Transcript Ch13 - E-Learning/An-Najah National University

Optimal Level of Product Availability
Chapter 13 of Chopra
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Outline
 Determining
optimal level of product availability
– Single order in a season
– Continuously stocked items
 Ordering
under capacity constraints
 Levers to improve supply chain profitability
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Example: Apparel Industry
How much to order? Parkas at L.L. Bean
Demand
D_i
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Probabability
p_i
.01
.02
.04
.08
.09
.11
.16
.20
.11
.10
.04
.02
.01
.01
Cumulative Probability of demand Probability of demand
being this size or less, F(.)
greater than this size, 1-F(.)
.01
.99
.03
.97
.07
.93
.15
.85
.24
.76
.35
.65
.51
.49
.71
.29
.82
.18
.92
.08
.96
.04
.98
.02
.99
.01
1.00
.00
Expected demand is 1,026 parkas.
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Parkas at L.L. Bean
Cost per parka = c = $45
Sale price per parka = p = $100
Discount price per parka = $50
Holding and transportation cost = $10
Salvage value per parka = s = 50-10=$40
Profit from selling parka = p-c = 100-45 = $55
Cost of overstocking = c-s = 45-40 = $5
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Optimal level of product availability
p = sale price; s = outlet or salvage price; c = purchase price
CSL = Probability that demand will be at or below reorder point
Raising the order size if the order size is already optimal
Expected Marginal Benefit =
=P(Demand is above stock)*(Profit from sales)=(1-CSL)(p - c)
Expected Marginal Cost =
=P(Demand is below stock)*(Loss from discounting)=CSL(c - s)
Define Co= c-s; Cu=p-c
(1-CSL)Cu = CSL Co
CSL= Cu / (Cu + Co)
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Order Quantity for a Single Order
Co = Cost of overstocking = $5
Cu = Cost of understocking = $55
Q* = Optimal order size
Cu
55
CSL  P( Demand  Q ) 

 0.917
Cu  Co 55  5
*
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Optimal Order Quantity
1.2
0.917
1
0.8
Cumulative
Probability
0.6
0.4
0.2
0
4 5 6 7 8 9 10 11 12 13 14 15 16 87
Optimal Order Quantity = 13(‘00)
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Parkas at L.L. Bean
Additional Expected
Expected
Expected Marginal
100s
Marginal Benefit Marginal Cost Contribution
11th
5500.49 = 2695 500.51 = 255 2695-255 = 2440
12th
5500.29 = 1595 500.71 = 355 1595-355 = 1240
13th
5500.18 = 990
500.82 = 410 990-410 = 580
14th
5500.08 = 440
500.92 = 460 440-460 = -20
15th
5500.04 = 220
500.96 = 480 220-480 = -260
16th
5500.02 = 110
500.98 = 490 110-490 = -380
17th
5500.01 = 55
500.99 = 495 55-495 = -440
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Revisit L.L. Bean as a Newsvendor Problem

Total cost by ordering Q units:
– C(Q) = overstocking cost
+
understocking cost
Q

0
Q
C (Q)  C o  (Q  x) f ( x)dx  C u  ( x  Q) f ( x)dx
dC (Q)
 Co F (Q)  Cu (1  F (Q))  F (Q)(Co  Cu )  Cu  0
dQ
Marginal cost of raising Q* - Marginal cost of decreasing Q* = 0
Cu
F (Q )  P( D  Q ) 
C o  Cu
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*
Show Excel to compute expected single-period cost curve.
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The Newsvendor Approach

Assumptions:
1. single period
2. random demand with known distribution
3. linear overage/shortage costs
4. minimum expected cost criterion

Examples:
– newspapers or other items with rapid obsolescence
– Christmas trees or other seasonal items
– capacity for short-life products
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Newsvendor Model Notation
X  demand (in units), a random variable.
G ( x)  P ( X  x), cumulative distribution function of demand
(assumed continuous.)
g ( x) 
d
G ( x)  density function of demand.
dx
co  cost (in dollars) per unit left over after demand is realized.
c s  cost (in dollars) per unit of shortage.
Q  production/order quantity (in units); this is the decision variable.
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Newsvendor Model
 Cost
Function:
Y ( x)  expected overage  expected shortage cost
 co E units over  c s E units short

Note: for any given
day, we will be either
over or short, not both.
But in expectation,
overage and shortage
can both be positive.

 co  max Q  x,0g ( x)dx  c s  max x  Q,0g ( x)dx
0
0
 co

Q
0
(Q  x) g ( x)dx  c s

 ( x  Q) g ( x)dx
Q
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Newsvendor Model (cont.)
 Optimal Solution: taking derivative of Y(Q) with respect to
Q, setting equal to zero, and solving yields:


G(Q * )  P X  Q * 
 Notes:
Q *  co
Q *  cs
cs
co  c s
Critical Ratio is
probability stock
covers demand
1
G(x)
cs
co  c s
Q*
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Newsvendor Example – T Shirts
 Scenario:
– Demand for T-shirts is exponential with mean 1000
(i.e., G(x) = P(X  x) = 1- e-x/1000). (Note - this is an
odd demand distribution; Poisson or Normal would
probably be better modeling choices.)
– Cost of shirts is $10.
– Selling price is $15.
– Unsold shirts can be sold off at $8.
 Model
Parameters:
cs = 15 – 10 = $5
co = 10 – 8 = $2
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Newsvendor Example – T Shirts (cont.)
 Solution:
G(Q )  1  e
*

