15.1 The basic model of Internet congestion control

This section presents the basic model of Internet congestion control.

Consider a network that consists of a set L = {1, …, M} of unidirectional links of capacity c_l, l ∈ L. The network is shared by a set I = {1, …, N} of sources. Let x_i be the rate of flow i and let X(t) = {x_i(t), i ∈ I} be the rate vector. Let R = (R_li, i ∈ I, l ∈ L) be the routing matrix, where R_li = 1 if flow i traverses link l, and 0 otherwise. By introducing the link rate vector Y (t) = {y_l(t), l ∈ L}, the link price vector P(t) = {p_l(t), l ∈ L}, and the source price vector Q(t) = {q_i(t), i ∈ I}, we then have

$\begin{array}{ll}\hfill Y(t)& =RX(t),\hfill \end{array}$

$\begin{array}{ll}\hfill Q(t)& =R^{T}P(t).\hfill \end{array}$

It is noted that the above two equations relate the individual source rate vector with the aggregate flow (at the links) rate vector and the individual link price vector with the source aggregate price vector.

At time t, we assume link l observes the aggregate source rate

$\begin{array}{l}\hfill y_l(t)=\sum _{i=1}^N{R}_{li}{x}_i(t-{\tau }_{li}^f),\end{array}$

and source i observes the aggregate price in its path

$\begin{array}{l}\hfill q_i(t)=\sum _{l=1}^M{R}_{li}{p}_l(t-{\tau }_{li}^b),\end{array}$

where ${\tau }_{li}^f$ is the forward feedback delay from source i to link l, and ${\tau }_{li}^b$ is the backward feedback delay from link l to source i. For simplicity, we assume that the feedback delays ${\tau }_{li}^f$ and ${\tau }_{li}^b$ are constants. Further, we consider propagation delay and neglect the queueing delay, thus treating round trip delay for any link l on the path of source i, i.e., ${\tau }_i={\tau }_{li}^f+{\tau }_{li}^b$, as a constant.

15.2 Internet congestion control: A multivariable control look

In this section, by describing the congestion control network model as a time-delayed multivariable control system, we propose a method to analyze the transfer functions between the individual source rate and the link aggregate rate, between the link price and the source aggregate price. Based on this, we look at the issue of feedback control design to relate the TCP source flow rate with the aggregate flow price, which is in a proportional (P) controller structure. This can further be utilized to analyze the important properties, such as stability, of the network systems.

Stability of the algorithm in Internet congestion control is especially important when it is implemented within a network. For example, to have an efficient throughput, to bound the delay and to minimize packet loss, the rate of data flow through its links should tend toward an equilibrium value, rather than continually oscillating between having bandwidth spare and being completely overloaded. Given the fact that the control system structure of Eqs. (15.5), (15.6) like the Smith-McMillan form governs their system performance, e.g., the stability, let us take a closer look.

Taking the Laplace transform of Eqs. (15.5), (15.6), respectively, and writing the delay routing matrices in the frequency domain as

$\begin{array}{l}\hfill R_f(s)=(R_{li}^f(s),i\in I,l\in L),\end{array}$

where

$\begin{array}{l}\hfill R_{li}^f(s)=\left\{\begin{array}{ll}{e}^{-{\tau }_{li}^{f}s},& \text{if flow}i\text{traverses link}l,\\ 0,& \text{otherwise},\end{array}\right.\end{array}$

and

$\begin{array}{l}\hfill R_b(s)=(R_{li}^b(s),i\in I,l\in L),\end{array}$

where

$\begin{array}{l}\hfill R_{li}^b(s)=\left\{\begin{array}{ll}{e}^{-{\tau }_{li}^{b}s},& \text{if flow}i\text{traverses link}l,\\ 0,& \text{otherwise},\end{array}\right.\end{array}$

we then have

$\begin{array}{ll}\hfill Y(s)& =R_f(s)X(s),\hfill \end{array}$

$\begin{array}{ll}\hfill Q(s)& ={(R_b(s))}^{T}P(s),\hfill \end{array}$

where X(s), Y (s), P(s), and Q(s) denote the Laplace transform of the vectors X(t), Y (t), P(t), and Q(t), respectively, and T denotes the transport of the matrix.

