piecewise linear approximations of the standard normal first … · 2020-01-28 · piecewise linear...
TRANSCRIPT
Piecewise linear approximations of the standardnormal first order loss function and an application
to stochastic inventory control
Dr. Roberto Rossi
The University of Edinburgh Business School,The University of Edinburgh, UK
Friday, June the 21th, 2013
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IntroductionWorking papers
This presentation illustrates results covered in the followingworking papers:
Roberto Rossi, S. Armagan Tarim, Brahim Hnich, and Steven D.Prestwich. Piecewise linear approximations of the standard normalfirst order loss function. Submitted to Applied Mathematics andComputation, arXiv:1307.1708, 2013
Roberto Rossi, Onur A. Kilic, and S. Armagan Tarim. Piecewiselinear approximations for the static-dynamic uncertainty strategy instochastic lot-sizing. Submitted to Omega, arXiv:1307.5942, 2013
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IntroductionRelated literature (sketch)
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilistic lot-sizing problem withservice-level constraints. Management Science, 34:1096–1108, 1988
S. A. Tarim and Brian G. Kingsman. The stochastic dynamic production/inventorylot-sizing problem with service-level constraints. International Journal of ProductionEconomics, 88(1):105–119, March 2004
S. Armagan Tarim and Brian G. Kingsman. Modelling and computing (Rn,Sn) policiesfor inventory systems with non-stationary stochastic demand. European Journal of
Operational Research, 174(1):581–599, October 2006
H. Tempelmeier. On the stochastic uncapacitated dynamic single-item lotsizingproblem with service level constraints. European Journal of Operational Research,181(1):184–194, August 2007
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IntroductionResearch questions
The investigation resorts to answering the following two keyquestions:
How can we produce “effective” piecewise linearisations of the firstorder loss function?
How can we employ these linearizations to model in a seamlessway a number of variants of the stochastic lot-sizing problemunder a static-dynamic uncertainty control policy, thus avoidingad-hoc solutions?
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A motivating exampleThe newsboy problem
time
inventory
0
single period
Newsboy problem
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The newsboy problemOrder quantity
time
inventory
ord
er
qu
an
tity
: Q
0
single period6/82
The newsboy problemDeterministic demand
time
inventory
ord
er
quantity
: Q
0
single period
inventory holding costs
deterministic demand: d
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The newsboy problemDeterministic demand
time
inventory
ord
er
quantity
: Q
0
single period
penalty costs
deterministic demand: d
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The newsboy problemDeterministic demand
time
inventory
ord
er
quantity
: Q
0
single period
deterministic demand: d
no cost
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The newsboy problemCost structure under deterministic demand
tota
l co
st:
g(Q
)
order quantity: Q
Q* = d
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The newsboy problemRandom demand
time
inventory
ord
er
qu
an
tity
: Q
0
single period
random demand: d
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The newsboy problemRandom demand
time
inventory
ord
er
qu
an
tity
: Q
0
single period
pmf: g(d) 0.8
random demand: d
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The newsboy problemRandom demand
time
inventory
ord
er
qu
an
tity
: Q
0
single period
pdf: g(d) 0.8
random demand: d
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The newsboy problemCost structure under random demand
time
inventory
ord
er
qu
an
tity
: Q
0
single period
inventory holding costs
random demand: d
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The newsboy problemCost structure under random demand
time
inventory
ord
er
qu
an
tity
: Q
0
single period
penalty costs
random demand: d
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The newsboy problemCost structure under random demand
exp
ecte
d t
ota
l co
st:
G(Q
)
order quantity: Q0 Q*
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The newsboy problemMathematical formulation
Consider
d: a one-period random demand that follows a probability
distribution f(d)
h: unit holding cost
p: unit penalty cost
Let I be the end of period inventory and
g(I) = hI+ + pI−,
where I+ = max(I, 0) and I− = −min(I, 0).
The expected total cost is G(Q) = E[g(Q− d)], where E[·]denotes the expected value.
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The newsboy problemMathematical formulation
Define:
E[I+] = E[max(Q− d, 0)]: complementary first order loss functionE[I−] = E[max(d−Q, 0)]: first order loss function
The expected total cost comprises two separable components
G(Q) = E[g(Q − d)] = hE[I+] + pE[I−]
5 10 15 20 25 30Q
10
20
30
40
50
cost
p E@I-D
h E@I+D
h E@I+D+p E@I-D
d = Normal(10, 5)h =$1p =$5
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The first order loss functionA graphical outlook
5 10 15 20Q
2
4
6
8
10
cost
E@I-D
E@I+D
E[I+]: complementary first order loss functionE[I−]: first order loss function
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The first order loss functionProperties
Consider a continuous random variable ω with support over R,probability density function gω(x) : R → (0, 1) and cumulativedistribution function Gω(x) : R → (0, 1).
