Transcript Document

Inventory Optimization under
Correlated Uncertainty
Abhilasha Aswal
G N S Prasanna,
International Institute of Information Technology –
Bangalore
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
1
Outline





Motivation
Optimizing with correlated demands
Generalized EOQ
Related work
Some Extensions:
 Generalized base stock
 Geman Tank
 Relational Algebra
 Conclusions
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
2
The EOQ model
 The EOQ model (Classical – Harris 1913)
Q* 





2CD
h
f*
Dh
2C
C: fixed ordering cost per order
h: per unit holding cost
D: demand rate
Q*: optimal order quantity
f*: optimal order frequency
Q*
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
3
Inventory optimization for multiple
products
EOQ(K)?
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
4
Motivation
 Inventory optimization example
Car type I
Car type II
Car type III
Automobile
store
Tyre type I
Supplies
Tyre type II
Petrol
Drivers
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
5
Motivation
 Ordering and holding costs
Product
Ordering Cost in Rs.
(per order)
Holding Cost in Rs.
(per unit)
Car Type I
1000
50
Car Type II
1000
80
Car Type III
1000
10
Tyre Type I
250
0.5
Tyre Type II
500 (intl shipment)
0.5
Petrol
600
1
Drivers
750
300
Abhilasha Aswal & G N S Prasanna
IIIT-B
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1 product versus 7 products
 Exactly Known Demands, no uncertainty
 EOQ solution and Constrained Optimization solution match exactly:
EOQ Solution
Product
Demand per
month
Car Type I
40
Car Type II
25
Car Type III
50
Tyre Type I
250
Tyre Type II
125
Petrol
300
0.5
600
Drivers
5
1
5
Constrained Optimization Solution
Order
Frequency
Order
Quantity
Cost
Order
Frequency
Order
Quantity
Cost
1
40
2000
1
40
2000
1
25
2000
1
25
2000
0.5
100
1000
0.5
500
250
0.25
500
250
600
0.5
600
600
1500
1
5
1500
UNREALISTIC!!!
0.5
100
1000
We cannot
know
the future
0.5
500
250
demands
0.25 exactly.
500
250
Total
7600
7600
 But…
Abhilasha Aswal & G N S Prasanna
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1 product versus 7 products
 Bounded Uncorrelated Uncertainty
 Assuming the range of variation of the demands is known, we can get
bounds on the performance by optimizing for both the min value and the
max value of the demands.
 EOQ solution and Constrained Optimization solution are almost the same.
EOQ solution
Product
Constrained Optimization
Order Frequency
Order Quantity
Order Frequency
Order Quantity
Min
Max
Min
Max
Min
Max
Min
Max
Car Type I
0.5
1
20
40
0.5
1
20
40
Car Type II
0
1
0
25
0
1
0
25
Car Type III
0.5
1
100
200
0.5
1
100
200
Tyre Type I
0.25
0.5
248.99
500
0.25
0.5
248
500
Tyre Type II
0.25
0.5
500
1000
0.25
0.5
500
1000
Petrol
0.25
0.5
300
600
0.25
0.5
300
600
Drivers
0.45
1
2.24
5
0.5
1
2
5
Abhilasha Aswal & G N S Prasanna
IIIT-B
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1 product versus 7 products
 Beyond EOQ: Correlated Uncertainty in Demand
 Considering the substitutive effects between a class of products (cars,
tyres etc.)
200 ≤ dem_tyre_1 + dem_tyre_2 ≤ 700
65 ≤ dem_car_1 + dem_car_2 + dem_car_3 ≤ 250
 Considering the complementary effects between products that track each
EOQ cannot incorporate such
other
forms of uncertainty.
5 ≤ (dem_car_1 + dem_car_2 + dem_car_3) – dem_petrol ≤ 20
5 ≤ dem_car_2 – dem_drivers ≤ 20
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
9
1 product versus 7 products
 Beyond EOQ: Correlated Uncertainty in Demand
 Min-Max solution for different scenarios:
EOQ
Order
Frequency
Products
Order
Quantity
With Substitutive
Constraints
With Complementary
Constraints
With both Substitutive and
Complementary constraints
Order
Frequency
Order
Quantity
Order
Frequency
Order
Quantity
Order
Frequency
Order
Quantity
1
40 I
Car Type
0.75
25
0.5
38
0.5
40
1
25II
Car Type
0.5
13
0.5
22
1
10
0.5
100
Car Type
III
0.75
125
0.75
121
0.5
180
0.5
500I
Tyre Type
500II
Tyre Type
0.25
362
0.75
250
0.75
200
0.75
500
0.75
373
0.5
400
600
Petrol
5
Drivers
0.5
400
0.5
208
0.5
222.5
0.5
5
0.5
2
0.5
3
0.25
0.5
1
7600
Cost (Rs.)
4590.438
Abhilasha Aswal & G N S Prasanna
4593.688
IIIT-B
4654.188
INFORMS 2010
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1 product versus 7 products
 Beyond EOQ: Correlated Uncertainty in Demand
 Comparison of different uncertainty sets
Scenario sets
Absolute Minimum Cost
Absolute Maximum Cost
Bounds only
3349.5
9187.5
Bounds and Substitutive
constraints
3412.5
9100
Bounds and Complementary
constraints
4469.5
8972.5
Bounds, Substitutive and
Complementary constraints
4482.5
8910
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
11
Optimizing with Correlated Demands
Mathematical Programming
Formalism
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
12
Optimal Inventory policy using “ILP”
Minimizedecision Max uncertain
Subject to:

ytp  htp Invtp1

T 1
 N  T-1

p
P
p 
 
It  C 
yt
 

t 0

 p 1  t  0
 

ytp   S tp Invtp1
I
I
p
t
p
t

 

