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**Inventory Control Models**

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

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**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

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**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?

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**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

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**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 % 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

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**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!

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**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

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**Lets 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 Cost assumes an average monthly inventory of trucks based on on hand data $3333

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**Relevant Costs (continued)**

Ordering Cost (or Production Cost). Includes both fixed and variable components slope = c K C(x) = K + cx for x > 0; 0 for x = 0.

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**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.

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**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.

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**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).

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**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.

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**Inventory Levels for the EOQ Model**

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**The Average Annual Cost Function G(Q)**

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Modeling Inventory:

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Subbing Q/ for T

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**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

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**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)

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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

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**Plugging and chugging:**

= 145*52 = 7540 In your teams: Compute Pr. 10, pg 201

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**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!

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**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 from now!) – be very careful!!!

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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:

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Sensitivity We find that this model is quite robust to Q errors if holding costs are relatively low We find, given a Q – error in ordering quantity that 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

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**Inventory Levels for Finite Production Rate Model**

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**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)

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**This is based on solving:**

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**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 and requires an expendable tool costing $25

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**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

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**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

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**All-Units Discount Order Cost Function**

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**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.

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**Incremental Discount Order Cost Function**

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**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).

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**All-Units Discount Average Annual Cost Function**

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**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)

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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

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**Lets 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)/2 Order 674 at a time!

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**Average Annual Cost Function for Incremental Discount Schedule**

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**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 Check if “Realizable” Compute G(Q*) for realizable Q*’s

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**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 = ((3900 – 2100)/Q), Q 600 or C(Q)/Q = (1800)/Q (orders beyond 601)

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**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]

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Differentiating G2(Q) Realizable!

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**Now Compute G(Q) for both and “cusp”**

Lowest cost – purchase 1783 about every 2 years!

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**Properties of the Optimal Solutions**

Lets jump back into our teams and do some! TRY 21b and 22b on Pg 211

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**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:

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**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.

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**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:

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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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**EOQ 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.

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**Working forward: This last statement means:**

(sj+(Qj/Pj)) T Of course: Qj = j*T So – with substitution: (sj+((j*T )/Pj) T Or: T((sj/(1- j/Pj)) = Tmin Finally, we must Choose T(actual cycle time) = MAX(T*,Tmin)

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**Given: 20 days/month and 12 month/year; $85/hr for setup**

Lets Try Problem 30 ITEM Mon Req’r Daily Req’r h = .2*c /P h’ Setup Time Setup Cost Unit Cost Daily Pr. Rate Mon. Pr. Rate J55R 125 6.25 4 .045 3.82 1.2 $102 $20 140 2800 H223 7 .032 6.78 0.8 $68 $35 220 4400 K-18R 45 2.25 0.6 .023 0.586 2.2 $187 $12 100 2000 Z-344 240 12 9 .073 8.34 3.1 $263.5 $45 165 3300 Given: 20 days/month and 12 month/year; $85/hr for setup

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**Compute the Following – in teams!:**

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