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5 Reorder Point: Determining When to Order

# 5 Reorder Point: Determining When to Order

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206

CHAPTER 6 • INVENTORY CONTROL MODELS

FIGURE 6.4
Reorder Point Graphs

Inventory
Level
Q

ROP
0
Lead time = L
ROP < Q

Time

Inventory
Level
On Order
Q

On hand

0
Lead time = L
ROP > Q

Time

Since the maximum on-hand inventory level is the order quantity of 400, an inventory position
of 480 would be:
Inventory position = 1Inventory on hand2 + 1Inventory on order2
480 = 80 + 400
Thus, a new order would have to be placed when the on-hand inventory fell to 80 while there
was one other order in-transit. The second graph in Figure 6.4 illustrates this type of situation.

6.6

EOQ Without the Instantaneous Receipt Assumption

The production run model
eliminates the instantaneous
receipt assumption.

When a firm receives its inventory over a period of time, a new model is needed that does not
require the instantaneous inventory receipt assumption. This new model is applicable when
inventory continuously flows or builds up over a period of time after an order has been placed or
when units are produced and sold simultaneously. Under these circumstances, the daily demand
rate must be taken into account. Figure 6.5 shows inventory levels as a function of time. Because
this model is especially suited to the production environment, it is commonly called the
production run model.
In the production process, instead of having an ordering cost, there will be a setup cost. This
is the cost of setting up the production facility to manufacture the desired product. It normally
includes the salaries and wages of employees who are responsible for setting up the equipment,
engineering and design costs of making the setup, paperwork, supplies, utilities, and so on. The
carrying cost per unit is composed of the same factors as the traditional EOQ model, although
the annual carrying cost equation changes due to a change in average inventory.

6.6

FIGURE 6.5
Inventory Control and
the Production Process

Inventory
Level

EOQ WITHOUT THE INSTANTANEOUS RECEIPT ASSUMPTION

Part of Inventory Cycle
During Which Production Is
Taking Place

207

There Is No Production
During This Part of the
Inventory Cycle

Maximum
Inventory

t

Solving the production run
model involves setting setup costs
equal to holding costs and solving
for Q.

Time

The optimal production quantity can be derived by setting setup costs equal to holding
or carrying costs and solving for the order quantity. Let’s start by developing the expression
for carrying cost. You should note, however, that making setup cost equal to carrying cost
does not always guarantee optimal solutions for models more complex than the production
run model.

Annual Carrying Cost for Production Run Model
As with the EOQ model, the carrying costs of the production run model are based on the average inventory, and the average inventory is one-half the maximum inventory level. However,
since the replenishment of inventory occurs over a period of time and demand continues during
this time, the maximum inventory will be less than the order quantity Q. We can develop the
annual carrying, or holding, cost expression using the following variables:
Q = number of pieces per order, or production run
Cs = setup cost
Ch = holding or carrying cost per unit per year
p = daily production rate
d = daily demand rate
t = length of production run in days
The maximum inventory level is as follows:
1Total produced during the production run2 - 1Total used during production run2
= 1Daily production rate21Number of days of production2
-1Daily demand21Number of days of production2

= 1pt2 - 1dt2
Since

total produced = Q = pt,
we know that
t =

Q
p

Maximum inventory level = pt - dt = p
The maximum inventory level
in the production model is less
than Q.

Q
Q
d
- d = Qa1 - b
p
p
p

Since the average inventory is one-half of the maximum, we have
Average inventory =

Q
d
a1 - b
p
2

(6-9)

208

CHAPTER 6 • INVENTORY CONTROL MODELS

and
Annual holding cost =

Q
d
a 1 - bCh
p
2

(6-10)

Annual Setup Cost or Annual Ordering Cost
When a product is produced over time, setup cost replaces ordering cost. Both of these are independent of the size of the order and the size of the production run. This cost is simply the number of orders (or production runs) times the ordering cost (setup cost). Thus,
Annual setup cost =

D
Cs
Q

(6-11)

Annual ordering cost =

D
C
Q o

(6-12)

and

Determining the Optimal Production Quantity
When the assumptions of the production run model are met, costs are minimized when the setup
cost equals the holding cost. We can find the optimal quantity by setting these costs equal and
solving for Q. Thus,
Annual holding cost = Annual setup cost
Q
d
D
a1 - bCh =
C
p
2
Q s
Here is the formula for the
optimal production quantity.
Notice the similarity to the basic
EOQ Model.

