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
Demand Uncertainty
Costs of Maintaining Control System
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 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


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



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
Holding Costs - Costs proportional to
the quantity of inventory held.
Includes:


1.
2.
3.
4.

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:



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)

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.
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 (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!

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.)

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.
Inventory Levels for the EOQ
Model
The Average Annual
Cost Function G(Q)
Modeling Inventory:
K  cQ 

G(Q) 
T
hQ 


2
K : setup cost
c: Unit cost

Q
T: cycle length T=
l

h: holding cost  % of unit cost 
Subbing Q/l for T


K

cQ


hQ


G (Q) 

 Q
 2
l 

Kl
hQ
G (Q) 
 lc 
Q
2
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:
Kl h
G (Q) 
 0
2
Q
2
'

Now, Solve for Q
Properties of the EOQ
(optimal) Solution
2K l
Q
h



Q is increasing with both K and l 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!






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:


2 * 35 * 7540
 513.7 514
h = $8*.25 = $2 Q 
2
l = 145*52 =
*
Q
T
 514
 .068 yr
7540
l
7540
or: 514
 3.54 wk
145
*
Annual Inventory Cost:
Then, In your
teams: Compute
Pr. 10, pg 201
35  7540 514  2
G Q  

 7540  8
514
2
G  Q   513.42  514  60320  $61347.42
Cost of Ordering & Holding:
513.42 + 514 = $1027.42
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
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
Let G(Q) be the average annual holding and setup cost function given by
G (Q)  K l / Q  hQ / 2
and let G* be the optimal average annual cost.
Then it can be shown that:
G (Q) 1  Q * Q 
 


G*
2  Q Q *
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
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:
2k l
Q
h'

Except that h’ is defined as h’= h(1- λ/P)
This is based on solving:
K hH
G (Q)  
T
2
K l  hQ

G (Q) 

 1  l / P 
Q
2
P is annual production rate
H is maximum on hand quantity

H  Q  1 l
P

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:



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

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
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
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
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 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
Q 
 2  300  900 
Q 
 2  300  900 
*
1
*
2
 0.34  6.50 
 495
 0.34  3.50 
 674
Lets Try one:



Both of these are Realizable (the value is ‘in
range’)
Compute G(Q) for both and breakpoint (600)
G(Q) = cl + (l*K)/Q + (h*Q)/2
G (495)  900  (6.50) 
 900  300 
495

 .34  6.5 495

 .34  3.5 674 
2
G (495)  $6942.43
G (674)  900  (3.50) 
 900  300 
674
2
G (674)  $3951.62
G (600)  900  (6.50) 
G (600)  $3957.00
 900  300 
600

 .34  6.5 600 
2
Order 674
at a time!
Average Annual Cost Function
for Incremental Discount Schedule
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):

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:

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) = cl + (l*K)/Q + (h*Q)/2]


 1800     300  900 
 1800    Q2
G2 (Q2* )  900   3.5  


0.34

3.5



 2



Q
Q
Q
2 
2
2 




 900  2100   Q   .34  3.5   .34  900
G2 (Q2* )  900  3.5 
2 
Q2
2 

Differentiating G2(Q)
d  G2  Q2  
   900  2100   .34  3.5 
0
0

2
dQ2
Q2
2


Set equal to 0 and solve for Q2
 900  2100 
2
2
Q

2
Q 
 0.595  0
Realizable!
 900  2100 
0.595
 1783
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 Solutions
Lets jump back into our
teams and do some!
TRY 22b and 23 on Pg 211
Resource Constrained MultiProduct 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:
c1Q1  c2Q2  ...  cnQn  C
Resource Constrained MultiProduct Systems

When the condition that:
c1 / h1  c2 / h2  ...  cn / hn

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

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:
n
lj
P
j 1
j
1
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
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:
n
T* 
2 K j
j 1
n
h
j 1

j
' l j
We must assure that this time allows for all setups
and all production times.
Working forward:

This last statement means:






(sj+(Qj/Pj))  T
sj is set up time
Of course: Qj = lj*T
So – with substitution: (sj+((lj*T )/Pj)  T
Or: T((sj/(1- lj/Pj)) = Tmin
Finally, we must Choose
T(chosen cycle time) = MAX(T*,Tmin)
Lets Try Problem 30
ITEM
Mon
Req’r
Daily
Req’r
h=
.2*c
l/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
140
7
7
.032
6.78
0.8
$68
$35
220
4400
K-18R
45
2.25
0.6
.023
0.58
6
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
Compute the Following – in
teams!:
n


2Kj 
j 1



T  n
  l j  h'j 
j 1
n
TMin 
s
j 1
j
  n lj
1   

  j 1 Pj
Topt  T  , TMin 

 

MAX
Lot Size: Topt  Demand Rate
Uptime; Drawdown Time; Utilization