American Journal of Operations Research, 2014, 4, 132-141
Published Online May 2014 in SciRes. http://www.scirp.org/journal/ajor
http://dx.doi.org/10.4236/ajor.2014.43013
Optimal Stopping Time to Buy an Asset
When Growth Rate Is a Two-State Markov
Chain
Pham Van Khanh
Military Technical Academy, Hanoi, Vietnam
Email: [email protected]
Received 13 March 2014; revised 13 April 2014; accepted 20 April 2014
Copyright © 2014 by author and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/
Abstract
In this paper we consider the problem of determining the optimal time to buy an asset in a position of an uptrend or downtrend in the financial market and currency market as well as other
markets. Asset price is modeled as a geometric Brownian motion with drift being a two-state
Markov chain. Based on observations of asset prices, investors want to detect the change points of
price trends as accurately as possible, so that they can make the decision to buy. Using filtering
techniques and stochastic analysis, we will develop the optimal boundary at which investors implement their decisions when the posterior probability process reaches a certain threshold.
Keywords
Optimal Stopping Time, Posterior Probability, Threshold
1. Introduction
In [1], the authors consider the problem of determining the optimal time to sell a property while price growth
rate is a random variable that takes the value of the given set. Under the assumptions of the problem considered
in [1], growth rate only gets one of the possible values that do not change from this value to other values, which
means that transition probability density is 0; but at a time I do not know the accuracy of price growth rate and
the probability of receiving a certain value of growth rate also changes over time.
The authors in [2] also have much optimal stopping time in mathematical finance, but this is the classical
problem and less common in practice because of its assumptions.
The authors in [3] consider the optimal stopping time problem when the growth rate of price process is not a
random variable but in many cases.
How to cite this paper: Van Khanh, P. (2014) Optimal Stopping Time to Buy an Asset When Growth Rate Is a Two-State
Markov Chain. American Journal of Operations Research, 4, 132-141. http://dx.doi.org/10.4236/ajor.2014.43013
P. Van Khanh
In [4] the authors consider the problem to find the optimal time to sell when the price growth rate is Markov
chain, however their approach is different from our method in this paper and the results are also different.
In this paper we consider the price process is described by geometric Brownian motion when its drift (price
growth) is the Markov chain with two states 0 and 1 (0: decrease, downtrend; 1: increase, uptrend).
There is a phenomenon known as the momentum of the stock price, which means that the stock price has
increased or is increasing, it will be increased in the future (usually near future), and a stock price has decreased
or is decreasing, it will be decreased in the future. The investment is based on the momentum of price process
that is called the momentum trading or trend following trading. By this way if an investor holding an asset wants
to sell the asset, he will have to wait until a bull appeared and continues to hold such assets to the next price
increase (in momentum) and when momentum is no longer available, i.e. prices going down or just starting
going down, then he decides to sell. Similarly, if a person wants to buy an asset, he will wait for the appearance
of an opportunity of going down in price and wait for prices to go down further (in momentum) until no longer
falling price, then he decides to buy. According to this way of investment, investors are expected to buy the
property at the bottom of market and sell the property at the highest point of the market.
In this paper, we analyze how to select buy and sell optimal strategies for a momentum trading. More
precisely, we seek to maximize the expected profit from a momentum trade.
The method we use to study in this paper is the martingale theory, change of measure and the optimal
stopping time is referred in the literature [2] [5] and [6].
2. Buying Asset Problem
Now we consider the case at , t ≥ 0 , is a Markov chain with two states 0 and 1 that
P ( a = ah ) = π 0 ; P ( a = al ) = 1 − π 0 and only capable of moving from state 0 to state 1 with transition density as
 −λ λ 
=
follows Q 
 (λ > 0)
 0 0
{
π t P=
a ah t X
At the time of t > 0 we put =
}
where
{ }
X
t
0 ≤ t ≤T
is the filter generated by X.
We now consider the problem: Find  X -stopping time τ , 0 ≤ τ ≤ T such that:
U = inf E e − rτ X τ 
0 ≤τ ≤T
(2.1)
According to the theorem 9.1 (view [5]), posterior probability process π t satisfying:
dπ t =λ (1 − π t ) dt + π t
ah X t −  ahπ t + al (1 − π t )  X t
σ Xt
a −a 
dWt =λ (1 − π t ) dt + π t (1 − π t )  h l  dWt
 σ 
where
=
dWt
=
(
and W ,  X
)
dX t − (1 − π t ) al + π t ah  X t dt
=
σ Xt
{a − (1 − π ) a + π a } X dt + σ X dW
t
{a − (1 − π ) a + π a } dt + dW
t
t
l
t
t
l
t
h
t
t
t
σ Xt
h
σ
t
is a Brownian motion and the filter generated by W coincides with filter  X .
Thus the process X t satisfies the equation:
dX t
=
E  at t X  dt + σ dWt =
(1 − π t ) al + π t ah  dt + σ dWt
Xt
Then the process X t and the posterior probability process π t satisfies the equation:
133
P. Van Khanh
 dX t
( al + π t ( ah − al ) ) dt + σ dWt
X =
 t

