Table Of Content8 Algorithm for solving optimization problems with
0
0 Interval Valued Probability Measure
2
n
a Phantipa Thipwiwatpotjana Weldon A. Lodwi
k
J
Department of Mathemati
al S
ien
es Department of Mathemati
al S
ien
es
4
2 University of Colorado Denver University of Colorado Denver
Phantipa.Thip
udenver.edu Weldon.Lodwi
k
udenver.edu
]
C
O
. Abstra
t
h
t
a We are
on
erned with three types of un
ertainties: probabilisti
, possibilitis-
m
ti
and interval. By using possibilityand ne
essitymeasuresasanInterval Valued
[ Probability Measure (IVPM), we present IVPM's interval expe
ted values whose
possibility distributions are in the form of polynomials. By working with interval
1
v expe
ted values of independent un
ertainty
oe(cid:30)
ients in a linear optimization
6 problemtogetherwithoperationssuggestedinLodwi
kandJamison[3℄,theprob-
1 lem after applying these operations be
omes a linear programming problem with
8
onstant
oe(cid:30)
ients. This is a
hieved by the appli
kation of two fun
tions. The
3 v : I → R I = {[a,b] |a≤b}
(cid:28)rst is applied to thke interval
oe(cid:30)
ients, , where .
. u : R → R
1 The se
ond is , applied to the produ
t we got from a previous fun
-
0
tion. Similar
on
epts hold for any types of optimization problems with linear
8
onstraints. Moreover, it implied that optimization problems
ontaining all three
0
: typesofun
ertaintiesinoneproblem
anbesolvedasordinaryoptimizationprob-
v
lems.
i
X
r
a
1 Introdu
tion
AnIntervalValuedProbability Measure(IVPM)whi
hisgeneratedfromthede(cid:28)nitions
provided by Wei
hselberger [8℄, is a tool that gives a partial representation for an
unknownprobability measure. Inthispaperweexpandtheideain[3℄ofusinganIVPM
toanoptimization problem withun
ertain
oe(cid:30)
ients. Thetypesoftheseun
ertainties
in this resear
h are probabilisti
, possibilisti
and interval un
ertainties. We provide
the ne
essary de(cid:28)nitions and explanations in the next se
tion.
Toapplyinanoptimizationproblem,wewillusetheintervalexpe
tedvaluedes
ribedin
se
tion2astherepresentative ofea
hun
ertain
oe(cid:30)
ient. We
onstru
tageneralform
of an interval expe
ted value whose IVPM
onstru
tion forms a polynomial possibility
density fun
tion.
In se
tion 3, by assuming independen
e, we apply interval expe
ted values to all un
er-
tain
oe(cid:30)
ientrandomvariables. We
reateanalgorithmforalinearprogram(LP)with
interval un
ertainty
oe(cid:30)
ients. Using this algorithm, we show that our un
ertainty
problem be
omes an ordinary LP. Similar details are given in se
tion 4. Examples,
on
lusion and further resear
h idea are provided.
2 Interval Valued Probability Measure (IVPM)
Before giving the de(cid:28)nition of an IVPM, we would like the readers to feel
omfortable
m˘
with the notation .
m¯; or
m˘ m˘ = m; or ;
De(cid:28)nition 2.1 De(cid:28)ne an un
ertain random variable as where
[m ,m ]
1 2
b
m¯ m m
is
orresponding to as a random variable with probabilisti
distribution and
m
means that is a random variable, with possibilisti
distribution. b
Inrealappli
ations, wemight notknow (with
ertaiint(yA))th=epir−o(bAab),ilii+ty(Am)easureforour
problems. Lodwi
k and Jamison [3℄ use an IVPM, m˘ m˘ m˘ , to measure
(cid:2) (cid:3)
a partial representation for an unknown probability measure. The original paper for
the idea of IVPM is adopted by Wei
hselberger [8℄. We use the following notation and
information throughout the paper unless stated otherwise:
•
The arithmeti
operations applied to intervals are those of interval arithmeti
[4℄.
