Reducing computation time of a wireless resource scheduler by exploiting temporal channel characteristics

Abstract

The long term evolution downlink scheduler must attain a low computation time as it performs scheduling decisions every 1 ms. 5G New Radio introduced mini-slots for the purpose of ultra-reliable and low-latency communications further shrinking the time to make scheduling decisions to 0.125 ms. Many optimal scheduling schemes have such high computation times that they are not suitable for implementation. Previous works generally attack this problem from a computational complexity theory perspective and devise alternative non-optimal problem formulations. Here, we tackle the problem from a practical point of view and propose to reduce the quantity of users and resources in the scheduling problem over a given time. We achieve this by scheduling relatively slow varying signal-to-noise ratio (SNR) users not as frequently but for relatively longer time durations. To evaluate the performance of our idea, we derive a novel correlated bivariate received SNR distribution. The derived distribution can also be applied to a signal-to-interference ratio limited system. We show that the number of operations it takes to make scheduling decisions can be reduced by 33% with confidence probability of 0.7 and by 58% with confidence probability of 0.4. We also evaluate the potential drawbacks of the proposed scheme in terms of efficiency and error rate.

Appendices

Appendix 1

1.1 Derivation of bivariate SNR density

Here, we derive bivariate SNR density by first converting joint amplitude density distribution to joint power density distribution before concluding with ratio of signal to noise. We start by joint signal and noise amplitude distribution density, Eq. (16), $d_2(r_1,r_2,n_1,n_2)=$

$$\begin{aligned}&\frac{r_1r_2n_1n_2\exp {\left( \frac{-\left( r_1^2+r_2^2\right) }{2\zeta _r(1-k^2)}\right) }\exp \left( {\frac{-(n_1^2+n_2^2)}{2\zeta _n}}\right) }{\zeta _r^2\zeta _n^2(1-k^2)} I_0 \left( \frac{kr_1r_2}{\zeta _r(1-k^2)}\right) , \nonumber \\&\quad 0\le r_1,r_2,n_1,n_2\le \infty \end{aligned}$$

(43)

$$\begin{aligned}&\beta _1=r_1^2, \beta _2=r_2^2, \beta _3=n_1^2, \beta _4=n_2^2\nonumber \\&\quad |J|=1/\left( 16\sqrt{\beta _1\beta _2\beta _3\beta _4}\right) \end{aligned}$$

(44)

$$\begin{aligned}&I_0(x)=\sum _{l=0}^{\infty }\left( \tfrac{x}{2}\right) ^{2l}\tfrac{1}{(l!)^2} \end{aligned}$$

(45)

$$\begin{aligned}&=\frac{\exp {\left( \frac{-\left( \beta _1+\beta _2\right) }{2\zeta _r(1-k^2)}\right) }\exp {\left( \frac{-(\beta _3+\beta _4)}{2\zeta _n}\right) \sum _{l=0}^{\infty }\left( \frac{k\sqrt{\beta _1}\sqrt{\beta _2}}{2\zeta _r(1-k^2)}\right) ^{2l}}}{16\zeta _r^2\zeta _n^2(1-k^2)(l!)^2} \end{aligned}$$

(46)

$$\begin{aligned}&\varphi _1=\frac{\beta _1}{\beta _3},\varphi _2=\frac{\beta _2}{\beta _4},\varphi _3=\beta _3,\varphi _4=\beta _4, \quad |J|=\varphi _3\varphi _4 \end{aligned}$$

(47)

$$\begin{aligned}&=\frac{\varphi _3\varphi _4\exp {\left( \frac{-\left( \varphi _1\varphi _3+\varphi _2\varphi _4\right) }{2\zeta _r(1-k^2)}\right) }\exp {\left( \frac{-(\varphi _3+\varphi _4)}{2\zeta _n}\right) }}{16\zeta _r^2\zeta _n^2(1-k^2)(l!)^2}\nonumber \times \\&\quad \sum _{l=0}^{\infty }\left( \frac{k\sqrt{\varphi _1\varphi _3}\sqrt{\varphi _2\varphi _4}}{2\zeta _r(1-k^2)}\right) ^{2l} \end{aligned}$$

