Transcript Managing Flow Variability: Safety Inventory Managing Flow Variability: Safety Inventory Forecasts Depend on: (a) Historical Data and (b) Market Intelligence. Demand Forecasts and Forecast.

Managing Flow Variability: Safety Inventory
Managing Flow Variability: Safety Inventory
1
Forecasts Depend on: (a) Historical Data and (b) Market
Intelligence.
Demand Forecasts and Forecast Errors
Safety Inventory and Service Level
Optimal Service Level – The Newsvendor Problem
Lead Time Demand Variability
Pooling Efficiency through Aggregation
Shortening the Forecast Horizon
Levers for Reducing Safety Inventory
Managing Flow Variability: Safety Inventory
Four Characteristics of Forecasts
2
Forecasts are usually (always) inaccurate (wrong). Because of
random noise.
Forecasts should be accompanied by a measure of forecast error.
A measure of forecast error (standard deviation) quantifies the
manager’s degree of confidence in the forecast.
Aggregate forecasts are more accurate than individual forecasts.
Aggregate forecasts reduce the amount of variability – relative to the
aggregate mean demand. StdDev of sum of two variables is less than
sum of StdDev of the two variables.
Long-range forecasts are less accurate than short-range forecasts.
Forecasts further into the future tends to be less accurate than those of
more imminent events. As time passes, we get better information, and
make better prediction.
Managing Flow Variability: Safety Inventory
Demand During Lead Time is Variable N(μ,σ)
3
Demand of sand during lead time has an average of 50 tons.
Standard deviation of demand during lead time is 5 tons
Assuming that the management is willing to accept a risk
no more that 5%.
Managing Flow Variability: Safety Inventory
Forecast and a Measure of Forecast Error
4
Forecasts should be accompanied by a measure of forecast
error
Managing Flow Variability: Safety Inventory
Demand During Lead Time
Inventory
5
Demand
during LT
Lead Time
Time
Managing Flow Variability: Safety Inventory
ROP when Demand During Lead Time is Fixed
6
LT
Managing Flow Variability: Safety Inventory
Demand During Lead Time is Variable
7
LT
Managing Flow Variability: Safety Inventory
Demand During Lead Time is Variable
8
Inventory
Time
Managing Flow Variability: Safety Inventory
Safety Stock
Quantity
9
A large demand
during lead time
Average demand
during lead time
ROP
Safety stock reduces risk of
stockout during lead time
Safety stock
LT
Time
Managing Flow Variability: Safety Inventory
Safety Stock
Quantity
10
ROP
LT
Time
Managing Flow Variability: Safety Inventory
Re-Order Point: ROP
11
Demand during lead time has Normal distribution.
If we order when the inventory on hand is
equal to the average demand during the lead
time;
then there is 50% chance that the demand
during lead time is less than our inventory.
However, there is also 50% chance that the
demand during lead time is greater than our
inventory, and we will be out of stock for a
while.
We usually do not like 50% probability of stock
out
We can accept some risk of being out of stock,
but we usually like a risk of less than 50%.
Managing Flow Variability: Safety Inventory
Safety Stock and ROP
12
Risk of a
stockout
Service level
Probability of
no stockout
RO
P
Average
demand
Quantity
Safety
stock
0
z
z-scale
Each Normal variable x is associated with a standard Normal Variable z
x is Normal (Average x , Standard Deviation x)  z is Normal (0,1)
Managing Flow Variability: Safety Inventory
z Values
13
Risk of a
stockout
Service level
Probability of
no stockout
RO
P
Average
demand
Quantity
Safety
stock
0
z
z-scale
SL
0.9
0.95
0.99
z value
1.28
1.65
2.33
There is a table for z which tells us
a) Given any probability of not exceeding z. What is the value of z
b) Given any value for z. What is the probability of not exceeding z
Managing Flow Variability: Safety Inventory
μ and σ of Demand During Lead Time
14
Demand of sand during lead time has an average of 50 tons.
Standard deviation of demand during lead time is 5 tons.
Assuming that the management is willing to accept a risk no
more that 5%. Find the re-order point.