• [0,1]
The set of all intervals
ontained in is denoted as
Int ≡ {[a,b] |0 ≤ a≤ b ≤ 1}.
[0,1]
• S A σ S S = R
denotes the universal set and is a -algebra of . Note that .
(S,A)
De(cid:28)nition 2.2 (Wei
hselberger [8℄)Given measurable spa
e , aninterval valued
i : A⊆ A → Int
m˘ [0,1]
fun
tion is
alled an R-probability if:
• i (A) = i− (A),i+ (A) ⊆ [0,1]
m˘ m˘ m˘ ,
(cid:2) (cid:3)
• ∃ Pr A ∀A∈ A Pr(A) ∈ i (A).
m˘
a probability measure, , on su
h that ,
i− i+
An R-probability from de(cid:28)nition 2.2 isan IVPM where m˘ and m˘ are
onstru
ted from
a possibility (fuzzy) density fun
tion.
2
p : S → [0,1]
De(cid:28)nition 2.3 A fun
tion is
alled a regular possibility distribution
fun
tion if
sup{p(x) |x ∈S} = 1.
Pos: S → [0,1]
Possibility distribution fun
tions (see [7℄) de(cid:28)ne a possibility measure,
where
Pos(A) = sup{p(x) | x∈ A}
and its dual ne
essity measure is
Nec(A)= 1−Pos(Ac),
sup{p(x) |x∈ ∅} = 0 n : S → [0,1]
where . A ne
essity distribution fun
tion
an be
de(cid:28)ned by setting
n(x) = 1−p(x)
and the
orresponding ne
essity measure
Nec(A)= inf{n(x)| x∈ Ac},
inf{n(x)| x∈ ∅} = 1
where .
In [2℄, it is shown that possibility distributions
an be
onstru
ted whi
h satisfy the
following
onsisten
y de(cid:28)nition.
p : S → [0,1]
De(cid:28)nition 2.4 Let be a regular possibility distribution fun
tion with
Pos Nec p
asso
iatedpossibilitymeasure andne
essitymeasure . Then issaidtobe
on-
X A Nec(A) ≤ Pr(X ∈ A)≤
sistent with random variable if for every measurable set ,
Pos(A)
.
i
m˘
The R-probability fun
tion in de(cid:28)nition 2.2 is used to de(cid:28)ne IVPMs. A possibility
i (A) = [Nec(A),Pos(A)]
m˘
and ne
essity pair, ,
onstru
ted by de(cid:28)nition 2.4 is able to
bound an unknown probability of interest. Therefore it
an be used to de(cid:28)ne an IVPM.
The reader
ould (cid:28)nd more explanations, examples and a
onstru
tion of an IVPM in
[3℄ .
2.1 The Interval Expe
ted Value Constru
ted From Polynomial Pos-
sibility Density Fun
tion
In this paper we
onsider the interval expe
ted value (de(cid:28)nition is given in [3℄),
∞ ∞
xdi = xd−(x)dx, xd+(x)dx ,
m˘
ZR (cid:20)Z−∞ Z−∞ (cid:21) (1)
of an IVPM
onstru
ted from possibility and ne
essity measures as the spe
i(cid:28)
upper
andlower
umulativeprobability distributionfun
tions, respe
tively. Whentheinterval
3
expe
ted value is
al
ulated as we will see, the lower
umulative distribution fun
tion
gives the right end-point of the interval expe
ted value, while the upper
umulative
distribution fun
tion de(cid:28)nes the left end-point. We give a formal de(cid:28)nition of the
interval expe
ted value in de(cid:28)nition 2.5
−
d (x)
De(cid:28)nition 2.5 The interval expe
ted value is de(cid:28)ned in (1), where refers to
the left possibility density fun
tion
orresponding to the upper
umulative distribution.
d+(x)
Similarly, is the right possibility density fun
tion
orresponding to the lower
umulative distribution.