The first order loss function can be rewritten as
L(x, ω) =
∫ ∞
−∞max(t− x, 0)gω(t) dt =
∫ ∞
x
(t− x)gω(t) dt. (1)
The complementary first order loss function can be rewritten as
L(x, ω) =
∫ ∞
−∞max(x− t, 0)gω(t) dt =
∫ x
−∞(x− t)gω(t) dt. (2)
LemmaL(x, ω) and L(x, ω) are convex in x.
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The first order loss functionProperties
There is a close relationship between the first order loss functionand the complementary first order loss function.
LemmaThe first order loss function L(x, ω) can also be expressed as
L(x, ω) = L(x, ω)− (x− ω) (3)
where ω = E[ω].
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The first order loss functionProperties
5 10 15 20Q
-10
-5
5
10
cost
L`Hx,ΩL
LHx,ΩL
-Hx-ΩL
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The first order loss functionProperties
LemmaThe first order loss function L(x, ω) can also be expressed as
L(x, ω) =
∫ ∞
x
(1−Gω(t)) dt (4)
LemmaThe complementary first order loss function L(x, ω) can also beexpressed as
L(x, ω) =
∫ x
−∞Gω(t) dt. (5)
These two results are not easily derived from each other!
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The first order loss functionProperties for symmetric distributions
LemmaIf the probability density function of ω is symmetric about a meanvalue ω, then
L(x, ω) = L(2ω − x, ω).
LemmaIf the probability density function of ω is symmetric about a meanvalue ω, then
L(x, ω) = L(2ω − x, ω) + (x− ω)
andL(x, ω) = L(2ω − x, ω)− (x− ω).
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The first order loss functionProperties for normal distribution
Let ζ be a normally distributed random variable with mean µ andstandard deviation σ.
LemmaThe complementary first order loss function of ζ can be expressedin terms of the standard Normal cumulative distribution function as
L(x, ζ) = σ
∫ x−µσ
−∞Φ(t) dt = σL
(x− µ
σ,Z
), (6)
where Z is a standard Normal random variable.
Unfortunately, no closed form expression exists for Φ(t).
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The first order loss functionNon-linear approximations
Several approximations have been discussed for Φ(t), see e.g.
Marvin Zelen and Norman C. Severo. Probability functions. InMilton Abramowitz and Irene A Stegun, editors, Handbook ofMathematical Functions, volume 5 of Applied Mathematics Series,pages 925–995. GPO, 1964
Approximation to L(x, ζ) have been recently discussed in
Steven K. De Schrijver, El-Houssaine Aghezzaf, and HendrikVanmaele. Double precision rational approximation algorithm forthe inverse standard normal first order loss function. AppliedMathematics and Computation, 219(3):1375–1382, October 2012
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The first order loss functionNon-linear approximations
DrawbacksExisting approximations are non-linear and cannot be easilyembedded in MILP models — ad-hoc strategies are needed.
Existing approximations do not provide upper and lower bounds forL(x, ζ) — it is hard to estimate the goodness of the solutionsobtained and to obtain optimality gaps.
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The first order loss functionPiecewise linear approximations
We introduce a well-known inequality from stochastic programming
Peter Kall and Stein W. Wallace. Stochastic Programming (WileyInterscience Series in Systems and Optimization). John Wiley &Sons, August 1994, p. 167.
Theorem (Jensen’s inequality)
Consider a random variable ω with support Ω and a functionf(x, s), which for a fixed x is convex for all s ∈ Ω, then
E[f(x, ω)] ≥ f(x,E[ω]).
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The first order loss functionThe newsboy problem & Jensen’s inequality
For a fixed Q, the total cost is convex for all values in the supportof d.
gQ(d) = g(Q− d) = hmax(Q− d, 0) + pmax(d−Q, 0))
5 10 15 20d
10
20
30
40
50
cost
gHQ-dL
Q = 10h =$1p =$5
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The first order loss functionThe newsboy problem & Jensen’s inequality
Define:
E[I+] = E[max(Q− d, 0)]: complementary first order loss functionE[I−] = E[max(d−Q, 0)]: first order loss function
The expected total cost can be bounded from below as follows.
hE[I+]+pE[I−] ≥ hmax(Q−E[d], 0)+pmax(E[d]−Q, 0) = g(Q−E[d])
5 10 15 20 25 30Q
10
20
30
40
50
cost
h maxHQ-E@dD,0L+p maxHE@dD-Q,0L
h E@I+D+p E@I-D
d = Normal(10, 5)E[d] = 10h =$1p =$5
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The first order loss functionBounding techniques
Define:
E[I+] = E[max(Q− d, 0)]: complementary first order loss function
The complementary first order loss function can be bounded frombelow as follows.
