In this section, we relate each of the scheduling criterion with the bounds on V-BLAST capacity. The best criterion technique is the one that is able to provide tighter bounds on

\underset{k}{\text{min}}\phantom{\rule{1pt}{0ex}}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}\phantom{\rule{1pt}{0ex}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}

where \underset{i}{\text{min}}\parallel {W}_{\text{ZF},n,i}^{k}{\parallel}^{2} is the minimum *i* th diagonal value of {\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1} of the *n* th layer of the *k* th user. In this section, we will find bounds that are independent of *n* and *i* for each scheduling technique.

### MaxMinSV

Choosing a user with maximum minimum singular value of **H**_{
k
} is the same as choosing a user with maximum minimum eigenvalue of {\mathbf{H}}_{k}{\mathbf{H}}_{k}^{H}. We have

\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}={\left[{\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1}\right]}_{i,i}.

This means that we are looking for the minimum *i* th diagonal value of {\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1}. We can write it in the form of the following inequality:

\begin{array}{l}{\lambda}_{\text{min}}^{k,n}\left({\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1}\right)\le \underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\\ \phantom{\rule{11.6em}{0ex}}\le {\lambda}_{\text{max}}^{k,n}\left({\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1}\right).\end{array}

(13)

Since **H**_{
n
}^{H}**H**_{
n
} is a squared Hermetian matrix, it can be written as

\begin{array}{l}{\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}=\mathbf{U}{\Lambda}_{k,n}{\mathbf{U}}^{H}\end{array}

(14)

where **U** is a unitary matrix with orthonormal eigenvectors and *Λ*_{k,n} is the diagonal matrix of eigenvalues \left[{\lambda}_{1},{\lambda}_{2},\dots ,{\lambda}_{{M}_{T}}\right]. Another approach is to use QR decomposition as done in [18].

Using simple matrix manipulations, the inverse is written as

\begin{array}{l}{\left[{\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right]}^{-1}=\mathbf{U}{\Lambda}_{k,n}^{-1}{\mathbf{U}}^{H}.\end{array}

(15)

The diagonal elements of {\Lambda}_{k,n}^{-1} are \left[\frac{1}{{\lambda}_{1}},\frac{1}{{\lambda}_{2}},\dots ,\frac{1}{{\lambda}_{{M}_{\text{T}}}}\right]. Therefore,

\begin{array}{ll}{\lambda}_{\text{min}}^{k,n}\left({\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1}\right)& =\frac{1}{{\lambda}_{\text{max}}^{k,n}\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}\phantom{\rule{2em}{0ex}}\end{array}

(16)

\begin{array}{ll}{\lambda}_{\text{max}}^{k,n}\left({\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}^{-1}\right)& =\frac{1}{{\lambda}_{\text{min}}^{k,n}\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right)}.\phantom{\rule{2em}{0ex}}\end{array}

(17)

For the sake of simplicity, we will henceforth refer to {\lambda}_{\text{max}}^{k,n}\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right) and {\lambda}_{\text{min}}^{k,n}\left({\mathbf{H}}_{k,n}^{H}{\mathbf{H}}_{k,n}\right) as {\lambda}_{\text{max}}^{k,n} and {\lambda}_{\text{min}}^{k,n}, respectively. We now have

\begin{array}{l}\frac{1}{{\lambda}_{\text{max}}^{k,n}}\le \underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \frac{1}{{\lambda}_{\text{min}}^{k,n}}\phantom{\rule{2em}{0ex}}\end{array}

(18)

\begin{array}{l}\frac{1}{{\text{min}}_{n}\phantom{\rule{1pt}{0ex}}{\lambda}_{\text{max}}^{k,n}}\phantom{\rule{1pt}{0ex}}\le \underset{n}{\text{max}}\underset{i}{\phantom{\rule{1pt}{0ex}}\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \frac{1}{\underset{n}{\text{min}}{\lambda}_{\text{min}}^{k,n}}.\phantom{\rule{2em}{0ex}}\end{array}

(19)

According to the inclusion principle for matrices, the minimum value of *λ*_{min} occurs at *n*=1 and the minimum vale of *λ*_{max} occurs at *n*=*M*_{T}.