Thisde(cid:28)nitionwillbe
ome
learwhenwepresent howto
al
ulatetheinterval expe
ted
value through a polynomial possibility density fun
tion.
n a/b/c/d
De(cid:28)nition 2.6 A polynomial degree fuzzy number , is a random number
a d
whosevalueisfuzzybetween and . The
orresponding polynomial fuzzymembership
f :R → [0,1]
fun
tion is de(cid:28)ned as
f (x) ; ∀x∈ [a,b)
L
1 ; ∀x∈ [b,c]
f(x) =
f (x) ; ∀x∈ (c,d]
R
0 ; otherwise,
f f f
L R
where and arede(cid:28)nedintable2.1. We
anuse asapolynomialpossibilitydensity
fun
tion when we have a
orresponding possibilisti
un
ertainty random variable.
n a/b/c/d [b,c] [a,d]
Consider a polynomial degree fuzzy number ,
ore , support , the
orresponding fuzzy membership fun
tion (or possibility density fun
tion) whi
h origi-
xn n = 1,2,3,...
nally
omes from a polynomial fun
tion where has the general form
as shown in table 2.1.
Table 2.1: General form of a polynomial possibility density fun
tion
n Left Density Fun
tion
f (x) = 1 (x−b)n +1
odd L (b−a)n
f (x) = − 1 (x−b)n+1
even L (b−a)n
n Right Density Fun
tion
f (x) = − 1 (x−c)n +1
odd R (d−c)n
f (x) = − 1 (x−c)n +1
even L (d−c)n
4
Ingeneral,theupper
umulative(Possibility measure)andthelower
umulative(Ne
es-
sity measure) distribution fun
tions generated by polynomial possibility density fun
-
Fu Fl
tion are and , respe
tively, where
0 ;x < a
Fu(x) = Pos(x) = fL(x) ;a ≤ x≤ b
1 ;b < x
and 0 ;x < c
Fl(x)= Nec(x) = 1−fR(x) ;c ≤ x≤ d
1 ;d < x.
n
For the odd number of , we
al
ulate the left and right density fun
tions respe
tively
as follows: d n (x−b)n−1 ;a ≤ x≤ b
d−(x) = [Fu(x)] = (b−a)n
dx (cid:26) 0 ;otherwise,
d n (x−c)n−1 ;c ≤ x≤ d
d+(x) = Fl(x) = (d−c)n
dx (cid:26) 0 ;otherwise.
h i
Theupper
umulative distributionprodu
esthelower integralandthelower
umulative
distribution produ
es the upper integral. Therefore
∞
b−a
−
xd (x)dx = a+
Z−∞ n+1
∞
and d−c
xd+(x)dx = d− .
Z−∞ n+1
Similar work
ould be done for the even degree. Thus, the interval expe
ted value of an
n n ∈ N
IVPM
onstru
ted by a polynomial possibility density fun
tion degree where
is b−a d−c
xdi = a+ ,d− .
m˘
Z (cid:20) n+1 n+1(cid:21)
R
c
Remark 2.1 The interval expe
ted value of an IVPM
onstru
ted from a
onstant
[a,b]
or an interval , (uniform regular possibilitydensity fun
tion), isthe
onstant or the
interval themselves.
a/b/c/d
Remark 2.2 Forany
ontinuouspossibilitydensityfun
tionwithfuzzynumber
[b,c] [a,d] xdi = [α,β] ⊆
whose
ore is and support is , the interval expe
ted value R m˘
[a,d] α∈ [a,b] β ∈ [c,d] R
, where and .
⋄
Proof: It is a property of an expe
ted value.
5
3 IVPM with Linear Programming
Consider a linear programming (LP) with some un
ertainty
oe(cid:30)
ients
n
max f(~x,a˘):= a˘ x
i i
(2)
Xi=1
n
s.t. g(~x,˘b,c˘):= ˘b x +c˘ = 0;j = 1,...,m
ji i ji
Xi=1
~0≤~x≤~t; where ~t ≥~0,
a˘ ˘b c˘
where some
omponents of , and
ould represent probabilisti
, possibilisti
, or
~x ∞
interval un
ertain random variables. The bound on ve
tor
ould be . In this paper
we
onsider the situation when these random variables are mutually independent.
