Chapter 4
Inventory Control Subject to
Known Demand
McGraw-Hill/Irwin
Copyright © 2005 by The McGraw-Hill Companies, Inc. All rights reserved.
Breakdown of the Total
Investment in Inventories in the U.S. Economy (1999)
4-2
4-3
Reasons for Holding Inventories






Economies of Scale
Uncertainty in delivery leadtimes
Speculation. Changing Costs Over Time
Smoothing.
Demand Uncertainty
Costs of Maintaining Control System
4-4
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?


Treatment of Excess Demand.
 Backorder all Excess Demand
 Lose all excess demand
 Backorder some and lose some
Inventory that changes over time
 perishability
 obsolescence
4-5
Relevant Costs

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%)
Note: Since inventory may be changing on a continuous
basis, holding cost is proportional to the area under the
inventory curve. (See examples.)
Inventory as a Function of Time
4-6
4-7
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.
4-8
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 bookeeping for backordered demands
Loss of goodwill for being unable to satisfy
demands when they occur.
Generally assume cost is proportional to
number of units of excess demand.
4-9
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).
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.
Inventory Levels for the EOQ Model
Saw structure is typical. First order when inventory is 0.
Reordering Q everytime when inventory is 0 must be optimal
4-10
4-11
The EOQ Model: Notation
D
= λ is the demand rate (in units per year)
c
= unit production cost, not counting setup or inventory costs
(in dollars per unit)
K
= setup costs (per placed order) in dollars
h
= holding cost (in dollars per unit per year), if the holding cost
consists entirely of interest on money tied up in inventory,
h = ic, where i – is an annual interest rate
Q
= lot size (order size) in units
T
= time between orders (cycle length)
G(Q) = average annual cost
T
Q
l
Q K cQ
G Q   h  
2 T
T
Inventory (I(t))
4-12
Relationships

Ordering Costs: (Order amount Q)
Assume Constant
Demand
slope = -l
Q
C(Q) = K + cQ
Time (t)
T
Instantaneous
Time between orders
Replenishment

Holding Cost:
h = Ic =(Interest Rate)(Cost of Inv.)

Average Inventory Size?
Under constant demand: Q/2

Time Between Orders:
l  Q/T
T = Q/l
Q
Rate of consumption
l
T
4-13
Total Costs

What is the average annual cost?
G(Q) = average order cost + average holding
cost
K  cQ hQ
G(Q) 

T
2
Average ordering
cost per time T
Average inventory
level at any time
4-14
Total Costs

What is the average annual cost?
K  cQ hQ
G (Q) 

T
2
K  cQ hQ


Q
2
l
hQ Kl


 lc
2
Q
The Average Annual Cost Function G(Q)
GQ   h
Q K cQ hQ Kl
 


 cl
2 T
T
2
Q
4-15
4-16
The Average Annual
Cost Function G(Q)
Minimize Annual Costs
4-17
GQ  non linear function of Q

Take the derivative of G(Q)
Kl
hQ
G (Q) 
 lc 
Q
2
 Kl h
G(Q) 

2
Q
2

Is this a minimum?
2Kl
G (Q)  3  0,Q  0
Q

EOQ:
YES!
 Kl h
2Kl
 0  Q* 
2 
Q
2
h
4-18
4-19
Properties of the EOQ 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 = Ic)
Properties of the EOQ Solution
hQ Kl
GQ  

 cl
2
Q




2 Kl
Q 
h

This formula is well-known economic order quantity, is also
known as economic lot size
This is a tradeoff between lot size and inventory
“Garbage in, garbage out” - usefulness of the EOQ formula for
computational purposes depends on the realism of input data
Estimating setup cost is not easily reduced to a single invariant
cost K
4-20
4-21
Example


Uvic requires 3600 gallons of paint annually for
scheduled maintenance of buildings. Cost of
placing an order is $16 and the interest rate
(annual) is 25%. Price of paint is $8 per gallon.
How much paint should be ordered, and how often?
2 Kl
2(16)(3600)
Q* 

 57,600  240
h
.25(8)
Q
240
 .07 years* (250 workingdays/year)
T 
l 3600
= 17.5 workingdays = 18 days
Order Point for the EOQ Model
4-22
Does it matter if
τ < T or τ > T ?
Keep track of
time left to zero
inventory or set
automatic reorder
at a particular
inventory level, R.
τ
τ
τ
τ
R = λ*τ, if τ < T
R = λ*MOD(τ/T),
if τ > T
Assumption: Delivery is immediate
There is no time lag between production and availability to satisfy
demand
Relax this assumption! Let the order lead time to be equal to τ
Sensitivity Analysis
Let G(Q) be the average annual
holding and set-up cost
function given by
hQ Kl
G Q  

2
Q
G  2hKl
4-23
independent
of Q
GQ  
hQ Kl

 cl
2
Q
Holding & Setup costs
2 Kl
Q 
h

6.000
5.000
and let G* be the optimal
average annual holding and
setup cost. Then it can be
shown that:
G (Q) 1  Q * Q 
 

