User Association in Massive MIMO HetNets

0.26 MB·English

by  Yi Xu

User Association in Massive MIMO HetNets
Yi Xu Student Member, IEEE and Shiwen Mao, Senior Member, IEEE
Abstract
5 Massive MIMO and small cell are both recognized as the key technologies for the future 5G wireless systems.
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In this paper, we investigate the problem of user association in a heterogeneous network (HetNet) with massive
0
2 MIMO and small cells, where the macro base station (BS)isequipped with amassive MIMO and the picocell BS’s

n are equipped with regular MIMOs. We first develop centralized user association algorithms with proven optimality,
a
consideringvariousobjectivessuchasratemaximization,proportionalfairness,andjointuserassociationandresource
J

4 allocation.WethenmodelthemassiveMIMOHetNetasarepeatedgame,whichleadstodistributeduserassociation
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algorithms with proven convergence to the Nash Equilibrium (NE). We demonstrate the efficacy of these optimal

] schemes by comparison with several greedy algorithms through simulations.
T
I
. Index Terms
s
c
[
Massive MIMO; small cells; heterogeneous networks (HetNet); user association; unimodularity; game theory.

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7 I. INTRODUCTION
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Overthepasttwodecades,MultipleInputMultipleOutput(MIMO)hasevolvedfromapuretheorytoapractical
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technology, and has greatly enhanced the wireless system capacity by offering many degrees of freedom (DoF)
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for wireless transmissions. However, due to the so-called “smartphone” revolution, mobile users are demanding
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increasinglyhigherdatarates forrichmultimediaapplications.Existingand futurewireless networksare facingthe
.
1 grand challenge of a 1000-time increase in mobile data in the near future [1]. There have been tremendousefforts
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made aiming to cater for this demand. For example, based on MIMO and OFDM, LTE-Advancedtargetsat a peak
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rate of 1 Gbps, but the average rate is still less than 100 Mbps. In the foreseeable future, such rates can hardly be
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satisfactory for data-hungry wireless users.
:
v To boost wireless capacity, two technologies have gained most attention from both industry and academia. The
i first one is massive MIMO (a.k.a., large-scale MIMO, full-dimension MIMO, or hyper MIMO) [2], [3]. The idea
X
is to equip a base station (BS) with hundreds, thousands, or even tens of thousands of antennas, hereby providing
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a an unprecedented level of DoF for mobile users. The massive MIMO concept has been successfully demonstrated
in recent works [4], [5]. The second technology is small cell. A great benefit of deploying small cells is that
the distance of the user-BS link can be effectively reduced, leading to reduced transmit power, higher data rate,
enhanced coverage, and better spatial reuse of spectrum. Both massive MIMO and small cells are recognized as
key technologies of the future 5G wireless systems [6].
In this paper, we consider a heterogeneous network (HetNet) with massive MIMO and small cells, where the
macrocell BS (MBS) is equipped with a massive MIMO and the picocell BS’s (PBS) are equipped with regular
MIMOs.TofullyharvestthebenefitspromisedbythesetwotechnologiesinanintegratedHetNetsystem,itiscritical
to investigate the user association problem, i.e., how to assign active users to the BS’s such that the system-wide
capacity can be maximized and users’ experience can be enhanced.
There are already several recent works pushing forward in this direction. In [7]–[10], the authors consider the
problem of user association in massive MIMO systems operated in the frequency-divisionduplexing (FDD) mode.
Y. Xu and S. Mao are with the Department of Electrical and Computer Engineering, Auburn University, Auburn, AL 36849-5201. Email:
[email protected], [email protected].
ShiwenMaoisthecorresponding author:[email protected], Tel:[phone], Fax:[phone].
These papers are focused on a macrocell without small cells. In [11], user association in time-division duplexing
(TDD) massive MIMO system is addressed, where factional user association is allowed. Bayat et al. in [12] model
the problem of user association in a femtocell HetNet as a dynamic matching game and derive the optimal user
association. However, massive MIMO is not considered in the system model. In [13], the authors investigate the
problem of user association with conventional MIMO BS’s and propose a simple bias based selection criterion to
