In the remainder of the paper, we will consider optimal resource allocation on the bounds obtained for the MBRC, that is, for hybrid wireless networks where the source node has access to distinct bands (standards) and a second node that overhears the source information relays to the destination using additional orthogonal bands. We consider the Gaussian case, where all the transmitted signals are corrupted by additive white Gaussian noises.

We have the input-output signal model given by (1) under source and relay power constraints:

where
and
are the total available power at the source and relay node.
and
are the nonnegative power allocation parameters for each orthogonal band at the source and relay node, and
. Unlike [5, 10], we do not have a total power constraint between the source and the relay and assume that each has its own battery.

We assume that the system has total bandwidth

. We define the received SNRs at the relay and the destination over channel

as

Note that the actual received SNR values are the scaled versions of (11) depending on the power and bandwidth allocation. For example, the actual received SNR at the relay from channel 1, which is allocated
fraction of the source power and
fraction of the bandwidth, simply is
. Given the received SNRs which are available at the source and relay, our aim is to find the optimal resource allocation parameters that maximize capacity lower bound in terms of the transmitted power and the total bandwidth for
-MBRC, which leads to optimally allocating the source power among
source bands, the relay power among
relay bands, and the total bandwidth among
bands. We can obtain the capacity lower and upper bounds of
-MBRC from Theorem 1 as follows.

Theorem 2.

The upper and lower bounds for the capacity of the Gaussian

-MBRC are

We omit the proof for Theorem 2 since the derivation for each mutual information follows directly from [15]. For each broadcast channel, if the relay node sees a higher received SNR than the destination node, then a superposition coding scheme [17] is used to convey independent information to the relay node, which cannot be decoded by the destination directly. The relay node then collects this information from all the channels where superposition coding is used, and transmits it to the destination at the appropriate rate.

Based on whether the relay node is utilized by a certain channel (band), we note that there are
possible schemes. We observe that these
schemes are not exclusive to each other, since a superposition coding scheme may be reduced to a direct source-to-destination transmission if no band is allocated to the relay-to-destination link. We also note that which scheme yields the largest rate is completely decided by the SNR relationship, namely, the componentwise relationship between the received SNRs of the source-to-relay links, that is,
and the received SNRs of the source-to-destination links, that is,
.

If
,
, then for any bandwidth allocation, the signal received by the relay over this broadcast channel can be viewed as a degraded version of the signal received by the destination. Therefore, direct link transmission should be used for this band, regardless of what scheme is used for the other bands. On the other hand, if
, then the relay node can always learn something more than the destination node over this band and uses the superposition code scheme, and although the superposition scheme may be reduced to a direct link transmission scheme, optimizing under this scheme does not incur any rate loss. Based on these observations, we conclude that there is no need to examine all the schemes to find the best rate and the corresponding resource allocation. That is, practically, the system checks the received SNRs and chooses one of
schemes satisfying the relationship of the received SNRs to communicate and the rate with optimized resource allocation for the chosen scheme is the maximum achievable rate, and the corresponding resource allocation is the globally optimal solution.

Next, we maximize the capacity lower bound in (12). To achieve this goal, we introduce the following general max-min optimization problem. We define

and

as any utility function with any resource allocation vector

over the convex set

:

Proposition 1.

If

and

are nonnegative and concave over

, there must exist

such that maximizing the following equation with respect to

is equivalent to (14):

Proof.

See Appendix A.

Note that the optimization problem in (14) corresponds to finding
and
maximizing the minimum of two end points in
. One possible technique to solve the max-min optimization problem in (14) is given by the following proposition [15], which we will also utilize.

Proposition 2 ([15, Proposition
]).

The relationship between optimal resource allocation parameters
and the corresponding optimal point
is given by the following.

Case 1:If
,

Case 2:If
,
.

Case 3:Neither case 1 nor 2 occurs; under this case, if
,
.

Now, one can restate our max-min optimization problem given in Theorem 2 as follows:

where

and

are the first and the second terms of max-min optimization problem in (12). Next, one needs to prove that

and

are concave over

in (16). Define

It is easy to see that
is continuous over
. Then, one has the following proposition.

Proposition 3.

is concave over
.

Proof.

First, note that due to the continuity of

, we only need to prove that

is concave over the

*interior* of the region, that is,

. This is done by examining the Hessian,

, of (17). The second-order derivatives of (17) with respect to

and

are

We note that
.

The Hessian is the

block diagonal matrix with the following matrix in its

th diagonal:

It is readily seen that
is singular. Since
for
from (18),
is the negative semidefinite. Thus,
is concave over
. Since
is continuous over
,
is concave over
.

We note that for any choice of set
,
corresponds to
in (17) with
,
,
,
,
,
,
,
, and
,
. For
, the Hessian is a
block diagonal matrix. Similarly,
corresponds to
with
,
,
,
,
,
, and
,
. For
, the Hessian is a
block diagonal matrix.

Remark 1.

Since
is concave over the set
, it is also concave over any convex subset of it. Thus,
and
are concave over
. (It is readily seen that the sum constraints define a convex set.) This establishes that the local optimal for (16) is also the global optimal [18, Theorem
, page 125-126].

Remark 2.