E[I+] ≥ hmax(Q− E[d], 0)
5 10 15 20 25 30Q
5
10
15
20
cost
maxHQ-E@dD,0L
E@I+D
d = Normal(10, 5)E[d] = 10h =$1p =$5
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The first order loss functionBounding techniques
Define:
E[I−] = E[max(d−Q, 0)]: first order loss function
The first order loss function can be bounded from below as follows.
E[I−] ≥ max(E[d]−Q, 0)
5 10 15 20 25 30Q
2
4
6
8
10
cost
maxHE@dD-Q,0L
E@I-D
d = Normal(10, 5)E[d] = 10h =$1p =$5
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The first order loss functionBounding techniques
Let gω(·) denote the probability density function of ω and considera partition of the support Ω of ω into N disjoint compactsubregions Ω1, . . . ,ΩN . We define, for all i = 1, . . . , N
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
Theorem
E[f(x, ω)] ≥
N∑
i=1
pif(x,E[ω|Ωi])
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The first order loss functionBounding techniques
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
pi = 0.5841376
E@Ω WiD=8.56168
-20 -10 10 20 30
0.02
0.04
0.06
0.08
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The first order loss functionBounding techniques
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
pi = 0.158624
E@Ω WiD=17.623
-20 -10 10 20 30
0.02
0.04
0.06
0.08
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The first order loss functionBounding techniques
For the (complementary) first order loss function (Llb(x, ω)) Llb(x, ω) the lowerbound
E[f(x, ω)] ≥N∑
i=1
pif(x,E[ω|Ωi])
is a piecewise linear function with N + 1 segments.
Consider the bound presented above and let f(x, ω) = max(x− ω, 0),
Llb(x, ω) =
N∑
i=1
pi max(x− E[ω|Ωi], 0)
this function is equivalent to
Llb(x, ω) =
0 −∞ ≤ x ≤ E[ω|Ω1]p1x − p1E[ω|Ω1] E[ω|Ω1] ≤ x ≤ E[ω|Ω2](p1 + p2)x − (p1E[ω|Ω1] + p2E[ω|Ω2]) E[ω|Ω2] ≤ x ≤ E[ω|Ω3]
.
.
.
.
.
.(p1 + p2 + . . . + pN )x − (p1E[ω|Ω1] + . . . + pNE[ω|ΩN ]) E[ω|ΩN−1] ≤ x ≤ E[ω|ΩN ]
which is piecewise linear in x with breakpoints at E[ω|Ω1],E[ω|Ω2], . . . ,E[ω|ΩN ].
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The first order loss functionBounding techniques
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
5 10 15 20 25 30Q
5
10
15
20
cost
piecewise-2
E@I+D
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The first order loss functionBounding techniques
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
pi = 0.5
E@Ω WiD=6.01058
-20 -10 10 20 30
0.02
0.04
0.06
0.08
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The first order loss functionBounding techniques
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
pi = 0.5
E@Ω WiD=13.9875
-20 -10 10 20 30
0.02
0.04
0.06
0.08
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The first order loss functionBounding techniques
pi = Prω ∈ Ωi =
∫
Ωi
gω(t) dt
E[ω|Ωi] =1
pi
∫
Ωi
tgω(t) dt
5 10 15 20 25 30Q
5
10
15
20
cost
piecewise-2
E@I+D
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The first order loss functionMinimax optimal linearisation parameters for a standard normal random variable
Piecewise linear approximation parametersSegments Error i 1 2 3 4 5 6 7 8 9 10
2 0.398942
bi ∞pi 1E[ω|Ωi] 0
3 0.120656
bi 0 ∞pi 0.5 0.5E[ω|Ωi] −0.797885 0.797885
4 0.0578441
bi −0.559725 0.559725 ∞pi 0.287833 0.424333 0.287833E[ω|Ωi] −1.18505 0 1.18505
5 0.0339052
bi −0.886942 0 0.886942 ∞pi 0.187555 0.312445 0.312445 0.187555E[ω|Ωi] −1.43535 −0.415223 0.415223 1.43535
6 0.0222709