Now we have an upper bound given by

\begin{array}{l}\underset{n}{\text{max}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \frac{1}{{\lambda}_{\text{min}}^{\left(k\right)}}\end{array}

(20)

or by taking the minimum over *K* users

\begin{array}{l}\underset{k}{\text{min}}\phantom{\rule{1pt}{0ex}}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \frac{1}{{\text{max}}_{k}\phantom{\rule{1pt}{0ex}}{\lambda}_{\text{min}}^{\left(k\right)}}.\end{array}

(21)

We will use this inequality to establish lower bounds on MaxMinSV.

For analytical tractability, we will focus on the case when *M*_{T}=*M*_{R}. The probability density function (pdf) of the smallest eigenvalue for the case of *M*_{T}=*M*_{R} is given by [19]

\begin{array}{l}{f}_{{\lambda}_{\text{min}}}\left(x\right)={M}_{\text{T}}{e}^{-x{M}_{\text{T}}}u\left(x\right).\end{array}

(22)

In this case, \mu =\frac{1}{{M}_{\text{T}}} and \sigma =\frac{1}{{M}_{\text{T}}}, where *μ* and *σ*^{2} represent the mean and variance. From chapter 10 of [20], we see that the maximum of *λ*_{min} will scale as \frac{1}{M}\text{log}K. Therefore, we have a new bound on the capacity of VBLAST as

\begin{array}{l}{C}_{\text{VBLAST}}^{\text{ZF}}\ge {M}_{\text{T}}{\text{log}}_{2}\left(1+\frac{\text{SNR}{\text{log}}_{2}K}{{M}_{\text{T}}^{2}}\right).\end{array}

(23)

### MinES

In this case, we start with the inequality

\begin{array}{l}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \frac{1}{{\lambda}_{\text{min}}^{k,n}}\le {\kappa}^{(k,n)}.\end{array}

(24)

Here *κ*^{(k,n)} is the condition number of the matrix {\mathbf{H}}_{k,n}{\mathbf{H}}_{k,n}^{H} defined as *λ*_{max}/*λ*_{min}. Now

\begin{array}{l}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \underset{n}{\text{max}}{\kappa}^{(k,n)}.\end{array}

(25)

The condition number *κ* is maximum when *n*=1, a lemma which follows from the inclusion principle of matrices. Therefore, we have

\begin{array}{ll}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}& \le {\kappa}^{\left(k\right)}\phantom{\rule{2em}{0ex}}\end{array}

(26)

\begin{array}{ll}\underset{k}{\text{min}}\phantom{\rule{1pt}{0ex}}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}& \le \underset{k}{\text{min}}{\kappa}_{\left(k\right)}.\phantom{\rule{2em}{0ex}}\end{array}

(27)

The pdf of *s*, the condition number of **H**_{
k
} for the case of *M*_{T}=*M*_{R}, is given by [19]

\begin{array}{l}{f}_{s/{M}_{\text{T}}}\left(x\right)=\frac{8}{{x}^{3}}{e}^{-\frac{4}{{x}^{2}}}u\left(x\right).\end{array}

(28)

As *κ*=*s*^{2}, we can show using a variable transformation argument that

\begin{array}{l}{f}_{\kappa}\left(x\right)=\frac{4{M}_{\text{T}}^{2}}{{x}^{2}}{e}^{-\frac{4{M}_{\text{T}}^{2}}{x}}u\left(x\right).\end{array}

(29)

Now we find the upper bound on min*k* *κ*^{(k)}. Gordon et al. [21] provide bounds on inequalities for the minimum of random variables. These inequalities exist if a random variable *ξ* satisfies the (*α*, *β*) condition defined as

\begin{array}{ll}P\left(\left|\xi \right|\le t\right)& \le \mathrm{\alpha t}\phantom{\rule{2em}{0ex}}\end{array}

(30)

\begin{array}{ll}P\left(\left|\xi \right|\ge t\right)& \le {e}^{-\mathrm{\beta t}}\phantom{\rule{2em}{0ex}}\end{array}

(31)

where *α*>0 and *β*>0.