Solving this for Q, we get the optimal production quantity 1Q*2:
Q* =

2DCs

(6-13)

d
Ch a1 - b
p
Q

It should be noted that if the situation does not involve production but rather involves the receipt
of inventory over a period of time, this same model is appropriate, but Co replaces Cs in the
formula.
Production Run Model
Annual holding cost =
Annual setup cost =

Q
d
a1 - bCh
p
2
D
C
Q s
2DCs

Optimal production quantity Q * =
Q

Ch a1 -

d
b
p

Brown Manufacturing Example
Brown Manufacturing produces commercial refrigeration units in batches. The firm’s estimated
demand for the year is 10,000 units. It costs about \$100 to set up the manufacturing process, and
the carrying cost is about 50 cents per unit per year. When the production process has been
set up, 80 refrigeration units can be manufactured daily. The demand during the production
period has traditionally been 60 units each day. Brown operates its refrigeration unit production
area 167 days per year. How many refrigeration units should Brown Manufacturing produce in

6.6

EOQ WITHOUT THE INSTANTANEOUS RECEIPT ASSUMPTION

209

each batch? How long should the production part of the cycle shown in Figure 6.5 last? Here is
the solution:
Annual demand = D = 10,000 units
Setup cost = Cs = \$100
Carrying cost = Ch = \$0.50 per unit per year
Daily production rate = p = 80 units daily
Daily demand rate = d = 60 units daily
2DCs

1. Q * =
R
2. Q * =

=

Ch a1 -

d
b
p

2 * 10,000 * 100
60
b
0.5a1 80
R
2,000,000

B 0.5 A 1>4 B

= 116,000,000

= 4,000 units
If Q* = 4,000 units and we know that 80 units can be produced daily, the length of each production cycle will be Q>p = 4,000>80 = 50 days. Thus, when Brown decides to produce refrigeration units, the equipment will be set up to manufacture the units for a 50-day time span.
The number of production runs per year will be D>Q = 10,000>4,000 = 2.5. This means that
the average number of production runs per year is 2.5. There will be 3 production runs in one
year with some inventory carried to the next year, so only 2 production runs are needed in the
second year.
USING EXCEL QM FOR PRODUCTION RUN MODELS The Brown Manufacturing production run

model can also be solved using Excel QM. Program 6.2A contains the input data and the Excel
formulas for this problem. Program 6.2B provides the solution results, including the optimal production quantity, maximum inventory level, average inventory level, and the number of setups.

PROGRAM 6.2A

Excel QM Formulas and Input Data for the Brown Manufacturing Problem

Enter the demand rate, setup cost, and
holding cost. Notice that the holding cost
is a fixed dollar amount rather than a
percentage of the unit price.
Enter daily production rate
and daily demand rate.

Calculate the optimal production quantity.

Calculate the
average number
of setups.

Calculate the maximum inventory.

Calculate the annual holding costs based on average inventory
and the annual setup cost based on the number of setups.

210

CHAPTER 6 • INVENTORY CONTROL MODELS

PROGRAM 6.2B
The Solution Results
for the Brown
Manufacturing Problem
Using Excel QM

IN ACTION

Fortune 100 Firm Improves Inventory
Policy for Service Vehicles

M

ost manufacturers of home appliances provide in-home
repair of the appliances that are under warranty. One such Fortune 100 firm had about 70,000 different parts that were used in
the repair of its appliances. The annual value of the parts inventory was over \$7 million. The company had more than 1,300 service vehicles that were dispatched when service requests were
received. Due to limited space on these vehicles, only about 400
parts were typically carried on each one. If a service person
arrived to repair an appliance and did not have necessary part
(i.e., a stockout occured), a special order was made to have the
order delivered by air so that the repair person could return and
fix the appliance as soon as possible.
Deciding which parts to carry was a particularly difficult
problem. A project was begun to find a better way to forecast the
demand for parts and identify which parts should be stocked on
each vehicle. Initially, the intent was to reduce the parts inventory
on each truck as this represented a significant cost of holding

6.7

the inventory. However, upon further analysis, it was decided that
the goal should be to minimize the overall cost—including the
costs of special deliveries of parts, revisiting the customer for
the repair if the part was not initially available, and overall customer satisfaction.
The project team improved the forecasting system used to
project the number of parts needed on each vehicle. As a result,
the actual number of parts carried on each vehicle increased.
However, the number of first-visit repairs increased from 86% to
90%. This resulted in a savings of \$3 million per year in the cost
of these repairs. It also improved customer satisfaction because
the problem was fixed without the service person having to
return a second time.
Source: Based on Michael F. Gorman and Sanjay Ahire. “A Major Appliance
Manufacturer Rethinks Its Inventory Policies for Service Vehicles,” Interfaces
36, 5 (September–October 2006): 407–419.

