1. Chapter 4 Inventory Control Subject to Known Demand
McGraw-Hill/Irwin

2. Reasons for Holding Inventories
Economies of Scale
Uncertainty in delivery leadtimes
Speculation. Changing Costs Over Time
Smoothing: seasonality, Bottlenecks
Demand Uncertainty
Costs of Maintaining Control System
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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?
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4. Characteristics of Inventory Systems
Treatment of Excess Demand.
Backorder all Excess Demand
Lose all excess demand
Backorder some and lose some
Inventory that changes over time
perishability
obsolescence
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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 % 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 20 – 30% of the inventory items sku’s) stored in small quantities and/or worth very little
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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 it!
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7. Relevant Costs a) Physical Cost of Space (3%)
Holding Costs - Costs proportional to the quantity of inventory held. Includes:
a) Physical Cost of Space (3%)
b) Taxes and Insurance (2 %)
c) Breakage Spoilage and Deterioration (1%)
*d) Opportunity Cost of alternative investment. (18%)
(Total: 24%)
h  .24*Cost of product
Note: Since inventory may be changing on a continuous basis, holding cost is proportional to the area under the inventory curve.
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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 = $
Average Annual Holding Cost
assumes an average monthly inventory of trucks based on on hand data
$3333
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9. Relevant Costs (continued)
Ordering Cost (or Production Cost).
Includes both fixed and variable components.
slope = c K
C(x) = K + cx for x > 0 and = 0 for x = 0.
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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 for lost demand
Costs of book-keeping for backordered demands
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11. Relevant Costs (continued)
Penalty or Shortage Costs. All costs that accrue when insufficient stock is available to meet demand. These include:
Loss of goodwill for being unable to satisfy demands when they occur.
Generally assume cost is proportional to number of units of excess demand.
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12. The Simple EOQ Model Assumptions:
1. Demand is fixed at l units per unit time.
2. Shortages are not allowed.
3. Orders are received instantaneously. (this will be relaxed later).
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13. Simple EOQ Model (cont.)
Assumptions (cont.):
4. Order quantity is fixed at Q per cycle. (can be proven optimal.)
5. Cost structure:
a) Fixed and marginal order costs (K + cx)
b) Holding cost at h per unit held per unit time.
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14. Inventory Levels for the EOQ Model
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15. The Average Annual Cost Function G(Q)
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16. Modeling Inventory:
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17. Subbing Q/ for T
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18. Finding an Optimal Level of ‘Q’ – the so-called EOQ
Take derivative of the G(Q) equation with respect to Q
Set derivative equals Zero:
Solve for Q
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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 = Ic)
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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 ans sells for $12.50/bx
Cost of ordering is $35
Determine EOQ (Q*)
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21. Plugging and chugging:
 = 145*52 = 7540
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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 trigger an order at a point before the delivery is needed to not have 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 units of lead time match units in T!
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23. But, Orders usually take time to arrive!
What happens when lead time exceeds T?
It is just as before (but we compute /T)
 is the lead time is similar units as T
Here, in weeks  = 6 weeks then:
/T = 6/3.545 = 1.69
Place order 1.69 cycles before we need product
Trip Point is then .69*Q* = .69*514 = 356 boxes
This trip point is not for the next stock out but the one after that (1.7 T from now!) – be very careful!!!
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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:
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25. Sensitivity
We find that this model is quite robust to Q errors if holding costs are relatively low
We find, given a Q
that Q* + Q has smaller error than Q* - Q
Error here mean storage costs penalty
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26. EOQ With Finite Production Rate
Suppose that items are produced internally at a rate P > λ. 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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27. This is based on solving
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28. Inventory Levels for Finite Production Rate Model
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29. Lets Try one: We work for Sam’s Active Suspensions
They sell after market kits for car “Pimpers”
They have an annual demand of 650 units
Production rate is 4/day (250 d/y)
Setup takes 2 techs working 45 and requires an expendible tool costing $25
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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)
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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
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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32. All-Units Discount Order Cost Function
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33. Incremental Discount Order Cost Function
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34. 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 succeeding 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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35. All-Units Discount Average Annual Cost Function
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36. To Find EOQ in ‘All Units’ discount case:
Compute Q* for each cost type
Check for Feasibility (the Q computed is applicable to the range) – “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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37. 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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38. 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!