bi −1.11507 −0.33895 0.33895 1.11507 ∞pi 0.132411 0.234913 0.265353 0.234913 0.132411E[ω|Ωi] −1.61805 −0.691424 0 0.691424 1.61805
7 0.0157461
bi −1.28855 −0.579834 0 0.579834 1.28855 ∞pi 0.0987769 0.182236 0.218987 0.218987 0.182236 0.0987769E[ω|Ωi] −1.7608 −0.896011 −0.281889 0.281889 0.896011 1.7608
8 0.0117218
bi −1.42763 −0.765185 −0.244223 0.244223 0.765185 1.42763 ∞pi 0.0766989 0.145382 0.181448 0.192942 0.181448 0.145382 0.0766989E[ω|Ωi] −1.87735 −1.05723 −0.493405 0 0.493405 1.05723 1.87735
9 0.00906529
bi −1.54317 −0.914924 −0.433939 0 0.433939 0.914924 1.54317 ∞pi 0.0613946 0.118721 0.152051 0.167834 0.167834 0.152051 0.118721 0.0613946E[ω|Ωi] −1.97547 −1.18953 −0.661552 −0.213587 0.213587 0.661552 1.18953 1.97547
10 0.00721992
bi −1.64166 −1.03998 −0.58826 −0.19112 0.19112 0.58826 1.03998 1.64166 ∞pi 0.0503306 0.0988444 0.129004 0.146037 0.151568 0.146037 0.129004 0.0988444 0.0503306E[ω|Ωi] −2.05996 −1.30127 −0.8004 −0.384597 0. 0.384597 0.8004 1.30127 2.05996
11 0.00588597
bi −1.72725 −1.14697 −0.717801 −0.347462 0. 0.347462 0.717801 1.14697 1.72725 ∞pi 0.0420611 0.0836356 0.110743 0.127682 0.135878 0.135878 0.127682 0.110743 0.0836356 0.0420611E[ω|Ωi] −2.13399 −1.39768 −0.9182 −0.526575 −0.17199 0.17199 0.526575 0.9182 1.39768 2.13399
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The first order loss functionApproximation error of Llb(x,Z) with up to eleven segments
4 6 8 10Number of segments
0.1
0.2
0.3
0.4
Absolute error
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The first order loss functionFive-segment piecewise Jensen’s bound for L(x, ζ), where µ = 0 and σ = 1
-2 -1 1 2x
0.5
1.0
1.5
2.0
L`Hx,ZL-L
`lbHx,ZL
L`
lbHx,ZL
L`Hx,ZL
Five-segment piecewise Jensen’s bound for L(x, Z), where Z is a standard normally distributed random variable.The maximum error is 0.0339052 and it is observed at x ∈ ±1.43535, ±0.415223.
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The first order loss functionFive-segment piecewise Jensen’s bound for L(x, ζ), where µ = 20 and σ = 5
Below we exploit the fact that the complementary first order loss function of ζ can be expressed in terms of thestandard Normal cumulative distribution function as
L(x, ζ) = σ
∫ x−µσ
−∞
Φ(t) dt = σL
(x − µ
σ, Z
),
where Z is a standard Normal random variable.
15 20 25 30x
2
4
6
8
10
L`Hx,ΖL-L
`lbHx,ΖL
L`
lbHx,ΖL
L`Hx,ΖL
Five-segment piecewise Jensen’s bound for L(x, ζ), where ζ is a normally distributed random variable with meanµ = 20 and standard deviation σ = 5. The maximum error is σ0.0339052 and it is observed atx ∈ σ(±1.43535) + µ, σ(±0.415223) + µ.
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The first order loss functionFive-segment piecewise linear upper bound for L(x, ζ), where µ = 20 and σ = 5
15 20 25 30x
2
4
6
8
10
L`Hx,ΖL-L
`ubHx,ΖL
L`
ubHx,ΖL
L`Hx,ΖL
Five-segment piecewise linear upper bound for L(x, ζ), where ζ is a normally distributed random variable withmean µ = 20 and standard deviation σ = 5. The maximum error is σ0.0339052 and it is observed atx ∈ ±∞, σ(±0.886942) + µ, µ.
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Stochastic lot-sizingGeneral framework
minE[TC] =
∫
d1
∫
d2
. . .
∫
dN
N∑
t=1
(aδt + hmax(It, 0) + vQt)×
g1(d1)g2(d2) . . . gN (dN ) d(d1)d(d2) . . . d(dN )
subject to, for t = 1, . . . N
It = I0 +
t∑
i=1
(Qi − di)
δt =
1 if Qt > 0,0 otherwise
Qi ≥ 0, δt ∈ 0, 1
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Stochastic lot-sizingα service level
minE[TC] =
∫
d1
∫
d2
. . .