dπ = λ (1 − π ) dt + π (1 − π )  ah − al
t
t
t 
1
 t
 σ

 dWt

(2.2)
π
Put Φ t = t . According to Ito formula we have:
1− πt
dΦ t
=
1
(1 − π )
2
dπ t +
t
=
(1 − π )
3
t

 ah − al
π t (1 − π t ) 
 σ

2

  dt

π
π2 2
λ1
 λ

dt + t ω dWt + t ω=
dt  1 + ω 2π t Φ t  dt + ωΦ t dWt
π
−
1− πt
1− πt
1− πt
1
t


{ }
Definition process Wt
1
as follows:
dWt = (ωπ t − σ ) dt + dWt
And the new measure P* satisfying:
T
T
 1T

1 T

dP*
2
2
exp
d
t
σ
ωπ
= exp − ∫ (σ − ωπ t ) dt + ∫ ( −σ + ωπ t ) dWt=
+
+
(
)
( −σ + ωπ t ) dWt 

 ∫
t
∫
dP
0
0
 20

2 0

where ω =
ah − al
σ
.
According to Girsanov theorem, Wt is a P* -Brownian motion.
We have:
λ


+ ω 2π t Φ t  dt + ωΦ t ( dWt − (ωπ t − σ ) dt ) = ( λ + ( λ + σω ) Φ t ) dt + ωΦ t dWt
dΦ t = 
Φt
 1 −

 1 + Φt

t




ω2 
−1
⇒ Φ=
Φ
+
=
+
−
Z
λ
Z
s
Z
λ
σω
d
;
exp


 t + ωWt 
t
t  0
∫ s  t
2 
0




dX t
=(1 − π t ) al + π t ah  dt + σ ( dWt − (ωπ t − σ ) dt ) =
( al + σ 2 ) dt + σ dWt
Xt


σ2 
=
⇒ X t X 0 exp  al +
 t + σ Wt 
2 


Consider the following process:
t
1 t

2
=
ηt exp  ∫ (σ − ωπ t ) dt − ∫ (σ − ωπ t ) dWt 
0
2 0

t
t
1 t

 1t

2
2
= exp  ∫ (σ − ωπ t ) dt − ∫ (σ − ωπ t ) ( dWt − (ωπ t − σ ) dt )  = exp − ∫ (σ − ωπ t ) dt − ∫ (σ − ωπ t ) dWt 
0
0
2 0

 20

dηt
Is a  X -martingale and =
ηt
(ωπ t − σ ) dWt
with η0 = 1.
a −λ t
e( l )
1
−λU t dt − σ U t dWt , we
have U t =
Let U t is process determined by dU t =
, U0
=
Xt
X0
Consider the following process:
(1 + Φt )U t
=
=
Yt
, Y 1. According to Ito formula we have:
(1 + Φ 0 )U 0 0
134
P. Van Khanh
=
dYt U t ( λ + ( λ + σω ) Φ t ) dt + ωΦ t dWt  + (1 + Φ t )  −λU t dt − σ U t dW t  − ωΦ t σ U t dt
Φt