approximate more complex selection rules. Bjo¨rnson, et al. in [14] consider the problem of improving the energy
efficiency without sacrificing the quality of service (QoS) of users in a massive MIMO and small cell HetNet.
Motivatedbytheseinterestingworks,weconsidertheuserassociationprobleminaTDDmassiveMIMOHetNet
inthispaper,takingintoconsiderationofthepracticalconstraints,suchasthelimitedloadcapacityateachBS,while
without allowing fractional user association. The main goal is to maximize the system capacity while enhancing
user experience.
More specifically, this paper contains two parts: (i) centralized user association and (ii) distributed user as-
sociation. For centralized user association, we investigate the problems of rate maximization, rate maximization
with proportional fairness, and joint resource allocation and user association. We prove the unimodularity of our
formulatedproblemand developoptimaluser association algorithmsto the problemsof rate maximizationand rate
maximizationwithproportionalfairness.We thenproposea seriesofprimaldecompositionanddualdecomposition
algorithms to solve the problem of joint resource allocation and user association and prove the optimality of the
proposed scheme. For distributed user association, we model the behavior and interaction between the service
provider, who owns the BS’s, and users as repeated games. We consider two types of operations: (i) the service
provider sets the price and the users decide which BS to connect to, and (ii) the users bid for the opportunity of
connection. We prove that in both cases the the proposed algorithms converge to the respective Nash Equilibrium
(NE).
In the reminder of this paper, Section II introduces the system model and preliminaries. Optimal centralized
and distributed user association schemes are presented in Sections III and IV, respectively. Section V presents the
simulation study and Section VI concludesthis paper. Throughoutthis paper,we use a boldfaceupper(lower)case
symboltodenoteamatrix(vector),andanormalsymboltodenoteascalar.(·)H denotestheHermitianofamatrix.
II. SYSTEMMODELAND PRELIMINARIES
The system considered in this paper includes K users and J BS’s, including an MBS with a massive MIMO
and (J −1) PBS’s, each equipped with a conventional MIMO. The channel model is h = g l , where
j,k,n j,k,n j,k
h is the channel of antenna n at BS j to user k, g represents the small scale fading coefficient between
j,k,n j,k,n
antenna n of BS j and user k, and l stands for the large scale fading coefficient between BS j and user k [16].
j,k
Concatenatingall the channelcoefficientsfromall the antennasofBS j, we obtainthe channelvectorh , as well
j,k
as the channel coefficient matrix for signals transmitted from BS j as H =[h ,h ,··· ,h ] .
j j,1 j,2 j,k
Let y denote the signals received by the users connecting to BS j, W the precoding matrix of BS j, and d
j j j
the data sent from BS j. We have
y =H W d +n , (1)
j j j j j
where n is the zero mean circulant symmetric complex Gaussian noise vector.
j
Each active user has the options to connect to either the MBS or a PBS. For a user k, define user association
index variable x as
kj
1, if user k is connected to BS j.
x = (2)
kj 
0, otherwise.
Let its achievable rate if connected to BSj be Rkj, ηkj =xkjRkj, and its actual data rate be ηk. We have
η = η = x R . (3)
k kj kj kj
j j
X X
For users connecting to a massive MIMO BS j (i.e., the MBS), their achievable rate can be approximated with
the following deterministic rate [11].
M −L +1 P l
R =log 1+ j j j j,k , (4)
kj Lj 1+ j′=jPj′lj′,k!
6
P
where M is the number of antennas at the BS, L is the prefixed load parameter of the BS indicating how many
j j
users it could serve, and P is transmitpower from the MBS. Note that there is no small scale fading factor in (4).
j
This approximation has been proven to be accurate [11].
ForaPBSwithaconventionalMIMO,weassumethattheinter-cellinterferenceisnegligibleamongthepicocells,
due to the small transmission powers and effective inter-cell interference coordination (ICIC) [15]. The achievable
rate of user k connecting to PBS j can be represented as follows.
2
P hH w
j j,k j,k
R =log 1+ , (5)
kj  (cid:12) (cid:12) 2
1+ k′=(cid:12)(cid:12)kPj hHj,k(cid:12)(cid:12)wj,k′
e  6 
 (cid:12) (cid:12) 
wherew is thek-thcolumnofBSj’sprecodingmPatrixW .(cid:12)Thereare(cid:12)manyprecodingdesignsforconventional
j,k j (cid:12) (cid:12)
MIMO BS’s, such as matched filter (MF) precoding, zero forcing (ZF) precoding, and regularized zero forcing
(RZF) precoding[16]. Withoutloss of generality,we adoptMF precodingin this paperwith W = 1 HH, where
j √ϕ j
ϕ is a power normalization factor. The signal received by all the users connecting to PBS j can be rewritten as
follows.
hH h d +hH h d +···+hH h d
j,1 j,1 1 j,1 j,2 2 j,1 j,k k
hH h d +hH h d +···+hH h d 
y = j,2 j,1 1 j,2 j,2 2 j,2 j,k k . (6)
j
 