What is the service level.
Service level = 1-risk of stockout = 1-.05 = .95
Find the z value such that the probability of a standard
normal variable being less than or equal to z is .95
Go to normal table, look inside the table. Find a probability
close to .95. Read its z from the corresponding row and
column.
Managing Flow Variability: Safety Inventory
z Value using Table
15
Given a 95% SL
95% Probability
Page 319: Normal table
0.05
The table will give you z
z
Second digit
after decimal
Up to
the first
digit
after
decimal
Probability
Z = 1.65
1.6
Managing Flow Variability: Safety Inventory
The standard Normal Distribution F(z)
16
z
F(z) = Prob( N(0,1) < z)
F(z)
0
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
0.00
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
0.8849
0.9032
0.9192
0.9332
0.9452
0.9554
0.9641
0.9713
0.9772
0.9821
0.9861
0.9893
0.9918
0.9938
0.9953
0.9965
0.9974
0.9981
0.9987
0.9990
0.9993
0.9995
0.01
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
0.8869
0.9049
0.9207
0.9345
0.9463
0.9564
0.9649
0.9719
0.9778
0.9826
0.9864
0.9896
0.9920
0.9940
0.9955
0.9966
0.9975
0.9982
0.9987
0.9991
0.9993
0.9995
0.02
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
0.8888
0.9066
0.9222
0.9357
0.9474
0.9573
0.9656
0.9726
0.9783
0.9830
0.9868
0.9898
0.9922
0.9941
0.9956
0.9967
0.9976
0.9982
0.9987
0.9991
0.9994
0.9995
0.03
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
0.8907
0.9082
0.9236
0.9370
0.9484
0.9582
0.9664
0.9732
0.9788
0.9834
0.9871
0.9901
0.9925
0.9943
0.9957
0.9968
0.9977
0.9983
0.9988
0.9991
0.9994
0.9996
0.04
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
0.8925
0.9099
0.9251
0.9382
0.9495
0.9591
0.9671
0.9738
0.9793
0.9838
0.9875
0.9904
0.9927
0.9945
0.9959
0.9969
0.9977
0.9984
0.9988
0.9992
0.9994
0.9996
0.05
0.5199
0.5596
0.5987
0.6368
0.6736
0.7088
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
0.8944
0.9115
0.9265
0.9394
0.9505
0.9599
0.9678
0.9744
0.9798
0.9842
0.9878
0.9906
0.9929
0.9946
0.9960
0.9970
0.9978
0.9984
0.9989
0.9992
0.9994
0.9996
0.06
0.5239
0.5636
0.6026
0.6406
0.6772
0.7123
0.7454
0.7764
0.8051
0.8315
0.8554
0.8770
0.8962
0.9131
0.9279
0.9406
0.9515
0.9608
0.9686
0.9750
0.9803
0.9846
0.9881
0.9909
0.9931
0.9948
0.9961
0.9971
0.9979
0.9985
0.9989
0.9992
0.9994
0.9996
0.07
0.5279
0.5675
0.6064
0.6443
0.6808
0.7157
0.7486
0.7794
0.8078
0.8340
0.8577
0.8790
0.8980
0.9147
0.9292
0.9418
0.9525
0.9616
0.9693
0.9756
0.9808
0.9850
0.9884
0.9911
0.9932
0.9949
0.9962
0.9972
0.9979
0.9985
0.9989
0.9992
0.9995
0.9996
0.08
0.5319
0.5714
0.6103
0.6480
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.9162
0.9306
0.9429
0.9535
0.9625
0.9699
0.9761
0.9812
0.9854
0.9887
0.9913
0.9934
0.9951
0.9963
0.9973
0.9980
0.9986
0.9990
0.9993
0.9995
0.9996
0.09
0.5359
0.5753
0.6141
0.6517
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
0.9177
0.9319
0.9441
0.9545
0.9633
0.9706
0.9767
0.9817
0.9857
0.9890
0.9916
0.9936
0.9952
0.9964
0.9974
0.9981
0.9986
0.9990
0.9993
0.9995
0.9997
Managing Flow Variability: Safety Inventory
Relationship between z and Normal Variable x
17
Risk of a
stockout
Service level
Probability of
no stockout
RO
P
Average
demand
Quantity
Safety
stock
0
z
z-scale
z = (x-Average x)/(Standard Deviation of x)
x = Average x +z (Standard Deviation of x)
μ = Average x
 x = μ +z σ
σ = Standard Deviation of x
Managing Flow Variability: Safety Inventory
Relationship between z and Normal Variable ROP
18
Risk of a
stakeout
Service level
Probability of
no stockout
RO
P
Average
demand
Quantity
Safety
stock
0
z
z-scale
LTD = Lead Time Demand
ROP = Average LTD +z (Standard Deviation of LTD)
ROP = LTD+zσLTD  ROP = LTD + Isafety
Managing Flow Variability: Safety Inventory
Demand During Lead Time is Variable N(μ,σ)
19
Demand of sand during lead time has an average of 50 tons.
