Inventory
MATERIAL FLOW SYSTEMS
Overview
DEMAND MODELLING
INVENTORY MODELS
Probability review
Newsvendor problem
◦ Binomial distribution
◦ Poisson distribution
◦ Normal distribution
Distribution fitting
EOQ model
General inventory strategy
◦ When to reorder
◦ How much to order
Demand Modelling
INVENTORY
Binomial distribution
The binomial distribution is appropriate for modeling demand when there are n
customers and each of them will purchase with probability p.
𝑃𝑃 𝐷𝐷 = 𝑥𝑥 =
𝑛𝑛 𝑥𝑥
𝑝𝑝 1 − 𝑝𝑝
𝑥𝑥
𝑛𝑛−𝑥𝑥
where
𝑛𝑛
𝑛𝑛!
.
=
𝑥𝑥! 𝑛𝑛−𝑥𝑥 !
𝑥𝑥
If there are 10 customers and p = 0.1, probability of selling 3 items is:
𝑃𝑃 𝐷𝐷 = 3 =
10!
× 0.13 × 0.97 = 0.0574
3! 7!
Q1-15
Binomial distribution
The binomial distribution is appropriate for modeling demand when there are n
customers and each of them will purchase with probability p.
𝜇𝜇 = 𝑛𝑛𝑛𝑛
and
𝜎𝜎 2 = 𝑛𝑛𝑛𝑛(1 − 𝑝𝑝)
See here for proof.
If there are 20 customers and p = 0.4:
◦ Expected demand is 20 × 0.4 = 8
◦ Demand variance is 20 × 0.4 × 0.6 = 4.8
Q16
Poisson distribution
The Poisson distribution can be appropriate for modeling demand when average
demand m is known.
𝑃𝑃 𝐷𝐷 = 𝑥𝑥 =
𝑒𝑒 −𝑚𝑚
𝑚𝑚 𝑥𝑥
𝑥𝑥!
If average daily demand is 10, probability of selling 7 items is:
𝑃𝑃 𝐷𝐷 = 7 =
𝑒𝑒 −10
107
= 0.0901
7!
Q17
Binomial to Poisson
The binomial distribution is
appropriate for modeling demand
when there are n customers and each
will purchase with probability p.
However, determining the value of n
to use can be tricky in practice.
As n increases, the binomial
distribution approaches the Poisson
distribution.
Suppose a store sells an average of 10
products each day. Furthermore, the
largest sale volume in a day is 20.
Under this setting, we could assume
𝑛𝑛 = 20 and 𝑝𝑝 = 0.5.
However, selling 100 products in a day
is also possible. Hence, one can also
assume 𝑛𝑛 = 100 and 𝑝𝑝 = 0.1.
Q18-21
Binomial to Poisson
The binomial distribution is appropriate for modeling demand when there are n
customers and each of them will purchase with probability p.
However, determining the value of n to use can be tricky in practice. In theory, n can be
very large in many cases.
As n increases, the binomial distribution approaches the Poisson distribution. Hence, the
Poisson distribution suitable for modeling demand when average demand m is known.
Under the Poisson distribution, mean and variance take the same value:
See here and here for proof.
𝜇𝜇 = 𝜎𝜎 2 = 𝑚𝑚
Q22
Normal distribution
The normal distribution can be appropriate for modeling demand when
average demand 𝜇𝜇 and demand variance 𝜎𝜎 2 are both known.
Similar to the Poisson distribution, the normal distribution
approximates a binomial distribution when n is sufficiently large (i.e.,
𝑛𝑛𝑛𝑛 > 5 and 𝑛𝑛 1 − 𝑝𝑝 > 5).
Q29
However, it is more flexible than the Poisson distribution in that it
allows for differing mean and variance values.