 ··· 
 
hHj,khj,1d1+hHj,khj,2d2+···+hHj,khj,kdk
 
 
Thus, the achievable rate for user k regarding to PBS j can be obtained as follows.
2
P x hH h
j kj j,k j,k
η =log 1+ . (7)
kj  (cid:12) (cid:12) 2
 1+ k′6=(cid:12)(cid:12)kPj xkj′hHj,(cid:12)(cid:12)khj,k′ 
 (cid:12) (cid:12) 
P (cid:12) (cid:12)
III. CENTRALIZEDUSE(cid:12)RASSOCIATI(cid:12)ON
In this section, we consider the problem of centralized user association. We assume that the BS’s have all the
channel state information (CSI) via uplink training. We adopt the following utility function for each user k with
achievable rate η .
k
ηk1−α/(1−α), if α>0,α6=1

U(ηk)=ηk, if α=0 (8)
log(η ), if α=1.
k
When α = 0, maximizing U(·) yields the maximization of the sum rate (but no fairness); when α → ∞, it leads
to the maximization of the worst-case rate (i.e, max-min fairness); when α=1, it yields the maximization of the
geometric mean rate (i.e., proportional fairness).
Our goal is to maximize the system utility by configuring the user-BS association. Typically, we consider the
cases when α=0 and α=1. In the case of α=1, we define U(0)=0.
A. Maximizing Sum-rate
We firstly investigate the problem of maximizing the system sum rate, i.e., α = 0 in (8) and U(η ) = η . The
k k
problem can be formulated as follows.
K
P1-1: max η (9)
k
{xkj}k=1
X
s.t. x ≤L ≤M , j =1,2,··· ,J
kj j j
k
X
x ≤1, k =1,2,··· ,K
kj
j
X
Constraints (2), (3), (4), (7).
Note that the second constraint requires the number of users connecting to a BS to be no more than its prefixed
load, which should in turn be no more than the number of antennas it has, since theoretically BS j can provide
at most M degrees of freedom (DoF). Assuming the L ’s are already chosen to satisfy L ≤ M , we drop this
j j j j
constraint in the remainder of this paper. The third constraint simply claims that each user can connect to at most
one BS at a time.
A key observation is that (7) can be rewritten as
2
P hH h
j j,k j,k
η =x log 1+ . (10)
kj kj  (cid:12) (cid:12) 2
 1+ k′6=k(cid:12)(cid:12)Pj xkj′h(cid:12)(cid:12)Hj,khj,k′ 
 (cid:12) (cid:12) 
Thus the R in (5) can be redefined as P (cid:12) (cid:12)
kj (cid:12) (cid:12)
2
e P hH h
j j,k j,k
R =log 1+ . (11)
kj  (cid:12) (cid:12) 2
e  1+ k′6=k(cid:12)(cid:12)Pj xkj′h(cid:12)(cid:12)Hj,khj,k′ 
 (cid:12) (cid:12) 
In(11),itcanbeseenthatRkj dependsonotherPusers’choi(cid:12)(cid:12)cesxkj′,fora(cid:12)(cid:12)llk 6=k′,aswell.Tomaketheproblem
tractable, we adopt the worst-case approximation by assuming the users within the coverage of BS j (denoted as
Gj) all connect to BS j with eperfect channels. This way, (11) can be approximated as
2
P hH h
j j,k j,k
R =log 1+ , (12)
kj  1+((cid:12)|G |−1)(cid:12)P 
(cid:12) j (cid:12) j
(cid:12) (cid:12)
e  
 