Standard deviation of demand during lead time is 5 tons
Assuming that the management is willing to accept a risk
no more that 5%. z = 1.65
Compute safety stock
Isafety = zσLTD
Isafety = 1.64 (5) = 8.2
ROP = LTD + Isafety
ROP = 50 + 1.64(5) = 58.2
Managing Flow Variability: Safety Inventory
Service Level for a given ROP
20
SL = Prob (LTD ≤ ROP)
LTD is normally distributed  LTD = N(LTD, sLTD ).
ROP = LTD + zσLTD  ROP = LTD + Isafety  I safety = z sLTD
At GE Lighting’s Paris warehouse, LTD = 20,000, sLTD = 5,000
The warehouse re-orders whenever ROP = 24,000
I safety = ROP – LTD = 24,000 – 20,000 = 4,000
I safety = z sLTD  z = I safety / sLTD = 4,000 / 5,000 = 0.8
SL= Prob (Z ≤ 0.8) from Appendix II  SL= 0.7881
Managing Flow Variability: Safety Inventory
Excel: Given z, Compute Probability
21
Managing Flow Variability: Safety Inventory
Excel: Given Probability, Compute z
22
Managing Flow Variability: Safety Inventory
μ and σ of demand per period and fixed LT
23
Demand of sand has an average of 50 tons per week.
Standard deviation of the weekly demand is 3 tons.
Lead time is 2 weeks.
Assuming that the management is willing to accept a risk
no more that 10%. Compute the Reorder Point
Managing Flow Variability: Safety Inventory
μ and σ of demand per period and fixed LT
24
R: demand rate per period (a random variable)
R: Average demand rate per period
σR: Standard deviation of the demand rate per period
L: Lead time (a constant number of periods)
LTD: demand during the lead time (a random variable)
LTD: Average demand during the lead time
σLTD: Standard deviation of the demand during lead time
Managing Flow Variability: Safety Inventory
μ and σ of demand per period and fixed LT
25
A random variable R: N(R, σR) repeats itself L times
during the lead time. The summation of these L random
variables R, is a random variable LTD
If we have a random variable LTD which is equal to
summation of L random variables R
LTD = R1+R2+R3+…….+RL
Then there is a relationship between mean and standard
deviation of the two random variables
LTD  LR
2
s LTD
 Ls R2
s LTD  Ls R
Managing Flow Variability: Safety Inventory
μ and σ of demand per period and fixed LT
26
Demand of sand has an average of 50 tons per week.
Standard deviation of the weekly demand is 3 tons.
Lead time is 2 weeks.
Assuming that the management is willing to accept a risk
no more that 10%. z = 1.28, R = 50, σR = 3, L = 2
LTD  LR
LTD  2(50)  100
s LTD  Ls R
s LTD  Ls R  2 (3)  4.24
Isafety = zσLTD = 1.28(4.24) = 5.43
ROP = 100 + 5.43
Managing Flow Variability: Safety Inventory
Lead Time Variable, Demand fixed
27
Demand of sand is fixed and is 50 tons per week.
The average lead time is 2 weeks.
Standard deviation of lead time is 0.5 week.