We further find that
is strictly concave over any convex subset of
,
, jointly when
, are held constant. Note that when all
, are held constant, that is,
, we have
as a function of
,
. In this case, it is easily seen that the Hessian is the
*diagonal* matrix in which
th diagonal term is given by
,
,
. Since now all of the diagonal terms are strictly negative when
,
,
is strictly concave over all
,
, jointly when all
,
, are held constant. Similarly,
is strictly concave over
,
, jointly when all
,
, are held constant. Since if a function is strictly concave over a set, it is also strictly concave over any convex subset of that set, the preceding argument implies that
is strictly concave over any convex subset of
, when all
, are held constant. This fact will be useful in the sequel.

Based on Proposition 1 and Proposition 3, the methodology given in Proposition 2 can be applied to our max-min optimization problem in (16) for an arbitrary
. That said, in the remainder of the paper, we will examine the optimal resource allocation for
-MBRC where the source has two bands and the relay has a single band available to communicate and uses its own full power
. We find this network model representative and meaningful because of the following two observations. First, if there is more than one band available for the link between the relay and the destination, then only the best band among them will be used. This can be seen by fixing the overall band for this link and performing joint power and bandwidth optimization. Therefore, as long as the relay-to-destination SNRs are different, which is usually the case in practice,
-MBRC will have the same resource allocation parameters as those of
-MBRC. Secondly, the case with
is similar to the case with
except that there are more schemes to choose from. Therefore, we focus on the
-MBRC in the sequel.

### 3.1. Maximization of Capacity Bounds for the Gaussian
-MBRC

For

-MBRC, there are four schemes to choose from. Let us label them Schemes I through IV. From Theorem 2, upper and lower bounds for the capacity of the Gaussian

-MBRC are

where
=
,
,
, and
for schemes I to IV, respectively. Each scheme materializes as a function of the received SNRs as follows.

Scheme
:
, the scenario where transmission from the source node over both links is decoded at the relay node. This scheme is chosen if
and
.

Scheme II:
, the scenario where transmission from the source node over band 1 is decoded at the relay node while band 2 is used for direct transmission only. This scheme is chosen if
and
.

Scheme III:
, the scenario where transmission from the source node over band 2 is decoded at the relay node while band 1 is used for direct transmission only. This scheme is chosen if
and
.

Scheme IV:
, the scenario where transmissions from the source node from both bands are used only for direct transmission. This scheme is chosen if
and
.

We define

as the optimal resource allocation parameters for (21).

and

are the first and second terms in (21). From (21), we note that the capacity for scheme IV is given by

In this case, the max-min optimization problem reduces to a maximization problem and it is readily shown that the optimal resource allocation for the rate of scheme IV is given by

For schemes I, II, III, once the appropriate scheme is decided upon, parameters
can be substituted accordingly and we can examine
for each of the cases in Proposition 2.

Case 1.

, and
maximizes
.

This case holds if the following condition is satisfied:

The received SNRs must satisfy

Proof.

See Appendix B.

Case 2.

, and
maximizes
.

This case holds if the following condition is satisfied:

Proof.

See Appendix B.

Remark 3.

By substituting the appropriate parameters for
for each scheme into (30), we observe that Case 2 does not ever materialize for schemes I, II, III.

Case 3.

, and
maximizes
for a fixed
.

This case occurs when (25) or (29) doES not hold. The closed form solution for this optimization problem does not exist. Thus, we have to rely on an iterative algorithm. We propose to use alternating maximization algorithm that calls for optimizing
=
in one stage, followed by optimizing
in the next stage. The iterations are obtained by finding KKT points of the corresponding optimization problem with the variable vector
or
. We note that the objective function is not differentiable at the boundary of the feasible region, that is, for
,
and the corresponding KKT points are not defined. Thus, we need to introduce a small positive value,
, and define a modified feasible region as illustrated in (B.3) and (B.4) that excludes the boundary point. Every time an iteration reaches the boundary of the new feasible region, we expand the feasible region by successively reducing
so that we can continue with the iterations until convergence. The detailed description of the following proposed iterative algorithm and proof of its convergence to the global optimal solution is given in Appendix C.

Step 1.

- (i)

- (ii)

- (iii)

- (iv)
Repeat step (ii) until the optimal
is found by
in (C.3).

Step 2.

If the iteration does not reach the boundary of the feasible region of (B.3) and (B.4), the algorithm terminates.

Step 3.

Otherwise, set
,
in (B.3) and (B.4) and repeat Steps (1) to (2) by using the KKT points from the previous iteration as the initial points. (For numerical results, we use
.)

We reiterate that based on the scheme at hand, we would substitute the correct parameters for
to find the optimal resource allocation strategy.

### 3.2. Upper Bound on Capacity

Recall that the upper bound given by (13) is obtained by the max-flow min-cut theorem. The maximization for the upper bound follows same steps to that of the lower bound, details of which we will omit here. In general, the upper bound is not tight. One exception is that for
-MBRC, since Case 2 for schemes I, II, and III is not possible, the optimal resource allocation parameters
maximize
(Case 1) or
(Case 3). There exists a
such that
if
, otherwise
. Since the first term of the upper bound in (22) is the same as
, we know that for
-MBRC,
maximizes
for
and the resulting optimized rate is the capacity of
-MBRC. A similar observation was made for the frequency division relay network, that is, when one band exists from the source in [11]. It is interesting to observe that the same observation extends to the multiband case.