where |·| for a set stands for the cardinality of the set.
Define auxiliary variables c as follows.
kj
R in (4), if BS j is the MBS;
c = kj (13)
kj 
Rkj in (12), if BS j is a PBS.
e
The sum rate maximization problem can be reformulated as
K J
P1-2: max x c (14)
kj kj
{xkj}k=1j=1
XX
s.t. x ≤L , j =1,2,··· ,J
kj j
k
X
x ≤1, k =1,2,··· ,K
kj
j
X
Constraints (2), (13).
Since the variablesx ’s are binary,problemP1-2 falls into the categoryof Multiple Knapsack Problems, which
kj
is one of Karp’s 21 NP-complete problems [17]. Although a greedy algorithm could be developed to compute
sub-optimal solutions, we show that problem P1-2 can actually be optimally solved by taking advantage of its
special structure.
Let X be a matrix with entries x , k = 1,2,··· ,K, j = 1,2,··· ,J. We could convert X to a vector x by
kj
concatenating the rows of X and taking a transpose as x = [x x ··· x ··· x ··· x ]T, and simplify
11 21 K1 1J KJ
thenotationasx=[x x ··· x ]T.We thenapplythesameconversiontothematrixcomprisingc andobtain
1 2 KJ kj
vector c. Problem P1-2 can be rewritten as
P1-3: maxcTx (15)
x
K
s.t. x ≤L , j =1,2,··· ,J
(j 1)K+k j
−
k=1
X
J
x ≤1, k=1,2,··· ,K
k+(j 1)K
−
j=1
X
Constraints (2), (13).
Ignoring constraints (2) and (13), define A as the constraint matrix of problem P1-3, with entries being the
coefficients of the first and second constraints. We next introduce an important definition and derive a key lemma.
Definition 1. A matrix A is called totally unimodular if the determinant of every square submatrix of A is either
0, +1 or −1 [18].
Lemma 1. The constraint matrix A of problem P1-3 is totally unimodular.
Proof: Inspecting the constraints in problem P1-3, we find that A is of the following form.
1 1···1 0 0···0 0 0···0
0 0···0 1 1···1 0 0···0
. . .
 . . . 
 . . . 
 
 
0 0···0 0 0···0 1 1···1
A =  . (16)
 
1 0···0 1 0···0 1 0···0
 
 
0 1···0 0 1···0 0 1···0
 
 ... ... ... 
 
 
0 0···1 0 0···1 0 0···1
 
 
We can divide A into blocks as follows.
A A ··· A
1 2 J
A = , (17)
B B ··· B 
1 2 J
 