Assuming that the management is willing to accept a risk
no more that 10%. Compute ROP and Isafety.
Managing Flow Variability: Safety Inventory
μ and σ of lead time and fixed Demand per period
28
L: lead time (a random variable)
L: Average lead time
σL: Standard deviation of the lead time
RL
R: Demand per period (a constant value)
LTD: demand during the lead time (a random variable)
LTD: Average demand during the lead time
σLTD: Standard deviation of the demand during lead time
R
L
Managing Flow Variability: Safety Inventory
μ and σ of demand per period and fixed LT
29
A random variable L: N(L, σL) is multiplied by a constant
R and generates the random variable LTD.
RL
If we have a random variable LTD which is equal to a
constant R times a random variables L
LTD = RL
Then there is a relationship between mean and standard
deviation of the two random variables
LTD  LR
2
s LTD
 R 2s L2
s LTD  Rs L
R
L
Managing Flow Variability: Safety Inventory
Variable R fixed L…………….Variable L fixed R
30
LTD  LR
2
s LTD
 Ls R2
s LTD  Ls R
R
R
R
R
R
+
R
+
R
+
R
+
R R
RL
LTD  LR
2
s LTD
 R 2s L2
s LTD  Rs L
R
L
L
Managing Flow Variability: Safety Inventory
Lead Time Variable, Demand fixed
31
Demand of sand is fixed and is 50 tons per week.
The average lead time is 2 weeks.
Standard deviation of lead time is 0.5 week.
Assuming that the management is willing to accept a risk no
more that 10%. Compute ROP and Isafety.
z = 1.28, L = 2 weeks, σL = 0.5 week, R = 50 per week
LTD  LR  50(2)  100
s LTD  Rs L  50(0.5)  25
Isafety = zσLTD = 1.28(25) = 32
ROP = 100 + 32
Managing Flow Variability: Safety Inventory
Both Demand and Lead Time are Variable
32
R: demand rate per period
R: Average demand rate
σR: Standard deviation of demand
LTD  LR
L: lead time
s LTD 
L: Average lead time
σL: Standard deviation of the lead time
Ls R  R s L
2
LTD: demand during the lead time (a random variable)
LTD: Average demand during the lead time
σLTD: Standard deviation of the demand during lead time
Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
33
How do we choose what level of service a firm should offer?
Cost of Holding
Extra Inventory
Improved
Service
Optimal Service Level under uncertainty
The Newsvendor Problem
The decision maker balances the expected costs of ordering
too much with the expected costs of ordering too little to
determine the optimal order quantity.
Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
34
Cost =1800, Sales Price = 2500, Salvage Price = 1700
Underage Cost = 2500-1800 = 700, Overage Cost = 1800-1700 = 100
Demand
100
110
120
130
140
150
160
170
180
190
200
Probability of Demand
0.02
0.05
0.08
0.09
0.11
0.16
0.2
0.15
0.08
0.05
0.01
What is probability of demand to be equal to 130?
What is probability of demand to be less than or equal to 140?
What is probability of demand to be greater than 140?
What is probability of demand to be equal to 133?
Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
35
Demand
100
101
102
103
104
105
106
107
108
109
Probability of Demand
0.002
0.002
0.002
0.002
0.002
0.002
0.002
0.002
0.002
0.002
Demand
110
111
112
113
114
115
116
117
118
119
Probability of Demand
0.005
0.005
0.005
0.005
0.005
0.005
0.005
0.005
0.005
0.005
What is probability of demand to be equal to 116?
What is probability of demand to be less than or equal to 160?
What is probability of demand to be greater than 116?
What is probability of demand to be equal to 13.3?
Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
36
Average Demand
100
110
120
130
140
150
160
170
180
190
200
Probability of Demand
0.02
0.05
0.08
0.09
0.11
0.16
0.2
0.15
0.08
0.05
0.01
What is probability of demand to be equal to
130?
What is probability of demand to be less than
or equal to 140?
What is probability of demand to be greater
than 140?
What is probability of demand to be equal to
133?