Then, the random variable *ξ* is bounded as

\begin{array}{l}\frac{1}{2\mathrm{\alpha K}}\le E\left[\underset{k}{\text{min}}{\xi}^{\left(k\right)}\right]\le \frac{1}{\mathrm{\beta K}},\\ \phantom{\rule{1em}{0ex}}\text{where}E\text{is the expected value.}\end{array}

(32)

Using the pdf of *κ*, we find that \alpha =\beta =\frac{1}{4{M}_{\text{T}}^{2}}. Now the upper bound on the minimum is given by

\begin{array}{l}E\underset{k}{\text{min}}{\xi}^{\left(k\right)}\le \frac{4{M}_{\text{T}}^{2}}{K}.\end{array}

(33)

Now we have

\begin{array}{l}\underset{k}{\text{min}}\phantom{\rule{1pt}{0ex}}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\le \frac{4{M}_{\text{T}}^{2}}{K}.\end{array}

(34)

### MaxSNR

We start with the established inequality

\begin{array}{l}\frac{1}{{\text{max}}_{k}\phantom{\rule{1pt}{0ex}}{\text{min}}_{n}\phantom{\rule{1pt}{0ex}}\text{trace}\left({\mathbf{H}}_{k,n}{\mathbf{H}}_{k,n}^{H}\right)}\\ \le \frac{1}{{\text{max}}_{k}\phantom{\rule{1pt}{0ex}}{\text{min}}_{n}\phantom{\rule{1pt}{0ex}}{\lambda}_{\text{max}}^{k,n}}\le \underset{k}{\text{min}}\phantom{\rule{1pt}{0ex}}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}.\end{array}

(35)

Again from the inclusion principle, the minimum of the trace occurs at *n*=*M*_{T}.

Since we know the last layer, the pdf of \text{trace}\left({\mathbf{H}}_{k,n}{\mathbf{H}}_{k,n}^{H}\right) for the last *n* becomes conditional. If we solve for the unordered VBLAST algorithm, we will get

\begin{array}{l}\frac{1}{{\text{max}}_{k}\text{trace}\left(\underset{k}{\overset{H}{\mathbf{h}}}{\mathbf{h}}_{k}\right)}\le \underset{k}{\text{min}}\phantom{\rule{1pt}{0ex}}\underset{n}{\text{max}}\phantom{\rule{1pt}{0ex}}\underset{i}{\text{min}}{\u2225{W}_{\text{ZF},n,i}^{k}\u2225}^{2}\end{array}

(36)

where **h**_{
k
} is a row vector containing *M*_{R} i.i.d complex Gaussian entries. In this case, the trace will be a chi-squared random variable with 2*M*_{R} degrees of freedom. The pdf in this case is given as

\begin{array}{l}{f}_{\mathit{\text{tr}}}\left(x\right)=\frac{{x}^{{M}_{\text{R}}-1}}{\left({M}_{\text{R}}-1\right)!}{e}^{-x}u\left(x\right).\end{array}

(37)

Now we use an upper bound on the maximum of the trace. Bertsimas et al. [22] provide us with a tight upper bound:

\begin{array}{l}E\left[{\text{max}}_{k}\phantom{\rule{1pt}{0ex}}\text{trace}\left({\mathbf{h}}_{k}^{H}{\mathbf{h}}_{k}\right)\right]\le \mu +\sigma \sqrt{K-1}\end{array}

(38)

where *μ* and *σ*^{2} are mean and variances, respectively. In this case, *μ*=2*M*_{R} and *σ*^{2}=4*M*_{R}. Now we have

\begin{array}{l}E\left[{\text{max}}_{k}\phantom{\rule{1pt}{0ex}}\text{trace}\left({\mathbf{h}}_{k}^{H}{\mathbf{h}}_{k}\right)\right]\le 2\sqrt{{M}_{\text{R}}}\left(\sqrt{{M}_{\text{R}}}+\sqrt{K-1}\right).\end{array}

(39)

Here we define an upper bound on capacity:

\begin{array}{l}{C}_{\text{VBLAST}}^{\text{ZF}-\text{unorder}}\le {M}_{\text{T}}{\text{log}}_{2}\left(1+\frac{2\text{SNR}\left(\sqrt{{M}_{\text{R}}}+\sqrt{K-1}\right)}{\sqrt{{M}_{\text{T}}}}\right).\end{array}