Asymptotics of insensitive load balancing and blocking phases

Abstract

We study a single class of traffic acting on a symmetric set of processor-sharing queues with finite buffers, and we consider the case where the load scales with the number of servers. We address the problem of giving robust performance bounds based on the study of the asymptotic behaviour of the insensitive load balancing schemes, which have the desirable property that the stationary distribution of the resulting stochastic network depends on the distribution of job-sizes only through its mean. It was shown for small systems with losses that they give good estimates of performance indicators, generalizing henceforth Erlang formula, whereas optimal policies are already theoretically and computationally out of reach for networks of moderate size. We characterize the response of symmetric systems under those schemes at different scales and show that three amplitudes of deviations can be identified according to whether $\rho < 1$, $\rho = 1$, or $\rho > 1$. A central limit scaling takes place for a sub-critical load; for $\rho =1$, the number of free servers scales like $n^{ {\theta \over \theta +1}}$ ($\theta$ being the buffer depth and n being the number of servers) and is of order 1 for super-critical loads. This further implies the existence of different phases for the blocking probability. Before a (refined) critical load $\rho _c(n)=1-a n^{- {\theta \over \theta +1}}$, the blocking is exponentially small and becomes of order $n^{- {\theta \over \theta +1}}$ at $\rho _c(n)$. This generalizes the well-known quality-and-efficiency-driven regime, or Halfin—Whitt regime, for a one-dimensional queue and leads to a generalized staffing rule for a given target blocking probability.

Proofs of results in Sect. 5

1.1 A.1 Proof of Theorem 8

Proof

We first prove a local convergence. Let $q = \hat{p}+ \beta /\sqrt{n}$, and let $c = \sum _k k q_k$, $\bar{\beta } = \sum _i i\beta _i$. Since $\sum _k q_k = \sum _k \hat{p}_k = 1$, we have

$$\begin{aligned} \sum _k\beta _k = 0, \ \ \ c = \hat{c} + \frac{\bar{\beta }}{\sqrt{n}}. \end{aligned}$$

(73)

We remind the reader that in order to simplify notation, we shall use p instead of p(c). Starting from (33),

$$\begin{aligned} \frac{\pi (q)}{\pi (\hat{p})}&\sim \left( \frac{\psi (c)}{\psi (\hat{c})}\right) ^n e^{\sqrt{n}\bar{\beta }}\prod _k\left( \frac{p_k}{q_k}\right) ^{nq_k} \end{aligned}$$

(74)

$$\begin{aligned}&= \left( \frac{\psi (c)}{\psi (\hat{c})}\right) ^n e^{\sqrt{n}\bar{\beta }}\prod _k\left( \frac{\hat{p}_k}{q_k} \frac{p_k}{\hat{p}_k}\right) ^{nq_k} \end{aligned}$$

(75)

$$\begin{aligned}&= e^{\sqrt{n}\bar{\beta }}\prod _k\left( \frac{\hat{p}_k}{q_k} \frac{p_k\psi (c)^{-1}}{\hat{p}_k\psi ({\hat{c}})^{-1}}\right) ^{nq_k} \end{aligned}$$

(76)

$$\begin{aligned}&= e^{\sqrt{n}\bar{\beta }}\prod _k\left( \frac{\hat{p}_k}{q_k}\right) ^{nq_k} \prod _k\left( \frac{p_k\psi (c)^{-1}}{\hat{p}_k\psi ({\hat{c}})^{-1}}\right) ^{nq_k}. \end{aligned}$$

(77)

We shall compute the asymptotics of the two products separately. The first product gives

$$\begin{aligned} \prod _k\left( \frac{\hat{p}_k}{q_k}\right) ^{nq_k}&= \prod _k\left( 1 + \frac{\beta _k}{\hat{p}_k\sqrt{n}}\right) ^{-n\hat{p}_k - \sqrt{n}\beta _k} \end{aligned}$$

(78)

$$\begin{aligned}&\sim \prod _k \exp \left( - \sqrt{n}\beta _k - \frac{\beta _k^2}{2\hat{p}_k}\right) \end{aligned}$$

(79)

$$\begin{aligned}&= \prod _k \exp \left( -\frac{\beta _k^2}{2\hat{p}_k}\right) , \end{aligned}$$