15.3 Padé approximations to the system (15.7) and (15.8)

Polynomials are not such a good class of functions if one wants to approximate functions with singularities because polynomials are entire functions without singularities. They are only useful up to the first singularity. Rational functions are the simplest ones with singularities. The idea is that the poles of the rational functions will move to the singularities of the function and hence the domain of convergence can be enlarged, and singularities of it may be discovered using the poles of the rational approximations.

The [m, n] Padé approximation of a function f in a point a is the rational function $\frac{Q_m}{P_n},$ with Q_m a polynomial of degree ≤ m and P_n a polynomial of degree ≤ n, for which we have the following interpolation condition at a

$\begin{array}{l}\hfill f(z)-\frac{Q_m(z)}{P_n(z)}=O({(z-a)}^{m+n+1}),z\to a.\end{array}$

The computation of the polynomials P_n and Q_m is not so straightforward in the above condition since one first has to compute the Taylor expansion of $\frac{Q_m}{P_n}$ and then equate the first m + n + 1 Taylor coefficients to the first m + n + 1 Taylor coefficients of f. Usually the Padé approximation is defined by linearizing the interpolation condition as

$\begin{array}{l}\hfill P_n(z)f(z)-Q_m(z)=O({(z-a)}^{m+n+1}),z\to a.\end{array}$

For Padé approximation near infinity to a function of the form

$\begin{array}{l}\hfill f(z)=\sum _{k=0}^{\infty }\frac{c_k}{z^{k+1}},\end{array}$

we let m = n − 1 and the interpolation condition is

$\begin{array}{l}\hfill P_{n}f(z)-Q_{n-1}(z)=O(z^{-n-1}),z\to \infty .\end{array}$

There is a degree of freedom since we can multiply both sides of Eq. (15.10) by a constant. Usually we normalize this by taking P_n monic, that is it is in the form xⁿ + ⋯, when this is possible. This can only be done if P_n is of the exact degree n. If we take P_n monic, then we can determine the n unknown coefficients a_k (k = 1, …, n) in

$\begin{array}{l}\hfill P_n=:\sum _{k=0}^n{a}_k{(z-a)}^{n-k},a_0=1,\end{array}$

by putting the coefficients of (z−a)^k for k = m + 1, m + 2, …, m + n in the Taylor expansion of P_nf equal to zero. The polynomial Q_m then corresponds to the Taylor polynomial of degree m of P_nf.

Hereafter we present another approach. Suppose f is analytic in a domain Ω that contains a. Again we take a contour Γ inside Ω encircling a once in the positive direction. Divide both sides of Eq. (15.10) by (z−a)^m+k+2 and integrate, to find

$\begin{array}{ll}\hfill & \frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{P_n(z)f(z)}{{(z-a)}^{m+k+2}}dz-\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{Q_m(z)}{{(z-a)}^{m+k+2}}dz\hfill \\ \hfill & =\sum _{j=m+n+1}^{\infty }{b}_{n,j}\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{(z-a)}^{j-m-k-2}dz,\hfill \end{array}$

where the b_{n, j}s are the coefficients in the expansion of P_nf − Q_m around a. The integral involving Q_m is zero for k ≥ 0 since it is proportional to the (m + k + 1)th derivative of Q_m, which is zero for k ≥ 0. The summation on the right-hand side has a contribution only when j = m + k + 1, but when 0 ≤ k ≤ n − 1 then j ≤ m + n, and such indices do not appear in the summation. Hence the right hand side also vanishes for k ≤ n − 1. Therefore Eq. (15.10) tells us that

$\begin{array}{l}\hfill \frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\frac{P_n(z)}{{(z-a)}^{m+k+2}}f(z)dz=0,k=0,1,\dots ,n-1.\end{array}$

