Lect 02

IEOR4000: Production Management Professor Guillermo Gallego
Lecture 2 September 9, 2004

Lecture Plan
1. The EOQ and Extensions
2. Multi-Item EOQ Model

1 The EOQ and Extensions
This section is devoted to the Economic Production Quantity (EPQ) model, its specialization to the
Economic Order Quantity (EOQ) model and its extension to allow backorders. The next section
deals with multiple products. The primary driver to hold inventories is the presence of a fixed pro-
duction/ordering cost that is incurred every time a positive number of units are produced/ordered.
We wish to determine the number of units to produce/order every time the fixed production/ordering
cost is incurred. Producing/ordering in large quantities reduce the average fixed production/ordering
cost. What prevents the production/ordering quantities from becoming too large is that it costs
money to hold inventories. Inventory costs include the cost of storage and insurance and the cost of
capital tied up in inventory. The EPQ balances the average fixed ordering cost (that decreases with
large orders) and the average inventory holding cost (that increases with large orders).
We will assume that demand is constant and continuous over an infinite horizon. We will also
assume a constant, continuous, production rate. The objective is to minimize the average cost.
Data:
K set up or ordering cost,
c unit production cost,
h holding cost,
λ demand rate,
µ production rate.
The holding cost is often model as h = Ic where I represents the inventory carrying cost per
unit per unit time and includes the cost of capital per unit per unit time. Let ρ = λ/µ. We assume
ρ ≤ 1, since otherwise the production capacity is not enough to keep up with demand.
Since we have assumed that production and demand rates are constant and costs are stationary,
it makes intuitive sense to produce/order in cycles of equal lengths and to start/end each cycle with
zero inventory. This last property is know as the Zero Inventory Property (ZIP) and can be formally
verified. Let T denote the cycle length. The cycle length is the time between consecutive orders; it
is also called the order/reorder interval. We want to express the average cost a(T ) as a function of
T and then select T to minimize a(T ). We could have also chosen the production/order quantity
Q = λT as the decision variable. The advantage of using a time variable is apparent when dealing
with multiple items, where time provides a unified variable.
Let I(t) denote the inventory at time t. We will assume that I(0) = 0, and by design we want
I(T ) = 0. At time zero we start production at rate µ. We need to determine when to stop production
so that I(T ) = 0. Since λT units are demanded over the interval [0, T ], it takes λT = ρT units of
µ
time to produce λT units at rate µ. Because demand is constant at rate λ, inventory accumulates
at rate µ − λ over the interval [0, ρT ] reaching a peak equal to (µ − λ)ρT = λ(1 − ρ)T at time ρT .
Since production stops at time ρT , inventory decreases at rate λ over the interval [ρT, T ] reaching
zero at time T .
Let us now compute the average inventory over the interval [0, T ]. Formally, this is equal to
T
1
I(t)dt.
T 0

The function I(t) forms a triangle that repeats ever T units of time. The base of each triangle is T
and the height is λ(1 − ρ)T . It follows that
T
1 11 1
I(t)dt = λ(1 − ρ)T 2 = λ(1 − ρ)T.
T 0 T2 2

IEOR4000: Production Management page 2 Professor Guillermo Gallego

Consequently, the average holding cost per unit time is equal to 1 hλ(1 − ρ)T .
2

Figure 1: Inventory as a function of time.

Since we incur the fixed ordering cost K once every T units of time, the average ordering cost is
simply K . Consequently, the average ordering and holding cost per unit time is:
T

1 K
a(T ) = hλ(1 − ρ)T + . (1)
2 T
Notice that the unit production cost, say c, is not included in (1) because it does not depend on T .
Thus a(T ) is the relevant function to study although cλ + a(T ) is the actual average cost per unit
of time.
By taking the first derivative of (1), setting it to zero, and solving for T , we find