= U t ωΦ t dW − (1 + Φ t ) σ U t dWt= (ωΦ t − (1 + Φ t ) σ ) U t dWt=  ω
− σ  (1 + Φ t ) U t dWt
+
Φ
1
t


d
Y
t
= (ωπ t − σ ) Yt dW ⇒
= (ωπ t − σ ) dWt
Yt
1 + Φ 0 ( al − λ − r )τ
e
(1 + Φτ ) .
X0
x
a − λ − r )τ
Denote X 0 = x, Φ 0 = φ then=
EP e − rτ X τ E=
E * e( l
η e− rτ X τ E=
η e− rτ X τ
(1 + Φτ )
P* T
P* τ
1+ φ P
Problem (2.1) is equivalent to the following problem:
rτ
rτ
From this we have Yt= ηt ∀t (a.s.), then e −=
ητ X τ e−=
Yτ X τ
inf EP* e(
=
F (φ )
=
Put Z t e(
al − λ − r )τ
0 ≤τ ≤T
(1 + Φt )
(2.3)
al − λ − r )τ
(1 + Φt ) . According to Ito formula we have:
dZ=
( al − λ − r ) e( a −λ − r )t (1 + Φt ) dt + e( a −λ − r )t dΦt
t
a − λ − r )t
=( al − λ − r ) e(
(1 + Φt ) dt + e( a −λ − r )t ( ( λ + ( λ + σω ) Φt ) dt + ωΦt dWt )
a − λ − r )t
( al − λ − r )(1 + Φ t ) + ( λ + ( λ + σω ) Φ t )  dt + e( a − λ − r )t ωΦ t dW
= e(
a − λ − r )t
a λ r t
( al − r ) + ( ah − r ) Φ t  dt + e( − − ) ωΦ t dW
= e(
a − λ − r )t
( al − r ) + ( ah − r ) Φ t  , it will be positive if
So, the drift of dU is e(
r − al
r − al
, and it will be negative if Φ t <
. This suggests
( al − r ) + ( ah − r ) Φt > 0 ⇔ Φt >
l
l
l
l
l
l
l
l
l
ah − r
optimal stopping time is the first time the process hits Φ t in area
satisfy the following free boundary problem:
to us that the
ah − r
[ A, ∞ ) with some A. Moreover pair ( A, F )
F + ( al − λ − r ) F= 0 0 < z < A

z≥A
1+ z
F ( z ) =

 F ′ ( A) = 1
F 0 + < ∞
 ( )
(2.4)
w2 z 2
F ′′ + ( λ + ( λ + ωσ ) z ) F ′ .
2
Differential Equation in (2.4) has the general solution as follows:
where  is infinitesimal operator =
F
=
F ( z ) C1 F1 ( z ) + C2 F2 ( z )
where
F1 ( z ) = x
−
ω2
2λ
eω
2
z
 −ω 2 + σω + λ
Whittaker W 
,
ω2

4ω 2 λ − 4ω 2 ( ah + al − 2r ) + ω 4 + 4 ( ah − al + λ )
−
F2 ( z ) = x
−ω 2 + 2σω + 2 λ
2
,−
2ω 2
−
−
−ω 2 + 2σω + 2 λ
ω2
2λ
eω
2
z

2λ 
;
ω 2 z 

 −ω 2 + σω + λ
Whittaker M 
,
ω2

4ω 2 λ − 4ω 2 ( ah + al − 2r ) + ω 4 + 4 ( ah − al + λ )
2ω 2
135
2
,−

2λ 
ω 2 z 

P. Van Khanh
and
−
z
Whittaker M
=
( a , b, z ) e 2 z
b+
1
2
−
z
1
Γ (b )
1
b − a −1 zt


M  b − a + ,1 + 2b, z  ; M
=
t a −1 (1 − t )
e dt
( a , b, z )
∫
2
Γ
b
−
a
Γ
a
(
) ( )0


Whittaker W=
( a, b, z ) e 2 z
b+
1
2
1
1 ∞ a −1
b − a −1 − zt


e dt
U  b − a + ,1 + 2b, z  ;U=
t (1 + t )
( a, b, z )
∫
2
Γ (a) 0


Changing variables and using some analytic transformations we obtain
=
F1 ( z )
∞
∫e
−
2λu
ω
2
u
1  2 λ 2σ
+
+
2  ω 2 ω
( β −1)2 −
8( al − λ − r )
ω2