If we use the expansion (15.11) then this gives

$\begin{array}{l}\hfill \sum _{j=0}^n{a}_j\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{(z-a)}^{n-j-m-k-2}f(z)dz=0,k=0,1,2,\dots ,n-1\end{array}$

and then

$\begin{array}{l}\hfill \frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}{(z-a)}^{n-j-m-k-2}f(z)dz=c_{m-n+k+j+1}.\end{array}$

Therefore we obtain the following system of equations

$\begin{array}{l}\hfill \left(\begin{array}{llll}{c}_{m-n+1}& c_{m-n+2}& \dots & c_{m+1}\\ c_{m-n+2}& c_{m-n+3}& \dots & c_{m+2}\\ ⋮& ⋮& \dots & ⋮\\ c_m& c_{m+1}& \dots & c_{m+n}\end{array}\right)\left(\begin{array}{l}{a}_0\\ a_1\\ ⋮\\ a_n\end{array}\right)=\left(\begin{array}{l}0\\ 0\\ ⋮\\ 0\end{array}\right).\end{array}$

There is one degree of freedom here since we have n + 1 unknowns and n (homogeneous) equations. The choice a₀ = 1 (if possible) gives the monic polynomial P_n, but sometimes other normalization will be used.

15.4 Analyses into system structure of congestion control of a simple network in frequency domain

Let us consider a simple network with three links and four flows as shown in Fig. 15.3. In this network, there are four flows: f1 traverses link L1, f2 traverses link L2, f3 traverses link L3, and f4 traverses link L1, link L2, and link L3. Flows f1, f2, and f3 represent one-hop short flow, which is dominant in the Internet: most peer-to-peer applications involve one-link flows between a data source and a receiver (downloader); while flow f4 represents long flows which traverse more than one hop. The routing structure of this network is described by the following routing matrix

$\begin{array}{l}\hfill R=\left[\begin{array}{llll}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill & \hfill \hfill \end{array}\right].\end{array}$

Letting x_i denote the source sending rate of flow fi, (i = 1, 2, 3, 4). The individual flow rate vector is then

$\begin{array}{l}\hfill X(t)=\left[\begin{array}{ll}\hfill x_1(t)\hfill & \hfill \hfill \\ \hfill x_2(t)\hfill & \hfill \hfill \\ \hfill x_3(t)\hfill & \hfill \hfill \\ \hfill x_4(t)\hfill \end{array}\right].\end{array}$

Letting y_l (l = 1, 2, 3) denote the aggregate flow rate at link l, the aggregate flow rate vector is then denoted by

$\begin{array}{l}\hfill Y(t)=\left[\begin{array}{ll}\hfill y_1(t)\hfill & \hfill \hfill \\ \hfill y_2(t)\hfill & \hfill \hfill \\ \hfill y_3(t)\hfill \end{array}\right].\end{array}$

Each link is suggesting a price (either delay or packet loss ratio), one has the link price vector

$\begin{array}{l}\hfill P(t)=\left[\begin{array}{ll}\hfill p_1(t)\hfill & \hfill \hfill \\ \hfill p_2(t)\hfill & \hfill \hfill \\ \hfill p_3(t)\hfill \end{array}\right].\end{array}$

The total price for flow fi is the sum of all the prices for all links it traverses and is denoted by q_i. We then denote the aggregate price vector as

$\begin{array}{l}\hfill Q(t)=\left[\begin{array}{ll}\hfill q_1(t)\hfill & \hfill \hfill \\ \hfill q_2(t)\hfill & \hfill \hfill \\ \hfill q_3(t)\hfill & \hfill \hfill \\ \hfill q_4(t)\hfill \end{array}\right].\end{array}$

The aggregate flow rate vector and the individual rate vector is associated by the routing matrix, and the aggregate price vector and the individual price vector are also associated by the transpose of the routing matrix. They are mathematically described by

$\begin{array}{ll}\hfill Y(t)& =RX(t),\hfill \end{array}$

$\begin{array}{ll}\hfill Q(t)& =R^{T}P(t).\hfill \end{array}$

In particular, at time t, the links observe the following corresponding aggregate source rate

