Managing Flow Variability: Safety Inventory
1
Operations Management
Session 23: Newsvendor Model
Managing Flow Variability: Safety Inventory
Uncertain Demand
2
Uncertain Demand: What are the relevant trade-offs?
– Overstock
• Demand is lower than the available inventory
• Inventory holding cost
– Understock
• Shortage- Demand is higher than the available inventory
– Why do we have shortages?
– What is the effect of shortages?
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
The Magnitude of Shortages (Out of Stock)
3
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
What are the Reasons?
4
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Consumer Reaction
5
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
What can be done to minimize shortages?
6
Better forecast
Produce to order and not to stock
– Is it always feasible?
Have large inventory levels
Order the right quantity
– What do we mean by the right quantity?
Session 23
Operations Management
6
Managing Flow Variability: Safety Inventory
Uncertain Demand
7
What is the objective?
– Minimize the expected cost (Maximize the expected profits).
What are the decision variables?
– The optimal purchasing quantity, or the optimal inventory
level.
Session 23
Operations Management
7
Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
8
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.
Session 23
Operations Management
8
Managing Flow Variability: Safety Inventory
News Vendor Model
9
Assumptions
–
–
–
–
–
–
–
Demand is random
Distribution of demand is known
No initial inventory
Set-up cost is equal to zero
Single period
Zero lead time
Linear costs:
•
•
•
•
Session 23
Purchasing (production)
Salvage value
Revenue
Goodwill
Operations Management
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Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
10
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?
Session 23
Operations Management
10
Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
11
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 116?
What is probability of demand to be greater than 116?
What is probability of demand to be equal to 113.3?
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Optimal Service Level: The Newsvendor Problem
12
Average Demand
100
110
120
130
140
150
160
170
180
190
200
Session 23
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 145?
What is probability of demand to be greater
than 145?
Operations Management
12
Managing Flow Variability: Safety Inventory
Compute the Average Demand
13
N
Average Demand   X i P( x  X i )
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
X
100
110
120
130
140
150
160
170
180
190
200
P(x=X)
0.02
0.05
0.08
0.09
0.11
0.16
0.2
0.15
0.08
0.05
0.01
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?
Session 23
Operations Management
13
Managing Flow Variability: Safety Inventory
Deamand (X) 100
Porbability
0.02
Prob(x ≥ X)
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
14
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  x ≥ 140
Prob(x ≥ 140) = 0.76
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
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Deamand (X) 100
Porbability
0.02
Prob(x ≥ X)
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
15
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 salvaged 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 salvaged  0.05(30)= 1.5
0.08 probability to sell 120  20 salvaged  0.08(20) = 1.6
0.09 probability to sell 130  10 salvaged  0.09(10) =0.9
0.8 + 1.5 + 1.6 + 0.9 = 4.8
Total number Sold 135.2 @ 700 = 94640
Total number Salvaged 4.8 @ -100 = -480
Expected Profit = 94640 – 480 = 94,160
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Cumulative Probabilities
16
X
100
110
120
130
140
150
160
170
180
190
200
Probabilities
P(x=X)
P(x<X)
P(x≥X)
0.02
0
1
0.05
0.02
0.98
0.08
0.07
0.93
0.09
0.15
0.85
0.11
0.24
0.76
0.16
0.35
0.65
0.2
0.51
0.49
0.15
0.71
0.29
0.08
0.86
0.14
0.05
0.94
0.06
0.01
0.99
0.01
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Number of Units Sold, Salvages
17
X
100
110
120
130
140
150
160
170
180
190
200
Probabilities
P(x=X)
P(x<X)
P(x≥X)
0.02
0
1
0.05
0.02
0.98
0.08
0.07
0.93
0.09
0.15
0.85
0.11
0.24
0.76
0.16
0.35
0.65
0.2
0.51
0.49
0.15
0.71
0.29
0.08
0.86
0.14
0.05
0.94
0.06
0.01
0.99
0.01
Session 23
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
Operations Management
Sold@700
Salvaged@-100
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Managing Flow Variability: Safety Inventory
Total Revenue for Different Ordering Policies
18
X
100
110
120
130
140
150
160
170
180
190
200
Probabilities
P(x=X)
P(x<X)
P(x≥X)
0.02
0
1
0.05
0.02
0.98
0.08
0.07
0.93
0.09
0.15
0.85
0.11
0.24
0.76
0.16
0.35
0.65
0.2
0.51
0.49
0.15
0.71
0.29
0.08