Normal distribution
Q23-28
The normal distribution is a continuous distribution but demand is often an
integer value. Hence continuity correction is applied when estimating
demand using a normal distribution:
𝑃𝑃 Demand = 5 = 𝑃𝑃 4.5 ≤ 𝑋𝑋 ≤ 5.5 = 𝑃𝑃
4.5 − 𝜇𝜇
5.5 − 𝜇𝜇
≤ 𝑍𝑍 ≤
𝜎𝜎
𝜎𝜎
Probabilities based on normal distributions can be obtained by looking up
the standard normal table:




𝑃𝑃
𝑃𝑃
𝑃𝑃
𝑃𝑃
𝑍𝑍
𝑍𝑍
𝑍𝑍
𝑍𝑍
≤ 𝑥𝑥
≤ 𝑥𝑥
≥ 𝑥𝑥
≥ 𝑥𝑥
for positive x
for negative x
for positive x
for negative x
Linear interpolation
The z-values in the standard normal table are only given up to 2 dec. pl.
Linear interpolation can be used to estimate the probability p
associated with a given z value that is larger and smaller than z1 and z2,
respectively:
𝑝𝑝 ≈
𝑝𝑝1 𝑧𝑧2 − 𝑧𝑧 + 𝑝𝑝2 𝑧𝑧 − 𝑧𝑧1
𝑧𝑧2 − 𝑧𝑧1
Interested students can find out more details from the URL below.
https://en.wikipedia.org/wiki/Linear_interpolation#Linear_interpolation_between_two_known_points
Linear interpolation
Suppose we are interested in 𝑃𝑃 𝑍𝑍 ≤ 0.231 .
Clearly:
𝑃𝑃 𝑍𝑍 ≤ 0.23 < 𝑃𝑃 𝑍𝑍 ≤ 0.231 < 𝑃𝑃 𝑍𝑍 ≤ 0.24
0.5910 < 𝑃𝑃 𝑍𝑍 ≤ 0.231 < 0.5948
By applying the linear interpolation formula given in the previous slide:
𝑃𝑃 𝑍𝑍 ≤ 0.231 ≈
0.5910 0.24 − 0.231 + 0.5948 0.231 − 0.23
= 0.5914
0.24 − 0.23
Normal distribution
Unlike binomial distribution, normal distribution does not require the
number of customers, which can be difficult to determine in practice.
Unlike Poisson distribution, normal distribution makes use of both
mean and variance information.
It is also easier to compute demand across multiple periods under the
normal distribution.
Demand across multiple periods
We are often interested in demand across multiple periods. For example, we want to
know the number of patients in a month.
Suppose daily demand is equally likely to be 0, 1 or 2. Demand across two days is:
Scenario
Demand
0, 0
0
0, 1
1
0, 2
2
1, 0
1
1, 1
2
1, 2
3
2, 0
2
2, 1
3
2, 2
4
P(Demand = 0) =
P(Demand = 1) =
P(Demand = 2) =
P(Demand = 3) =
P(Demand = 4) =
1
9
2
9
3
9
2
9
1
9
Analyzing demand
across 30 days is
challenging.
Demand across multiple periods
We are often interested in demand across multiple periods. For example, we want to
know the number of patients in a month.
Suppose daily demand is normally distributed with mean 100 and variance 200. Demand
across 30 days is also normally distributed with:
mean: 30 × 100 = 3000
and
variance 30 × 200 = 6000
More generally, if daily demand is normally distributed with mean 𝜇𝜇 and variance 𝜎𝜎 2 , then
demand across 𝑘𝑘 days is also normally distributed with mean 𝑘𝑘𝑘𝑘 and variance 𝑘𝑘𝜎𝜎 2 .
The above assumes daily demand is independent.
Distribution fitting
The normal distribution is defined by its mean and variance. Hence we estimate
these values from the data available.
Suppose the following 12 demand data are available: 85, 55, 150, 93, 65, 197,
54, 84, 116, 70, 54 and 69.
Estimated 𝜇𝜇 =
Estimated
𝜎𝜎 2
=
85+55+⋯+69
12
= 91
85−91 2 + 55−91 2 +⋯+ 69−91 2
12−1
= 1915.1
Q30
Distribution fitting
Suppose the following 12 demand data are available: 85, 55, 150, 93, 65, 197,
54, 84, 116, 70, 54 and 69.