(80)

where the last equality follows from (73). For the second product, from (27),

$$\begin{aligned} \log \left( \frac{p_k\psi (c)}{\hat{p}_k\psi ({\hat{c}})}\right)&= \log \left( \left( \frac{\theta -\hat{c}-\bar{\beta }/\sqrt{n}}{\rho }\right) ^{\theta -k} \left( \frac{\theta -\hat{c}}{\rho }\right) ^{-(\theta -k)}\right) \end{aligned}$$

(81)

$$\begin{aligned}&\sim \log \left( 1 - \frac{(\theta -k)\bar{\beta }}{(\theta -\hat{c})\sqrt{n}}+ \frac{(\theta -k)(\theta -k-1)}{2}\frac{\bar{\beta }^2}{(\theta -\hat{c})^2 n}\right) \end{aligned}$$

(82)

$$\begin{aligned}&\sim \frac{-(\theta -k)\bar{\beta }}{(\theta -\hat{c})\sqrt{n}} - \frac{(\theta -k)}{2}\frac{\bar{\beta }^2}{(\theta -\hat{c})^2 n}. \end{aligned}$$

(83)

Thus,

$$\begin{aligned} \prod _k\left( \frac{p_k\psi (c)^{-1}}{\hat{p}_k\psi ({\hat{c}})^{-1}}\right) ^{nq_k}&\sim \exp \left( (n\hat{p}_k + \sqrt{n}\beta _k)\left( \frac{-(\theta -k)\bar{\beta }}{(\theta -\hat{c})\sqrt{n}} - \frac{(\theta -k)\bar{\beta }^2}{2(\theta -\hat{c})^2 n}\right) \right) \end{aligned}$$

(84)

$$\begin{aligned}&\sim -\sqrt{n}\bar{\beta } +\frac{\bar{\beta }^2}{2(\theta -\hat{c})}, \end{aligned}$$

(85)

where the equalities (73), $\sum _k\beta _k = \bar{\beta }$ and $\sum _k\hat{p}_k = \hat{c}$ helped in the simplification.

Substituting the asymptotics of the two products in (77), we get

$$\begin{aligned} \frac{\pi (q)}{\pi (p)}&= \exp \left( \frac{\bar{\beta }^2}{2(\theta -\hat{c})}\right) \prod _k \exp \left( -\frac{\beta _k^2}{2\hat{p}_k}\right) . \end{aligned}$$

(86)

Consider the exponent on the RHS. Since $\sum _i\beta _i = 0$, we have $\bar{\beta } = \sum _i i \beta _i = \sum _{i=0}^{\theta -1} i \beta _i - \theta \sum _{i=0}^{\theta -1}\beta _i = -\sum _i(\theta - i)\beta _i$. Therefore,

$$\begin{aligned} \frac{\bar{\beta }^2}{2(\theta -\hat{c})} - \frac{1}{2p_k}\sum _k \beta _k^2&= \frac{1}{2(\theta -\hat{c})}\left( \sum _{k=0}^{\theta -1}(\theta - k)\beta _k\right) ^2 - \frac{1}{2\hat{p}_k}\sum _{k=0}^{\theta -1}\beta _k^2 - \frac{1}{2\hat{p}_\theta }\left( \sum _{i=0}^{\theta -1}\beta _i\right) ^2 \end{aligned}$$

(87)

$$\begin{aligned}&= \frac{1}{2(\theta -\hat{c})}\left( \sum _{k=0}^{\theta -1}(\theta - k)\beta _k\right) ^2 - \frac{1}{2\hat{p}_k}\sum _{k=0}^{\theta -1}\beta _k^2 - \frac{\psi }{2}\left( \sum _{i=0}^{\theta -1}\beta _i\right) ^2. \end{aligned}$$

(88)

Since the multivariate Gaussian distribution has exponent $-\frac{1}{2}\beta \varSigma ^{-1}\beta$, we can deduce from the above equation the inverse of the covariance matrix to be that stated in the theorem and the local convergence of $\frac{\pi (q)}{\pi (p)}$ to the Gaussian density.

