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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

Inventory Control Models

Ch 4 (Known Demands)

R. R. Lindeke

IE 3265, Production And Operations Management


Reasons for holding inventories
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


Characteristics of inventory systems
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?


Characteristics of inventory systems1
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


  • Real inventory systems abc ideas
    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


    Real inventory systems abc ideas and control
    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!


    Relevant inventory costs
    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


    Lets try one
    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


    Relevant costs continued
    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.


    Relevant costs continued1
    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.


    Relevant costs continued2
    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.


    The simple eoq model the most fundamental of all
    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).


    Simple eoq model cont
    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.



    The average annual cost function g q
    The Average Annual Cost Function G(Q)



    Subbing q for t
    Subbing Q/ for T


    Finding an optimal level of q the so called eoq
    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


    Properties of the eoq optimal solution
    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)


    Try one
    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


    Plugging and chugging
    Plugging and chugging:

    • h = $8*.25 = $2

    •  = 145*52 = 7540

    Then, In your teams: Compute Pr. 10, pg 201


    But orders usually take time to arrive
    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!


    But orders usually take time to arrive1
    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!!!


    Sensitivity analysis
    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:


    Sensitivity
    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


    Inventory levels for finite production rate model
    Inventory Levels for Finite Production Rate Model


    Eoq with finite production rate
    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)



    Lets try one1
    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


    Continuing
    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


    Quantity discount models
    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


    All units discount order cost function
    All-Units Discount Order Cost Function


    Quantity discount models1
    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.


    Incremental discount order cost function
    Incremental Discount Order Cost Function


    Properties of the optimal solutions
    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).


    All units discount average annual cost function
    All-Units Discount Average Annual Cost Function


    To find eoq in all units discount case
    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)


    Lets try one2
    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


    Lets try one3
    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!


    Average annual cost function for incremental discount schedule
    Average Annual Cost Function for Incremental Discount Schedule


    In an incremental case
    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


    Trying the previous problem but as incremental case
    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)


    Trying the previous but as incremental case
    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]


    Differentiating g 2 q
    Differentiating G2(Q)

    Realizable!


    Now compute g q for both and cusp
    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!


    Properties of the optimal solutions1

    Properties of the Optimal Solutions

    Lets jump back into our teams and do some!

    TRY 22b and 23 on Pg 211


    Resource constrained multi product systems
    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:


    Resource constrained multi product systems1
    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.


    Eoq models for production planning
    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:


    Issues
    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


    Eoq models for production planning1
    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 all production times.


    Working forward
    Working forward:

    • This last statement means:

      • (sj+(Qj/Pj))  T

      • sj is set up time

      • 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(chosen cycle time) = MAX(T*,Tmin)


    Lets try problem 30
    Lets Try Problem 30

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



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