Demand is modelled using a normal distribution defined by the estimated mean
and variance obtained (i.e., 91 and 1915.1)
Under the above assumption:
𝑃𝑃 𝐷𝐷 ≤ 101 = 𝑃𝑃 𝑍𝑍 ≤
101.5 − 91
1915.1
= 𝑃𝑃 𝑍𝑍 ≤ 0.24 = 0.59
Q30
Distribution fitting
Suppose the following 12 demand data are available: 85, 55, 150, 93, 65, 197,
54, 84, 116, 70, 54 and 69.
We estimate 𝑃𝑃 𝐷𝐷 ≤ 101 by assuming that demand is normally distributed
with mean 91 and variance 1915.1.
Next, we check if this assumption is reasonable. If the assumption is true:
◦ 𝑃𝑃 Demand ≤ 91 = 𝑃𝑃 𝑍𝑍 ≤
91.5−91
1915.1
= 𝑃𝑃 𝑍𝑍 ≤ 0.0114 = 0.505
◦ 𝑃𝑃 Demand > 91 = 1 − 𝑃𝑃 Demand ≤ 91 = 0.495
Distribution fitting
Assumption may
not be reasonable
Suppose the following 12 demand data are available: 85, 55, 150, 93, 65, 197,
54, 84, 116, 70, 54 and 69.
We estimate 𝑃𝑃 𝐷𝐷 ≤ 101 by assuming that demand is normally distributed
with mean 91 and variance 1915.1.
Q31-37
Next, we check if this assumption is reasonable. If the assumption is true:
◦ Expected observations ≤ 91 = 12 × 0.505 = 6.06
◦ Expected observations > 91 = 12 × 0.495 = 5.94
Actual observations is 8
Actual observations is 4
Data transformation
If data is not normally distributed, one can try to make the data
“normal” by applying some transformation.
Log transformation:
Power transformation:
𝑌𝑌 = ln(𝑋𝑋)
𝑌𝑌 = 𝑋𝑋 𝜆𝜆 where 𝜆𝜆 ≠ 0
Commonly used 𝜆𝜆 values are:
-3, -2, -1, -0.5, 0.5, 1, 2 and 3
Data transformation
If data is not normally distributed, one can try to make the data
“normal” by applying some transformation.
Demand
ln(Demand)
85
4.443
55
4.007
150
5.011
93
4.533
65
4.174
197
5.283
54
3.989
84
4.431
116
4.754
70
4.248
54
3.989
69
4.234
• Estimated mean =
4.443+4.007+⋯+4.234
12
• 6 data points are above 4.425
= 4.425
• 6 data points are below 4.425
• Data appears to fits the properties of the
normal distribution better after transformation.
Q38-39
Data transformation
If data is not normally distributed, one can try to make the data
“normal” by applying some transformation.
Demand
ln(Demand)
85
4.443
55
4.007
150
5.011
93
4.533
65
4.174
197
5.283
54
3.989
84
4.431
116
4.754
70
4.248
54
3.989
69
4.234
Estimated mean =
4.443+4.007+⋯+4.234
12
Estimated variance = 0.171
𝑃𝑃 Demand ≤ 101 = 𝑃𝑃 𝑍𝑍 ≤
= 4.425
ln(101.5) − 4.425
= 𝑃𝑃 𝑍𝑍 ≤ 0.472
This probability
estimate is better since
ln(demand) is better = 0.68
described by the
normal distribution.
0.171
Q40-43
Checking model fit
Previously, we subjectively concluded that the fit is poor based on observations:
◦ At most 91:
8 (actual) versus 6.06 (expected)
◦ Greater than 91: 4 (actual) versus 5.94 (expected)
The Pearson's chi-squared test (goodness-of-fit test) is a more objective way for assessing fit:
𝑘𝑘
observed𝑖𝑖 − expected𝑖𝑖 2
2
𝜒𝜒 = �
expected𝑖𝑖
𝑖𝑖=1
where 𝑘𝑘 is the number of bins used. As a guide, expected𝑖𝑖 should be at least 5.
Interested students can find out more about Pearson's chi-squared test here.
For assessing normality, can use Shapiro-Wilk normality test. See online tool and description.
Demand modeling procedure
1. Check data for normality.
2. If fit is poor, perform transformation to improve fit.
3. Estimate mean and variance from data.
4. Use estimated mean and variance to determine
probabilities.