0.86
0.14
0.05
0.94
0.06
0.01
0.99
0.01
Session 23
Units
Sold
Salvaged
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
Operations Management
Sold
70000
76860
83370
89320
94640
99190
102620
104650
105630
106050
106120
Revenue
Salvaged
0
20
90
240
480
830
1340
2050
2910
3850
4840
Total
70000
76840
83280
89080
94160
98360
101280
102600
102720
102200
101280
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Managing Flow Variability: Safety Inventory
Example 2: Denim Wholesaler
19
The demand for denim is:
– 1000 with probability 0.1
Cost parameters:
– 2000 with probability 0.15
– 3000 with probability 0.15
– 4000 with probability 0.2
Unit Revenue (r ) = 30
Unit purchase cost (c )= 10
Salvage value (v )= 5
Goodwill cost (g )= 0
– 5000 with probability 0.15
– 6000 with probability 0.15
– 7000 with probability 0.1
How much should we order?
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
20
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
21
Marginal analysis: What is the value of an additional unit?
Suppose the wholesaler purchases 1000 units
What is the value of the 1001st unit?
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
22
Wholesaler purchases an additional unit
Case 1: Demand is smaller than 1001 (Probability 0.1)
– The retailer must salvage the additional unit and losses $5 (10 – 5)
Case 2: Demand is larger than 1001 (Probability 0.9)
– The retailer makes and extra profit of $20 (30 – 10)
Expected value = -(0.1*5) + (0.9*20) = 17.5
Session 23
Operations Management
22
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
23
What does it mean that the marginal value is positive?
– By purchasing an additional unit, the expected profit
increases by $17.5
The dealer should purchase at least 1,001 units.
Should he purchase 1,002 units?
Session 23
Operations Management
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Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
24
Wholesaler purchases an additional unit
Case 1: Demand is smaller than 1002 (Probability 0.1)
– The retailer must salvage the additional unit and losses $5 (10 – 5)
Case 2: Demand is larger than 1002 (Probability 0.9)
– The retailer makes and extra profit of $20 (30 – 10)
Expected value = -(0.1*5) + (0.9*20) = 17.5
Session 23
Operations Management
24
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
25
Assuming that the initial purchasing quantity is between 1000 and
2000, then by purchasing an additional unit exactly the same
savings will be achieved.
Conclusion: Wholesaler should purchase at least 2000 units.
Session 23
Operations Management
25
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
26
What is the value of the 2001st unit?
Wholesaler purchases an additional unit
Case 1: Demand is smaller than 2001 (Probability 0.25)
– The retailer must salvage the additional unit and losses $5 (10 – 5)
Case 2: Demand is larger than 2001 (Probability 0.75)
– The retailer makes and extra profit of $20 (30 – 10)
Expected value = -(0.25*5) + (0.75*20) = 13.75
Session 23
Operations Management
26
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
27
Why does the marginal value of an additional unit decrease, as
the purchasing quantity increases?
– Expected cost of an additional unit increases
– Expected savings of an additional unit decreases
Session 23
Operations Management
27
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
28
We could continue calculating the marginal values
Demand
1000
2000
3000
4000
5000
6000
7000
Session 23
Probability
0.1
0.15
0.15
0.2
0.15
0.15
0.1
Cumlative
Probability
0.1
0.25
0.4
0.6
0.75
0.9
1
Expected
marginal cost
0.5
1.3
2.0
3.0
3.8
4.5
5.0
Operations Management
Expected marginal
savings
18.0
15.0
12.0
8.0
5.0
2.0
0.0
Marginal Value
17.50
13.75
10.00
5.00
1.25
-2.50
-5.00
28
Managing Flow Variability: Safety Inventory
Example 2: Marginal Analysis
29
What is the optimal purchasing quantity?
– Answer: Choose the quantity that makes marginal value: zero
Marginal value
17.5
13.75
10
5
1.3
Quantity
-2.5
1000
2000
3000 4000
5000
6000
7000 8000
-5
Session 23
Operations Management
29
Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
30
Net Marginal Benefit:
MB = p – c
MB = 30 - 10 = 20
Net Marginal Cost:
MC = c - v
MC = 10-5 = 5
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( D ≤ Q).
The expected marginal cost =MC× P( D ≤ 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(D>Q) = 1- P( D≤ Q).
The expected marginal benefit = MB× [1-Prob.( D ≤ Q)]
Session 23
Operations Management
30
Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
31
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(D ≤ Q*) ≥ MB× [1 – Prob(D ≤ Q*)]
MB
Prob(D ≤ Q*) ≥
MB  MC
MB = p – c = Underage Cost = Cu
MC = c – v = Overage Cost = Co
MB
P( D  Q ) 
MB  MC
*
Session 23
cu