$\begin{array}{ll}\hfill y_1(t)& =x_1(t-{\tau }_{11}^f)+x_4(t-{\tau }_{14}^f),\hfill \end{array}$

$\begin{array}{ll}\hfill y_2(t)& =x_2(t-{\tau }_{22}^f)+x_4(t-{\tau }_{24}^f),\hfill \end{array}$

$\begin{array}{ll}\hfill y_3(t)& =x_3(t-{\tau }_{33}^f)+x_4(t-{\tau }_{34}^f),\hfill \end{array}$

where ${\tau }_{li}^f$ (l = 1, 2, 3;i = 1, 2, 3, 4) denotes the feed-forward delay of the data flow fi by the traveling link l. This delay is determined by the propagation delay and the forward queuing delay at the link. The propagation delay is determined by the distance between the source and the destination while the forward queuing delay is determined by the queue length and the leaking capacity at the link.

Similarly, the sources observe the following corresponding aggregate price along their path

$\begin{array}{ll}\hfill q_1(t)& =p_1(t-{\tau }_{11}^b),\hfill \end{array}$

$\begin{array}{ll}\hfill q_2(t)& =p_2(t-{\tau }_{22}^b),\hfill \end{array}$

$\begin{array}{ll}\hfill q_3(t)& =p_3(t-{\tau }_{33}^b),\hfill \end{array}$

$\begin{array}{ll}\hfill q_4(t)& =p_1(t-{\tau }_{14}^b)+p_2(t-{\tau }_{24}^b)+p_3(t-{\tau }_{34}^b),\hfill \end{array}$

where ${\tau }_{li}^b$ (l = 1, 2, 3;i = 1, 2, 3, 4) denotes the feed-backward delay of flow fi. These backward delays are actually experienced by the acknowledgment (ack) packets for their corresponding data flows. They are all related to the distance between the destination and the source and the capacity (in the opposite direction) correspondingly.

By taking the Laplace transform of Eqs. (15.23), (15.24), (15.25), and denoting the Laplace transformation of X(t) and Y (t) as X(s) and Y (s), respectively, one has

$\begin{array}{l}\hfill Y(s)=R_f^P(s)X(s),\end{array}$

where the transfer function $R_f^P(s)$ is given by

$\begin{array}{l}\hfill R_f^P(s)=\left[\begin{array}{llll}{e}^{-{\tau }_{11}^{f}s}& 0& 0& e^{-{\tau }_{14}^{f}s}\\ & & & \\ 0& e^{-{\tau }_{22}^{f}s}& 0& e^{-{\tau }_{24}^{f}s}\\ & & & \\ 0& 0& e^{-{\tau }_{33}^{f}s}& e^{-{\tau }_{34}^{f}s}\end{array}\right].\end{array}$

The second order Padé approximation to the above transfer function is

$\begin{array}{l}\hfill R_f^P(s)=\left[\begin{array}{llll}\frac{12-6{\tau }_{11}^{f}s+{({\tau }_{11}^f)}^2{s}^2}{12+6{\tau }_{11}^{f}s+{({\tau }_{11}^f)}^2{s}^2}& 0& 0& \frac{12-6{\tau }_{14}^{f}s+{({\tau }_{14}^f)}^2{s}^2}{12+6{\tau }_{14}^{f}s+{({\tau }_{14}^f)}^2{s}^2}\\ & & & \\ 0& \frac{12-6{\tau }_{22}^{f}s+{({\tau }_{22}^f)}^2{s}^2}{12+6{\tau }_{22}^{f}s+{({\tau }_{22}^f)}^2{s}^2}& 0& \frac{12-6{\tau }_{24}^{f}s+{({\tau }_{24}^f)}^2{s}^2}{12+6{\tau }_{24}^{f}s+{({\tau }_{24}^f)}^2{s}^2}\\ & & & \\ 0& 0& \frac{12-6{\tau }_{33}^{f}s+{({\tau }_{33}^f)}^2{s}^2}{12+6{\tau }_{33}^{f}s+{({\tau }_{33}^f)}^2{s}^2}& \frac{12-6{\tau }_{34}^{f}s+{({\tau }_{34}^f)}^2{s}^2}{12+6{\tau }_{34}^{f}s+{({\tau }_{34}^f)}^2{s}^2}\end{array}\right].\end{array}$