where each A , j ∈ [1,J], is a submatrix of A of size J ×K; and each B , j ∈ [1,J], is an identity matrix of
j j
size K×K.
Let S denote an arbitrary square submatrix of matrix A of size n. For any submatrix of A of size n=1, it is
n
trivial to see that the determinant of this submatrix is either 0 or +1. So we only need to consider the case where
the size of the square submatrix is greater than or equal to 2.
Case 1: S is taken entirely from one of the submatrices A or B , j ∈ [1, J]. We can see from the structure
n j j
that at least one row of A is all zero. So if the square submatrixis entirely taken fromA , the determinantof the
j j
submatrixis zero. Since matrix B , for all j, is simply an identity matrix, it is straightforwardthat the determinant
j
of any square submatrix of B is either 0 or +1.
j
Case 2: S is not entirely taken from any one of the submatrices A or B , j ∈[1,J]. In this case, the square
n j j
submatrix must be taken from 2n (n = 1,··· ,J) submatrices of the submatrix set (A ∪B , j ∈ 1,··· ,J). We
j j
next proceed with our proof by applying induction method.
For the base case n = 1, the square submatrix to be examined is of size 2. Since the entries can only be 0 or
+1, the determinant can only be 0, +1 or −1.
Now assuming that any square submatrix of size (n−1) has determinant 0, +1 or −1, we need to check if the
same conclusion holds for any square submatrix of size n.
We first notice that each column of A has exactly two +1s. Moreover, exactly one of them is in A , and the
j
other in B . Let q = argmin S , where S is the (i,q)-th entry of S . That is, column q has the
j ∗ q i ni,q ni,q n ∗
minimum number of 1s among all the columns of S .
n
Let ζq∗ =minq iSni,q. ζq∗ cPan only be 0, 1, or 2.
If ζq∗ =0, then allthe entriesof the q∗-th columnof Sn are 0, which resultsin det(Sn)=0, where det is short
for determinant. P
If ζq∗ = 1, then we could calculate det(Sn) by expanding the q∗-th column and obtain det(Sn)=det(S(n 1)).
Since det(S ) is 0, 1 or −1 by our induction hypothesis, we conclude det(S ) is 0, 1 or −1. −
(n 1) n
Ifζq∗ =2,−we couldfirstly negatealltheentriestakenfromBj,andthenaddalltherowsin Bj toanynon-zero
row in A . After this procedure, if that non-zero row in A is still non-zero, add that row to any other non-zero
j j
row in A . Repeat this process until we get a zero row in A . The reason why this process always give us a
j j
all-zero row is that we have equal number of +1s in A and B . Since any basic row operation does not change
j j
the determinant and we finally get a all-zero row, we have det(S )=0. That completes our induction.
n
Fact 1. For a linear programming problem, if its constraint matrix satisfies totally unimodularity, then its has all
integral vertex solutions [18].
Fact 2. For a linear programming problem, if it has feasible optimalsolutions, then at least one of them occurs at
a vertex of the polyhedron define by its constraints [19].
Given the facts and Lemma 1, we have the following theorem. The proof is straightforward and omitted.
Theorem1. TheoptimalsolutionofproblemP1canbeobtainedbysolvingarelaxedproblemwhere thevariables
x are allowed to take real values between [0,1].
kj
Given the above theorem, we could obtain the optimal solution of P1 by solving the relaxed problem, termed
NP1, using common LP solvers [18].
B. Proportional Fairness
In this section, we take proportional fairness among user achievable rates into consideration. The problem can
be formulated as follows.
K J
P2-1: max log x c (18)
{xkj}k=1 j=1 kj kj
X X
 
s.t. same constraints as problem P1-2.
Problem P2-1 is a nonlinear integer programming problem, which is generally NP-hard. To get a better under-
standing of the problem, we examine its equivalent problem as follows.
K J
P2-2: max x c (19)
xkj k=1j=1 kj kj
Y X
 
s.t. same constraints as problem P1-2.
ProblemP2-2isageometricprogrammingproblem,withbinaryvariables.Theobjectivefunctionisaposynomial
function with JK terms. Conventionally, to solve geometric programming problems we need to introduce new
variables such as y = log(x) so that geometric programming can be solved via convex programming. However,
here x ’s are binary. Since log(0) = −∞, we could not apply these techniques. Another heuristic scheme is to
kj
firstly sort these JK coefficients, and then find L maximal coefficients for each BS. However, even sorting these
j
JK coefficients could be computationally prohibitive even for a small system, which requires O(JKlog(JK))
operations.
A key observation about the logarithm function is that log( τ ) ≤ log(τ ), for all τ ≥ 2. Therefore, in
i i i i i
practice,1 the optimal value of problem P2-1 is upper bounded by that of the following problem.
P P
K J
NP2: max x log(c ) (20)
xkj k=1j=1 kj kj
XX
s.t. same constraints as problem P1-2.
We have the following results for the transformed problems.
Lemma 2. Problems P2-1 and NP2 are equivalent.
Proof: Recall that if η = 0, we define U(η ) = 0. The second constraint J x ≤ 1 imposes that
k k j=1 kj
each user could only connect to one BS. Consequently, x log(c )=log( x c ). Furthermore, we have
j kj kj j kPj kj
x log(c )= log( x c ).
k j kj kj k j kj kj P P
Comparing problems NP2 to P1-2, we find they are actually equivalent. Thus we can obtain the optimal value
P P P P
of P2-1 by applying the same technique used to solve problem P1-2. We then have the following lemma.
Lemma 3. Sum rate maximization in Section III-A also achieves proportional fairness.
Lemma 4. The optimal value of problem P2-1 is upper bounded by UB = K max log (c ).
1 k=1 j 2 kj
Proof: Denote m=max{ln(c ),ln(c ),··· ,ln(c )}, we have P
k1 k2 kJ
J J em
log2 xkjckj ≤ log2 emeln(ckj)
j=1 j=1
X X
   