2K 1
T∗ = . (2)
hλ 1−ρ
It is easy to see, by looking at the sign of the second derivative of a(T ), that (1) is convex and
consequently (2) is the optimal order interval. As a consequence, the economic production quantity
(EPQ) is equal to
2Kλ 1
Q∗ = λT ∗ = . (3)
h 1−ρ
Substituting (2) into (1), we find that optimal average cost is given by
√
a∗ ≡ a(T ∗ ) = 2Khλ 1 − ρ (4)

Example: Suppose K = 100, h = 5, λ = 200, µ = 1000. Then ρ = 0.2, the optimal order interval
is T ∗ = 0.50, the optimal order quantity is Q∗ = 100, and the optimal average cost is a∗ = $400.
We now explore how T ∗ and a∗ change with ρ. Notice that T ∗ increases and a∗ decreases as ρ
increases. Intuitively, inventories accumulate at a slower rate when ρ increases. As a result, we can
use longer reorder intervals and reduce costs. It is interesting to explore what happens in the limit
as ρ → 1. In this case, the production rate and demand rate are in perfect balance. Thus, after an
initial set up, we simply produce to meet demand and keep doing that forever. Thus, T ∗ → ∞, and
a∗ → 0. Notice that for fixed µ < ∞ the average cost is zero when λ = 0 or λ = µ. The average cost


is the highest when λ = µ/2, i.e., when only half of the production capacity is utilized. Thus, when
selecting the production rate µ to satisfy demand λ, it is better to avoid µ = 2λ as lower average
costs will result with either a lower or a higher production rate µ.
Example: Suppose K = 100, h = 5, λ = 500, µ = 1000. Then ρ = 0.5, the optimal order interval
is T ∗ = 0.40, the optimal order quantity is Q∗ = 200, and the optimal average cost is a∗ = $500.
The average cost a∗ is increasing concave in the setup cost K. This means that the savings that
accrue from reducing the fixed cost K become larger the more we reduce K. Table 4 makes this
clear. This is an example of changing the givens. Japanese manufacturers, such as Toyota, were
quick to recognize the value of reducing setup costs.

K a∗ savings
100 500.00 0
90 474.34 25.66
80 447.21 27.13
70 418.33 28.88
60 387.30 31.03
50 353.55 33.74
40 316.23 37.33
30 273.86 42.37
20 223.61 50.25
10 158.11 65.49
0 0 158.11

Table 1: Savings from reducing fixed costs.

1.1 The Economic Order Quantity (EOQ) Model
The infinite production rate case is often used to model the situation where the product is ordered
rather than produced and corresponds to ρ = 0. The infinite production rate case is also known as
the Economic Order Quantity (EOQ) model. This optimal order interval for this model is given by

2K
T∗ = . (5)
hλ
The optimal order quantity for this model is known as the EOQ and is given by

2Kλ
Q∗ = λT ∗ = . (6)
h
Finally, the optimal average cost is given by
√
a∗ ≡ a(T ∗ ) = 2Khλ (7)

1.2 Sensitivity Analysis.
Using the results from previous sections, we can write the average cost as
K
a(T ) = HT + .
T
Here K is the fixed ordering cost, and H = 1 hλ(1−ρ) for the finite production rate case and H = 1 h
2 2
for the infinite production rate case.


K
Notice that HT ∗ = T∗ = 1 a(T ∗ ), so we can write
2

T 1 T
HT = HT ∗ = a(T ∗ ) ∗
T∗ 2 T
and
K K T∗ 1 T∗
= ∗ = a(T ∗ ) .
T T T 2 T
Consequently,
K
a(T ) = HT +
T
∗ T K T∗
= HT ∗
+ ∗
T T T
1 T T∗
= a(T ∗ ) +
2 T∗ T
T
= a(T ∗ )e
T∗

where e(x) ≡ 1 (x + 1/x).
2
This allow us to write
a(T ) T
=e . (8)
a(T ∗ ) T∗
Notice that e(x) ≥ 1 and that it achieves its minimum at x = 1
There are two immediate applications of this sensitivity result.