Quantity Discount Models
In developing the EOQ model, we assumed that quantity discounts were not available. However,
many companies do offer quantity discounts. If such a discount is possible, but all of the other
EOQ assumptions are met, it is possible to find the quantity that minimizes the total inventory
cost by using the EOQ model and making some adjustments.

6.7

TABLE 6.3
Quantity Discount
Schedule

QUANTITY DISCOUNT MODELS

211

DISCOUNT
NUMBER

DISCOUNT
QUANTITY

DISCOUNT
(%)

DISCOUNT
COST (\$)

1

0 to 999

0

5.00

2

1,000 to 1,999

4

4.80

3

2,000 and over

5

4.75

When quantity discounts are available, the purchase cost or material cost becomes a relevant cost, as it changes based on the order quantity. The total relevant costs are as follows:
Total cost = Material cost + Ordering cost + Carrying cost
Q
D
Co + Ch
Total cost = DC +
Q
2

(6-14)

where
D
Co
C
Ch

=
=
=
=

annual demand in units
ordering cost of each order
cost per unit
holding or carrying cost per unit per year

Since holding cost per unit per year is based on the cost of the items, it is convenient to express
this as
Ch = IC
where
I = holding cost as a percentage of the unit cost 1C2

The overall objective of the
quantity discount model is to
minimize total inventory costs,
which now include actual
material costs.

For a specific purchase cost (C), given the assumptions we have made, ordering the EOQ will
minimize total inventory costs. However, in the discount situation, this quantity may not be large
enough to qualify for the discount, so we must also consider ordering this minimum quantity for
the discount. A typical quantity discount schedule is shown in Table 6.3.
As can be seen in the table, the normal cost for the item is \$5. When 1,000 to 1,999 units
are ordered at one time, the cost per unit drops to \$4.80, and when the quantity ordered at one
time is 2,000 units or more, the cost is \$4.75 per unit. As always, management must decide
when and how much to order. But with quantity discounts, how does the manager make these
decisions?
As with other inventory models discussed so far, the overall objective will be to minimize
the total cost. Because the unit cost for the third discount in Table 6.3 is lowest, you might be
tempted to order 2,000 units or more to take advantage of the lower material cost. Placing an
order for that quantity with the greatest discount cost, however, might not minimize the total
inventory cost. As the discount quantity goes up, the material cost goes down, but the carrying
cost increases because the orders are large. Thus, the major trade-off when considering quantity
discounts is between the reduced material cost and the increased carrying cost.
Figure 6.6 provides a graphical representation of the total cost for this situation. Notice the
cost curve drops considerably when the order quantity reaches the minimum for each discount.
With the specific costs in this example, we see that the EOQ for the second price category
11,000 … Q … 1,9992 is less than 1,000 units. Although the total cost for this EOQ is less than
the total cost for the EOQ with the cost in category 1, the EOQ is not large enough to obtain this
discount. Therefore, the lowest possible total cost for this discount price occurs at the minimum
quantity required to obtain the discount 1Q = 1,0002. The process for determining the minimum cost quantity in this situation is summarized in the following box.

212

CHAPTER 6 • INVENTORY CONTROL MODELS

FIGURE 6.6
Total Cost Curve for the
Quantity Discount Model

Total
Cost
\$

TC Curve for Discount 3
TC Curve for
Discount 1

TC Curve for Discount 2

EOQ for Discount 2

0

1,000

2,000
Order Quantity

Quantity Discount Model
2DCo
.
B IC
2. If EOQ < Minimum for discount, adjust the quantity to Q = Minimum for discount.
Q
D
3. For each EOQ or adjusted Q, compute Total cost = DC + Co + Ch.
Q
2
4. Choose the lowest-cost quantity.
1. For each discount price (C), compute EOQ =

Brass Department Store Example
Let’s see how this procedure can be applied by showing an example. Brass Department Store
stocks toy race cars. Recently, the store was given a quantity discount schedule for the cars; this
quantity discount schedule is shown in Table 6.3. Thus, the normal cost for the toy race cars is \$5.