− 3 

1 
(1 + zu ) 2 
( β −1)2 −
8( al − λ − r ) 2 λ 2σ 
−
−
+1
ω2
ω2 ω 
du
0
and F2 ( z )
1
z
∫
2λu
2
eω
u


2
1 2  λ
  2λ
  8( al − λ − r ) 
−3
 +σ  +   +σ  −1 −

2  ω  ω
 ωω
 
ω2



2
1  2  λ
  8( al − λ − r ) 2  λ

−  +σ  +1
  +σ  −1 −
ωω
 
 
ωω
ω2
(1 − zu ) 2 
du
0
Denote
α=
2λ
ω
> 0, β =α +
2
2σ
ω
> 0, γ =
( β − 1)2 +
8 ( λ + r − al )
ω2
> β −1 .
We have
∞
F1 ( z ) =
∫e u
−α u
( β + γ − 3) 2
(1 + zu )
( γ − β +1) 2
1
z
α u ( β + γ − 3) 2
du and F2 ( z ) =
(1 − zu )
∫e u
0
( γ − β +1) 2
du
0
Now we consider the property of the function
F2 ( z )
=
1
z
α u ( β + γ − 3) 2
∫e
u
(1 − zu )(
γ − β +1) 2
du .
0
First, we calculate the derivative of F2 ( z ) .
F2′ ( z ) = lim
F2 ( z + h ) − F2 ( z )
h →0
h
1


z
( γ − β +1) 2
1

( γ − β +1) 2
α u ( β + γ − 3) 2
α u ( β + γ − 3) 2
1
d
e
1
d
z
h
u
u
u
zu
u
= lim  ∫ e u
−
+
−
−
(
)
(
)
(
)
∫

h →0 h
0
 0



1


( γ − β +1) 2
1  z+h

( γ − β +1) 2
du 
= lim  ∫ eα u u ( β +γ −3) 2 (1 − ( z + h ) u )
− eα u u ( β +γ −3) 2 (1 − zu )
h →0 h
 0



1
z+h
)
(
 1

1  z α u ( β + γ − 3) 2

( γ − β +1) 2
1
d
zu
u
− lim  ∫ e u
−
(
)

h →0 h
 1

 z+h

1


( γ − β +1) 2
1  z + h α u ( β + γ − 3) 2

( γ − β +1) 2
du 
= lim  ∫ e u
− (1 − zu )
(1 − ( z + h ) u )
h →0 h
 0



1


1  z α u ( β + γ − 3) 2

( γ − β +1) 2
du 
− lim  ∫ e u
(1 − zu )
h →0 h
 1

 z+h

)
(
136
P. Van Khanh
= lim
1
z+h
h →0
∫
eα u u ( β +γ −3) 2
(1 − ( z + h ) u )
( γ − β +1) 2
− (1 − zu )
( γ − β +1) 2
du
h
0
1
γ − β +1 z
α u ( β + γ −1) 2
=
−
(1 − zu )
∫e u
2
0
( β − 1)2 +
Because γ=
8 ( λ + r − al )
( γ − β −1) 2
du
> β − 1 so F2′ ( z ) < 0 .
ω2
1
According to average integral theorem we have F=
e
2 (z)
α u∗
(1 − zu )
∗
( γ − β +1) 2 z
∫u
( β + γ − 3) 2
du
0
1
 1
where u ∗ ∈  0,  . Moreover, with small enough z we have zu ∗ < z ⇔ z <
 z
u∗
( )
1
(
F2 ( z ) > 1 − z
)
( γ − β +1) 2 z
du
∫ u=
0
because 0 < z < 1 and −
=
F1′( z )
We have
2
( β + γ − 3) 2
β + γ −1
4
β + γ −1
(1 − z )
( γ − β +1) 2 −
( β +γ −1)
>εz
4
z
−
2
. Then we have:
( β +γ −1)
4
→ +∞
when
z → 0+
<0.
γ − β + 1 ∞ −α t ( β +γ −1) 2
γ − β −1 2
(1 + zt )( ) dt > 0 because r > al so
∫e t
2
0
( β − 1)2 +
γ=
8 ( λ + r − al )
ω2
> β −1
And then F1 ( z ) is increasing function.
=
Moreover F1′′( z )
γ − β +1 γ − β −1 ∞
2
2
4 × 2 ( λ + σω )
−α t
t ( β +γ +1) 2 (1 + zt )
( γ − β − 3) 2
dt < 0 because
0
( β − 1)2 +
γ=
So γ < ( β − 1) +
∫e
2
8 ( λ + r − al )
ω
2
<
( β − 1)2 +
8 ( λ + ah − al )
ω2
= ( β − 1) + 4 β =β + 1 and F1′′( z ) < 0 means that the function F1 ( z )
2
ω
is increasing and convex function on ( 0, ∞ ) . Figure 1 shows the graph of function F1 ( z ) , we can check the
2
increase and convex properties of it. The graph of F2 ( z ) is shown in Figure 2, we can see that it tends to infinite when z as 0+.
∞
But F ( 0 + ) < ∞ , then =
C2 = 0 so F ( z ) C1 ∫ e −α u u ( β +γ −3) 2 (1 + zu )
( γ − β +1) 2
du
0
 ∞ −α u ( β + γ − 3 ) 2
γ − β +1 2
1+ A
(1 + Au )( ) du =
C1 ∫ e u
 0
By (2.4) we have 
∞
C γ − β + 1 e −α t u ( β +γ −1) 2 1 + Au (γ − β −1) 2 du =
1
(
)
1
∫

