Based on the above network model and assumptions, our objective is to derive an optimal association scheduling method that maximizes the throughput of a target vehicle. In this section, we show that computing the offline optimal schedule can be casted into a linear programming problem, and thus, we have the optimal solution in polynomial time. In addition to the optimal solution that maximizes the throughput, our framework can be extended to obtain the optimality of association that minimizes the frequency of handoffs occurring in vehicles (see Appendix).
4.1 Objective function
The objective of the association scheduling is to maximize the total transmitted bits during the time intervals I = {1,2,…,m} for the given available APs and effective bit rate information. Computing the optimal association scheduling is equivalent to finding the optimal values for x
ij
,i∈I andj∈J
i
by optimizing the trade-off between the throughput gain from a handoff and the resulting re-association overhead c.
Therefore, we formulate the offline optimal association control scheduling as a combinatorial optimization problem that finds the optimal values for x
ij
(i∈I and j∈J
i
) with the following objective function:
(1)
subject to
(2)
The objective function given in Equation 1 consists of two terms. The first term denotes the total amount of transmitted bits. The re-association overhead is reflected by the second term. Note that the re-association overhead should be considered only when a handoff occurs. In other words, if the AP j used in interval i-1 is continuously used in interval i (i.e., xi-1j = x
ij
), then there is no re-association overhead. On the contrary, if the mobile user selects a new AP j′ which was not used in interval i-1 (i.e., xi-1j = 1,x
ij
= 0, yielding xi-1j-x
ij
=1), the vehicle cannot transmit data during c seconds due to the handoff overhead. Thus, we express the re-association overhead as
(4)
Also, Equation 2 is provided since the vehicle can associate with at most one AP among the available APs according to the system model given above.
4.2 Computing the optimal scheduling problem
The objective function given in Equation 1 is a non-linear integer programming problem with binary integer decision variable x
ij
. Although the optimal solution of the problem is a subset of all combinatorial and thus is solvable by brute-force enumerative algorithms [21], its calculation complexity is too high to be practical.
To address this challenging issue, we transform the above non-linear function into a linear function by exploiting the properties of the binary decision variable.
Since the decision variable x
ij
is a binary value from which it is given , we rewrite the non-linear component in the objective function of Equation 1 as
(5)
Then, by setting z
ij
=x
ij
·xi-1j, we view z
ij
as another binary variable. This yields two additional constraints in the objective function that reflect the same handoff costs:
(6)
Thus, the optimization problem is rewritten as
(7)
subject to
However, it is still non-trivial to solve the above integer program (IP) because of the integrality constraints. We will discuss the property of the objective function and show that its solution can be obtained in polynomial time.
Let us consider a IP max and its linear programming (LP) relaxation max . A linear program in real variables is said to be integral if it has at least one optimal solution which is integral. Likewise, a polyhedron P={x:A x≤b} is said to be integral if the linear program has an optimum solution x∗ with integer coordinates. According to the LP theory [22], solving an IP over an integral polyhedron can be done in polynomial time through LP relaxation of integral constraints, and then the LP has an integral optimal solution.
One common way of proving that a polyhedron is integral is to show that its constraint matrix is TU [21], where a matrix is said to be totally unimodular if the determinant of each square submatrix is 0,+1,or -1. This leads to the following proposition.
Proposition
1
If a matrix A is totally unimodular and b is an integral vector, then the polyhedron P = {x:A x≤b} is integral.
Proof
See [21].
We now show that the constraint matrix of the objective function given in Equation 7 satisfies TU. At this point, however, it is not easy to recognize that the constraint matrix in Equation 7 is TU by directly using the definition of TU. Thus, we consider a general sufficient condition with the following theorem.
Theorem
1
A matrix A is totally unimodular if
-
(i)
a
pq
∈{0,+1,-1} for all p, q, and
-
(ii)
for any subset M of the rows of A, there exists a partition (M 1, M 2) of M such that each column q satisfies
Proof.
See [22].
With a more general condition in Theorem 1, we will show that the constraint matrix of the linear integer problem given in Equation 7 is totally unimodular. We first rewrite the constraints of the linear integer problem given in Equation 8 in the form of the standard formation as
(8a)
(8c)
Then, one can easily express the above constraint as the form of A x≤b, where A denotes the constraint matric, defined as , such that b
q
≥0. Then, we have the following theorem.