IN ACTION

L

Lucent Technologies Develops Inventory
Requirements Planning System

ucent Technologies has developed an inventory requirements
planning (IRP) system to determine the amount of safety stock
(buffer stock) to carry for a variety of products. Instead of looking
only at the variability of demand during the lead time, the company looked at both the supply and demand of the products during the lead time. The focus was on the deviations between the
forecast demand and the actual supply. This system was used
both for products with independent demand and for products
with dependent demand.
A modified ABC classification system was used to determine
which items received the most careful attention. Items were considered both for the dollar volume and the criticality. In addition

to the Class A, B, and C categories, a D category was created to
include items that were both low dollar volume and low criticality. A simple two-bin system was used for these items.
In order to gain acceptance of the IRP system, business managers from all functions were involved in the process, and the system was made transparent so that everyone understood the
system. Because of the IRP system, overall inventory was reduced
by \$55 million, and the service level was increased by 30%. The
success of the IRP system helped Lucent receive the Malcolm
Baldrige Award in 1992.
Source: Based on Alex Bangash, et al. “Inventory Requirements Planning at
Lucent Technologies,” Interfaces 34, 5 (September–October 2004): 342–352.

6.8

TABLE 6.4

USE OF SAFETY STOCK

213

Total Cost Computations for Brass Department Store

ORDER
QUANTITY (Q)

ANNUAL
MATERIAL
COST (\$) = DC

ANNUAL
ORDERING
D
COST (\$) = Co
Q

ANNUAL
CARRYING
Q
COST (\$) = Ch
2

TOTAL (\$)

DISCOUNT
NUMBER

UNIT
PRICE (C)

1

\$5.00

700

25,000

350.00

350.00

25,700.00

2

4.80

1,000

24,000

245.00

480.00

24,725.00

3

4.75

2,000

23,750

122.50

950.00

24,822.50

For orders between 1,000 and 1,999 units, the unit cost is \$4.80, and for orders of 2,000 or more
units, the unit cost is \$4.75. Furthermore, the ordering cost is \$49 per order, the annual demand
is 5,000 race cars, and the inventory carrying charge as a percentage of cost, I, is 20% or 0.2.
What order quantity will minimize the total inventory cost?
The first step is to compute EOQ for every discount in Table 6.3. This is done as follows:
EOQ 1 =
EOQ values are computed.

EOQ 2 =
EOQ 3 =

EOQ values are adjusted.

12215,00021492

B 10.2215.002

= 700 cars per order

B 10.2214.802

= 714 cars per order

B 10.2214.752

= 718 cars per order

12215,00021492
12215,00021492

The second step is to adjust those quantities that are below the allowable discount range. Since
EOQ1 is between 0 and 999, it does not have to be adjusted. EOQ2 is below the allowable range
of 1,000 to 1,999, and therefore, it must be adjusted to 1,000 units. The same is true for EOQ3;
it must be adjusted to 2,000 units. After this step, the following order quantities must be tested
in the total cost equation:
Q1 = 700
Q2 = 1,000
Q3 = 2,000

The total cost is computed.

Q* is selected.

The third step is to use Equation 6-14 and compute a total cost for each of the order quantities.
This is accomplished with the aid of Table 6.4.
The fourth step is to select that order quantity with the lowest total cost. Looking at Table 6.4,
you can see that an order quantity of 1,000 toy race cars minimizes the total cost. It should be recognized, however, that the total cost for ordering 2,000 cars is only slightly greater than the total
cost for ordering 1,000 cars. Thus, if the third discount cost is lowered to \$4.65, for example, this
order quantity might be the one that minimizes the total inventory cost.
USING EXCEL QM FOR QUANTITY DISCOUNT PROBLEMS As seen in the previous analysis, the

quantity discount model is more complex than the inventory models discussed so far in this
chapter. Fortunately, we can use the computer to simplify the calculations. Program 6.3A shows
the Excel formulas and input data needed for Excel QM for the Brass Department Store problem. Program 6.3B provides the solution to this problem, including adjusted order quantities and
total costs for each price break.

6.8

Use of Safety Stock

Safety stock helps in avoiding
stockouts. It is extra stock kept
on hand.

When the EOQ assumptions are met, it is possible to schedule orders to arrive so that stockouts
are completely avoided. However, if the demand or the lead time is uncertain, the exact demand
during the lead time (which is the ROP in the EOQ situation) will not be known with certainty.
Therefore, to prevent stockouts, it is necessary to carry additional inventory called safety stock.

214

CHAPTER 6 • INVENTORY CONTROL MODELS

PROGRAM 6.3A

Excel QM’s Formulas and the Input Data for the Brass Department Store
Quantity Discount Problem
Enter demand rate,
setup cost, and
holding cost.

Enter the quantity discount
schedule of quantities and unit
prices for each price break.
Compute the order quantities for
each price break and adjust
them upward if necessary.
Compute holding,
setup, and unit cost
for each price break.