∫
dN
N∑
t=1
(aδt + hmax(It, 0) + vQt)×
g1(d1)g2(d2) . . . gN (dN ) d(d1)d(d2) . . . d(dN )
subject to, for t = 1, . . . N
It = I0 +
t∑
i=1
(Qi − di)
δt =
1 if Qt > 0,0 otherwise
PrIt ≥ 0 ≥ α
Qi ≥ 0, δt ∈ 0, 1
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Stochastic lot-sizingPenalty cost
minE[TC] =
∫
d1
∫
d2
. . .
∫
dN
N∑
t=1
(aδt + hmax(It, 0) + pmax(−It, 0) + vQt)×
g1(d1)g2(d2) . . . gN (dN ) d(d1)d(d2) . . . d(dN )
subject to, for t = 1, . . . N
It = I0 +
t∑
i=1
(Qi − di)
δt =
1 if Qt > 0,0 otherwise
Qi ≥ 0, δt ∈ 0, 1
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Stochastic lot-sizingβcyc service level
H. Tempelmeier. On the stochastic uncapacitated dynamic single-item lotsizingproblem with service level constraints. European Journal of Operational Research,181(1):184–194, August 2007
minE[TC] =
∫
d1
∫
d2
. . .
∫
dN
N∑
t=1
(aδt + hmax(It, 0) + vQt)×
g1(d1)g2(d2) . . . gN (dN ) d(d1)d(d2) . . . d(dN )
subject to, for t = 1, . . . N
It = I0 +t∑
i=1
(Qi − di)
δt =
1 if Qt > 0,0 otherwise
1− maxi=1,...,m
[E
Total backorders in replenishment cycle i
Total demand in replenishment cycle i
]≥ βcyc
Qi ≥ 0, δt ∈ 0, 1
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Stochastic lot-sizingβ service level
minE[TC] =
∫
d1
∫
d2
. . .
∫
dN
N∑
t=1
(aδt + hmax(It, 0) + vQt)×
g1(d1)g2(d2) . . . gN (dN ) d(d1)d(d2) . . . d(dN )
subject to, for t = 1, . . . N
It = I0 +
t∑
i=1
(Qi − di)
δt =
1 if Qt > 0,0 otherwise
1− E
Total backorders within the planning horizon
Total demand within the planning horizon
≥ β
Qi ≥ 0, δt ∈ 0, 150/82
Problem parameters
Normally distributed demand with constant coefficient of variation
c =σt
µt
51/82
Static uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
Q4
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Static uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
Q4
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Static uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
Q4
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Static uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
Q4
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Dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
S9
S10
s9 s
10
56/82
Dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
S9
S10
s9 s
10
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Dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
S9
S10
s9 s
10
58/82
Dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
S9
S10
s9 s
10
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Static-dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
R2
S2
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Static-dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
R2
S2
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Static-dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
R2
S2
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Static-dynamic uncertainty
J. H. Bookbinder and J. Y. Tan. Strategies for the probabilisticlot-sizing problem with service-level constraints. ManagementScience, 34:1096–1108, 1988
1 2 3 4 5 6 7 8 9 10 11 12Period0
100
200
300
400Inventory level
R2
S2
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Static-dynamic uncertaintyMILP model under α service level
S. A. Tarim and Brian G. Kingsman. The stochastic dynamic production/inventory lot-sizing problem withservice-level constraints. International Journal of Production Economics, 88(1):105–119, March 2004
E[TC] = −vI0 + vN∑
t=1
dt + minN∑
t=1
(aδt + hIt) + vIN (7)
subject to, for t = 1, . . . N
It + dt − It−1 ≥ 0
It + dt − It−1 ≤ δtMt
It ≥t∑
j=1
G−1
dj...t(α) −
t∑
k=j
dk
Pjt
t∑
j=1
Pjt = 1
Pjt ≥ δj −t∑
k=j+1
δk j = 1, . . . , t
Pjt ∈ 0, 1 j = 1, . . . , t
δt ∈ 0, 1
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Static-dynamic uncertaintyEnhanced MILP model under α service level
We introduce two new sets of decision variables: Ilbt and Iubt for t = 1, . . . , N . These represent, respectively, a
lower and an upper bound to the true value of E[max(It, 0)].
We introduce the following constraints in the model
Ilbt ≥ It
i∑
k=1
pk −
t∑
j=1
i∑
k=1
pkEZ|Ωi
Pjtσdj...tt = 1, . . . , N; i = 1, . . . ,W
where σdj...tdenotes the standard deviation of dj + . . . + dt and Ilbt ≥ 0.