(48)

Integrate over $\varphi _3$:

$$\begin{aligned}&\int _0^{\infty }\varphi _3^{l+1}\exp {\left( \frac{-\varphi _1\varphi _3}{2\zeta _r(1-k^2)}\right) }\exp {\left( \frac{-\varphi _3}{2\zeta _n}\right) }d\varphi _3\nonumber \\&\quad = \frac{2^{2+l}\Gamma {(2+l)}}{\left( \tfrac{1}{\zeta _n}+\tfrac{\varphi _1}{(1-k^2)\zeta _r}\right) ^{2+l}} \end{aligned}$$

(49)

Integration over $\varphi _4$ yields similar expression. The SNR distribution density, $\xi (\varphi _1,\varphi _2)$, becomes:

$$\begin{aligned}&=\sum _{l=0}^{\infty }\frac{k^{2l}\varphi _1^l\varphi _2^l2^{2l+4}\Gamma ^2(2+l)}{2^{2l}16\zeta _r^{2l}(1-k^2)^{2l}\zeta _r^2\zeta _n^2(1-k^2)(l!)^2\left( \tfrac{1}{\zeta _n}+\tfrac{\varphi _1}{(1-k^2)\zeta _r}\right) ^{2+l}\left( \tfrac{1}{\zeta _n}+\tfrac{\varphi _2}{(1-k^2)\zeta _r}\right) ^{2+l}} \end{aligned}$$

(50)

$$\begin{aligned}&=\sum _{l=0}^{\infty } \frac{\Gamma ^2(2+l)\left( k\sqrt{\varphi _1}\sqrt{\varphi _2}\right) ^{2l}\zeta ^2_r\zeta _n^{2+2l}(1-k^2)^3}{{(\varphi _1\varphi _2\zeta _n^2+\zeta _n\zeta _r\varphi _1(1-k^2)+} {\zeta _n\zeta _r\varphi _2(1-k^2)+\zeta _r^2(1-k^2)^2)^{2+l}(l!)^2}}\nonumber , \\&\quad 0\le \varphi _1, \varphi _2 \le \infty \end{aligned}$$

(51)

Appendix 2

1.1 Derivation of trivariate distribution density of Rayleigh random variables

Joint density function, $w_2(z_{I_1},z_{Q_1},z_{I_2},z_{Q_2},z_{I_3},z_{Q_3})$ is given by

$$\begin{aligned} =\frac{\exp { \left( {-\frac{\mathbf {Z}{^T}\mathbf {K}^{-1}\mathbf {Z}}{2}}\right) }}{(2\pi )^3(\det \mathbf {K})^{(1/2)}} \end{aligned}$$

(52)

where

$$\begin{aligned} \mathbf {Z}= \begin{bmatrix} z_{I_1} \\ z_{Q_1} \\ z_{I_2} \\ z_{Q_2} \\ z_{I_3} \\ z_{Q_3} \end{bmatrix} \quad \mathbf {K}=\zeta \begin{bmatrix} 1&0&k_1&0&k_2&0 \\ 0&1&0&k_1&0&k_2\\ k_1&0&1&0&k_1&0 \\ 0&k_1&0&1&0&k_1 \\ k_2&0&k_1&0&1&0\\ 0&k_2&0&k_1&0&1\\ \end{bmatrix} \end{aligned}$$

(53)