Q
1000

cs
5

 0.714
co  cs 2  5
Q*  1,253
 Sensitivity: If co = $10 (i.e., shirts must be discarded) then
G(Q* )  1  e

Q
1000

cs
5

 0.333
co  cs 10  5
Q*  405
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Newsvendor Model with Normal
Demand
 Suppose
demand is normally distributed with
mean  and standard deviation . Then the
critical ratio formula reduces to:
3.00
cs
 Q *  
G (Q * )  

   co  c s
Q * 

cs
 z where  ( z ) 
co  c s
Q*    z
(z)
0.00
1
7
13
19
25
31
37
43
49
55
61
67
73
79
85
0
91
97 103 109 115 121 127 133 139 145 151 157
z
Note: Q* increases in both
 and  if z is positive (i.e.,
if ratio is greater than 0.5).
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Multiple Period Problems
 Difficulty: Technically, Newsvendor model is for a single
period.
 Extensions: But Newsvendor model can be applied to
multiple period situations, provided:
– demand during each period is iid, distributed according
to G(x)
– there is no setup cost associated with placing an order
– stockouts are either lost or backordered
 Key: make sure co and cs appropriately represent overage and
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shortage cost.
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Example
 Scenario:
–
–
–
–
GAP orders a particular clothing item every Friday
mean weekly demand is 100, std dev is 25
wholesale cost is $10, retail is $25
holding cost has been set at $0.5 per week (to reflect
obsolescence, damage, etc.)
 Problem: how should they set order amounts?
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Example (cont.)
 Newsvendor
Parameters:
c0 = $0.5
cs = $15
15
 0.9677
0.5  15
 Solution:
 Q  100 

  0.9677
 25 
Q  100
 1.85
25
Q  100  1.85(25)  146
G (Q * ) 
Every Friday, they should
order-up-to 146, that is, if
there are x on hand, then
order 146-x.
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Newsvendor Takeaways
•
Inventory is a hedge against demand uncertainty.
•
Amount of protection depends on “overage” and
“shortage” costs, as well as distribution of
demand.
•
If shortage cost exceeds overage cost, optimal
order quantity generally increases in both the
mean and standard deviation of demand.
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Impact of Improving Forecasts
EX: Demand is Normally distributed with a mean of R = 350 and standard
deviation of R = 150
Purchase price = $100 , Retail price = $250
Disposal value = $85 , Holding cost for season = $5
How many units should be ordered as R changes?
Price=p=250; Salvage value=s=85-5=80; Cost=c=100
Understocking cost=p-c=250-100=$150,
Overstocking cost=c-s=100-80=$20
Critical ratio=150/(150+20)=0.88
Optimal order quantity=Norminv(0.88,350,150)=526 units
Expected profit? Expected profit differs from the expected cost by a constant.
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Computing the Expected Profit with Normal Demands
(p : price; s : salvage value; c : cost) per unit.
x : demand with pdf f(x); Q : order quantity.
(p - c)x - (c - s)(Q - x) if x  Q
Profit(x, Q)  

(p
c)Q
if
x

Q



Expected Profit 
 Profit(x, Q) f(x) dx

Suppose that the demand is Normal with mean μ and standard deviation σ
Expected Profit  (p - s) μ normdist(( Q - μ)/σ,0,1,1)
- (p - s) σ normdist(( Q - μ)/σ,0,1,0)
- (c - s) Q normdist(Q , μ, σ,1)
 (p  c) Q (1  normdist(Q , μ, σ,1))
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Impact of Improving Forecasts
R
Q*
150
526
120
491
149.3
6.9
$48,476
90
456
112.0
5.2
$49,482
60
420
74.7
3.5
$50,488
30
385
37.3
1.7
$51,494
0
350
0
0
$52,500
Expected Expected Expected
Overstock Understock Profit
186.7
8.6
$47,469
Where is the trade off? Expected overstock vs. Expected understock.
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Expected profit vs. ?????
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Cost or Profit; Does it matter?
(p : price; s : salvage value; c : cost) per unit.
x : demand with pdf f(x); Q : order quantity.
(p - c)x - (c - s)(Q - x) if x  Q
Profit(x, Q)  

(p
c)Q
if
x

Q


 (c - s)(Q - x) if x  Q 
Cost(x, Q)  

(p
c)(x
Q)
if
x

Q


(p - c)x if x  Q 
Profit(x, Q)  Cost(x, Q)  
  (p  c)x
(p - c)x if x  Q
E[Profit(x , Q)]  E[Cost(x, Q)]  (p - c)E(Demand )  Constant in Q
Max E[Profit(x , Q)] and Min E[Cost(x, Q)] are equivalent ; they yield the same optimal.
Q
Q
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Learning Objectives
Optimal order quantities are obtained by
trading off cost of lost sales and cost of excess
stock
 Levers for improving profitability

–
–
–
–
Increase salvage value and decrease cost of stockout
Improved forecasting
Quick response with multiple orders
Postponement
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