Consider a replenishment cycle covering periods j, . . . , t and associated order-up-to-level S. We aim to enforce
Ilbt ≥ σLilb
((S − µdj...t
)/σdj...t, Z
)for all i = 1, . . . ,W , where µdj...t
is the expected value and
σdj...tthe standard deviation of the demand over periods j, . . . , t. Observe that S − µdj...t
= It , the above
expression follows immediately.
Iubt ≥ It
i∑
k=1
pk −t∑
j=1
i∑
k=1
pkEZ|Ωi
Pjtσdj...t+
t∑
j=1
eW
Pjtσdj...t
t = 1, . . . , N,i = 1, . . . ,W ;
where Iubt ≥ eW and eW denotes the maximum approximation error associated with a partition comprising W
regions.
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Static-dynamic uncertaintyEnhanced MILP model under α service level
Finally, the objective function then becomes
E[TC] = −vI0 + vN∑
t=1
dt + minN∑
t=1
(aδt + hIlbt ) + vIN (8)
if our aim is to compute a lower bound for the cost of an optimal plan, or
E[TC] = −vI0 + vN∑
t=1
dt + minN∑
t=1
(aδt + hIubt ) + vIN (9)
if our aim is to compute an upper bound for the cost of an optimal plan.
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Static-dynamic uncertaintyNumerical example under α service level
We demonstrate our approach on an instance originally discussed in S. A. Tarim and Brian G. Kingsman. Thestochastic dynamic production/inventory lot-sizing problem with service-level constraints. International Journal ofProduction Economics, 88(1):105–119, March 2004. The instance comprises N = 10 periods in the planninghorizon. Demand dt in period t is normally distributed with mean µt and standard deviation σt.
t 1 2 3 4 5 6 7 8 9 10µt 200 50 100 300 150 200 100 50 200 150σt 60 15 30 90 45 60 30 15 60 45
Inventory holding costs are set to h = 1 setup costs are set to a = 2500; we target an α service level of 0.95.We ignore unit costs, i.e. v = 0.
Piecewise linear approximation (2 seg.) - E[TC]∈ [9989.07, 10314.00]t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 0 0 1 0 0 0 0St 1000.46 - - - - 867.35 - - - -
Piecewise linear approximation (11 seg.) - E[TC]∈ [9993.66, 9998.46]t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 0 0 1 0 0 0 0St 1000.46 - - - - 867.35 - - - -
The expected total cost estimated by Tarim and Kingsman’s model is 9989.07. We simulated this policy andestimated its expected total cost with a margin of error of ±0.001% at 95% confidence; the resulting cost is9993.74 ± 0.1.
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Static-dynamic uncertaintyMILP model under penalty cost
S. Armagan Tarim and Brian G. Kingsman. Modelling and computing (Rn,Sn) policies for inventory systems withnon-stationary stochastic demand. European Journal of Operational Research, 174(1):581–599, October 2006
We introduce two new sets of variables Blbt and Bub
t for t = 1, . . . , N , which represent a lower and upperbound, respectively, for the true value of E[−min(It, 0)] and thus allow us to compute lower and upper boundsfor the expected backorders in each period.
Blbt ≥ −It + It
i∑
k=1
pk −
t∑
j=1
i∑
k=1
pkEZ|Ωi
Pjtσdj...t
t = 1, . . . , N,i = 1, . . . ,W ;
where Bubt ≥ −It and
Bubt ≥ −It+It
i∑
k=1
pk−t∑
j=1
i∑
k=1
pkEZ|Ωi
Pjtσdj...t+
t∑
j=1
eW
Pjtσdj...t
t = 1, . . . , N,i = 1, . . . ,W ;
where Bubt ≥ −It + eW .
The objective function then becomes
E[TC] = −vI0 + v
N∑
t=1
dt + min
N∑
t=1
(aδt + hIlbt + bB
lbt ) + vIN (10)
if our aim is to compute a lower bound for the cost of an optimal plan, or
E[TC] = −vI0 + vN∑
t=1
dt + minN∑
t=1
(aδt + hIubt + bB
ubt ) + vIN (11)
if our aim is to compute an upper bound for the cost of an optimal plan.
