Dr. Ron Lembke
Random Demand
 Don’t know how many we will sell
 Sales will differ by period
 Average always remains the same
 Standard deviation remains constant
How would our policies change?
 How would our order quantity change?
 EOQ balances ordering vs holding, and is unchanged
 How would our reorder point change?
 That’s a good question
Constant Demand vs Random
•Steady demand
•Always buy Q
•Reorder at R=dL
•Sell dL during LT
•Inv = Q after arrival
•Random demand
•Always buy Q
•Reorder at R=dL + ?
•Sell ? during LT
•Inv = ? after arrival
Inv
Q
R
L
Q
Q
Q
R
L
Random Demand
 Reorder when on-hand inventory is equal to the
amount you expect to sell during LT, plus an extra
amount of safety stock
 Assume daily demand has a normal distribution
 If we want to satisfy all of the demand 95% of the time,
how many standard deviations above the mean should
the inventory level be?
 Just considers a probability of running out, not the
number of units we’ll be short.
Demand over the Lead Time
R = Expected Demand over LT + Safety Stock
R  d L  z L
d = Average demand per day
L = Lead Time in days
 L = st deviation of demand over Lead Time
z from normal table, e.g. z.95 = 1.65
Random Example
 What should our reorder point be?
 Demand averages 50 units per day, L = 5 days
 Total demand over LT has standard deviation of 100
 want to satisfy all demand 90% of the time
 To satisfy 90% of the demand, z = 1.28
R  d L  z L
R  50 * 5  1.28 *100
R  250  128  378
Random Demand
•Random demand
•Always buy Q
•Reorder at R=dL + SS
•Sometimes use SS
•Sometimes don’t
•On average use 0 SS
Inv
R
dL
SS
L
L
L
L
St Dev of Daily Demand
 What if we only know the average daily demand, and
the standard deviation of daily demand?
 Lead time = 4 days,
 daily demand = 10,
 Daily demand has standard deviation = 5,
 What should our reorder point be, if z = 3?
St Dev Over LT
 If the average each day is 10, and the lead time is 4
days, then the average demand over the lead time must
be 40.
d * L  10 * 4  40
 What is the standard deviation of demand over the
lead time?
 Std. Dev. ≠ 5 * 4
St Dev Over Lead Time
 We can add up variances, not standard deviations
 Standard deviation of demand over LT =
 L  Ldays  day
 45  10
R  d L  z L  d L  z Ldays  day
Demand Per Day
R  d L  z * L * D
Or, same
thing:
 L  L * D
R  d L  z L
L = Lead time in days
d = average demand per day
 D = st deviation of demand per day
z from normal table, e.g. z.95 = 1.65
Random Demand
Fixed Order Quantity
 Demand per day averages 40 with standard deviation
15, lead time is 5 days, service level of 90%
L = 5 days
d = 40
R  d L  z * L * D
z = 1.30
R  200  43.6  243.6
 D = 15
R  40 * 5  1.30 * 5 *15
Fixed-Time Period Model
 Place an order every, say, week.
 Time period is fixed, order quantity will vary
 Order enough so amount on hand plus on order gets
up to a target amount
 Q = S – Inv
 Order “up to” policies
S  d L  T   z * L  T *  D
Fixed Order Period
•Order every T days
•Q = S – Inv
•Order comes in L days
after being placed
S
•Amount on hand after arrival differs
•Order today, next order comes T+L
days later
T
L
Inv
On hand
On hand
+ on order
L
T
Service Level Criteria
 Type I: specify probability that you do not run out
during the lead time
 Chance that 100% of customers go home happy
 Type II: (Fill Rate) proportion of demands met from
stock
 100% chance that this many go home happy, on average
Two Types of Service
Cycle
1
2
3
4
5
6
7
8
9
10
Sum
Demand
180
75
235
140
180
200
150
90
160
40
1,450
Stock-Outs
0
0
45
0
0
10
0
0
0
0
55
Type I:
8 of 10 periods
80% service
Type II:
1,395 / 1,450 =
96%
Summary
 Fixed Order Quantity – always order same
 Random demand – reorder point needs to change
 Standard Deviation over the LT is given
 Standard Deviation per day is given
 Fixed Time Period
 Always order once a month, e.g.
 Amount on hand plus on order will add up to S
 Different service metrics: Type I, Type II