J
= log2(em)+log2 eln(ckj)−m
j=1
X
 
≤ mlog (e)+log (J) (21)
2 2
The first inequality is because x ≤ 1. The second inequality is due to the fact that m is the largest one among
kj
all the xkjckj and eln(xkjckj)−m ≤1.
On the other hand, it follows the constraint J x ≤ 1 that log J x c ≤ log (em), which is a
j=1 kj 2 j=1 kj kj 2
better bound than (21). We thus have UB1 = PKk=1maxjlog2(ckj). (cid:16)P (cid:17)
P
1Recall that ckj is the achievable rate of user k connecting to BS j. ckj ≥ 2 is generally satisfied in current wireless systems with a
sufficiently largebandwidth andhightransmissionpower.
Algorithm 1: Greedy Algorithm 1 for User Association
1 Initialize K={1,2,··· ,K}, Lj,∀j∈J and xkj to be an all-zero matrix ;
2 for k=1 to K do
3 for j =1 to J do
4 Compute ckj as in (13) ;
5 end
6 end
7 while∃j, Lj 6=0 do
8 Find (k∗,j∗)=argmaxk,j{ck,j} ;
9 if Lj∗ 6=0 then
10 xkj∗∗ =1 ;
11 Lj∗ =Lj∗ −1 ;
12 K=K\k∗ ;
13 end
14 end
Algorithm 2: Greedy Algorithm 2 for User Association
1 Initialize K={1,2,··· ,K}, Lj,∀j∈J and xkj to be an all-zero matrix ;
2 for k=1 to K do
3 for j =1 to J do
4 Compute ckj as in (13) ;
5 end
6 end
7 for j =1 to J do
8 whileLj 6=0 do
9 Find (k∗,j)=argmaxk{ck,j} ;
10 xk∗ =1 ;
j
11 Lj =Lj−1 ;
12 K=K\k∗ ;
13 end
14 end
For comparison purpose, we propose two sub-optimal greedy algorithms, i.e., Algorithms 1 and 2, as bench-
marks. They can be directly used for comparison with problem P1-1. To compare with problem P2-1, in Al-
gorithm 1, we need to change Steps 7 and 8 as “while ∃j, L 6= 0 & max log(c ) > 0 do” and “Find
j k,j k,j
(k ,j )=argmax {log(c )},” respectively.In Algorithm2, we need to changeStep 8 and9 as “while L 6=0
∗ ∗ k,j k,j j
& max log(c )>0” and “Find (k ,j)=argmax log(c ),” respectively.
k k,j ∗ k k,j
C. Joint Resource Allocation and User Association
Inthissection,wetakeresourceallocationintoaccount.ConsideramassiveMIMOOFDMAHetNet.InOFDMA
systems, such as LTE, the time-frequency resource is divided into resource blocks (RB). A typical RB consists of
12 subcarriers (180kHz) in the frequency domain and 7 OFDMA symbols in the time domain (0.5 ms). So the
system mayhaveupto severalhundredsof RBs. We normalizeitto bea unitnumber.A userk connectingtoa BS
j gets a portion β of the overall resource. The goal is to maximize the system utility considering both resource
kj
allocation and user association.
Considering the logarithm rate utility and defining Φ = k |x =1 , the problem is formulated as follows.
j kj
K (cid:8) J (cid:9)
P3-1: max log x c β (22)
{xkj,βkj}k=1 j=1 kj kj kj
X X
 