Cu  C o
Operations Management
pc
pc


pccv pv
31
Managing Flow Variability: Safety Inventory
Marginal Value: The General Formula
32
P(D ≤ Q*) ≥ Cu / (Co+Cu)
Cu / (Co+Cu) = (30-10)/[(10-5)+(30-10)] = 20/25 = 0.8
Order until P(D ≤ Q*) ≥ 0.8
P(D ≤ 5000) ≥ = 0.75 not > 0.8 still order
P(D ≤ 6000) ≥ = 0.9 > 0.8 Stop
Order 6000 units
Session 23
Operations Management
Demand
1000
2000
3000
4000
5000
6000
7000
Cum
Prob
0
0
0
0
0
0
Probability
0.1
0.15
0.15
0.2
0.15
0.15
0.1
32
Managing Flow Variability: Safety Inventory
Analytical Solution for the Optimal Service Level
33
In Continuous Model where demand for example has
Uniform or Normal distribution
MB
cu
P( D  Q ) 

MB  MC
Cu  C o
*
Session 23
pc

pv
Operations Management
33
Managing Flow Variability: Safety Inventory
Marginal Value: Uniform distribution
34
Suppose instead of a discreet demand of
Demand
1000
2000
3000
4000
5000
6000
7000
Probability
0.1
0.15
0.15
0.2
0.15
0.15
0.1
Cumlat
Probab
0.1
0.25
0.4
0.6
0.75
0.9
1
We have a continuous demand uniformly distributed between
1000 and 7000
1000
7000
Pr{D ≤ Q*} = 0.80
How do you find Q?
Session 23
Operations Management
34
Managing Flow Variability: Safety Inventory
Marginal Value: Uniform distribution
35
Q-l = Q-1000
?
0.80
l=1000
1/6000
u=7000
u-l=6000
(Q-1000)*1/6000=0.80
Q = 5800
Session 23
Operations Management
35
Managing Flow Variability: Safety Inventory
Type-1 Service Level
36
What is the meaning of the number 0.80?
F(Q) = (30 – 10) / (30 – 5) = 0.8
–
Pr {demand is smaller than Q} =
–
Pr {No shortage} =
–
Pr {All the demand is satisfied from stock} = 0.80
It is optimal to ensure that 80% of the time all the demand is
satisfied.
Session 23
Operations Management
36
Managing Flow Variability: Safety Inventory
Marginal Value: Normal Distribution
37
Suppose the demand is normally distributed with a mean of 4000
and a standard deviation of 1000.
What is the optimal order quantity?
Notice: F(Q) = 0.80 is correct for all distributions.
We only need to find the right value of Q assuming the normal
distribution.
Session 23
Operations Management
37
Managing Flow Variability: Safety Inventory
Marginal Value: Normal Distribution
38
0.00045
0.0004
Probability of
excess inventory
0.00035
Probability of
shortage
4841
0.0003
0.00025
Series1
0.0002
0.00015
0.80
0.0001
0.00005
0.20
0
0
Session 23
1000
2000
3000
4000
5000
Operations Management
6000
7000
8000
38
Managing Flow Variability: Safety Inventory
Type-1 Service Level
39
Recall that:
F(Q) = Cu / (Co + Cu) = Type-1 service level
Session 23
Operations Management
39
Managing Flow Variability: Safety Inventory
Type-1 Service Level
40
Is it correct to set the service level to 0.8?
Shouldn’t we aim to provide 100% serviceability?
Session 23
Operations Management
40
Managing Flow Variability: Safety Inventory
Type-1 Service Level
41
What is the optimal purchasing quantity?
0.00045
Probability of
excess inventory
0.0004
0.00035
Probability of
shortage
0.0003
0.00025
5282
0.0002
Series1
0.90
0.00015
0.0001
0.00005
0
0
Session 23
1000
2000
3000
4000
5000
6000
Operations Management
7000
8000
41
Managing Flow Variability: Safety Inventory
Type-1 Service Level
42
How do you determine the service level?
For normal distribution, it is always optimal to have:
Mean + z*Standard deviation
µ + zs
The service level determines the value of Z
zs is the level of safety stock
m +zs is the base stock (order-up-to level)
Session 23
Operations Management
42
Managing Flow Variability: Safety Inventory
Type-1 Service Level
43
Given a service level, how do we calculate z?
From our normal table or
From Excel
– Normsinv(service level)
Session 23
Operations Management
43
Managing Flow Variability: Safety Inventory
Additional Example
44
Your store is selling calendars, which cost you $6.00 and sell for
$12.00 You cannot predict demand for the calendars with certainty.
Data from previous years suggest that demand is well described by a
normal distribution with mean value 60 and standard deviation 10.
Calendars which remain unsold after January are returned to the
publisher for a $2.00 "salvage" credit. There is only one opportunity
to order the calendars. What is the right number of calendars to
order?
Session 23
Operations Management
44
Managing Flow Variability: Safety Inventory
Additional Example - Solution
45
MC= Overage Cost = Co = Unit Cost – Salvage = 6 – 2 = 4
MB= Underage Cost = Cu = Selling Price – Unit Cost = 12 – 6 = 6
P( D  Q* ) 
Cu
6