1.2.1 Powers-of-Two Policies
P OT
Let T denote the minimizer a(T ) subject to T ∈ {2k β : k ∈ Z}, where β is a base planning
period and Z denotes the set of integers. It can be shown that

a(T P OT ) √
∗)
≤ e( 2) 1.06.
a(T

This tells us that if we restrict the order interval to be a power-of-two multiple of a base planning
period, the resulting level of sub-optimality is at most 6%.

1.2.2 Parameter Estimation
Consider an inventory system where the holding cost parameter H has been estimated correctly but
where the ordering cost parameter K is incorrectly estimated as K . Let T be the “optimal” order
size resulting from using K . What is the actual average cost as a function of K /K? From equation
(8) we have
a(T ) T
=e =e K /K .
a(T ∗ ) T∗
From this we can see that the cost of overestimating K is lower than that of underestimating K.
To see this more clearly, suppose we overestimate K by 40%, then K = 1.40K resulting e(1.40) =
1.057.Conversely, if we underestimate K by 40%, then K = 0.60K resulting in e(0.60) = 1.133.


1.3 The Backorder Case.
Backorders can be completely avoided in a deterministic setting. However, if customers are willing
to accept backorders, and backorders are charged at a linear rate, say b, per unit per unit time, then
it is optimal to incur some backorders during each cycle. Thus, in addition to a tradeoff with setup
costs, there is a tradeoff between holding and backorder costs. You can think of backorder cost rate
b as the cost rate, per unit of time, of keeping a customer waiting for one unit of inventory. On the
other hand, you can think of the holding cost rate h as the cost rate, per unit of time, of keeping
a unit of inventory waiting for a customer. It turns out that the optimal balance between holding
and backorder costs results in us holding inventory 100b/(b + h)% of the time. Thus, if b = 9h, then
100b/(b + h)% = 90% so it is optimal to keep inventories 90% of the time and incur backorders 10%
of the time. Thus, customers will find inventory on hand 90% of the time. This interpretation will
be useful later when we discuss models with random demands.
Because we hold less inventory when backorders are allowed, the costs will be lower. We will
later show that allowing backorders is mathematically equivalent to reducing the holding cost rate
by the factor b/(b + h) so that the modified holding cost rate is equal to hb/(h + b). Using this
modified holding cost rate results in the optimal order interval

2K b+h 1
T∗ = , (9)
hλ b 1−ρ
and optimal average cost
√ b
a∗ = 2Khλ 1 − ρ. (10)
b+h
It is instructive to view the optimal order interval (9) and the optimal average cost (10) as
functions of ρ and b. In particular it is interesting to compare (9) and (10) to the case where
production rates are infinite, ρ = 0, and backorders are not allowed (b = ∞). Note that

b+h 1
T ∗ (ρ, b) = T ∗ (0, ∞) (11)
b 1−ρ
b
a∗ (ρ, b) = a∗ (0, ∞) 1−ρ (12)
b+h
2K
√
Here T ∗ (0, ∞) = ∗
hλ and a (0, ∞) = 2Khλ are respectively the optimal cycle length and
the optimal average cost for the EOQ, i.e., the case of infinite production run and no backorders
allowed.
Notice that with finite production rates and finite backorder costs, the cycle length increases
while the average cost decreases, always satisfying

a∗ (ρ, b)T ∗ (ρ, b) = 2K. (13)
That is, the optimal cost per cycle is independent of the production rate, of the backorder cost,
and of the holding cost; and is always equal to twice the ordering or setup cost.

1.3.1 Derivation of the Backorder Cost Formula
Remember that without backorders the function I(t) was always non-negative and drops to zero at
the time of replenishment. When backorders are allowed, we expect the inventory to be negative
during some portion of the cycle. We can find the optimal proportion of time that the inventory
should be negative by shifting the inventory curve down by some s ≥ 0 to I(t) − s and by optimizing
over s for a fixed order interval T . With this convention we can write the average cost per unit time
as
T T
1
a(T, s) = K +h (I(t) − s)+ dt + b (I(t) − s)− dt , (14)
T 0 0


where x+ = max(x, 0) and x− = max(−x, 0).
The integrals in (14) represent the holding and backorder costs. Clearly a(T, s) is convex in s
for fixed T ; equating the partial derivative, of a(T, s) with respect to s, to zero, we obtain
T
1 b
1{t : I(t) > s}dt = . (15)
T 0 b+h