s.t. β ≤1, j =1,2,··· ,J
kj
kX∈Φj
same constraints as problem P1-2.
TosolveproblemP3-1,weneedto:(i)selectusersforeachBStoserveand(ii)allocateresourcestotheassociated
users at each BS. We next propose a series of primal decomposition and dual decomposition to solve the problem
optimally.
Itisworthnotingthattheproblemcanalsobeformulatedinadifferentway,bysubstitutingconstraint J x ≤
j=1 kj
1,k = 1,2,··· ,K with a new constraint J x = 1,k =1,2,··· ,K. We call this problem P3-2. Comparing
j=1 kj P
these two formulations, we have the following observations.
P
1) Problem P3-1 does not require that every user must be connected,while problem P3-2 requires each user be
connected, even under unfavorable conditions.
2) Problem P3-2 has a more stringent requirement than problem P3-1. Therefore the optimal value of problem
P3-2 is upper bounded by that of problem P3-1.
3) Since problem P3-1 offers more choices of user association, problem P3-1 is slower in convergence than
problem P3-2.
We focus on the harder problem P3-1. Given the algorithm to solve problem P3-1, problem P3-2 can be readily
solved.Duetointegervariablesx andrealvariablesβ ,problemP3-1isamixedintegernonlinearprogramming
kj kj
problem(MINLP),whichisgenerallyNP-hard.However,nextweproposeanalgorithmtoobtainitsoptimalsolution.
Sincex ’stakebinaryvaluesand J x ≤1,wehave K log J x c β = K J x log(c β ).
kj j=1 kj k=1 j=1 kj kj kj k=1 j=1 kj kj kj
Recall that if Jj=1xkj =0, the logParithmic utility is 0. ThuPs problem(cid:16)PP3-1 can be refo(cid:17)rmuPlated aPs
P K J
P3-3: max x log(c β ) (23)
{xkj,βkj}k=1j=1 kj kj kj
XX
s.t. same constraints as problem P3-1.
The choices of β rely on the values of x . Given these coupled variables, we first apply the Primal Decom-
kj kj
position method [20] to decompose problem P3-3 to the following two levels of problems. Fixing variables x ’s,
kj
we have the lower level problem as
K J
max x log(c β ) (24)
{βkj} k=1j=1 kj kj kj
XX
s.t. β ≤1,j =1,2,··· ,J.
kj
kX∈Φj
When the β ’s are fixed, the higher level problem (or, the master problem) is given by
kj
K J
max x log(c β ) (25)
{xkj} k=1j=1 kj kj kj
XX
s.t. same constraints as problem P1-2.
Since there are no couplingsamong the subproblems,the lower levelproblem(24) can be furtherdecomposedinto
L subproblems as follows.
K
max x log(c β ) (26)
{βkj} k=1 kj kj kj
X
K
s.t. β ≤1,j =1,2,··· ,J.
kj
kX∈Φj
Defining Lagrange multiplier λ, the Lagrangian of problem (26) is defined as
K K
L= x log(c β )+λ 1− β . (27)
kj kj kj kj
!
k=1 k=1
X X
Applying KKT conditions [21], the optimal solution can be obtained as follows.
x
β = kj . (28)
kj K x
k=1 kj
Substituting (28) into the master problem, the objectPive function becomes
K J
c
x log kj . (29)
Xk=1Xj=1 kj Kk=1xkj!
Note that we have dropped one x term in (29), since duPe to the definition (2), we have (x )2 = x . Since
kj kj kj
K x is in the denominator, problem (29) has coupled objectives. The main idea of addressing the coupled
k=1 kj
objective is to introduce auxiliary variables and additionalequality constraints so that the couplingin the objective
fPunction is transferred to coupling in the constraint [20]. We thus introduce a new variable, which is defined as:
K
Ξ = x . (30)
j kj
k=1
X
To solve the aboveproblem,we relaxx to a realnumberin [0,1].However,we will show later thatevenif we
kj
have relaxed the variables, we could still find the optimal solution to the originalproblem. The relaxed problem to
be solved is
K J
c
max x log kj (31)
{xkj} k=1j=1 kj (cid:18)Ξj (cid:19)
XX
s.t. Ξ ≤L , j =1,2,··· ,J
j j
J
x ≤1, k =1,2,··· ,K
kj
j=1
X
0≤x ≤1,for allk,j
kj
Constraints (13), (30).
Problem (31) is a convex optimization problem. Defining Lagrange multipliers for the equality constraints (30),
problem (31) can be solved with the dual decomposition method. Alternatively, we propose Algorithm 3 to obtain