If we set all the forward delays as ${\tau }_{11}^f=1,{\tau }_{22}^f=2,{\tau }_{33}^f=6,{\tau }_{14}^f=3,{\tau }_{24}^f=4,{\tau }_{34}^f=9,$ by denoting

$\begin{array}{ll}\hfill a_1^f& =12-6{\tau }_{11}^{f}s+{({\tau }_{11}^f)}^2{s}^2=12-6s+s^2\hfill \\ \hfill a_2^f& =12-6{\tau }_{22}^{f}s+{({\tau }_{22}^f)}^2{s}^2=12-12s+4s^2\hfill \\ \hfill a_3^f& =12-6{\tau }_{33}^{f}s+{({\tau }_{33}^f)}^2{s}^2=12-36s+36s^2\hfill \\ \hfill a_4^f& =12-6{\tau }_{14}^{f}s+{({\tau }_{14}^f)}^2{s}^2=12-18s+9s^2\hfill \\ \hfill a_5^f& =12-6{\tau }_{24}^{f}s+{({\tau }_{24}^f)}^2{s}^2=12-24s+16s^2\hfill \\ \hfill a_6^f& =12-6{\tau }_{34}^{f}s+{({\tau }_{34}^f)}^2{s}^2=12-54s+81s^2\hfill \\ \hfill b_1^f& =12+6{\tau }_{11}^{f}s+{({\tau }_{11}^f)}^2{s}^2=12+6s+s^2\hfill \\ \hfill b_2^f& =12+6{\tau }_{22}^{f}s+{({\tau }_{22}^f)}^2{s}^2=12+12s+4s^2\hfill \\ \hfill b_3^f& =12+6{\tau }_{33}^{f}s+{({\tau }_{33}^f)}^2{s}^2=12+36s+36s^2\hfill \\ \hfill b_4^f& =12+6{\tau }_{14}^{f}s+{({\tau }_{14}^f)}^2{s}^2=12+18s+9s^2\hfill \\ \hfill b_5^f& =12+6{\tau }_{24}^{f}s+{({\tau }_{24}^f)}^2{s}^2=12+24s+16s^2\hfill \\ \hfill b_6^f& =12+6{\tau }_{34}^{f}s+{({\tau }_{34}^f)}^2{s}^2=12+54s+81s^2\hfill \end{array}$

then we have the following approximated transfer function

$\begin{array}{l}\hfill R_f^P(s)=\left[\begin{array}{llll}\frac{a_1^f}{b_1^f}& 0& 0& \frac{a_4^f}{b_4^f}\\ & & \\ 0& \frac{a_2^f}{b_2^f}& 0& \frac{a_5^f}{b_5^f}\\ & & \\ 0& 0& \frac{a_3^f}{b_3^f}& \frac{a_6^f}{b_6^f}\\ & & \\ \end{array}\right].\end{array}$

Using d^f(s) to denote the least common multiple (lcm) of the denominators of all the elements of $R_f^P(s)$, one has

$\begin{array}{l}\hfill d^f(s)=b_1^f\times b_2^f\times b_3^f\times b_4^f\times b_5^f\times b_6^f.\end{array}$

Subsequently the Smith form of $R_f^P(s)$ is S_f(s), which is given by

$\begin{array}{l}\hfill S_f(s)=\left[\begin{array}{llll}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b_1^f\times b_2^f\times b_3^f\times b_4^f\times b_5^f\times b_6^f\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b_1^f\times b_2^f\times b_3^f\times b_4^f\times b_5^f\times b_6^f\hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right].\end{array}$