Using the approximation in (32), combined with the blocking probability estimates obtained in Theorem 5, it can be easily seen that

$$\begin{aligned} \pi (\hat{p}) \sim n^{- \theta /2}, \end{aligned}$$

which in turn implies that, for any $q =\hat{p}+ \beta /\sqrt{n}$,

$$\begin{aligned} \pi (q) n^{- \theta /2} \rightarrow \exp \left( \frac{1}{2(\theta -\hat{c})}\left( \sum _{k=0}^{\theta -1}(\theta - k)\beta _k\right) ^2 - \frac{1}{2\hat{p}_k}\sum _{k=0}^{\theta -1}\beta _k^2 - \frac{\psi }{2}\left( \sum _{i=0}^{\theta -1}\beta _i\right) ^2 \right) . \end{aligned}$$

Generalizing slightly the previous computations, the same would hold for any $q =\hat{p}+ (\beta + \varepsilon _n)/\sqrt{n}$, with $\varepsilon _n$ vanishing when n goes to infinity. Hence, to derive a global convergence result of the distribution function as stated in the theorem, we can now appeal to a variant of Scheffé’s lemma (see, for instance, Theorem 1.29 in [35] with $\delta _i(n)= {1 \over \sqrt{n}}, i =1, \ldots , k$ and $k= \theta$). $\square$

1.2 A.2 Proof of Theorem 9

Proof

Instead of defining q according to a predefined scaling as in the previous proof, we shall this time define it with an arbitrary scaling which shall be made precise later. Let $q = \hat{p}+ {\beta }^{(n)}$, where again we have $\sum _k {\beta }^{(n)}_k = 0$. For $\rho = 1$, $\hat{p}_0 = \cdots = \hat{p}_{\theta -1} = 0$, $\hat{p}_\theta = 1$, so we shall assume that ${\beta }^{(n)}_k \ge 0$ for $k < \theta$. Also, for $\rho = 1$, we have $\hat{c}= \theta$ and $\psi (\hat{c})=1$ so that

$$\begin{aligned} c = \theta + {\bar{\beta }}^{(n)}, \; \psi (c) = \sum _{j=0}^\theta \frac{(-{\bar{\beta }}^{(n)})^j}{j!}, \end{aligned}$$

(89)

where ${\bar{\beta }}^{(n)} = \sum _k k{\beta }^{(n)}_k <0$.

Our starting point is again (33), which for the present case reduces to

$$\begin{aligned} \frac{\pi (q)}{\pi (\hat{p})}&\sim \psi (c)^n e^{n{\bar{\beta }}^{(n)}}\prod _k\left( \frac{p_k}{q_k}\right) ^{nq_k} \end{aligned}$$

(90)

$$\begin{aligned}&=\left( \frac{\psi (c)}{q_\theta }\right) ^{nq_\theta }e^{n{\bar{\beta }}^{(n)}}\prod _{k=0}^{\theta -1}\left( \frac{p_k\psi (c)}{{\beta }^{(n)}_k}\right) ^{nq_k} \end{aligned}$$

(91)

$$\begin{aligned}&=\left( \frac{\psi (c)}{q_\theta }\right) ^{nq_\theta }e^{n{\bar{\beta }}^{(n)}}\prod _{k=0}^{\theta -1}\left( \frac{1}{(\theta -k)!}\frac{(\theta - \hat{c}- {\bar{\beta }}^{(n)})^{\theta -k}}{{\beta }^{(n)}_k}\right) ^{nq_k}\end{aligned}$$

(92)

$$\begin{aligned}&=\left( \frac{\psi (c)}{q_\theta }\right) ^{nq_\theta }e^{n{\bar{\beta }}^{(n)}}\prod _{k=0}^{\theta -1}\left( \frac{1}{(\theta -k)!}\frac{(-{\bar{\beta }}^{(n)})^{\theta -k}}{{\beta }^{(n)}_k}\right) ^{nq_k}. \end{aligned}$$

(93)