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39. Average Annual Cost Function for Incremental Discount Schedule
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40. In an Incremental Case:
Cost is strictly a 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 Equation
Check if “Realizable”
Compute G(Q*) for realizable Q*’s
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41. Trying the previous but as Incremental Case:
Cost Function: Basically states that we pay 6.50 for each up to 600 then 3.50 for any more:
C(Q) = 6.5(Q), Q  600
C(Q) = 3.5(Q – 600) , Q > 600
C(Q)/Q = 6.5, Q  600
C(Q)/Q = (1800/Q), Q > 600
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42. Trying the previous but as Incremental Case:
For the First Interval: Q* = [(2*300*900)/(.34*6.50)] = 495 (realizable)
Finding Q* is a process of writing a G(Q) equation for this range and then differentiation :
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43. Differentiating G2(Q)
Realizable!
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44. Now Compute G(Q) for both and “cusp”
Lowest cost – purchase 1783 about every 2 years!
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45. Properties of the Optimal Solutions
Lets jump back into our teams and do some!
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46. 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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47. 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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48. 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 we must have that
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49. Issues:
We are interested in the Family MAKESPAN (we wish to produce all products within the chosen cycle time)
Underlying Assumptions:
Setup Cost are not Sequence Dependent (this assumption is not accurate as we will later see)
Plant uses a “Rotation” Policy that produces a single ‘batch’ of each product each cycle
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50. 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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51. Working forward: This last statement means:
(sj+(Qj/Pj)  T
Since: Qj = j*T
Subbing:
(sj+((j*T )/Pj)  T
T(sj/(1- j/Pj) = Tmin
We must Choose T(planned cycle time) = MAX(T*,Tmin)
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52. Given: 20 days/month and 12 month/year; $85/hr for setup
Lets Try Problem 30
ITEM
Mon Reqr
Daily Reqr
h = .2*c
/P h’
Setup Time
Setup Cost
U. Cost
Daily Pr. Rate
Mon. Pr. Rate
J55R
125
6.25 4
.045
3.82
1.2
$120
$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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53. Compute:
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54. Lets do a ‘QUICK’ Exploration of Stochastic Inventory Control (Ch 5)
We will examine underlying ideas –
We base our approaches on Probability Density Functions (means & std. Deviations)
We are concerned with two competing ideas: Q and R
Q (as earlier) an order quantity and R a stochastic estimate of reordering time and level
Finally we are concerned with Servicing ideas – how often can we supply vs. not supply a demand (adds stockout costs to simple EOQ models)
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55. The Nature of Uncertainty
Suppose that we represent demand as
D = Ddeterministic + Drandom
If the random component is small compared to the deterministic component, the models of chapter 4 will be accurate. If not, randomness must be explicitly accounted for in the model.
In this chapter, assume that demand is a random variable with cumulative probability distribution F(t) and probability density function f(t).
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56. The Newsboy Model
At the start of each day, a newsboy must decide on the number of papers to purchase. Daily sales cannot be predicted exactly, and are represented by the random variable, D.
Costs: co = unit cost of overage
cu = unit cost of underage
It can be shown(*see over) that the optimal number of papers to purchase is the fractile of the demand distribution given by F(Q*) = cu / (cu + co).
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57. Determination of the Optimal Order Quantity for Newsboy Example
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58. Computing the Critical Fractile:
We wish to minimize competing costs (Co & Cu):
G(Q,D) = Co*MAX(0, Q-D) + Cu*MAX(0, D-Q) ---- D is actual (potential) Demand
G(Q) = E(G(Q,D)) ---- expected value
Therefore:
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59. Applying Leibniz’s Rule:
d(G(Q))/dQ = CoF(Q) – Cu(1 – F(Q))
F(Q) is a cumulative Prob. Density Function (as earlier)
G’(Q*) = (Cu)/(Co + Cu)
This is the critical fractile
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60. Lets see about this: Prob 5 pg 241
Observed sales given as a number purchased during a week (grouped)
Lets assume some data was supplied:
Make Cost: $1.25
Selling Price: $3.50
Salvageable Parts: $0.80
Co = overage cost = $ $0.80 = $0.45
Cu = underage cost = $ $1.25 = $2.25
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61. Continuing: Compute Critical Ratio:
CR = Cu/(Co + Cu) = 2.25/( ) = .8333
If we assume a continuous Pr. D. Function (lets choose a normal function):
Z(CR)  when F(Z) = (from Std. Normal Tables!)
