### 3.1 Problem formulation

For a given capacity of the feedback channel, we have characterized the distortion in Section 2. With the quantized downlink CSI, the resource allocation can be carried out for a given performance measure. From this, we can formulate the resource allocation with capacity constraints of the feedback channels. Toward this end, in this subsection, we introduce the outage throughput as the performance measure.

Given the quantized CSI, the outage probability on the *n*-th subcarrier to the *k*-th user is defined as

where *γ*_{
n
} is the input signal error ratio (SNR) of the *n*-th subcarrier and *R* is the transmission rate. From Equation 3, the maximum transmission rate *R* that can maintain the outage probability *ε* is

where . Thus, the expected rate of information successfully decoded at user *k* on subcarrier *n* is

It is possible to maximize by choosing *ε*,

Here, the throughput is termed as the outage throughput. Setting , we obtain

where . Substituting Equation 2 yields

The optimal *x* that maximizes is given by the following theorem:

*Theorem* 2. There exists a unique globally optimal *x* that maximizes in Equation 6, which is given by

where *W*(*x*) is the *Lambert-W* function, which is the solution to the equation *W*(*x*)*e* ^{W(x)}= *x*.

*Proof* See Appendix Appendix B.

Thus, for each given transmit power *γ*_{
n
}, quantized power gain and quantization error *v*_{
k, n
}, we can evaluate the outage throughput of the *k*-th user on the *n*-th subcarrier in Equation 5 by Theorem 2. The overall outage throughput conditioned on the quantized CSI is represented as

where *ρ*_{
k, n
} is the subcarrier allocation indicator: if the *n*-th subcarrier is assigned to the *k*-th user, then *ρ*_{
k, n
} = 1; otherwise *ρ*_{
k, n
} = 0. Here, the BS decides *γ*_{
n
} and *ρ*_{
k, n
} with the knowledge of quantized CSI . To emphasize this, we denote the input SNR and the allocation indicator as functions of by and , respectively. The average outage throughput is thus given by

Now, we can formulate the outage throughput maximization under feedback capacity constraints:

where the first constraint is the feedback capacity constraint, the second constraint ensures that each subcarrier is assigned to one user exclusively, and the third constraint is for total transmit power, denoted by *γ*_{
T
}.

By Theorem 1, for each *R*_{
k
} (*D*_{
k
}), there exists a test channel that achieves *R*_{
k
} (*D*_{
k
}). Thus, maximizing the downlink throughput under feedback capacity constraints is equivalent to maximizing the downlink throughput under the corresponding test channel. It can also be observed that to maximize *T*°, we can maximize the conditional outage throughput for each realization of under the conditional probability density function given in Equation 2. That is,

To make the problem in Equation 10 tractable, we consider a suboptimal solution by breaking the problem into two steps: subcarrier allocation and power allocation. In the first step, subcarriers are assigned to users under the assumption that the transmit power is identical over all subcarriers; in the second step, power is loaded on the subcarriers assigned in the first step.

### 3.2 Subcarrier allocation

Under the assumption of *γ*_{
n
} = *γ*_{
T
}/*N*, the optimization problem in Equation 10 reduces to

It implies that the subcarriers should be assigned based on the following criterion:

The above criterion requires to evaluate *KN* values of the rate given in Equation 5. However, we can simplify this criterion in the case where on subcarrier *n*, the mean quantization error *v*_{
k, n
} is identical among all users *k*. We state the following theorem:

*Theorem* 3. For any given *v*_{
k, n,
}, the throughput defined Equation 5 is monotonically increasing in if in Equation 5 is monotonically increasing in .

*Proof* By assumption, we have for . Thus,

It follows that

It can be shown that given in Equation 6 is monotonically increasing in . Thus, by Theorem 3, in the case of *v*_{
k', n
} = *v*_{
k, n
} for *k* ≠ *k*', the subcarrier allocation reduces to

When a tie occurs, we can select users in random fashion.

### 3.3 Power allocation

Denote by *k*_{
n
} the selected user on the *n*-th subcarrier, i.e., *k*_{
n
} = arg max _{
k
} *ρ*_{
k, n
}. Given the subcarrier allocation, the problem 10 becomes

From the Equations 6 and 7, we can observe that is not concave in *γ*_{
n
}. Hence, the problem 12 is not a convex optimization problem. However, we can employ a dual approach to obtain a suboptimal solution.

The dual problem is

where

where *μ* is the Lagrangian multiplier of the first constraint in Equation 12. Given *μ*, the optimal power allocation on the *n*-th subcarrier is

We can use a derivative-free line search method, such as the golden section method to find the *γ*_{
n
} for a given Lagrangian multiplier *μ* [19].

The Lagrangian dual problem 13 has been shown to be a convex optimization problem in *μ* [20]. Thus, we can use the bisection method to find the optimal global multiplier *μ* [19]. The bisection method requires to evaluate the first derivative of *g*(*μ)* with respect to *μ*. Although g(*μ)* is not continuously differentiable due to the max function, we can use the subgradient instead [21],

where *γ*_{
n
} is obtained from Equation 14.

Using the dual optimization approach, it is possible that the final power allocation may not satisfy . We can multiply the final power allocation on each subcarrier by a constant to arrive a feasible solution.

**Complexity**: in the first step, assigning subcarriers requires to find the maximum among *K* users for each subcarrier *n*, and thereby, the complexity of subcarrier allocation is *O*(*KN*). In the power allocation, in each iteration for *μ* in Equation 13, we need to compute *N* power allocation values given by Equation 14. Each power allocation value requires a search routine, which is assumed to converge within *I*_{
γ
} iterations. Assuming that *I*_{
μ
} iterations are required to find the optimal *μ*, the overall complexity of the suboptimal algorithm is *O*(*KN* + *I*_{
μ
} *I*_{
γ
}*N*). Ignoring the constants *I*_{
μ
} and *I*_{
γ
}, the complexity is just *O*(*KN*).