 0.6
Cu  Co 6  4
Look for P(Z ≤ z) = 0.6 in Standard Normal table or for
NORMSINV(0.6) in excel  0.2533
Q*  m
Q*  m
P( Z 
)  0.6 
 0.2533
s
s
Q*  m  0.2533s  60  10(0.2533)  62.533  63
By convention, for the continuous demand distributions, the results are
rounded to the closest integer.
Session 23
Operations Management
45
Managing Flow Variability: Safety Inventory
Additional Example - Solution
46
Suppose the supplier would like to decrease the unit cost in order to
have you increase your order quantity by 20%. What is the minimum
decrease (in $) that the supplier has to offer.
Qnew = 1.2 * 63 = 75.6 ~ 76 units
P( D  Q* )  P( D  76)  P( z 
76  60
)  P( Z  1.6)
10
Look for P(Z ≤ 1.6) = 0.6 in Standard Normal table or for
NORMSDIST(1.6) in excel  0.9452
P( D  Q* )  0.9452 
Cu
pc
12  c 12  c



Cu  Co p  c  c  v 12  2
10
12  c  9.452  c  2.55
Session 23
Operations Management
46
Managing Flow Variability: Safety Inventory
Additional Example
47
On consecutive Sundays, Mac, the owner of your local newsstand, purchases a
number of copies of “The Computer Journal”. He pays 25 cents for each copy
and sells each for 75 cents. Copies he has not sold during the week can be
returned to his supplier for 10 cents each. The supplier is able to salvage the
paper for printing future issues. Mac has kept careful records of the demand
each week for the journal. The observed demand during the past weeks has the
following distribution:
Quantity Q
Probability p(D=Q)
4
0.04
5
0.06
6
0.16
7
0.18
8
0.20
9
0.10
10
0.10
11
0.08
12
0.04
13
0.04
What is the optimum order quantity for Mac to minimize his cost?
Session 23
Operations Management
47
Managing Flow Variability: Safety Inventory
Additional Example - Solution
48
Overage Cost = Co = Unit Cost – Salvage = 0.25 – 0.1 = 0.15
Underage Cost = Cu = Selling Price – Unit Cost = 0.75 – 0.25 = 0.50
Cu
0.50
P( D  Q*) 

 0.77
Cu  Co 0.50  0.15
P( D  Q* )  0.77
The critical ratio, 0.77, is between Q = 9 and Q = 10.
Q
4
5
6
7
8
9
10
11
12
13
Probability
p(D=Q)
0.04
0.06
0.16
0.18
0.20
0.10
0.10
0.08
0.04
0.04
Cumulative
Probability
F(Q)
0.04
0.10
0.26
0.44
0.64
0.74
0.84
0.92
0.96
1.00
Remember from the marginal analysis explanation that the results are rounded up.
Because at 9 still it is at our benefit to order one more.
So Q* = 10
Session 23
Operations Management
48