We change coordinates to polar, $z_{I_1}=r_1\cos (\theta _1)$, $z_{Q_1}=r_1\sin (\theta _1)$, $z_{I_2}=r_2 \cos (\theta _2)$, $z_{Q_2}=r_2\sin (\theta _2)$, $z_{I_3}=r_3 \cos (\theta _3)$, $z_{Q_3}=r_3\sin (\theta _3)$, $|J|=r_1r_2r_3$. Density, $w_3(r_1,r_2,r_3,\theta _1,\theta _2,\theta _3)=$

$$\begin{aligned} \frac{r_1r_2r_3\exp {-\left( \frac{r_1^2\sigma _1+r_2^2\sigma _4+r_3^2\sigma _1+2r_1r_2\sigma _2\cos (\theta _1-\theta _2)+2r_1r_3\sigma _3\cos (\theta _1-\theta _3)+2r_2r_3\sigma _2\cos (\theta _2-\theta _3)}{2\zeta (k_2-1)(2k_1^2-k_2-1)}\right) }}{{(2\pi )^3\zeta ^3(k_2-1)(2k_1^2-k_2-1)}} \end{aligned}$$

(54)

where $\sigma _1=1-k_1^2$, $\sigma _2=k_1(k_2-1)$, $\sigma _3=k_1^2-k_2$, $\sigma _4=1-k_2^2$. Next, we make a change of variables $x_1=\theta _1-\theta _2$, $x_2=\theta _2-\theta _3$, and $x_3=(\theta _1+\theta _2+\theta _3)/3$. $|J|=1$. Integration over $x_i$’s would yield the density in terms of magnitude. Therefore,

$$\begin{aligned}&\int _0^{2\pi }\int _0^{2\pi }\int _0^{2\pi }e^{-(A\cos (x_1)+B\cos (x_2)+C\cos (x_1+x_2))}dx_1dx_2dx_3\nonumber \\&\quad =2\pi \int _0^{2\pi }\int _0^{2\pi }\left( I_0(A)+2\sum _{k=1}^{\infty }I_k(A)\cos (kx_1)\right) \nonumber \times \\&\qquad \left( I_0(B)+2\sum _{l=1}^{\infty }I_l(B)\cos (lx_2)\right) \nonumber \times \\&\qquad \left( I_0(C)+2\sum _{j=1}^{\infty }I_j(C)\cos (j(x_1+x_2))\right) dx_1dx_2 \end{aligned}$$

(55)

where

$$\begin{aligned} e^{-(A\cos (x_1))}=I_0(A)+2\sum _{k=1}^{\infty }I_k(A)\cos (kx_1) \end{aligned}$$

(56)

as given in [35] (9.6.34) and $I_k(A)$ represents the modified Bessel function. We are not providing all of the steps due to space limitations here. Since the region of integration is periodic, all terms integrate to zero but of this form:

$$\begin{aligned}&\sum _{k=1}^{\infty }\sum _{j=1}^{\infty }I_k(A)I_j(C)\int _0^{2\pi }\cos (kx_1)\cos (jx_1)dx_1 \nonumber \\&\quad =\pi \sum _{k=1}^{\infty }I_k(A)I_k(C) \end{aligned}$$

(57)

The solution to Eq. (55) and the density are given in Eqs. (58) and (59) respectively.

$$\begin{aligned}&(2\pi )^3\left( I_0(A)I_0(B)I_0(C)+2\sum _{k=1}^{\infty }I_k(A)I_k(B)I_k(C)\right) \end{aligned}$$

(58)

$$\begin{aligned}&w_3(r_1,r_2,r_3)=\frac{r_1r_2r_3\exp {-\left( \frac{r_1^2\sigma _1+r_2^2\sigma _4+r_3^2\sigma _1}{2D\zeta }\right) }}{\zeta ^3 D}\nonumber \times \\&\quad \left( I_0(A)I_0(B)I_0(C)+2\sum _{k=1}^{\infty }I_k(A)I_k(B)I_k(C)\right) \end{aligned}$$

(59)

where $A=2r_1r_2\sigma _2/2\zeta D$, $B=2r_2r_3\sigma _2/2\zeta D$, $C=2r_1r_3\sigma _3/2\zeta D$, $D=(k_2-1)(2k_1^2-k_2-1)$.