Theorem 2.
Let A be the constraint matrix of the linear integer problem given in Equation 7 (i.e., in the form of A x≤b), then matrix A is totally unimodular.
Proof.
We first introduce a partitioning method that obtains a partition (M1,M2) for an arbitrary subset M of the rows of A. Let B
i
be a subset of A such that B
i
includes all the rows corresponding to the coefficients of x
ij
in the constraints given in Equation 8. Let , , and denote the submatrices of A containing all the rows corresponding to each constraint in Equations 8a, 8b, and 8c for interval i, respectively. Then, B
i
is given as . For an arbitrary subset M, Algorithm 1 generates a partition (M1,M2) by iteratively testing every row in B
i
∩M and assigning the row into M1 or M2 for i∈I. It is straightforward to show that this portioning rule obtains a partition (M1,M2) of M correctly.
The basic principle of how to assign each testing row is as follows. For two different rows p and p′ which are both correspondent to interval i, we perform the partition according to two elements a
pq
and in the same column q. For the case that both a
pq
and are 1 (or -1), we assign such rows p, and p′ to different partitions M1 and M2, respectively. If both a
pq
and are non-zero but their signs are different, i.e., (a
pq
, ) = {1,-1} or {-1,1}, then we assign such rows to the same partition M1 or M2. Since B
i
∩M and Bi-1∩M can have non-zero values in the same column due to the third constraint (8.c), we consider the rows in M1 and M2 already assigned for interval i-1 when assigning the rows of B
i
∩M for interval i. Thus, we assign the rows of B
i
∩M into M1 or M2 depending on whether any non-zero value in the same column was already assigned in both M1(or M2) and B
i
∩M or not. The partitioning rule is described in Algorithm 1.
For a given arbitrary subset M of rows of A=[a
pq
], we can have its partition (M1,M2) based on Algorithm 1. Since condition (i) of Theorem 1 is already satisfied in Equation 8, we only need to show that the partition (M1,M2) will satisfy condition (ii) of Theorem 1.
We can easily see that the number of non-zero entries a
pq
on column q for all p∈M is at most three from Equation 8. Therefore, the number of non-zero entries on the same column in M1 and M2 is at most three. Let N
q
denote the number of non-zero entries either in M1 or in M2 for column q.
-
(i)
N
q
=0,1: nothing to prove.
-
(ii)
N
q
= 2: There are three subcases for where p≠p ′:
-
(ii-1)
: In this case, and (or and ) for ∃i. This happens only if B
i
∩M≠∅ by the definition of B
i
. Thus, it is belonging to the case 1-2 in Algorithm 1; the two rows p and p′ were partitioned to different partitions. Thus, holds.
-
(ii-2)
(ii-2) {1,-1} (or {-1,1}): Since for ∃i, and this is belonging to case 1-1 in Algorithm 1, the two rows p and p′ in B
i
∩M should be assigned to the same partition M1 or M2. Thus, holds.
-
(ii-3)
{1,1}: This can only happen for a column with respect to z
ij
for ∃i(Bi-1and B
i
). Then, the columns of M1(or M2) and B
i
∩M are belonging to the case 1-2 in Algorithm 1; thus, the two rows p and p′ are assigned to different partitions. Thus, holds.
-
(iii)
N
q
= 3: In this case, among three non-zero a
pq
, two of them should be -1 and the other one is +1 where p∈M. The partition (M 1, M 2) is performed according to case 1-1 in Algorithm 1, and the summation of 1,-1,-1 is -1. Thus, holds.
In all cases, non-zeros in a column for the partition (M1,M2) obtained from Algorithm 1 satisfy condition (ii) of Theorem 1. Therefore, the constraint matrix A is totally unimodular.
Solving an IP over an integral polyhedron P can be done in polynomial time. Therefore, from the results of Theorems 1 and 2, we can obtain the solution of the integer program given in Equation 8 in polynomial time.
Remark
In addition to the optimal scheduling problem that maximizes throughput, our framework can be extended to obtain the optimality of association that minimizes the frequency of handoffs occurring in vehicles. In the Appendix, we formulate the optimal handoff minimization scheduling problem as a non-linear integer programming and show that this problem also holds the TU property, and thus is solvable in polynomial time.