G*
2  Q Q * 
4.000
3.000
2.000
1.000
0.000
0
2
4
6
8
10
Cost penalties are quite small
Finite Replenishment Rate:
Economic Production Quantity (EPQ)
4-24
Assumptions for EOQ:
Production is instantaneous
There is no capacity constraint, and entire lot is produced simultaneously
Delivery is immediate
There is no time lag between production and availability to satisfy demand
Example:
Parts produced at the same factory –
production rate is P (P > λ),
arriving continuously.
Inventory Levels for Finite
Production Rate Model
4-25
The EPQ Model: Notation
D
c
K
h
Q
T
= λ is the demand rate (in units per year)
= unit production cost, not counting setup or inventory
costs (in dollars per unit)
= setup costs (per placed order) in dollars
= holding cost (in dollars per unit per year), if the
holding
cost consists entirely of interest on money tied up in
inventory, h=ic, where i is an annual interest rate
= size of each production run (order) in units
= time between initiation of orders arrival (cycle length)
T1 = production (replenishment) time
T2 = downtime
T  T1  T2
H = maximum on-hand inventory
G(Q) = average annual setup & holding cost
4-26
4-27
The EPQ Model: Formula
T  T1  T2
H K
G Q   h 
2 T
Q
T1 
P
Q Q
T2  T  T1  
l P
Pl
H  P  l T1 
Q
P
2 Kl
H K
h  l  Kl
 l


Q

,
where
h

h
G Q   h   Q 1   
1  
H
 slope  P  l
T1
2
For EOQ:
T
2
P
h
Q
hQ Kl
G Q  

2
Q
2 Kl
Q 
h


P
Quantity Discount Models
4-28

One of the most severe assumptions: the unit variable cost c did not
depend on the replenishment quantity

In practice: quantity discounts exist based on the purchase price or
transportation costs – take advantage of these can result in
substantial savings

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 in Ch. 4.7

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
All-Units Discount Order Cost Function
C 499  $149.70
C (500)  $145.00
C (516)  $149.64
0.30Q

C Q   0.29Q
0.28Q

for 0  Q  500
for 500  Q  1,000
for 1,000  Q
4-29
All-Units Discount
Average Annual Cost Function
G(Q)
0.30Q

C Q   0.29Q
0.28Q

4-30
for 0  Q  500
for 500  Q  1,000
for 1,000  Q
G0(Q)
G1(Q)
G2(Q)
Gmin(Q)
500
1,000
Q
Incremental Discount Order Cost Function
for 0  Q  500
0.30Q

C Q   150  0.29Q  500  5  0.29Q
for 500  Q  1,000
295  0.28Q  1,000  15  0.28Q for 1,000  Q

4-31
4-32
Average Annual Cost Function
for Incremental Discount Schedule
Properties of the Optimal Solutions

For all units discounts, the optimal will occur
at the minimum point of one of the cost
curves or at a discontinuity point


One compares the cost at the largest realizable
EOQ and all of the breakpoints succeeding it
For incremental discounts, the optimal will
always occur at a realizable EOQ value.

Compare costs at all realizable EOQ’s.
4-33
4-34
Example

Supplier of paint to the maintenance department
has announced new pricing:
$8 per gallon if order is < 300 gallons
$6 per gallon if order is ≥ 300 gallons

Data remains as before:


K = 16, I = 25%, l = 3600
Is this a case of all units or incremental
discount?
4-35
Solution

Step 1: For Price 1:
(1)
Q 
2Kl

Ic1
2(16)(3600)
 240 gallons
(.25)(8)

Step 2: As Q(1) < 300, EOQ is realizable.

Step 3: Price 2:

2Kl
2(16)(3600)
Q 

 277 gallons
Ic2
(.25)(6)
Step 4: As Q(2) < 300, EOQ is not realizable.
(2)
4-36
Cost Function
C(Q)
Realizable
G(Q|p1)
G(Q|p2)
Not
Realizable
240 277
300
Q
4-37
C(Q)
Cost Function
Only possible solutions G(Q|p )
1
G(Q|p2)
240 277
300
Q
4-38
Solution

Step 5: Compare costs of possible
solutions.
lK Ic jQ
G(Q)  lc j 

Q
2

For $8 price, Q=240:
(3600)(16) (.25)(8)(240)
G(240)  (3600)(8) 

 $29,280 per year
240
2

For $6 price, Q=300:
(3600)(16) (.25)(6)(300)
G(300)  (3600)(6) 

 $22,017 per year
300
2

Q=300 is the optimal quantity.
Q 300
1
Q*  300 and T  

year
l 3600 12
Resource Constrained
Multi-Product Systems
Classic EOQ model is for a single item. Setup plan for n items.
Option A:
Treat one system with multiple items as multiple
systems with one item
Works if:
There are no interactions among items, such as
sharing common resources – budget, storage
capacity, or both
Option B:
Modify classic EOQ to insure no violation of the
resource constraints
Works if:
Have not made any mistakes and know how to use
Lagrange multipliers 
4-39
Resource Constrained
Multi-Product Systems
4-40
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 budget constraint on the system
n
, where Qi is the order size for product i
c Q
i 1
i
i
n
w Q
i
i 1
i
W
, where wi is the volume occupied by product i
Q Kl
For EOQ: G Q   h 
2 Q
Minimize
s.t.
C
 Qi K i li 