Since ${\beta }^{(n)} \sim 0$ and ${\bar{\beta }}^{(n)} < 0$, the value of $k < \theta$ that makes the largest contribution is $\theta -1$. For all other values of k, $\frac{({\bar{\beta }}^{(n)})^{\theta -k}}{{\beta }^{(n)}_k} \rightarrow 0$ with respect to this fraction for $k = \theta - 1$. That is, fluctuations under this scaling will be visible only in ${S}^{(n)}_{\theta -1}$ and ${S}^{(n)}_\theta$ and not in lower values of k. In other words, given the number in the system, we can deduce the configuration to be ${S}^{(n)}_{\theta -1} = n\theta - nc$ and ${S}^{(n)}_\theta = n-{S}^{(n)}_{\theta -1}$. Therefore, there is only one vector p in the set ${\mathcal {P}}^{(n)}_c$. As a consequence, the only possible value of q in (90) is p, which then leads to

$$\begin{aligned} \frac{\pi (q)}{\pi (\hat{p})} \sim \psi (c)^n e^{n{\bar{\beta }}^{(n)}}. \end{aligned}$$

(94)

Consider $q_{\theta -1} = {\beta }^{(n)} \ge 0$, where ${\beta }^{(n)}$ is a scalar from now on. Since $q_\theta = 1 - {\beta }^{(n)}$, we have ${\bar{\beta }}^{(n)} = -{\beta }^{(n)}$. Let us compute the asymptotics of the term with $\psi$:

$$\begin{aligned} n\log (\psi (c))&= n\log \left( \sum _{j=0}^{\theta }\frac{{{\beta }^{(n)}}^j}{j!}\right) \sim n\left( {\beta }^{(n)} - \frac{{{\beta }^{(n)}}^{\theta +1}}{(\theta +1)!}\right) , \end{aligned}$$

(95)

where the last asymptotic form is a consequence of Lemma 4. Substituting the above relation back in (94), we get

$$\begin{aligned} \frac{\pi (q)}{\pi (\hat{p})} \sim \exp \left( -n\frac{{{\beta }^{(n)}}^{\theta +1}}{(\theta +1)!}\right) , \end{aligned}$$

(96)

where we have used the identity ${\bar{\beta }}^{(n)} = -{\beta }^{(n)}$ which was noted previously.

Consequently, the right scaling for ${\beta }^{(n)}$ is $z n^{-1/(\theta +1)}$, for $z > 0$, which means that ${S}^{(n)}_{\theta - 1} = n{\beta }^{(n)}$ lives on a scale of $n^{\theta /(\theta +1)}$. As for the proof of the central limit theorem, we can pass from local to global convergence by combining (96), the estimate of the blocking probabilities given in Theorem 7, and Theorem 1.29 in [35] with $\delta _1(n)= {1 \over \sqrt{n}}$ and $k= 1$. $\square$

1.3 A.3 Proof of Theorem 10

Proof

Following the same steps as in the proof of Theorem 9 until (93), we can arrive at the conclusion that ${S}^{(n)}_{\theta -1}$ and ${S}^{(n)}_\theta$ will be nonzero. Note that the only difference with the $\rho = 1$ case is that now

$$\begin{aligned} \psi (c) = \sum _{j=0}^\theta \frac{(-{\bar{\beta }}^{(n)})^j}{\rho ^j j!}, \end{aligned}$$

(97)

and $p_k$ has a factor $\rho ^{-(\theta -k)}$. Going further until (96) leads us to

$$\begin{aligned} \frac{\pi (q)}{\pi (\hat{p})} \sim \exp \left( -n{\beta }^{(n)} + n\frac{{\beta }^{(n)}}{\rho } -n\frac{{{\beta }^{(n)}}^{\theta +1}}{\rho ^{\theta +1} (\theta +1)!}\right) . \end{aligned}$$

(98)

The only possible scaling is, thus, ${\beta }^{(n)} = zn^{-1}$, which means that the fluctuations of ${S}^{(n)}_\theta$ around $n\theta$ are O(1).

We cannot carry on from this stage along the same line as that in the proof of Theorem 10, because to arrive at (94) we had assumed that the nonzero fluctuations were increasing with n. (This was needed to apply Stirling’s approximation.) So, we shall work directly with the stationary distribution. From (12),

$$\begin{aligned} {\mathbb {P}}({S}^{(n)}_{\theta -1} = s)&= {B}^{(n)}_\theta s! \frac{n!}{s! (n-s)!}(n\rho )^{-s} \end{aligned}$$

(99)

$$\begin{aligned}&= {\mathbb {P}}({S}^{(n)}_{\theta -1} = 0) \frac{n!}{(n-s)!}(n\rho )^{-s} \end{aligned}$$

(100)

$$\begin{aligned}&\sim {\mathbb {P}}({S}^{(n)}_{\theta -1} = 0) \rho ^{-s}, \end{aligned}$$

(101)

which is a consequence of Stirling’s approximation. $\square$

1.4 A.4 Concavity of R

Lemma 2

The function $R:{\mathbb {R}}_+ \rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} R(t) = \log \left( \sum _{k=0}^{\theta }\frac{t^k}{k!} \right) - \rho t \end{aligned}$$

(102)

is concave.

