Solution to Warehouse Problem1
A company currently has two warehouses. Each warehouse services half of the
company’s demand, and the annual demand serviced by each warehouse is
normally distributed with mean 10,000 and standard deviation 1000. The lead
time for meeting demand is 1/10 year. The company wants to meet 95% of all
demand on time. Assume that each warehouse uses the EOQ formula to
determine its order quantity, and that this leads to Q = 2000 for each warehouse.
Important Inventory Formulas for Q, R Policies
We’ll use these symbols to represent the parameters of inventory systems:
Q = Number of units per order
Q* = Optimum value for Q (EOQ)
D = Annual demand in units
S = Setup or ordering cost per order
H = Holding or carrying cost per unit per year
I = Percent holding cost
P = Price per unit
d = Daily demand
L = Lead time in days
Optimal Order Quantity
Q*
2 DS
H
2 * 1000 * 62.5

0.5
 500

Based on 13-50 (p. 761) in Practical Management Science (2nd ed., Winston and Albright, 2001
Duxbury Press). Solution by David Juran, 2002.
1
Reorder Point for a Service Level of 1 - 
R
 dL  z  dL
Where dL is the expected demand during the lead time, z is a number of
standard deviations above dL, and dL is the standard deviation of demand
during the lead time. (Our textbook uses the letter k instead of z to symbolize
the number of standard deviations.)
 is the probability that a customer order will be backordered, and 1 -  is the
probability that a customer order can be filled from inventory (commonly
referred to as the service level).
The first part of the reorder point formula is expected demand, and the second
part of the formula is safety stock. Safety stock, by definition, is inventory held in
case of unexpected demand.
(a) How much safety stock must be held at each warehouse?
We could try to use the safety stock formula given above, but unfortunately it
isn’t set up to deal with a service level target like this one. (Notice that
management wants to meet 95% of all demand on time, whereas the safety stock
formula is set up in terms of demand during the lead time only.)
It is useful here to consider the inventory system as having a cycle that repeats
itself, which allows us to study one cycle and then project our conclusions to the
system as a whole.
The order quantity is 2000, as given in the problem, so we can look at cycles of
2000 units. Approximately five times per year, each warehouse places an order
for 2000 units, waits during the 0.10-year lead time, and then receives 2000 units.
Over the long run, supply and demand will be the same, so we can think in
terms of a cycle of 2000 customer orders, 95% management wants to meet out of
inventory. This implies that management wants to have 2000 * (1 – 95%) = 100
backorders every cycle. This, in turn, implies that management wants 100
backorders during each lead time, because the only time a backorder can occur is
during the lead time.
If we had a way to set a desired level for expected backorders, and then solve for
the appropriate z-value, we would have all the ingredients to answer this
question. The problem is that calculating the expected number of backorders is
hard.
B60.2350
2
Prof. Juran
There is a nice trick using Excel’s normal distribution functions that can do this
easily (see p. 742 in the text). We set up a spreadsheet as shown:
1
2
3
4
5
A
B
Analysis for each of the two warehouses
Mean annual demand
StDev of annual demand
Lead time
E(DL)
6 StDev(DL)
7 Q
8 k
9 k*L
Order quantity (given)
(safety stock in terms of z)
10
11 R
Expected shortage per cycle
Reorder point
(safety stock in units)
C
D
10000
1000
1/10
1000.000
=C4*C2
=SQRT(C4)*C3
316.228
2000
1.00
E
F
G
H
Service level constraint
% of demand satisfied from "in stock"
Obtained
Required
98.68% =
95%
Total Safety Stock
632.5
=2*C9
=C8*C6
316.2
=C6*(NORMDIST(C8,0,1,0)-C8*(1-NORMSDIST(C8)))
=C5+C8*C6
26.35
1316.2
We use this spreadsheet to analyze the safety stock requirements of one
warehouse. Then, in cell E6, we calculate the total safety stock for both
warehouses. The trick here is in cell C10, which does all the nasty calculus
involved in calculating the expected number of shortages. Currently, the
spreadsheet is using 1 standard deviation’s worth of safety stock (k = 1), which
results in too few expected shortages.
We can find the right value of k by using one of our oldest tools, Goal Seek.
B60.2350
3
Prof. Juran
1
2
3
4
5
A
B
Analysis for each of the two warehouses
Mean annual demand
StDev of annual demand
Lead time
E(DL)
6 StDev(DL)
7 Q
8 k
9 k*L
Order quantity (given)
(safety stock in terms of z)
10
11 R
Expected shortage per cycle
Reorder point
C
D
10000
1000
1/10
1000.000
=C4*C2
=SQRT(C4)*C3
316.228
2000
0.18
(safety stock in units)
E
F
G
H
Service level constraint
% of demand satisfied from "in stock"
Obtained
Required
95.00% =
95%
Total Safety Stock
112.6
=2*C9
=C8*C6
56.3
=C6*(NORMDIST(C8,0,1,0)-C8*(1-NORMSDIST(C8)))
=C5+C8*C6
100.00
1056.3
Each warehouse needs to hold 56.3 units in safety stock.
(b) Show that if the company had only one warehouse, it would hold less safety
stock than it does when it has two warehouses.
We’ll use the same spreadsheet again, changing the data to reflect the combined
demand.
Expected Annual Demand:
D
 2*D
 2 * 10 ,000
 20 ,000
Expected Lead Time Demand:
dL
 0.1 * 20,000
 20 ,000
B60.2350
4
Prof. Juran
Standard Deviation of Annual Demand:
 D
 