From (15), we see that s should be selected so the proportion of time the inventory is positive is
b/(b + h). This is a result that holds for more general functions1 I(t), not just the triangle shaped
one we have been discussing.
By similar triangles
h
s∗ (T ) = λ(1 − ρ)T (16)
b+h
solves (15). Substituting s∗ (T ) into a(T, s) we obtain

1 hb K
a(T ) = λ(1 − ρ)T + . (17)
2h+b T
Clearly, a(T ) is convex in T so the optimal cycle length is given by equation (9) and the optimal
cost by equation (10).

1.4 Some useful relationships
It can be shown that the optimal average cost is equal to the holding cost rate times the maximum
inventory, say Imax reached through the cycle, and also equal to the backorder cost rate times the
absolute value of the minimum inventory, say Imin reached through the cycle.
These relationships are useful since it is often easier to collect information on Imax, Imin and
h, that it is on K and b. Suppose that inventories are managed optimally. Given Imax, Imin and
h, the imputed values for K and b are given by

Imax
b = h (18)
Imin
b+h
K = |Imax Imin |. (19)
2(1 − ρ)

2 Multi-item EOQ Models
Our concern here is with inventory problems involving a variety of items. We start with a discussion
of exchange curves for a family of items with similar ordering and inventory carrying costs, next we
study a heuristic that calls for using the same reorder interval for each item. Finally, we discuss
the joint replenishment problem where a major setup cost is incurred every time an order is placed
regardless of the number of different items in the order.

2.1 Exchange Curves for a Family of Products
Consider n items, where item i has unit cost ci and fixed cost Ki , and demand per unit time equal
to λi i = 1, . . . , n. Assume further that the Ki cannot be determined explicitly, but that it is
reasonable to assume that Ki is proportional to the number of standard hours devoted to the setup
of the order. That is, Ki = si a where si is the number of standard hours and a is the rate in dollars
per setup hour. We also assume that the holding cost rate hi is of the form hi = Ici where I is the
1 It is enough for the function to be integrable and nowhere flat for this result to hold.


per dollar inventory carrying charge. In what follows, we assume for simplicity the case of infinite
production rates with no backorders allowed, but the arguments to be made can be easily extended
to encompass those cases.
The economic order interval for item i is thus
2si a
Ti∗ = i = 1, . . . , n.
Ici λi
Then, the average cycle stock (ACS), i.e., the average number of dollars invested in inventory is
equal to
n n
1 a 1
ACS = ci λi Ti∗ = √ ci λi si , (20)
2 i=1 I 2 i=1
while the average number of setup hours per year is
n n
1 I 1
N= si = √ ci λi si . (21)
i=1
Ti a 2 i=1

Notice that the product of the number of setup hours per year times the average cycle stock is
a constant, so
n √
( i=1 ci λi si )2
ACS = (22)
2N
is a hyperbola and is independent of a and I. This tells us that if the family of items is managed
according to the EOQ logic then the number of orders per year and the average cycle stock should
lie on the hyperbola given by (22). This surprising result allow us to determine whether or not the
items are managed effectively without knowing the values of a and I. All we need to determine if
the items are managed effectively is the unit cost, the demand rate and the setup times. As we shall
see, the hyperbola represents different efficient points depending on the ratio of a to I.
Example: Consider the following inventory system consisting of 4 items:

λ c s
7200 $2.00 1
4000 0.90 2
500 5.00 1
100 0.81 3

Table 2: Data for Exchange Curve Example

Then the curve of efficient points is given by ACS = 36569/N . A few points of this curve are
given by the following table:
A manager can determine the average cycle stock and the number of hours spent in setups under
the current policy and then see whether or not the point lies on the hyperbola. Suppose, for example,
that the items are ordered once a week or 52 times a year. Then N = 208 and ACS = $396 which
is suboptimal. Suppose that the manager does not know the ratio a/I but feels that it would be a
great improvement to move to the point N = 200 and ACS = $182.85 in the curve. How can he
efficiently obtain the order intervals to achieve this goal? To answer this question first notice that
by dividing ACS (20) by N (21) we find
ACS a
= . (23)
N I
This equation makes intuitive sense. The hyperbola gives us the set of efficient points for different
values of a/I. Equation (23) tell us where in the hyperbola we should be for each value of a/I. If a/I