Proof

Recall that $g_\theta (t) = \sum _{k=0}^{\theta }\frac{t^k}{k!}.$ Rewrite $g_\theta (t)$ in terms of the incomplete gamma function using the following steps:

$$\begin{aligned} g_\theta (t)&= \frac{1}{\varGamma (\theta +1,0)}\int _{0}^{\infty }\left( t + u\right) ^\theta e^{-u}du \end{aligned}$$

(103)

$$\begin{aligned}&= e^{t}\tilde{\varGamma }(\theta +1, t), \end{aligned}$$

(104)

where $\tilde{\varGamma }$ is the normalized incomplete gamma function, that is, $\tilde{\varGamma }(m,x) = \frac{\varGamma (m,x)}{\varGamma (m,0)}$.

To show the concavity of R, we shall show that its second derivative is negative. Note that $g^\prime _\theta (t) = g_{\theta -1}(t)$ so that

$$\begin{aligned} R^\prime (t) = \frac{g_{\theta -1}(t)}{g_\theta (t)} - \rho \end{aligned}$$

(105)

and

$$\begin{aligned} R^{\prime \prime }(t)&= \frac{g_{\theta }(t)g_{\theta -2}(t) - g_{\theta -1}(t)^2}{g_\theta (t)^2} \end{aligned}$$

(106)

$$\begin{aligned}&= \frac{\tilde{\varGamma }(\theta +1,t)\tilde{\varGamma }(\theta -1, t) - \tilde{\varGamma }(\theta , t)^2}{\tilde{\varGamma }(\theta +1, t)^2}. \end{aligned}$$

(107)

It is shown in [1] that $\tilde{\varGamma }$, viewed as a function of $\theta$, is log-concave for all $t > 0$. We can thus infer that R is concave in $(0,\infty )$. $\square$

1.5 A.5 Generating functions

Let

$$\begin{aligned} {\mathcal {M}}^{(n)}(\mathbf {z}) = \sum _{s}\pi (s)\prod _{k=0}^\theta z_k^{s_k} \end{aligned}$$

(108)

be the moment generating function of $S^{(n)}$.

Theorem 12

$$\begin{aligned} {\mathcal {M}}^{(n)}(\mathbf {z}) = B_\theta n\rho \int _0^\infty \left( \sum _{k=0}^{\theta }\frac{1}{k!} t^k z_{\theta -k}\right) ^{n}e^{-tn\rho } dt. \end{aligned}$$

(109)

Proof

From (12) and using the facts that $x! = \int _0^\infty t^x e^{-t} dt$ and $(n\theta - \bar{s}) = \sum _k(\theta -k)s_k$,

$$\begin{aligned} \bar{ {\mathcal {M}}}^{(n)}(z)&= B_\theta \sum _s \int _0^\infty t^{\sum _k(\theta -k)s_k} e^{-t} dt \left( {\begin{array}{c}n\\ s\end{array}}\right) \prod _{k=0}^{\theta }\left( \frac{(n\rho )^{-(\theta -k)}z_k}{(\theta -k)!}\right) ^{s_k} \end{aligned}$$

(110)

$$\begin{aligned}&=B_\theta \sum _s \left( {\begin{array}{c}n\\ s\end{array}}\right) \int _0^\infty \prod _{k=0}^{\theta }\left( \frac{((n\rho )^{-1}t)^{(\theta -k)}z_k}{(\theta -k)!}\right) ^{s_k}e^{-t} dt \end{aligned}$$

(111)

$$\begin{aligned}&=B_\theta \int _0^\infty \sum _s \left( {\begin{array}{c}n\\ s\end{array}}\right) \prod _{k=0}^{\theta }\left( \frac{((n\rho )^{-1}t)^{(\theta -k)}z_k}{(\theta -k)!}\right) ^{s_k}e^{-t} dt \end{aligned}$$