Managing Flow Variability: Safety Inventory
Compute the Average Demand
37
N
AverageDemand  ri P( R  ri )
i 1
Average Demand =
+100×0.02 +110×0.05+120×0.08
+130×0.09+140×0.11 +150×0.16
+160×0.20 +170×0.15 +180×0.08
+190×0.05+200×0.01
Average Demand = 151.6
r
100
110
120
130
140
150
160
170
180
190
200
How many units should I have to sell 151.6 units (on average)?
How many units do I sell (on average) if I have 100 units?
P( R =r)
0.02
0.05
0.08
0.09
0.11
0.16
0.2
0.15
0.08
0.05
0.01
Managing Flow Variability: Safety Inventory
Deamand (r)
Porbability
Prob(R ≥ r)
100
0.02
1.00
110
0.05
0.98
120
0.08
0.93
130
0.09
0.85
140
0.11
0.76
150
0.16
0.65
160
0.20
0.49
170
0.15
0.29
180
0.08
0.14
190
0.05
0.06
200
0.01
0.01
38
Suppose I have ordered 140 Unities.
On average, how many of them are sold? In other words, what is
the expected value of the number of sold units?
When I can sell all 140 units?
I can sell all 140 units if  R≥ 140
Prob(R≥ 140) = 0.76
The the expected number of units sold –for this part- is
(0.76)(140) = 106.4
Also, there is 0.02 probability that I sell 100 units 2 units
Also, there is 0.05 probability that I sell 110 units5.5
Also, there is 0.08 probability that I sell 120 units 9.6
Also, there is 0.09 probability that I sell 130 units 11.7
106.4 + 2 + 5.5 + 9.6 + 11.7 = 135.2
Managing Flow Variability: Safety Inventory
Deamand (r)
Porbability
Prob(R ≥ r)
100
0.02
1.00
110
0.05
0.98
120
0.08
0.93
130
0.09
0.85
140
0.11
0.76
150
0.16
0.65
160
0.20
0.49
170
0.15
0.29
180
0.08
0.14
190
0.05
0.06
200
0.01
0.01
39
Suppose I have ordered 140 Unities.
On average, how many of them are salvaged? In other words,
what is the expected value of the number of sold units?
0.02 probability that I sell 100 units.
In that case 40 units are salvaged  0.02(40) = .8
0.05 probability to sell 110  30 salvage 0.05(30)= 1.5
0.08 probability to sell 120  30 salvage 0.08(20) = 1.6
0.09 probability to sell 130  30 salvage 0.09(10) =0.9
0.8 + 1.5 + 1.6 + 0.9 = 4.8
Total number Solved 135.2 @ 700 = 94640
Total number Salvaged 4.8 @ -100 = -480
Expected Profit = 94640 – 480 = 94,160
Managing Flow Variability: Safety Inventory
Cumulative Probabilities
40
100
110
120
130
140
150
160
170
180
190
200
Probabilities
P(R=r) P(R≤r) P(R≥r)
0.02
0.02
1
0.05
0.07
0.98
0.08
0.15
0.93
0.09
0.24
0.85
0.11
0.35
0.76
0.16
0.51
0.65
0.2
0.71
0.49
0.15
0.86
0.29
0.08
0.94
0.14
0.05
0.99
0.06
0.01
1
0.01
Units
Sold Salvage
100
0
109.8
0.2
119.1
0.9
127.6
2.4
135.2
4.8
141.7
8.3
146.6
13.4
149.5
20.5
150.9
29.1
151.5
38.5
151.6
48.4
Salvage
70000
76860
83370
89320
94640
99190
102620
104650
105630
106050
106120
Revenue
Sales
0
20
90
240
480
830
1340
2050
2910
3850
4840
Total
70000
76840
83280
89080
94160
98360
101280
102600
102720
102200
101280
Managing Flow Variability: Safety Inventory
Number of Units Sold, Salvages
41
100
110
120
130
140
150
160
170
180
190
200
Probabilities
P(R=r) P(R≤r) P(R≥r)
0.02
0.02
1
0.05
0.07
0.98
0.08
0.15
0.93
0.09
0.24
0.85
0.11
0.35
0.76
0.16
0.51
0.65
0.2
0.71
0.49
0.15
0.86
0.29
0.08
0.94
0.14
0.05
0.99
0.06
0.01
1
0.01
Units
Sold Salvage
100
0
109.8
0.2
119.1
0.9
127.6
2.4
135.2
4.8
141.7
8.3
146.6
13.4
149.5
20.5
150.9
29.1
151.5
38.5
151.6
48.4
Salvage
70000
76860
83370
89320
94640
99190