2
0

So A is the solution of equation
∞
2 ∫ e −α u u ( β +γ −3) 2 (1 + Au )
( γ − β +1) 2
∞
du = (1 + A )( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 (1 + Au )
0
0
137
( γ − β −1) 2
dt
(2.5)
P. Van Khanh
Figure 1. Graph of the function F1 ( z ) .
Figure 2. Graph of the function F2 ( z ) .
Lemma 2.1. Equation (2.5) has unique solution.
Proof: Equation (2.5) is equivalent to
∞
2 ∫ e −α u u ( β +γ −3) 2 (1 + xu )
( γ − β +1) 2
∞
du − (1 + x )( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 (1 + xu )
0
( γ − β −1) 2
du =
0
0
∞
=
Put: f ( x ) 2 ∫ e −α u u ( β +γ −3) 2 (1 + xu )
( γ − β +1) 2
∞
du − (1 + x )( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 (1 + xu )
0
( γ − β −1) 2
du
0
∞
∞
0
0
=
We
have f ( 0 + ) 2 ∫ e −α u u ( β +γ −3) 2 du − ( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 du
For integrals
∞
−α u ( β + γ − 3 ) 2
du
I1 2=
=
∫e u
0
=
∞
 −α u ( β +γ −1) 2
4
e −α u du ( β +γ −1) 2
=
e u
∫
β + γ −1 0
β + γ −1 
4
4α ∞ −α u ( β +γ −1) 2
e u
du
β + γ − 1 ∫0
138
∞
0
∞

+ α ∫ e −α u u ( β +γ −1) 2 du 
0

 4α + ( β − 1)2 − γ 2  ∞ −α u ( β +γ −1) 2
 4α
∞
γ −1) 2
du 
du
f (0 +) 
=
+ ( β − 1 − γ )  ∫ e −α u u ( β +=
∫e u
β −1+ γ

 β −1+ γ
0
 0
 8λ 8 ( λ + r − al ) 
So:
 2−
∞
8 ( r − al ) ∞ −α u ( β +γ −1) 2
ω2
ω
 ∫ e −α u u ( β +γ −1) 2 du =
e u
du < 0
=
− 2
β −1+ γ
ω ( β − 1 + γ ) ∫0