 Fixed costs and
breakpoints: nonconvexities that
preclude strong-duality
from being achieved.
 No breakpoints or fixed
costs: min-max
optimization  QP
 M  S tp

 1 M  S tp
Invtp1  Invtp  S tp  Dtp  0
(CP) D  E
S tp  0
Dtp  0
Abhilasha Aswal & G N S Prasanna
 Min-max optimization,
not an LP.
 Duality??
 Heuristics have to be
used in general.
IIIT-B
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Optimal Inventory policy by Sampling
 A simple statistical sampling heuristic
Begin
for i = 1 to maxIteration
{
parameterSample = getParameterSample(constraint Set)
bestPolicy = getBestPolicy(parameterSample)
findCostBounds(bestPolicy)
}
chooseBestSolution()
End
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
14
Optimizing with Correlated Demands:
Analytical Formulation: Generalized
EOQ(K)
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
15
Classical EOQ model
 Per order fixed cost = f(Q)
 holding cost per unit time = h(Q)
C  Q   h  Q   f  Q  D / Q 
Q*  2 fD / h;C *  Q   2 fDh
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
16
EOQ(K) with multiple products,
uncertain demands
 Additive SKU costs
Case with 2 commodities, generalized to n commodities
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
17
EOQ(K) with multiple products,
uncertain demands
 Holding cost linear, ordering cost fixed
Q1*  2 f1 D1 / h1 ;C1*  D1   2 f1 D1h1
Q2*  2 f 2 D2 / h2 ;C2*  D2   2 f 2 D2 h2
C *  D1 , D2   C1*  D1   C2*  D2   2 f1 D1h1  2 f 2 D2 h2
Cmax  max  D1 , D2 CP  2 f1 D1h1  2 f 2 D2 h2 
Cmin  min  D1 , D2 CP  2 f1 D1h1  2 f 2 D2 h2 
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
18
Analytical solution: Substitutive
constraints
 Holding cost linear, ordering cost fixed
 Under a substitutive constraint D1 + D2 <= D
C * ( D1 , D2 )  C1* ( D1 )  C 2* ( D2 )  2 f1 D1h1  2 f 2 D2 h2
D1  D2  D

 f1h1 D
f 2 h2 D 
  2 D f1h1  f 2 h2 
C max  C * 
,
 f1h1  f 2 h2 f1h1  f 2 h2 


C min  min C * 0, D , C * D,0  2 D  min f1h1 , f 2 h2 
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
19
Analytical solution: Substitutive
constraints - Example
 2 products, demands D1 & D2
 Costs:
h1 = 2/unit
h2 = 3/unit
f1 = 5/order
f2 = 5/order
 D1 + D2 = D = 100
 Maximum cost
C max  2 D f1h1  f 2 h2 
 2  100 10  15  70.71
 Minimum cost


2  100 15   44.72
C min  min 2 D f1h1 , 2 D f 2 h2 

 min 2  100 10,
Abhilasha Aswal & G N S Prasanna
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INFORMS 2010
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Analytical solution: Complementary
constraints
 Holding cost linear, ordering cost fixed
 Under a complementary constraint D1 – D2 <= D, with D1 and D2
limited to Dmax
Cmax  C * Dmax  D , Dmax 


Cmin  min C D,0, C 0, D 
Abhilasha Aswal & G N S Prasanna
*
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*
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Analytical solution: Complementary
constraints - Example
 2 products, demands D1 & D2
 Costs:
h1 = 2/unit
h2 = 3/unit
f1 = 5/order
f2 = 5/order
 Demand constraints:
D1 - D2 = K = 20
D1 <= Dmax = 50
D2 <= Dmax = 50
 Maximum cost
C max  2( Dmax  K ) f1 h1  2 Dmax f 2 h2
 2  30  10  2  50  15  45.83
 Minimum cost