Accordingly the Smith-McMillan form of $R_f^P(s)$ is M_f(s)

$\begin{array}{l}\hfill \begin{array}{c}{M}_f(s)=\left[\begin{array}{llll}\hfill \frac{1}{b_1^f\times b_2^f\times b_3^f\times b_4^f\times b_5^f\times b_6^f}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right]\\ =\left[\begin{array}{ll}\hfill \hfill & \hfill \frac{1}{(s+3+\sqrt{3}i)\times (s+3-\sqrt{3}i)}\times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{2}\right)\times \left(s+\frac{3-\sqrt{3}i}{2}\right)}\hfill \\ \hfill \hfill & \hfill \times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{6}\right)\times \left(s+\frac{3-\sqrt{3}i}{6}\right)}\times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{3}\right)\times \left(s+\frac{3-\sqrt{3}i}{3}\right)}\hfill \\ \hfill \hfill & \hfill \times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{4}\right)\times \left(s+\frac{3-\sqrt{3}i}{4}\right)}\times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{9}\right)\times \left(s+\frac{3-\sqrt{3}i}{9}\right)}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right].\end{array}\end{array}$

Similarly, by taking the Laplace transform of Eqs. (15.26), (15.27), (15.28), (15.29), and denoting the Laplace transformation of P(t) and Q(t) as P(s) and Q(s), respectively, one has

$\begin{array}{l}\hfill Q(s)={(R_f^P(s))}^{T}X(s),\end{array}$

where the transfer function ${(R_f^P(s))}^T$ is given by

$\begin{array}{l}\hfill {(R_b^P(s))}^T=\left[\begin{array}{lll}\frac{12-6{\tau }_{11}^{b}s+{({\tau }_{11}^b)}^2{s}^2}{12+6{\tau }_{11}^{b}s+{({\tau }_{11}^b)}^2{s}^2}& 0& 0\\ & & \\ 0& \frac{12-6{\tau }_{22}^{b}s+{({\tau }_{22}^b)}^2{s}^2}{12+6{\tau }_{22}^{b}s+{({\tau }_{22}^b)}^2{s}^2}& 0\\ & & \\ 0& 0& \frac{12+6{\tau }_{33}^{b}s+{({\tau }_{33}^b)}^2{s}^2}{12+6{\tau }_{33}^{b}s+{({\tau }_{33}^b)}^2{s}^2}\\ & & \\ \frac{12-6{\tau }_{14}^{b}s+{({\tau }_{14}^b)}^2{s}^2}{12+6{\tau }_{14}^{b}s+{({\tau }_{14}^b)}^2{s}^2}& \frac{12-6{\tau }_{24}^{b}s+{({\tau }_{24}^b)}^2{s}^2}{12+6{\tau }_{24}^{b}s+{({\tau }_{24}^b)}^2{s}^2}& \frac{12-6{\tau }_{34}^{b}s+{({\tau }_{34}^b)}^2{s}^2}{12+6{\tau }_{34}^{b}s+{({\tau }_{34}^b)}^2{s}^2}\end{array}\right].\end{array}$

If the backward delays are set as ${\tau }_{11}^b=11,{\tau }_{22}^b=5,{\tau }_{33}^b=7,{\tau }_{14}^b=14,{\tau }_{24}^b=8,{\tau }_{34}^b=12,$ by denoting

