 + (300*900)/495 + .34((6.5*495)/2) = $6942.43 • G(600) = 900*6.5 +(300*900/600) + .34((6.5*600)/2) = $6963.00 • G(1763) = 900*(3.5 +(1800/1783)) + (300*900)/1783 + .34*(3.5 +(1800/1783))*(1783/2) = $5590.67 Lowest cost – purchase 1783 about every 2 years!

  46. Properties of the Optimal Solutions Lets jump back into our teams and do some! TRY 22b and 23 on Pg 211

  47. Resource 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:

  48. Resource 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.

  49. EOQ 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:

  50. Issues: • 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

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