2  20  15   20
C min  min 2 K  f1 h1 , 2 K  f 2 h2 

 min 2  20  10,
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
22
Both substitutive & complementary
constraints
 Holding cost linear, ordering cost fixed
 Under both substitutive and complementary constraints
C *  D1 , D2   C1*  D1   C2*  D2   2 f1D1h1  2 f 2 D2 h2
 Dmin  D1  D2  Dmax
CP : 
   D1  D2  
Cmax  max  D1 , D2 CP  2 f1 D1h1  2 f 2 D2 h2 
Cmin  min  D1 , D2 CP  2 f1 D1h1  2 f 2 D2 h2 
 Convex optimization techniques are required for this
optimization.
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
23
Both substitutive & complementary
constraints - Optimization
 Objective function: concave
 Minimization: HARD!
 Envelope based bounding schemes
 Heuristics to find upper bound.
 Simulated annealing based
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
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Both substitutive & complementary
constraints - Example
 2 products, demands D1 & D2
 Costs:
h1 = 2/unit
h2 = 3/unit
f1 = 5/order
f2 = 5/order
 Demand constraints:
150 <= D1 + D2 <= 200
-20 <= D1 – D2 <= 20
 Maximum cost: 99.88
 Minimum cost
 Enumerating all vertices (exact)
85.39
 Error: 0.111247 %
 Simulated annealing heuristic
85.48499
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
25
Both substitutive & complementary
constraints – Example (contd)
 5 products, demands
D1, D2, D3, D4 & D5
 Costs:
 Demand constraints:
h1 = 2/unit
h2 = 3/unit
h3 = 4/unit
h4 = 5/unit
h5 = 6/unit
f1, f2, f3, f4, f5 = 5/order
Abhilasha Aswal & G N S Prasanna
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D1 + D2 + D3 + D4 + D5 <= 1000
D1 + D2 + D3 + D4 + D5 >= 500
2 D1 - D2 <= 400
2 D1 - D2 >= 100
5 D5 - 2 D4 <= 900
5 D5 - 2 D4 >= 150
D2 + D4 <= 400
D2 + D4 >= 250
D1 <= 350
D1 >= 100
D3 >= 150
D3 <= 300
D4 >= 75
D4 <= 200
INFORMS 2010
26
Both substitutive & complementary
constraints – Example (contd)
 Maximum cost: 436.6448
 Minimum cost:
 Enumerating all vertices (exact)
323.5942
 Simulated annealing heuristic
324.4728
 Error: 0.271505 %
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
27
Inventory constraints
 Constrained Inventory Levels
 If the inventory levels Qi and demands Di, are constrained as
 Q1 , Q2 , D1 , D2   0
 The vector constraint above can incorporate constraints like
 Limits on total inventory capacity (Q1+Q2 <= Qtot)
 Balanced inventories across SKUs (Q1-Q2) <= ∆
 Inventories tracking demand (Q1-D1<=Dmax)
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
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Inventory constraints
 Constrained Inventory Levels
C1  Q1 , D1   h1  Q1   f1  Q1  D1 / Q1 
C2  Q2 , D2   h2  Q2   f 2  Q2  D2 / Q2 
C  Q1 , Q2 , D1 , D2   C1  Q1   C2  Q2 
[ D1 , D2 ]  CP
 Q1 , Q2 , D1 , D2   0
C *  D1 , D2   min Q1 ,Q2 C  Q1 , Q2 , D1 , D2 
Cmax  max[ D1 , D2 ]CP C *  D1 , D2  
Cmin  min[ D1 , D2 ]CP C *  D1 , D2  
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
29
Related Work
McGill (1995)
Inderfurth (1995)
Dong & Lee (2003)
Stefanescu et. al.
(2004)
Abhilasha Aswal & G N S Prasanna
Bertsimas, Sim,
Thiele et. al.
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Related work
 Bertsimas, Sim, Thiele - “Budget of uncertainty”




 Uncertainty: aij  aij , aij  aij 


zij 
 Normalized deviation for a parameter:
aij  aij

aij
n
 Sum of all normalized deviations limited:
z
j 1
ij
 i , i
 N uncertain parameters  polytope with 2N sides
 In contrast, our polyhedral uncertainty sets:
 More general
 Much fewer sides
Abhilasha Aswal & G N S Prasanna
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31
Extensions:
Generalized basestock
German Tank
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
32
Basestock with correlated inventory
Abhilasha Aswal & G N S Prasanna
IIIT-B
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33
The German Tank Problem
Classical German Tank
 Biased estimators
Generalization
 Maximum likelihood
 Unbiased estimators
 Minimum Variance
unbiased estimator
(UMVU)
 Maximum Spacing
estimator
 Bias-corrected maximum
likelihood estimator
Abhilasha Aswal & G N S Prasanna
IIIT-B
 Given correlated data
samples, drawn from a
uniform distributionestimating the bounded
region formed by correlated
constraints enclosing the
samples.
 Estimating the constraints
without bias and with
minimum variance.
INFORMS 2010
34
Information Theory and Relational
Algebra
 Uncertainty can be identified with Information.
 Information  polyhedral volume
 Relational algebra between alternative
constraint polyhedra
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
35
Conclusions
 Generalized EOQ to Correlated Demands
 Analytical Solutions
 Computational Solutions
 Enumerative versus Simulated Annealing
 Extensions of formulations
 Generalized Basestock
 German Tank
 Information Theory and Relational Algebra
Abhilasha Aswal & G N S Prasanna
IIIT-B
INFORMS 2010
36
Thank you
Abhilasha Aswal & G N S Prasanna
IIIT-B
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