$\begin{array}{ll}\hfill a_1^b& =12-6{\tau }_{11}^{b}s+{({\tau }_{11}^b)}^2{s}^2=12+66s+121s^2\hfill \\ \hfill a_2^b& =12-6{\tau }_{22}^{b}s+{({\tau }_{22}^b)}^2{s}^2=12+30s+25s^2\hfill \\ \hfill a_3^b& =12-6{\tau }_{33}^{b}s+{({\tau }_{33}^b)}^2{s}^2=12+42s+49s^2\hfill \\ \hfill a_4^b& =12-6{\tau }_{14}^{b}s+{({\tau }_{14}^b)}^2{s}^2=12+84s+196s^2\hfill \\ \hfill a_5^b& =12-6{\tau }_{24}^{b}s+{({\tau }_{24}^b)}^2{s}^2=12+48s+64s^2\hfill \\ \hfill a_6^b& =12-6{\tau }_{34}^{b}s+{({\tau }_{34}^b)}^2{s}^2=12+72s+144s^2\hfill \\ \hfill b_1^b& =12+6{\tau }_{11}^{b}s+{({\tau }_{11}^b)}^2{s}^2=12+66s+121s^2\hfill \\ \hfill b_2^b& =12+6{\tau }_{22}^{b}s+{({\tau }_{22}^b)}^2{s}^2=12+30s+25s^2\hfill \\ \hfill b_3^b& =12+6{\tau }_{33}^{b}s+{({\tau }_{33}^b)}^2{s}^2=12+42s+49s^2\hfill \\ \hfill b_4^b& =12+6{\tau }_{14}^{b}s+{({\tau }_{14}^b)}^2{s}^2=12+84s+196s^2\hfill \\ \hfill b_5^b& =12+6{\tau }_{24}^{b}s+{({\tau }_{24}^b)}^2{s}^2=12+48s+64s^2\hfill \\ \hfill b_6^b& =12+6{\tau }_{34}^{b}s+{({\tau }_{34}^b)}^2{s}^2=12+72s+144s^2,\hfill \end{array}$

we then have

$\begin{array}{l}\hfill {(R_b^P(s))}^T=\left[\begin{array}{lll}\frac{a_1^b}{b_1^b}& 0& 0\\ & & \\ 0& \frac{a_2^b}{b_2^b}& 0\\ & & \\ 0& 0& \frac{a_3^b}{b_3^b}\\ & & \\ \frac{a_4^b}{b_4^b}& \frac{a_5^b}{b_5^b}& \frac{a_6^b}{b_6^b}\end{array}\right].\end{array}$

Further, if d^b(s) is the least common multiple (lcm) of the denominators of all the elements in ${(R_b^P(s))}^T$, that is

$\begin{array}{l}\hfill d^b(s)=b_1^b\times b_2^b\times b_3^b\times b_4^b\times b_5^b\times b_6^b,\end{array}$

then the Smith form of ${(R_b^P(s))}^T$ (denoted by S_b(s)) is given by

$\begin{array}{l}\hfill S_b(s)=\left[\begin{array}{lll}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill b_1^b\times b_2^b\times b_3^b\times b_4^b\times b_5^b\times b_6^b\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill b_1^b\times b_2^b\times b_3^b\times b_4^b\times b_5^b\times b_6^b\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right].\end{array}$

Therefore the Smith-McMillan form of ${(R_b^P(s))}^T$ (denoted by M_b(s)) is given by

$\begin{array}{l}\hfill \begin{array}{c}{M}_b(s)=\left[\begin{array}{llll}\hfill \frac{1}{b_1^b\times b_2^b\times b_3^b\times b_4^b\times b_5^b\times b_6^b}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right]\\ =\left[\begin{array}{ll}\hfill \hfill & \hfill \frac{1}{\left(s+\frac{3+\sqrt{3}i}{11}\right)\times \left(s+\frac{3-\sqrt{3}i}{11}\right)}\times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{5}\right)\times \left(s+\frac{3-\sqrt{3}i}{5}\right)}\hfill \\ \hfill \hfill & \hfill \times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{7}\right)\times \left(s+\frac{3-\sqrt{3}i}{7}\right)}\times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{14}\right)\times \left(s+\frac{3-\sqrt{3}i}{14}\right)}\hfill \\ \hfill \hfill & \hfill \times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{8}\right)\times \left(s+\frac{3-\sqrt{3}i}{8}\right)}\times \frac{1}{\left(s+\frac{3+\sqrt{3}i}{12}\right)\times \left(s+\frac{3-\sqrt{3}i}{12}\right)}\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill \hfill \\ \hfill \hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill \hfill \end{array}\right].\end{array}\end{array}$