De(cid:28)nition 3.1 (Lodwi
k and JX˘amisonY˘[3℄) The IVPM
onstru
ted from two un
ertain
independent random variables, and , is de(cid:28)ned as
i (A×B) ≡ i (A)i (B).
X˘×Y˘ X˘ Y˘
Unlike in the ordinary gL(P~x,,˘bf,oc˘r)t=his0 paper a prgoblem might have no feasible region.
Instead, the
onstraints mean that
an
ome as
lose to zero as possible.
Therefore,itisreasonabletousepenaltystrategiesforthistypeofproblem. Forexample
in [3℄
max f(~x,a˘) := 8x +7x
1 2
s.t. g (~x,˘b,c˘) := 3x +[1,3]x +4 = 0
1 1 2
g (~x,˘b,c˘) := ˆ2x +5x +1 = 0
2 1 2
~x∈ [0,2] ,
ˆ2 = 1/2/2/3
where ∗ . It is easy to see that this problem has no feasible set. Therefore,
~x
the solution for a modi(cid:28)ed problem does not need to satisfy the
onstraints of the
original one.
Lodwi
kandJamison[3℄presentanideatodealwithproblem(2)whi
hinvolvesinterval
types of un
ertainties by using the IVPM and the operations in the order below. The
explanation of these steps follows:
~p >0 g(~x,˘b,c˘)
1. apply a penalty
ost (determined by a de
ision maker) to (cid:12) (cid:12) ,
(cid:12) (cid:12)
(cid:12) (cid:12)
2.
al
ulate the interval expe
ted value
h ~x,a˘,˘b,c˘ di x
R (cid:16) (cid:17) a˘×˘b×c˘ (
ombining all liked terms i'stogether or not isdepend-
R
h ~x,a˘,˘b,c˘ = f(~x,a˘)−~p· g(~x,˘b,c˘) ,
ing on the user), where (cid:16) (cid:17) (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
6
v : [α,β] → Rk [α,β]
[α,β]
3. apply the ordered fun
tion to the interval
oe(cid:30)
ient of
h ~x,a˘,˘b,c˘ di
R (cid:16) (cid:17) a˘×˘b×c˘ from step 2,
R
u v ([α,β]) u :Rk → R
[α,β] [α,β] [α,β]
4. determine where ,
(cid:0) (cid:1)
p~ > 0 g(~x,˘b,c˘)
By assuming penalty
ost ve
tor to the
onstraints (cid:12) (cid:12), the problem
(cid:12) (cid:12)
be
omes (cid:12) (cid:12)
max h ~x,a˘,˘b,c˘ := f(~x,a˘)−p~· g(~x,˘b,c˘)
(cid:16) (cid:17) (cid:12) (cid:12) (3)
(cid:12) (cid:12)
g(~x,˘b,c˘) (cid:12) (cid:12)
For now, we ignore the fa
t that (cid:12) (cid:12) is a non smooth fun
tion and
onsider it
(cid:12) (cid:12)
as a linear fun
tion, so that we
a(cid:12)n explain(cid:12)
learly ea
h step of the operations above.
Next, by using the assumption that all un
ertain random variables are independent, we
i
form an IVPM a˘×˘b×c˘ then
al
ulate the interval expe
ted value with respe
t to this
IVPM. We get
h ~x,a˘,˘b,c˘ di .
ZR (cid:16) (cid:17) a˘×˘b×c˘ (4)
i ≡ i i i a˘ ˘b c˘
Sin
e a˘×˘b×c˘ a˘ ˘b c˘ ( , and are independent), the equation (4) is a triple integral
i i i x
with respe
t to a˘, ˘b and c˘. We
onsider i as a
onstant while (cid:28)nding the interval
expe
ted value for the IVPM. We get a linear fun
tion with interval
oe(cid:30)
ients as a
result for this step.
f g
Noting that if we (cid:28)nd the interval expe
ted value for ea
h fun
tion and before
applying the violation
ost, the result up to this point is the same as working with
h
violation
ost then the interval expe
ted value of fun
tion . So far we a
hieve a linear
un
onstrained obje
tive fun
tion with interval
oe(cid:30)
ients.