N ACS
50 731.38
100 365.69
150 243.79
200 182.85
250 146.28
300 121.90
350 104.48
400 91.42
450 81.26
500 73.14

Table 3: A Few Data Points of the Exchange Curve

is high, then the line will intersect the curve at a point where ACS is high and N is low. Conversely,
if a/I is low, the line will intersect the curve at a point where ACS is low and N is high.
Returning to our example, if the manager wants to operate at the point N = 200 and ACS =
$182.85 all he needs to do is compute the ratio ACS/N = 182.85/200 = 0.91423 and use this value
√
for a/I in the formulas for the reorder intervals, e.g., Ti∗ = 2si /ci λi .91423, i = 1, . . . , 4. The
resulting reorder intervals will be such that N = 200 and ACS = $182.5. See the Exchange-curve.xls
ﬁle.

2.2 A Simple Inventory Coordination Heuristic
Our previous analysis suggests that we should operate on the exchange curve by ordering items
according to their economic order interval. For management, however, it may be easier and more
convenient to order all the items in the family at the same time. We now investigate, in the worst
case sense, the relative loss of optimality that results from using a common reorder interval for a
family of items rather than using the economic reorder interval for each item.
Let
Ki 1
ai (T ) = + h i λi T
T 2
be the average cost of using reorder interval T for item i. Then, the average cost of using a common
reorder interval T for all the items in the group is given by
n
a(T ) = ai (T ).
i=1

The common reorder interval that minimizes a(T ) is
n
2 i=1 Ki
T∗ = n ,
i=1 hi λi

and the overall average cost is
n n n
a(T ∗ ) = ai (T ∗ ) = 2 Ki h i λi .
i=1 i=1 i=1

In contrast, if we individually minimize each item’s average cost
Ki 1
ai (Ti ) = + hi λi Ti
Ti 2


by using the reorder intervals
2Ki
Ti = , i = 1, . . . , n,
hi λi
the overall average cost is given by
n n
ai (Ti∗ ) = 2Ki hi λi .
i=1 i=1

Now, let us consider the ratio
n
i=1 ai (T ∗ )
n
i=1 ai (Ti∗ )
of the average cost of using a the best common reorder interval for all the items to the average cost
of using the optimal reorder interval for each item. Obviously the ratio is greater or equal to one,
and a number much larger than one would indicate that using a common order interval is a bad
idea. On the other hand, if the ratio is very close to one, say 1.03, we can have the benefits of order
coordination at a relatively small cost.
In order to obtain an upper bound on the cost ratio we order the items according to the ratios
Ki
hi λi , so that items with small ratios come first. Let

K n h 1 λ1
r=
K 1 h n λn
be the ratio of the largest to the smallest ratio. It is possible to show that
n ∗
i=1 a(T ) 1 √ √
n ∗
≤ ( 4 r + 1/ 4 r).
i=1 ai (Ti ) 2

Thus, if r ≤ 4, then the cost ratio is at most 1.06. This analysis indicates that using a common
order interval is a good idea as long as the items are not too different as measured by r.
4
For the following data set, we found that r = 3.19, so the ratio of a(T ∗ ) to i=1 ai (Ti∗ ) is at
√ √ ∗
most 1 ( 4 3.19+1/ 4 3.19) = 1.042. Actual calculations show that 4a(T ) ∗ = 1.018 so the relative
2 ai (Ti )
i=1
cost of using a common reorder interval to jointly managing the items is less than 2%.

λ h K
1550 0.4 7
1725 0.35 9
1875 0.5 14
1110 0.25 10

Table 4: Data for Coordinating Heuristic.

Lect 02

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