0


∞
f ′ ( x ) = ( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 (1 + xu )
( γ − β −1) 2
∞
du − ( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 (1 + xu )
0
− (1 + x )
P. Van Khanh
( γ − β −1) 2
du
0
γ − β −1
∞
(γ − β + 1) ∫ e−α u u ( β +γ +1) 2 (1 + xu )(
γ − β − 3) 2
du
2
0
∞
γ − β −1
γ − β −3 2
=− (1 + x )
(γ − β + 1) ∫ e−α u u ( β +γ +1) 2 (1 + xu )( ) du
2
0
Because
γ < ( β − 1) +
2
4 × 2 ( λ + σω )
ω
2
( β − 1)2 + 4β = β + 1; γ =
=
( β − 1)2 +
8 ( λ + r − al )
ω2
> β −1
and
so f ′ ( x ) =− (1 + x )
γ − β −1
2
∞
(γ − β + 1) ∫ e−α u u ( β +γ +1) 2 (1 + xu )(
γ − β − 3) 2
du > 0, ∀x > 0
0
We will prove lim f ( x ) = +∞
x →+∞
Indeed, for large enough x we have 1 + xu ∼ xu and
∞
=
f ( x ) 2 ∫ e −α u u ( β +γ −3) 2 (1 + xu )
( γ − β +1) 2
∞
du − (1 + x )( γ − β + 1) ∫ e −α u u ( β +γ −1) 2 (1 + xu )
0
( γ − β −1) 2
du
0
∞
∞
∼ 2 x(γ − β +1) 2 ∫ e −α u u ( β +γ −3) 2 u (γ − β +1) 2 du − (1 + x )( γ − β + 1) x(γ − β −1) 2 ∫ e −α u u ( β +γ −1) 2 u (γ − β −1) 2 du
= 2x
( γ − β +1) 2
0
∞
∫e
0
−α u
u
( β + γ − 3) 2 ( γ − β +1) 2
u
du − ( γ − β + 1) x
0
( γ − β +1) 2
∞
∫e
−α u
u γ −1du
0
∞
= (1 + β − γ ) x(γ − β +1) 2 ∫ e −α u u γ −1du > 0
0
Thus, lim f ( x ) < 0, lim f ( x ) = +∞ và f ( x ) is increasing function so f ( x ) = 0 have unique expex →0 +
x →+∞
rience. So the theorem is proved. 
The graph of f ( x ) is shown in Figure 3, it is an increase function and lim f ( x ) < 0, lim f ( x ) = +∞ .
x →0 +
x →+∞
τ A inf {t ≥ 0 : Φ t ≥ A} is the optimal stopping time for (2.1).
Theorem 2.2. Stopping time =
Proof:
According to the general theory of optimal stopping time is the point of making the expression:
(  + ah − λ − r ) (1 + φ ) =
ah − r + ( al − r ) φ
=
C {φ : F (φ ) > 1 + φ } .
is positive will be under the continuation area:
And stopping time will be optimal if it is the first time that the {Φ t }t ≥ 0 hits the stopping regions
D = {φ : F (φ ) = 1 + φ } or =
τ A inf {t ≥ 0 : Φ t ≥ A} . 
In Figure 4, we see a simulation of processes: Asset Price process, process Φ ( t ) , threshold probability,
posterior probability process and the optimal stopping time with
al =
−0.1; ah =
0.2; σ =
0.4 ; r =
0.05; λ =
0.3 .
139
P. Van Khanh
Figure 3. Graph of the function f ( x ) .
Figure 4. A simulation of asset price process, the posterior probability process, process Φ ( t ) , the threshold probability and the optimal stopping time (the optimal time to buy).
X ( 0 ) . In Figure 5, we simuWe find that τ * = 0.63 and the asset price (discounted) to buy is 1.24 < 1.5 =
late 4 processes: Asset Price process, process Φ ( t ) , threshold probability posterior probability process and the
optimal stopping time with parameters
al =
0.2; σ =
0.4 ; r =
0.05; λ =
0.3 .
−0.1; ah =
X (0) .
We find that τ = 0.58 and the asset price (discounted) to buy is 1.27 < 1.5 =
*
140
P. Van Khanh
Figure 5. A simulation of asset price process, the posterior probability process, process Φ ( t ) , the threshold probability and
the optimal stopping time (the optimal time to buy).
3. Conclusion
In this paper, we consider the problem of buying an asset when the asset price is modeled by the geometric
Brownian motion which has a change point, where price growth rate is the Markov process with two states that
describes the decreasing and increasing of asset prices process on the market. For buying problem we assume
that the price will be a shift from decreasing to increasing prices and a buying decision is made when the probability of decreasing state surpassed some certain threshold. We have to simulate the price process with a number
of parameters and conduct numerical solution to the experimental buying threshold. In the next study we will
consider the problem with assumptions that are closer to reality and more constraints.
Acknowledgements
This research is funded by Vietnam National Foundation for Science and Technology Development
(NAFOSTED) under grant number 10103-2012.17
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