The next question is how to deal with these
oe(cid:30)
ients to keep as mu
h information
about theintervalaspossible. Thisinformation isdependent onthede
isionmaker. He
might use the midpoint as his priority, or he might want to keep tra
k on the length of
theinterval. Sothede
isionmaker needstoputhispriorities intheorder. Forexample,
he might use the midpoint as his (cid:28)rst priority sin
e it is the best estimate for the true
value. Then his se
ond priority
ould be the length of this interval be
ause together
with the midpoint, he will be able to get ba
k to his interval easily. His third priority
ould be theright end point ofthe interval be
ausehedid not want to ex
essthatlimit,
and so on. The de
ision maker also
an have di(cid:27)erent orders and (or) methods for ea
h
of intervals.
v : [α,β] → Rk k
[α,β]
The de
ision maker will write down the fun
tion to represent his
k = 3
priorities. For example when ,
α+β
2
v ([α,β]) = β−α .
[α,β]
β
7
[µ,ν] v
[µ,ν]
For the di(cid:27)erent interval
oe(cid:30)
ient , the
orresponding fun
tion might have
v
[α,β]
di(cid:27)erent orders from .
Now, the de
ision maker might weigh these priorities equally or might have some fan
y
[α,β]
strategy to deal with them. Again, these strategies depend on the interval and
the opinion of the de
ision maker. For example, given equal importan
e, he
an de(cid:28)ne
u : Rk → R
[α,β]
the fun
tion as
α+β
2 1 α+β 1 1
u β −α = + (β −α)+ β.
[α,β]
3 (cid:18) 2 (cid:19) 3 3
β,
The bottom line is that now the interval
oe(cid:30)
ients be
ome a real number by using the
u v [α,β]
[α,β] [α,β]
fun
tions and on the interval
oe(cid:30)
ient , i.e.
u v ([α,β]) ∈ R.
[α,β] [α,β]
(cid:0) (cid:1)
Therefore, we transform our original problem to a linear un
onstrained obje
tive fun
-
~x
tion problem with real
oe(cid:30)
ients. Together with the bound on we get a solution for
[α,β] u v
our transformation problem, (drop the subs
ript from fun
tions and ),
max u v h(~x,a˘,˘b,c˘) di .
x (cid:18) (cid:18)Z a˘×˘b×c˘(cid:19)(cid:19)
R
h(~x,a˘,˘b,c˘)
Unfortunately, we have to deal with the non smooth fun
tion, . Also we will
notbethat lu
kytogetalinearun
onstrained obje
tivefun
tion aftertheseoperations.
2
By looking at step suggested above
arefully, we
an
al
ulate the interval expe
ted
g f ~p
i
value of ea
h
onstraint and obje
tive fun
tion before applying a penalty
ost to
them (without
hanging the result at the end of step 4). So, there will be some
hanges
in the operations stated above. The operations we use in this paper are in the following
steps.
Algorithm 3.1 IVPM with interval un
ertainty
oe(cid:30)
ients LP.
g (~x,a˘,˘b,c˘) f(~x,a˘)
i
1. Calg
u(~xla,t˘be,c˘t)hdeiinterval exip=e
1t,e.d..v,amlue of ea
fh(~x,a˘) di , and , i.e., (cid:28)nd
R i a˘×˘b×c˘ for and R a˘×˘b×c˘. For
onvenien
e, we
R R
store the result of this step as
f(~x) ← f(~x,a˘)di
Z a˘×˘b×c˘
R
g (~x) ← g (~x,˘b,c˘)di ; i = 1,...,m .
i Z i a˘×˘b×c˘
R
8
v : [α,β] → Rk [α,β]
[α,β]
2. Apply the ordered fun
tion to the interval
oe(cid:30)
ient of
f(~x) g (~x)
i
and re
eived from step 1. Again, store the result as
f(~x) ← v f(~x,a˘) di
(cid:18)Z a˘×˘b×c˘(cid:19)
R
g (~x) ← v g (~x,˘b,c˘) di ;i = 1,...,m.