Z = (Q* - )/ (we compute mean = 9856; StDev = )
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62. Continuing: Q* = Z +  = 4813.5*.967 + 9856 = 14511
Our best guess economic order quantity is 14511
We really should have done it as a Discrete problem
Taking this approach we would find that Q* is only 12898
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63. Lot Size Reorder Point Systems
Assumptions
Inventory levels are reviewed continuously (the level of on-hand inventory is known at all times)
Demand is random but the mean and variance of demand are constant. (stationary demand)
There is a positive leadtime, τ. This is the time that elapses from the time an order is placed until it arrives.
The costs are:
Set-up each time an order is placed at $K per order
Unit order cost at $c for each unit ordered
Holding at $h per unit held per unit time ( i. e., per year)
Penalty cost of $p per unit of unsatisfied demand
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64. Describing Demand
The response time of the system in this case is the time that elapses from the point an order is placed until it arrives. Hence, the uncertainty that must be protected against is the uncertainty of demand during the lead time. We assume that D represents the demand during the lead time and has probability distribution F(t). Although the theory applies to any form of F(t), we assume that it follows a normal distribution for calculation purposes.
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65. Decision Variables
For the basic EOQ model discussed in Chapter 4, there was only the single decision variable Q. The value of the reorder level, R, was determined by Q. In this case, we treat Q and R as independent decision variables. Essentially, R is chosen to protect against uncertainty of demand during the lead time, and Q is chosen to balance the holding and set-up costs. (Refer to Figure 5-5)
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66. Changes in Inventory Over Time for Continuous-Review (Q, R) System
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67. The Cost Function The average annual cost is given by:
Interpret n(R) as the expected number of stockouts per cycle given by the loss integral formula. The standardized loss integral values appear in Table A-4. The optimal values of (Q,R) that minimizes G(Q,R) can be shown to be:
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68. Solution Procedure
The optimal solution procedure requires iterating between the two equations for Q and R until convergence occurs (which is generally quite fast). A cost effective approximation is to set Q=EOQ and find R from the second equation. (A slightly better approximation is to set Q = max(EOQ,σ) where σ is the standard deviation of lead time demand when demand variance is high).
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69. Service Levels in (Q,R) Systems
In many circumstances, the penalty cost, p, is difficult to estimate. For this reason, it is common business practice to set inventory levels to meet a specified service objective instead. The two most common service objectives are:
Type 1 service: Choose R so that the probability of not stocking out in the lead time is equal to a specified value.
Type 2 service. Choose both Q and R so that the proportion of demands satisfied from stock equals a specified value.
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70. Computations
For type 1 service, if the desired service level is α then one finds R from F(R)= α and Q=EOQ.
Type 2 service requires a complex interative solution procedure to find the best Q and R. However, setting Q=EOQ and finding R to satisfy n(R) = (1-β)Q (which requires Table A-4) will generally give good results.
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71. Comparison of Service Objectives
Although the calculations are far easier for type 1 service, type 2 service is generally the accepted definition of service. Note that type 1 service might be referred to as lead time service, and type 2 service is generally referred to as the fill rate. Refer to the example in section 5-5 to see the difference between these objectives in practice (on the next slide).
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72. Comparison (continued)
Order Cycle Demand Stock-Outs
For a type 1 service objective there are two cycles out of ten in which a stockout occurs, so the type 1 service level is 80%. For type 2 service, there are a total of 1,450 units demand and 55 stockouts (which means that 1,395 demand are satisfied). This translates to a 96% fill rate.
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73. (s, S) Policies
The (Q,R) policy is appropriate when inventory levels are reviewed continuously. In the case of periodic review, a slight alteration of this policy is required. Define two levels, s < S, and let u be the starting inventory at the beginning of a period. Then
(In general, computing the optimal values of s and S is much more difficult than computing Q and R.)
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