GQ1 ,..., Qn     hi

2
Qi 
i 1 
and n
 Qi K i li 

GQ1 ,..., Qn     hi

2
Qi 
i 1 
n
n
 wi Qi  W
 ciQi  C
i 1
i 1
n
Resource Constrained
Multi-Product Systems
4-41
 Qi K i li 

Minimize GQ1 ,..., Qn     hi

2
Qi 
i 1 
n
Budget constraint
 ciQi  C
n
s.t.
i 1
n
w Q
i 1
i
i
Space constraint
W
Lagrange multipliers method: relax one or more constraints
Minimize
n
n
 Qi K i li 






GQ1 ,..., Qn , 1 , 2     hi 
 1  C   ci Qi   2 W   wi Qi 

2
Qi 
i 1 
i 1
i 1




n
G
G

0
,
 0 for i  1,..., n; j  1,2
by solving necessary conditions:
Qi
 j
Resource Constrained Multi-Product Systems:
Steps to Find Optimal Solution
4-42
Single constraint:
1.
Solve the unconstrained problem. If constraint is satisfied, this
solution is the optimal one.
2.
If the constraint is violated, rewrite objective function using Lagrange
multipliers
3.
Obtain optimal Qi* by solving (n+1) equations
G
G
 0,
 0 for i  1,..., n;
Qi

Resource Constrained Multi-Product Systems:
Steps to Find Optimal Solution
4-43
Double constraints:
1.
Solve the unconstrained problem. If both constraints are satisfied, this
solution is the optimal one.
2.
Otherwise rewrite objective function using Lagrange multipliers by including
one of the constraints, say budget, and solve one-constraint problem to find
optimal solution. If the space constraint is satisfied, this solution is the
optimal one.
3.
Otherwise repeat the process for the only space constraint.
4.
If both single-constraint solutions do not yield the optimal solution, then
both constraints are active, and the Lagrange equation with both
constraints must be solved.
n
n
 Qi K i li 






GQ1 ,..., Qn , 1 , 2     hi 
 1  C   ci Qi   2 W   wi Qi 

2
Qi 
i 1 
i 1
i 1




n
5.
Obtain optimal Qi* by solving (n+2) equations
G
G
 0,
 0 for i  1,..., n; j  1,2
Qi
 j
EOQ Models for Production Planning
4-44
Problem: determine optimal procedure for producing n products on a
single machine
Consider n items with known demand rates
, production rates
,
l j is to minimize the
holding costs , and set-up costs . The objective
Pj cost of holding and
h j setups, and to have noKstock-outs.
For the problem
j
to be feasible we must have that
n
j
l
P
j 1
 1.
j
Assumption: rotation cycle policy – exactly one setup for each product in
each cycle; production sequence stays the same in each next cycle
4-45
The method of solution is to express the average
annual cost function in terms of the cycle time, T to
assure no stock-outs. The optimal cycle time has the
following mathematical form, where sj is a setup time
n
n
T* 
2 K j
j 1
n
h
j 1
j
'lj
Tmin 
s
j 1
j
n
lj
j 1
Pj
1 
And the optimal production quantities are given by:

Q j  l jT , where T = max {T*, Tmin},
see pp.216-217
Homework:

Read Ch. 4


Problems
4.5, 4.12, 4.15, 4.16
4.17, 4.18, 4.22, 4.24, 4.25
4.26, 4.27, 4.28, 4.30

Work on appendix 4-A,


4-46
4-47
References

Presentations by McGraw-Hill/Irwin and by Wilson,G.R.

“Production & Operations Analysis” by S.Nahmias

“Factory Physics” by W.J.Hopp, M.L.Spearman

“Inventory Management and Production Planning and
Scheduling” by E.A. Silver, D.F. Pyke, R. Peterson

“Production Planning, Control, and Integration” by D. Sipper
and R.L. Bulfin Jr.
Reorder Point Calculation for Example 4.1
4-48
Reorder Point Calculation for Lead Times
Exceeding One Cycle
4-49
4-50
Sensitivity Analysis
Let G(Q) be the average annual holding and
cost function given by
set-up
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 * 
4-51
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:
2k l
Q
h'
Except that h’ is defined as h’= h(1- λ/P)
4-52
Inventory Levels for Finite
Production Rate Model
4-53
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.
All-Units Discount
Order Cost Function
4-54
4-55
Incremental Discount
Order Cost Function
4-56
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).
All-Units Discount Average
Annual Cost Function
4-57
4-58
Average Annual Cost Function
for Incremental Discount Schedule
4-59
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:
c1the
Q1 condition
 c2Q2 that
 ...  cnQn
When
C
c / h  c / h  ...  c / h
is met,
is straightforward.
If the condition is not
1 the1solution
2 procedure
2
n
n
met, one must use an iterative procedure involving Lagrange
Multipliers.
4-60
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
n
lj
P
j 1
j
 1.
4-61
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
'lj
And the optimal production quantities are given by:
Q j *  l jT *