Summary (demand modeling)
Demand modeling
 Binomial distribution: n customers, each purchase with probability p.
 Poisson distribution: only average demand m is known.
 Normal distribution: average demand 𝜇𝜇 and demand variance 𝜎𝜎 2 are known.
Advantages of modeling demand using the normal distribution:
 Uses information regarding average demand and demand variance
 Demand across multiple periods easy to analyze
If modeling based on historical data, check if data is normally
distributed. Data transformation should be used if data is not normally
distributed.
Inventory Models
INVENTORY
Newsvendor problem
Suppose demand for newspaper in a given day is as follows:
Demand
Probability
20
0.2
21
0.2
22
0.2
23
0.2
24
0.2
Furthermore, each paper sells for $1.00 and the cost price
of each paper is $0.30.
How many papers should be ordered?
Newsvendor problem
Suppose we order 20 newspapers:
𝐸𝐸[Profit] = 0.2 $20 + $20 + $20 + $20 + $20 − 20 $0.3 = $14
Suppose we order 21 newspapers:
𝐸𝐸[Profit] = 0.2 $20 + $21 + $21 + $21 + $21 − 21 $0.3 = $14.5
Newsvendor problem
Suppose we order 22 newspapers:
𝐸𝐸[Profit] = 0.2 $20 + $21 + $22 + $22 + $22 − 22 $0.3 = $14.8
Suppose we order 23 newspapers:
𝐸𝐸[Profit] = 0.2 $20 + $21 + $22 + $23 + $23 − 23 $0.3 = $14.9
Suppose we order 24 newspapers:
highest expected profit
𝐸𝐸[Profit] = 0.2 $20 + $21 + $22 + $23 + $24 − 24 $0.3 = $14.8
Q1-6
Newsvendor solution
Optimal order quantity: 𝑄𝑄 ∗ = max 𝑄𝑄: 𝑃𝑃 Demand ≥ 𝑄𝑄 ≥
Demand
Probability
𝑃𝑃(𝐷𝐷 ≥ 𝑄𝑄)
20
0.2
1.0
21
0.2
0.8
22
0.2
0.6
cost
price
23
0.2
0.4
24
0.2
0.2
Each paper sells for $1.00 and the cost price of each paper is $0.30.
The optimal order quantity is 23.
Q7-12
Newsvendor solution explained
Suppose we order 20 newspapers:
𝐸𝐸[Profit] = 0.2 $20 + $20 + $20 + $20 + $20 − 20 $0.3 = $14
Suppose we order 21 newspapers:
𝐸𝐸[Profit] = 0.2 $20 + $21 + $21 + $21 + $21 − 21 $0.3 = $14.5
Benefit of ordering 21 instead of 20:
$1 × 0.2 + 0.2 + 0.2 + 0.2 − 1 × $0.3 = $0.5
Newsvendor solution explained
Benefit of ordering 21 instead of 20:
1 × 0.2 + 0.2 + 0.2 + 0.2 − 1 × 0.3 = 0.5
Benefit of ordering Q instead of Q – 1:
price × 𝑃𝑃 Demand ≥ 𝑄𝑄 − cost ≥ 0
price × 𝑃𝑃 Demand ≥ 𝑄𝑄 ≥ cost
𝑃𝑃(Demand ≥ 𝑄𝑄) ≥
cost
price
Economic order quantity model
EOQ model studies a simple multi-period inventory problem.
Assumptions:
◦
◦
◦
◦
◦
◦
Demand is deterministic and uniform over time
Unit cost of each item is constant
Ordering cost is constant
Inventory holding cost is linear with respect to inventory
Lead time is constant
All demand must be satisfied, no stock-out allowed
Economic order quantity model
EOQ model studies a simple multi-period inventory problem.
Inventory
D: daily demand
Q
Q: order quantity
L: lead time (in days)
R=DL
𝑄𝑄
− 𝐿𝐿
𝐷𝐷
𝑄𝑄
𝐷𝐷
Time
R: reorder point
How much should we order?