i (cid:18)Z i a˘×˘b×c˘(cid:19)
R
u v ([α,β]) ∈ R v ([α,β]) f(~x)
[α,β] [α,β] [α,β]
3. Determine to the
oe(cid:30)
ient of and
g (~x) (cid:0) (cid:1) u : Rk → R
i [α,β]
re
eived from step 2, where . Store the result as
f(~x) ← u v f(~x,a˘) di
(cid:18) (cid:18)Z a˘×˘b×c˘(cid:19)(cid:19)
R
g (~x) ← u v g (~x,˘b,c˘) di .
i (cid:18) (cid:18)Z i a˘×˘b×c˘(cid:19)(cid:19)
R
f g
i
Now, the
oe(cid:30)
ients of and are all
onstants and we approa
h the following
LP problem
max f(~x)
s.t. g (~x)= 0; i= 1,...,m
i
(5)
~0 ≤~x ≤~t; where ~t ≥~0,
p~> 0 g
4. Applyapenalty
ost tothefun
tionve
tor re
eivedfromstep3. De(cid:28)nethe
h(~x) = f(~x)−pT |g(~x)|
penalty fun
tion . The problem be
omes an un
onstrained
h(~x)
obje
tive fun
tion to maximize the fun
tion .
|g(~x)|
5. Use modeling te
hniques to get rid of non-smooth absolute fun
tion, .
p~
In general, the penalty ve
tor is not a (cid:28)xed ve
tor. It depends upon the ex
ess or
g 0
shortage of fun
tion from . Moreover, one of the modeling di(cid:30)
ulty is that we
ould
notknowinadvan
ewhi
h
onstraint willla
korex
essthebalan
ezero. Letusdenote
e s g (~x)
i i i
$ and $ as the
ost penalty for ea
h ex
ess and shortage unit of from zero,
g(~x,˘b,c˘) g
respe
tively. Noting that (cid:12) (cid:12) , the absolute of the fun
tion ve
tor , is a non
(cid:12) ψi :(cid:12)= max {0,gi(~x)} ζi := max {0,−gi(~x)}
smooth fun
tion. So we d(cid:12)e(cid:28)ne (cid:12) and , for
i = 1,...,m
. Then the un
onstrained problem got form step 4 of Algorithm 3.1
max h(~x) := f(~x)−pT |g(~x)|
9
an be remodeled as
max h(~x,ψ~,ζ~) := f(~x)− m e ψ − m s ζ
i=1 i i i=1 i i
P P
(6)
ψ ≥ g (~x)
i i
ζi ≥ −gi(~x)
i = 1,...,m;
ψ ≥ 0
i
s.t.
ζ ≥ 0
i
0≤~x≤~t; ~t ≥ 0.
From the explanation above we provide the
on
lusion in the following lemma.
Lemma 3.1 Consider an LP (2), after working through the algorithm 3.1, the remod-
eled problem (6) be
omes an LP problem.
Proof: It is
lear from the algorithm 3.1.
Example 3.1 Consider the problem
max f(~x,a˘) := 2x −¯3x +[3,5]x
1 2 3
s.t. g (~x,˘b,c˘) := 4x +[1,5]x −2x −[0,2] = 0
1 b 1 2 3
g (~x,˘b,c˘) := 6x −¯2x +9x −9 = 0
2 b 1 2 3
g (~x,˘b,c˘) := −2x −[1,4]x −8x +¯5 = 0
3 1 2 3
0 ≤ x ≤ 3
1 b
0 ≤ x ≤ 2
2
0 ≤ x ≤ 2,
3
4 [1,5] −2
T
a˘ = 2,−¯3,[3,5] ˘b = 6 −¯2 9 c˘= [−[0,2],−9,¯5]T
where h i , b and .
−2 −[1,4] −8
b
f g
3
Note: The fun
tions and involve 3 types ofbun
ertainties.
These
oe(cid:30)
ients have possibility (or probability) polynomial density fun
tions as
shown in table 3.2. Apply Algorithm 3.1 step by step, we have:
10