102620
104650
105630
106050
106120
Revenue
Sales
0
20
90
240
480
830
1340
2050
2910
3850
4840
Total
70000
76840
83280
89080
94160
98360
101280
102600
102720
102200
101280
Managing Flow Variability: Safety Inventory
Total Revenue for Different Ordering Policies
42
100
110
120
130
140
150
160
170
180
190
200
Probabilities
P(R=r) P(R≤r) P(R≥r)
0.02
0.02
1
0.05
0.07
0.98
0.08
0.15
0.93
0.09
0.24
0.85
0.11
0.35
0.76
0.16
0.51
0.65
0.2
0.71
0.49
0.15
0.86
0.29
0.08
0.94
0.14
0.05
0.99
0.06
0.01
1
0.01
Units
Sold Salvage
100
0
109.8
0.2
119.1
0.9
127.6
2.4
135.2
4.8
141.7
8.3
146.6
13.4
149.5
20.5
150.9
29.1
151.5
38.5
151.6
48.4
Sales
70000
76860
83370
89320
94640
99190
102620
104650
105630
106050
106120
Revenue
Salvage
0
20
90
240
480
830
1340
2050
2910
3850
4840
Total
70000
76840
83280
89080
94160
98360
101280
102600
102720
102200
101280
Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
43
Net Marginal Benefit:
MB = p – c
MB = $2,500 - $1,800 = $700
Net Marginal Cost:
MC = c - v
MC = $1,800 - $1,700 = $100
Suppose I have ordered Q units.
What is the expected cost of ordering one more units?
What is the expected benefit of ordering one more units?
If I have ordered one unit more than Q units, the probability of not selling that extra
unit is if the demand is less than or equal to Q. Since we have P( R ≤ Q).
The expected marginal cost =MC× P( R ≤ Q)
If I have ordered one unit more than Q units, the probability of selling that extra
unit is if the demand is greater than Q. We know that P(R>Q) = 1- P( R ≤ Q).
The expected marginal benefit = MB× [1-Prob.( R ≤ Q)]
Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
44
As long as expected marginal cost is less than expected marginal profit we buy the
next unit. We stop as soon as: Expected marginal cost ≥ Expected marginal profit
MC×Prob(R ≤ Q*) ≥ MB× [1 – Prob(R ≤ Q*)]
MB
Prob(R ≤ Q*) ≥
MB  MC
In a continuous model: SL* = Prob(R ≤ Q*) =
MB
MB  MC
MB
$700
SL* 

 0.875
MB  MC $700  $100
If we assume demand is normally distributed, What quantity corresponds to this
service level ?
Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
45
Q*  R  z s R
Probability Less than Upper Bound is 0.87493
0.4
0.35
0.3
Density
0.25
0.2
z = 1.15
0.15
0.1
0.05
0
-4
-3
-2
-1
0
Critical Value (z)
1
2
3
4
Q*  R  z s R  151.6  1.15  22.44  177.41
Managing Flow Variability: Safety Inventory
Aggregate Forecast is More Accurate than Individual Forecasts
46
Demand forecast error for 5 products
Month P1 P2 P3 P4 P5
Total
1
2 4 -3 4 -2
5
2
0 0 5 0 1
6
3
2 5 -2 1 0
6
4
-1 2 -4 1 -2
-4
5
1 -4 -4 4 3
0
6
-2 -1 4 5 1
7
7
4 -5 0 5 -5
-1
8
-2 2 2 0 -4
-2
9
-4 1 1 4 3
5
10
-1 -1 -4 -2 4
-4
11
4 1 3 2 0
10
12
2 0 -4 -3 2
-3
13
-4 -1 2 -1 -4
-8
14
0 -5 4 5 -2
2
15
-5 4 4 0 -4
-1
16
4 2 2 1 3
12
17
-3 2 -2 5 -3
-1
18
3 1 -3 -1 3
3
19
2 3 -3 -5 3
0
20
0 -3 1 -1 -4
-7
21
-3 4 -4 0 5
2
22
4 1 -5 4 0
4
23
0 -4 -5 0 -5
-14
24
-2 0 -2 -4 -2
-10
25
2 5 -2 5 -1
9
26
4 5 5 0 1
15
27
2 1 4 1 -1
7
28
0 -3 3 0 -2
-2
29
3 -2 -1 -1 -4
-5
30
-2 1 -4 -4 0
-9
5
3
1
-1
-3
-5
15
10
5
0
15
-5
10
-10
5
-15
0
Managing Flow Variability: Safety Inventory
Physical Centralization
47
Physical Centralization: the firm consolidates all its warehouses in one
location from which is can serve all customers.