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Static-dynamic uncertaintyNumerical example under penalty cost
We demonstrate our approach on an instance originally discussed in Charles R. Sox. Dynamic lot sizing withrandom demand and non-stationary costs. Operations Research Letters, 20(4):155–164, May 1997. The instancecomprises N = 8 periods in the planning horizon. Demand dt in period t is normally distributed with mean µt
and standard deviation σt.
t 1 2 3 4 5 6 7 8µt 110 40 10 62 12 80 122 130σt 22 8 2 12.4 2.4 16 24.4 26vt 5.6 4.2 3.0 2.0 1.2 0.6 0.2 0
Piecewise linear approximation - E[TC]∈ [1024.70, 1034.24]t 1 2 3 4 5 6 7 8 9 10δt 1 1 0 1 0 1 1 1
Subt 130.2 57.072 - 85.597 - 102.363 156.103 185.484
Slbt 130.2 57.072 - 85.597 - 102.363 156.103 185.484
Tarim and Kingsman - E[TC]=1031 (simulated: 1036.30)t 1 2 3 4 5 6 7 8 9 10δt 1 1 0 1 0 1 1 1St 128.5 56.9 - 84.6 - 101.9 155.4 165.6
Inventory holding costs are set to h = 0.5; setup costs are set to a = 48; penalty costs are set to b = 12; finally,unit costs vt vary from period to period. The initial inventory is set to 98 units.
The expected total cost estimated by Tarim and Kingsman’s model is 1031. We simulated this policy and estimatedits expected total cost with a margin of error of ±0.01% at 95% confidence; the resulting cost is 1036.30 ± 0.1.
Policy parameters obtained via our MILP approximation converge for eleven segments; the optimality gap ishowever 0.92%, reflecting the fact that the actual cost of this policy lies somewhere between 1024.70 and 1034.24.We simulated this policy and estimated its expected total cost with a margin of error of ±0.01% at 95%confidence; the resulting cost is 1034.14 ± 0.1.
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Static-dynamic uncertaintyMILP model under βcyc service level as defined in Tempelmeier (2007)
We introduce constraints
Blbt ≤ (1 − β
cyc)
t∑
j=1
Pjtµdj...tt = 1, . . . , N, (12)
if our aim is to compute a lower bound for the cost of an optimal plan; or with
Bubt ≤ (1 − β
cyc)
t∑
j=1
Pjtµdj...tt = 1, . . . , N, (13)
if our aim is to compute an upper bound for the cost of an optimal plan. Finally, the objective function becomes
E[TC] = −vI0 + vN∑
t=1
dt + minN∑
t=1
(aδt + hIlbt ) + vIN (14)
if our aim is to compute a lower bound for the cost of an optimal plan, or
E[TC] = −vI0 + vN∑
t=1
dt + minN∑
t=1
(aδt + hIubt ) + vIN (15)
if our aim is to compute an upper bound for the cost of an optimal plan.
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Static-dynamic uncertaintyNumerical example under βcyc service level as defined in Tempelmeier (2007)
We solved the same instance discussed for the case of an α service level, however we now enforced a βcyc servicelevel of 0.95. By using eleven segments in the linearisation, the optimality gap is very narrow, i.e. 0.23%. Wesimulated both the policies obtained and estimated their expected total cost with a margin of error of ±0.001% at95% confidence; the resulting costs are 8347.71 ± 0.08 and 8361.31 ± 0.08, respectively.
Piecewise linear approximation - E[TC]∈ [8347.40, 8367.03]t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 1 0 0 0 0 0 0
Subt 373.95 - - 1150.85 - - - - - -
Slbt 372.84 - - 1149.17 - - - - - -
Tempelmeier - E[TC]=8348 (simulated: 8347.10)t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 1 0 0 0 0 0 0St 373 - - 1149 - - - - - -
It should be noted that an order-up-to-level of 1149, which is suggested in Tempelmeier’s work for the secondreplenishment cycle, does not strictly meet the prescribed cycle service level, since it provides a cycle fill ratestrictly lower than 0.95.
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Static-dynamic uncertaintyNumerical example under βcyc service level as defined in Tempelmeier (2007)
We consider the same instance, but now the cycle fill rate is set to βcyc = 0.6 and the setup costs are reduced toa = 1000.
Piecewise linear approximation - E[TC]∈ [2773.63, 2781.10]t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 1 0 0 0 0 0 0
Subt 210.71 - - 694.84 - - - - - -
Slbt 210.29 - - 690.00 - - - - - -
Tempelmeier - E[TC]=2776 (simulated: 2776.81)t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 1 0 0 0 0 0 0St 211 - - 690 - - - - - -
These results suggest that the model in Tempelmeier (2007) constitutes an excellent approach to the dynamiclot-sizing problem under non stationary stochastic demand and cycle fill rate constraints.