Q22-23
Costs in EOQ model
Costs in EOQ model
◦ Holding cost: Cost for storage, insurance, opportunity cost
◦ Ordering cost: Cost of placing an order
Q13-18
Objective: Minimize total cost (ordering + holding)
𝐷𝐷
𝑄𝑄
Ordering Cost = 𝐶𝐶
𝐶𝐶: Unit ordering cost
Holding Cost = 𝐻𝐻
𝐻𝐻: Unit holding cost
𝑄𝑄
2
Costs in EOQ model
Costs in EOQ model
◦ Holding cost: Cost for storage, insurance, opportunity cost
◦ Ordering cost: Cost of placing an order
Objective: Minimize total cost (ordering + holding)
𝐷𝐷
𝑄𝑄
𝑇𝑇𝑇𝑇 = 𝐶𝐶 + 𝐻𝐻
𝑄𝑄
2
𝑑𝑑
−𝐷𝐷𝐷𝐷
𝐻𝐻
𝑇𝑇𝑇𝑇 =
+
𝑑𝑑𝑑𝑑
𝑄𝑄 2
2
−𝐷𝐷𝐷𝐷
𝐻𝐻
+
=0
𝑄𝑄 ∗ 2
2
𝑄𝑄∗ =
2𝐷𝐷𝐷𝐷
𝐻𝐻
Q19-20
EOQ model example
C = $50 per order
D = 365 per year
H = $2.30 per year
L = 5 days
Find the optimal order quantity Q* and reorder point R:
𝑄𝑄∗
=
2𝐷𝐷𝐷𝐷
=
𝐻𝐻
𝑇𝑇𝐶𝐶125 =
𝑇𝑇𝐶𝐶126
2 × 365 × $50
= 125.97 units => 126 units
$2.30
365
125
$50 +
$2.30 = $289.75
2
125
365
126
=
$50 +
$2.30 = $289.74
2
126
lower total cost
Q21
EOQ model example
C = $50 per order
D = 365 per year
H = $2.30 per year
L = 5 days
Find the optimal order quantity Q* and reorder point R:
What if D = 354 per year?
𝑅𝑅 = 𝐷𝐷𝐷𝐷 =
365
× 5 = 5 units
365
Q22-23
General inventory policy
In practice, demand is uncertain and most products are not highly
perishable.
Here we discuss the multi-period inventory problem with uncertain
demand.
There are two parameters to determine:
◦ r: reorder point => when do we place an order
◦ Q: order quantity => how much do we order
We study this first
Order quantity, Q
Suppose objective is to minimize total cost (ordering + holding):
𝑇𝑇𝑇𝑇 =
𝐷𝐷
𝑄𝑄
𝐶𝐶 + 𝐻𝐻
𝑄𝑄
2
𝑑𝑑
−𝐷𝐷𝐷𝐷
𝑇𝑇𝑇𝑇 =
𝑑𝑑𝑑𝑑
𝑄𝑄2
−𝐷𝐷𝐷𝐷
𝐻𝐻
=0
+
𝑄𝑄∗ 2
2
𝑄𝑄∗ =
2𝐷𝐷𝐷𝐷
𝐻𝐻
𝐸𝐸 𝑇𝑇𝑇𝑇 =
Approximately
true if demand
is stationary
𝐸𝐸 𝐷𝐷
𝑄𝑄
𝐶𝐶 + 𝐻𝐻
𝑄𝑄
2
𝑑𝑑
−𝐸𝐸 𝐷𝐷 𝐶𝐶
𝐻𝐻
𝐸𝐸 𝑇𝑇𝑇𝑇 =
+
𝑑𝑑𝑑𝑑
𝑄𝑄2
2
−𝐸𝐸 𝐷𝐷 𝐶𝐶
𝐻𝐻
=0
+
𝑄𝑄∗ 2
2
𝑄𝑄∗ =
2𝐸𝐸 𝐷𝐷 𝐶𝐶
𝐻𝐻
In practice, an
approximate
solution suffices.
Order quantity, Q
C = $50 per order
𝑄𝑄 ∗ =
H = $2 per year
=
Daily demand ∼ N(100,1000)
2 × 𝐸𝐸 𝐷𝐷 × 𝐶𝐶
𝐻𝐻
2 × (365 × 100) × $50
$2
= 1350.9
𝑇𝑇𝑇𝑇1350 =
𝑇𝑇𝑇𝑇1351 =
365 × 100
1350
$50 +
$2 = $2701.852
2
1350
365 × 100
1351
$50 +
$2 = $2701.851
2
1351
𝑸𝑸∗ = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
Reorder point, r
Ideally, inventory should be zero when the next batch of goods arrives.