Example: Two warehouses. Demand in the two ware houses are independent.
Both warehouses have the same distribution for their lead time demand.
LTD1: N(LTD, σLTD )
LTD2: N(LTD, σLTD )
Both warehouses have identical service levels
To provide desired SL, each location must carry Isafety = zσLTD
z is determined by the desired service level
The total safety inventory in the decentralized system is
I
D
Safety
 2 zs LTD
Managing Flow Variability: Safety Inventory
Independent Lead time demands at two locations
48
LTDC = LTD1 + LTD2  LTDC = LTD + LTD = 2 LTD
C
C
VarLTD
 s 2 LTD  s 2 LTD  2s 2 LTD  s LTD
 2s LTD
I
C
Safety
 z 2s LTD
D
I Safety
 2 zs LTD
Centralization reduced the safety inventory by a factor of 1/√2
GE lighting operating 7 warehouses. A warehouse with average lead
time demand of 20,000 units with a standard deviation of 5,000 units
and a 95% service level needs to carry a safety inventory of
Isafety = 1.65×5000= 8250
I
D
safety
 7  8,250  57,750
C
I safety
 1.65  7  5000  21,827
Decrease in safety inventory by a factor of
7  2.65
Managing Flow Variability: Safety Inventory
independent Lead time demands at N locations
49
Centralization of N locations:
I
c
safety
 z N s LTD
Independent demand in N locations: Total safety inventory to
provide a specific SL increases not by N but by √N
If centralization of stocks reduces inventory, why doesn’t everybody do it?
– Longer response time
– Higher shipping cost
– Less understanding of customer needs
– Less understanding of cultural, linguistics, and regulatory barriers
These disadvantages my reduce the demand.
Managing Flow Variability: Safety Inventory
Dependent Demand
50
Does centralization offer similar
benefits when demands in
multiple locations are correlated?
120
100
80
60
40
LTD1 and LTD2 are statistically
identically distributed but correlated
with a correlation coefficient of ρ .
20
0
0
10
20
30
40
50
60
70
90
No Correlation: ρ close to 0
C
(s LTD
) 2  s 2 LTD  s 2 LTD  2s LTDs LTD
C
2
(s LTD
) 2  s 2 LTD  s 2 LTD  2s LTD
 2(1   )s 2 LTD
C
s LTD
 2(1   )s LTD
80
C
I safety
 z  2(1   )s LTD
100
Managing Flow Variability: Safety Inventory
+ Correlation, + Perfect Correlation
51
100
120
90
100
80
70
80
60
50
60
40
40
30
20
20
10
0
0
0
10
20
30
40
50
60
70
80
90
100
Positive Correlation: ρ close to 1
0
10
20
30
40
50
60
70
80
90
100
Perfect Positive Correlation: ρ = +1
120
100
90
100
80
70
80
60
50
60
40
40
30
20
20
10
0
0
0
10
20
30
40
50
60
70
80
90
100
Negative Correlation: ρ close to -1
0
10
20
30
40
50
60
70
80
90
100
Perfect Negative Correlation: ρ = -1
Managing Flow Variability: Safety Inventory
Correlation
52
C
I safety
 2(1   )  z  s LTD
D
I safety
 2  z  s LTD
The safety inventory in the two-location decentralized system is larger
than in the centralized system by a factor of
2/
2(1   ) 
2 /(1   )
If demand is positively fully correlated, ρ = 1, centralization
offers no benefits in the reduction of safety inventory
Benefits of centralization increases as the demand on the two
locations become negatively correlated. The best case is  = -1,
where we do not need safety inventory at all
Managing Flow Variability: Safety Inventory
Principle of Aggregation and Pooling Inventory
53
Inventory benefits  due to principle of aggregation.