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Static-dynamic uncertaintyMILP model under β service level
We introduce two new set of nonnegative variables Clbt and Cub
t for t = 0, . . . , N . These variables express the
expected total backorders within the replenishment cycle that ends at period t, if there is one. Hence, Clbt (resp.
Cubt ) should be equal to Blb
t (resp. Bubt ), if t is the last period of a replenishment cycle; otherwise Clb
t (resp.
Cubt ) should be equal to 0. We enforce this fact as follows. For convenience, we set
Blb0 = Bub
t = Clb0 = Cub
t = I0 , then we enforce
Clbt ≥ B
lbt − δt+1
t∑
k=1
dt t = 0, . . . , N − 1, (16)
Cubt ≥ B
ubt − δt+1
t∑
k=1
dt t = 0, . . . , N − 1. (17)
Finally, we must ensure that ClbN = Blb
N and CubN = Bub
N .
We then use these new variables to build constraint
N∑
t=1
Clbt ≤ (1 − β)
N∑
t=1
dt (18)
which will replace (12), if our aim is to compute a lower bound for the cost of an optimal plan; and constraint
N∑
t=1
Cubt ≤ (1 − β)
N∑
t=1
dt (19)
which will replace (13), if our aim is to compute an upper bound for the cost of an optimal plan.
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Static-dynamic uncertaintyNumerical example under β service level
We solved the same instance discussed for the case of an α service level, however we now enforced a β service levelof 0.95.
Piecewise linear approximation - E[TC]∈ [8313.48, 8335.38]t 1 2 3 4 5 6 7 8 9 10δt 1 0 0 1 0 0 0 0 0 0
Subt 413.12 - - 1129.21 - - - - - -
Slbt 413.12 - - 1126.71 - - - - - -
We simulated both the policies and estimated their expected total cost with a margin of error of ±0.001% at 95%confidence; the resulting costs are 8315.59 ± 0.08 and 8331.20 ± 0.08, respectively. This represents a 0.4%cost reduction with respect to the policies obtained via the model operating under a βcyc service level.
Note that a control policy that orders up to 413.12 in period 1 and up to 1129.21 in period 4 is infeasible accordingto a cycle β service level constraint. The expected number of unit short in the second replenishment cycle amountsto 68.08 unit, that is 5.92% of the expected demand for this cycle, which amounts to 1150 units. However, theexpected number of unit short over the planning horizon is 74.75 units, that is 6.67 units over the firstreplenishment cycle and 68.08 units over the second one. This represents 4.98% of the expected demand over thewhole planning horizon, which amounts to 1500 units. Therefore this policy satisfies a classical β service levelconstraint.
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Static-dynamic uncertaintyNumerical example under β service level
We consider the same instance, but now the cycle fill rate is set to βcyc = 0.6 and the setup costs are reduced toa = 1000.
Piecewise linear approximation - E[TC]∈ [2602.58, 2612.69]t 1 2 3 4 5 6 7 8 9 10δt 0 0 0 1 0 0 0 0 0 0
Subt - - - 903.49 - - - - - -
Slbt - - - 902.43 - - - - - -
We simulated both the policies in Table 1 and estimated their expected total cost with a margin of error of±0.01% at 95% confidence; the resulting costs are 2603.63 ± 0.26 and 2609.11 ± 0.26, respectively. For thisinstance, the cost reduction with respect to the policies obtained under a βcyc service level is substantial andamounts to 6.4%.
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Computational experienceInstances
We generated a total of 810 instances.
10 demand patternsordering cost [500,1000,2000]unit cost [2,5,10]coefficient of variation [0.10,0.20,0.30]penalty cost [2,5,10]
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Computational experienceDemand patterns
Period
Exp
ecte
d d
em
an
d
02
04
06
08
01
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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ecte
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d
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0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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ecte
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d
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04
06
08
01
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ecte
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d
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ecte
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Exp
ecte
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ecte
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30
05
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70
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ecte
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em
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d
01
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ecte
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Computational experienceα service level
0.00
50.
050
0.50
05.
000
Segments
Opt
imal
ity g
ap %
1 2 3 4 5 6 7 8 9 10
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Computational experiencePenalty cost
0.02
0.10
0.50
2.00
10.0
0
Segments
Opt
imal
ity g
ap %
1 2 3 4 5 6 7 8 9 10
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Computational experienceCycle β service level
0.1
10.0
1000
.0
Segments
Opt
imal
ity g
ap %
1 2 3 4 5 6 7 8 9 10
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Computational experienceβ service level
0.05
0.50
5.00
50.0
0
Segments
Opt
imal
ity g
ap %
1 2 3 4 5 6 7 8 9 10
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ConclusionsQuestions
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