Piecewise linear approximations of the standard normal first … · 2020-01-28 · Piecewise linear approximations of the standard normal ﬁrst order loss function and an application

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

[email protected]

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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time

inventory

ord

er

quantity

: Q

0

single period

penalty costs


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time

inventory

ord

er

quantity

: Q

0

single period


no cost

9/82

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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time

inventory

ord

er

qu

an

tity

: Q

0

single period

pmf: g(d) 0.8

random demand: d

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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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time

inventory

ord

er

qu

an

tity

: Q

0

single period

penalty costs

random demand: d

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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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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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5 10 15 20Q

-10

-5

5

10

cost

L`Hx,ΩL

LHx,ΩL

-Hx-ΩL

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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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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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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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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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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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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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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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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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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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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




It = I0 +

t∑

i=1

(Qi − di)

δt =


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)×



It = I0 +

t∑

i=1

(Qi − di)

δt =


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




It = I0 +t∑

i=1

(Qi − di)

δt =


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




It = I0 +

t∑

i=1

(Qi − di)

δt =


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

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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

52/82

Static uncertainty


1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

Q4

53/82

Static uncertainty


1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

Q4

54/82

Static uncertainty


1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

Q4

55/82

Dynamic uncertainty


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


1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

S9

S10

s9 s

10

57/82

Dynamic uncertainty


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


1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

S9

S10

s9 s

10

59/82

Static-dynamic uncertainty


1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

R2

S2

60/82



1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

R2

S2

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1 2 3 4 5 6 7 8 9 10 11 12Period0

100

200

300

400Inventory level

R2

S2

62/82



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)


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)


E[TC] = −vI0 + vN∑

t=1

dt + minN∑

t=1

(aδt + hIubt + bB

ubt ) + vIN (11)


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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)


E[TC] = −vI0 + vN∑

t=1

dt + minN∑

t=1

(aδt + hIubt ) + vIN (15)


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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.


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.


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.


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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77/82

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

78/82

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

79/82

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

80/82

Computational experienceβ service level

0.05

0.50

5.00

50.0

0

Segments

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imal

ity g

ap %

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ConclusionsQuestions

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