Statistics: Standard deviation of sum of random variables is less than
the sum of the individual standard deviations.
Physical consolidation is not essential, as long as available inventory is
shared among various locations  Pooling Inventory
–
Virtual Centralization
–
Specialization
–
Component Commonality
–
Delayed Differentiation
–
Product Substitution
Managing Flow Variability: Safety Inventory
Virtual Centralization
54
Virtual Centralization: inventory pooling in a network of locations is
facilitated using information regarding availability of goods and subsequent
transshipment of goods between locations to satisfy demand.
Location A
Exceeds Available stock
Location B
Less than Available stock
1. Information about product demand and availability must be available
at both locations
2. Shipping the product from one location to a customer at another
location must be fast and cost effective
Pooling is achieved by keeping the inventories at decentralized locations.
Managing Flow Variability: Safety Inventory
Specialization, Substitution
55
Demand for both products exist in both locations. But a large
portion of demand for P1 is in location A, while a large portion
of demand for P2 is in location B.
Location A
Location B
Product P1
Product P2
Both locations keep average inventory.
Safety inventory is kept only in the specialized warehouse
One other possibility to deal with variability is product substitution.
Managing Flow Variability: Safety Inventory
Component Commonality
56
Up to now we have discussed aggregating demand across various geographic
locations, either physical or virtual
Aggregating demand across various products has the same benefits.
Computer manufacturers: offer a wide range of models, but few components,
CPU, RMA, HD, CD/DVD drive, are used across product lines.
Replace Make-to-stock with make Make-to-Order
Commonality + MTO:
Commonality: Safety inventory of the common components much less than
safety inventory of unique components stored separately.
MTO: Inventory cost is computed in terms of WIP cost not in terms of
finished good cost (which is higher).
Managing Flow Variability: Safety Inventory
Postponement (Delayed Differentiation)
57
Forecasting Characteristic: Forecasts further into the future tends to be
less accurate than those of more imminent events.
Since shorter-range forecasts are more accurate, operational decisions
will be more effective if supply is postponed closer to the point of
actual demand.
Two Alternative processes (each activity takes one week)
 Alternative A: (1) Coloring the fabric, (2) assembling T-shirts
 Alternative B: (1) Assembling T-shirts, (2) coloring the fabric
No changes in flow time. Alternative B postponed the color difference
until one week closer to the time of sale. Takes advantage of the
forecasting characteristic: short-Range forecast more accurate.
Managing Flow Variability: Safety Inventory
Postponement (Delayed Differentiation)
58
Two advantages: Taking advantage of two demand forecasting
characteristics
 Commonality Advantage: At week 0; Instead of forecast for each
individual item, we forecast for aggregates item – uncolored Tshirt. Forecast for aggregate demand is more accurate than forecast
for individual item. It is easier to more accurately forecast total
demand for different colored T-shirts for next week than the week
after the next.
 Postponement Advantage: Instead of forecasting for each individual
items two weeks ahead, we do it at week 1. Shorter rang forecasts
are more accurate. It is easier to more accurately forecast demand
for different colored T-shirts for next week than the week after the
next.
Managing Flow Variability: Safety Inventory
Lessons Learned
59
Levels for Reducing Safety Capacity
 Reduce demand variability through improved forecasting
 Reduce replenishment lead time
 Reduce variability in replenishment lead time
 Pool safety inventory for multiple locations or products
 Exploit product substitution
 Use common components
 Postpone product-differentiation processing until closer to the
point of actual demand