When demand is deterministic, the reorder point is the inventory level that
is sufficient to meet demand over the order lead time.
When demand is uncertain, time when inventory hits zero is unclear.
Low reorder point (too late):
Delayed delivery
High reorder point (too early):
Positive inventory
Reorder point, r
NEWSVENDOR PROBLEM
Q too low: loss sales
REORDER POINT
r too low:
price – cost
Q too high: wastage
delay penalty, S
r too high: positive inventory
cost
cost
𝑄𝑄∗ = max 𝑄𝑄: 𝑃𝑃 Demand ≥ 𝑄𝑄 ≥
price
inventory holding cost, H
H
𝑟𝑟 = max 𝑟𝑟: 𝑃𝑃 Demand ≥ 𝑟𝑟 ≥
H+S
∗
Reorder point, r
What does it mean when a company’s inventory policy is such that:
𝑃𝑃 Demand ≥ 𝑟𝑟 = 0.05
It implies that:
𝐻𝐻
𝐻𝐻 + 𝑆𝑆
𝑆𝑆
≈ 0.05 ⇒
≈ 20 ⇒
≈ 19
𝐻𝐻 + 𝑆𝑆
𝐻𝐻
𝐻𝐻
Delaying the delivery of one item to customer by one day cost 19 times more then
the cost of keeping one item in inventory each day.
Reorder point, r
H = $2 per item per year
S = $18 per item per year
Daily demand ∼ N(100,1000)
L = 5 days
𝑟𝑟 ∗ = max 𝑟𝑟: 𝑃𝑃 Demand ≥ 𝑟𝑟 ≥
= max 𝑟𝑟: 𝑃𝑃 Demand ≥ 𝑟𝑟 ≥
H
H+S
$2
$2+$18
= max 𝑟𝑟: 𝑃𝑃 Demand ≥ 𝑟𝑟 ≥ 0.1
= max 𝑟𝑟: 𝑃𝑃 Demand < 𝑟𝑟 ≤ 0.9
= max 𝑟𝑟: 𝑃𝑃 𝑍𝑍 ≤
= max 𝑟𝑟:
𝑟𝑟 − 0.5 − 5 × 100
5 × 1000
𝑟𝑟 − 0.5 − 5 × 100
5 × 1000
≤ 1.28
= max 𝑟𝑟: 𝑟𝑟 ≤ 591.0097 ⇒ 𝟓𝟓𝟓𝟓𝟓𝟓
≤ 0.9
Summary (inventory model)
Newsvendor problem: Single period with uncertain demand
 Loss sales vs. wastage:
𝑄𝑄 ∗ = max 𝑄𝑄: 𝑃𝑃 Demand ≥ 𝑄𝑄 ≥
EOQ model: Multi-period with deterministic uniform demand
 Ordering cost vs. holding cost:
𝑄𝑄 ∗ =
2𝐷𝐷𝐷𝐷
𝑄𝑄 ∗ =
2×𝐸𝐸 𝐷𝐷 ×𝐶𝐶
and
𝐻𝐻
cost
price
𝑅𝑅 = 𝐷𝐷𝐷𝐷
Multi-period with uncertain demand
 Ordering cost vs. “regular” holding cost:
 Delay cost vs. “excess” holding cost:
𝐻𝐻
𝑟𝑟 ∗ = max 𝑟𝑟: 𝑃𝑃 Demand ≥ 𝑟𝑟 ≥
H
H+S
Factory dynamics
The next 3 weeks will be online learning. It is assumed that you have some
mathematics, probability, statistics and MS Excel background.
For students lacking the necessary background, the following videos might be useful:
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MS Excel: Relative and absolute cell references
Uniform distribution
Exponential distribution
Sample variance
Finding inverse of a function
FD-EXP (Week 7 E-Learning).xlsx is available for download from LumiNUS.
The next lecture is on 12 Mar 2020.