Determine the optimal order
quantity by finding the order
quantity that minimizes total
costs.

Compute the total cost
for each price break.

PROGRAM 6.3B
Excel QM’s Solution to
the Brass Department
Store Problem

When demand is unusually high during the lead time, you dip into the safety stock instead
of encountering a stockout. Thus, the main purpose of safety stock is to avoid stockouts when
the demand is higher than expected. Its use is shown in Figure 6.7. Note that although stockouts
can often be avoided by using safety stock, there is still a chance that they may occur. The
demand may be so high that all the safety stock is used up, and thus there is still a stockout.
One of the best ways to implement a safety stock policy is to adjust the reorder point. In the
EOQ situation where the demand and lead time are constant, the reorder point is simply the

6.8

FIGURE 6.7
Use of Safety Stock

USE OF SAFETY STOCK

215

Inventory
on
Hand

Time
Stockout
Inventory
on
Hand

Safety
Stock, SS

Stockout Is Avoided

0 Units
Time

amount of inventory that would be used during the lead time (i.e., the daily demand times the
lead time in days). This is assumed to be known with certainty, so there is no need to place an
order when the inventory position is more than this. However, when the daily demand or the lead
time fluctuate and are uncertain, the exact amount of inventory that will be used during the
lead time is uncertain. The average inventory usage during the lead time should be computed
and some safety stock should be added to this to avoid stockouts. The reorder point becomes
ROP = 1Average demand during lead time2 + 1Safety stock2
ROP = 1Average demand during lead time2 + SS

(6-15)

where
Safety stock is included in the
ROP.

SS = safety stock
How to determine the correct amount of safety stock is the only remaining question. Two important factors in this decision are the stockout cost and the holding cost. The stockout cost usually involves lost sales and lost goodwill, which results in loss of future sales. If holding cost is
low while stockout cost is high, a large amount of safety stock should be carried to avoid stockouts as it costs little to carry this, while stockouts are expensive. On the other hand, if stockout
cost is low but holding cost is high, a lower amount of safety stock would be preferred, as having a stockout would cost very little, but too much safety stock will result in much higher annual
holding costs.

216

CHAPTER 6 • INVENTORY CONTROL MODELS

How is the optimum stock level determined? If demand fluctuates while lead time is constant, and if both the stockout cost per unit and the holding cost per unit are known, the use of a
payoff/cost table might be considered. With only a small number of possible demand values during the lead time, a cost table could be constructed in which the different possible demand levels
would be the states of nature, and the different amounts of safety stock as the alternatives. Using
the techniques discussed in Chapter 3, the expected cost could be calculated for each safety
stock level, and the minimum cost solution could be found.
However, a more general approach is to determine what service level is desired and then to
find the safety stock level that would accomplish this. A prudent manager will look at the holding cost and the stockout cost to help determine an appropriate service level. A service level
indicates what percentage of the time customer demand is met. In other words, the service level
is the percentage of time that stockouts are avoided. Thus,
Service level = 1 - Probability of a stockout
or
Probability of a stockout = 1 - Service level
Once the desired service level is established, the amount of safety stock to carry can be found
using the probability distribution of demand during the lead time.
SAFETY STOCK WITH THE NORMAL DISTRIBUTION Equation 6-15 provides the general formula

for determining the reorder point. When demand during the lead time is normally distributed,
the reorder point becomes
ROP = 1Average demand during lead time2 + ZsdLT

(6-16)

where
Z = number of standard deviations for a given service level
␴dLT = standard deviation of demand during the lead time
Thus, the amount of safety stock is simply ZsdLT. The following example looks at how to determine the appropriate safety stock level when demand during the lead time is normally distributed and the mean and standard deviation are known.
HINSDALE COMPANY EXAMPLE The Hinsdale Company carries a variety of electronic inventory

items, and these are typically identified by SKU. One particular item, SKU A3378, has a demand that is normally distributed during the lead time, with a mean of 350 units and a standard
deviation of 10. Hinsdale wants to follow a policy that results in stockouts occurring only 5% of
the time on any order. How much safety stock should be maintained and what is the reorder
point? Figure 6.8 helps visualize this example.
FIGURE 6.8
Safety Stock and the
Normal Distribution
5% Area of Normal Curve
SS
μ = 350

X =?

μ = Mean demand = 350
σ = Standard deviation = 10
X = Mean demand + Safety stock
SS = Safety stock = X – μ = Zσ
Z =

X–μ
σ