2 * 1000 
 2 *  D2

2
 1414.2
Standard Deviation of Lead Time Demand:
   0.10 * 
dL
2
D
 0.10 * 1414.2 2
 447.2
Order Quantity:
Recall that for each warehouse,
Q

2 DS
H
EOQ

22 D S
H
Therefore,
 2
2 DS
H
 2 EOQ 
 2 2000 
 2828.4
We plug these values into our spreadsheet:
1
2
3
4
5
A
B
Analysis for a single warehouse
Mean annual demand
StDev of annual demand
Lead time
E(DL)
6 StDev(DL)
7 Q
8 k
9 k*L
Order quantity (given)
(safety stock in terms of z)
10
11 R
Expected shortage per cycle
Reorder point
(safety stock in units)
C
D
E
F
G
H
Service level constraint
% of demand satisfied from "in stock"
=SQRT(1000^2+1000^2)
Obtained
Required
95.00% =
95%
=C4*C2
20000
1414.2
1/10
2000.000
=SQRT(C4)*C3
447.214
2828.4
0.18
=C8*C6
79.6
=C6*(NORMDIST(C8,0,1,0)-C8*(1-NORMSDIST(C8)))
=C5+C8*C6
141.42
2079.6
A single warehouse would only need to hold 79.6 units of safety stock to meet
the same 95% service level target (as compared to 112.6 with two warehouses).
B60.2350
5
Prof. Juran
(c) A young MBA argues, “By having one central warehouse, I can reduce the
total amount of safety stock needed to meet 95% of all customer demands on
time. Therefore, we can save money by having only one central warehouse
instead of several branch warehouses.” How might this argument be
rebutted?
He might be right, but there are other things to consider besides safety stock. A
single warehouse is likely to involve longer shipping times, distances, and costs.
It might turn out to be less expensive overall to maintain 2 warehouses and have
more safety stock.
1-Warehouse Inventory System
4000
3500
3000
Inventory
2500
2000
1500
1000
500
0
-500
0
50
100
150
200
250
300
350
250
300
350
-1000
Days
2-Warehouse Inventory System
4000
3500
3000
Inventory
2500
2000
1500
1000
500
0
-500
0
50
100
150
200
-1000
Days
B60.2350
6
Prof. Juran