(112)

$$\begin{aligned}&=B_\theta \int _0^\infty \left( \sum _{k=0}^{\theta }\frac{1}{k!}t^k(n\rho )^{-k} z_{\theta -k}\right) ^{n}e^{-t} dt, \end{aligned}$$

(113)

where the last identity is a consequence of the multinomial theorem followed by a relabelling of the index inside the sum. Finally, making the transformation $t \mapsto tn\rho$ inside the integral completes the proof. $\square$

Let

$$\begin{aligned} \bar{ {\mathcal {M}}}^{(n)}(z) = \sum _{j}\left( \sum _{s: \sum _k ks_k = j}\pi (s)\right) z^j \end{aligned}$$

(114)

be the moment generating function of the number of tasks in steady state.

Lemma 3

$$\begin{aligned} \bar{ {\mathcal {M}}}^{(n)}(z) = {\mathcal {M}}^{(n)}\left( z^0, z^1, \ldots ,z^\theta \right) . \end{aligned}$$

(115)

Proof

From its definition,

$$\begin{aligned} \bar{ {\mathcal {M}}}^{(n)}(z)&= \sum _{j}\left( \sum _{s: \sum _k ks_k = j}\pi (s)\right) z^j \end{aligned}$$

(116)

$$\begin{aligned}&= \sum _{j}\left( \sum _{s: \sum _k ks_k = j}\pi (s)z^{k s_k}\right) \end{aligned}$$

(117)

$$\begin{aligned}&= \sum _{j}\left( \sum _{s: \sum _k ks_k = j}\pi (s)(z^k)^{s_k}\right) \end{aligned}$$

(118)

$$\begin{aligned}&= \sum _s\pi (s)(z^k)^{s_k} \end{aligned}$$

(119)

$$\begin{aligned}&= {\mathcal {M}}^{(n)}\left( z^0, z^1, \ldots ,z^\theta \right) . \end{aligned}$$

(120)

$\square$

Theorem 13

$$\begin{aligned} \bar{ {\mathcal {M}}}^{(n)}(z) = B_\theta n\rho \int _0^\infty \left( \sum _{k=0}^{\theta }\frac{1}{k!} t^k z^{\theta -k}\right) ^{n}e^{-tn\rho } dt. \end{aligned}$$

(121)

Proof

The result is a direct consequence of Theorem 12 and Lemma 3. $\square$

1.6 A.6 Miscellaneous results

Lemma 4

For $\theta \ge 1$,

$$\begin{aligned} \log \left( \sum _{i=0}^\theta \frac{t^i}{i!}\right) = t - \frac{1}{(\theta +1)!} t^{\theta +1} + o\left( t^{\theta +1}\right) ,\quad \mathrm{as}\quad t\rightarrow 0. \end{aligned}$$

(122)

Proof

Let $h_\theta (t)=\log \left( \sum _{i=0}^\theta \frac{t^i}{i!}\right)$. The proof is based on computing the coefficients in the Taylor series expansion of h around 0. The first derivative of h is

$$\begin{aligned} h^{(1)}_{\theta }(t) = 1 - \frac{t^\theta }{\theta !}g_\theta (t)^{-1}, \end{aligned}$$

(123)

where $g_\theta (t) = \sum _{i=0}^\theta \frac{t^i}{i!}$, which gives the coefficient of t as 1.

For $k\le \theta$, taking the kth derivative of $h^{(1)}$ and evaluating it using the product rule for higher-order derivatives, we obtain the $(k+1)$th derivative of h as

$$\begin{aligned} h^{(k+1)}_\theta (t) = -\sum _{j=0}^k \left( {\begin{array}{c}k\\ j\end{array}}\right) \frac{t^{\theta -j}}{(\theta -j)!}\left( g_\theta (t)^{-1}\right) ^{(k-j)}, \end{aligned}$$

(124)

where $(g_\theta (t)^{-1})^{(k-j)}$ is the $(k-j)$th derivative of $g_\theta (t)^{-1}$. Assuming that the derivatives of $g_\theta (t)^{-1}$ do not go to $\infty$ at $t=0$ (which can be seen to be true), at $t=0$, the only nonzero derivative is obtained for $k=\theta$ and $j=k$. That is,

$$\begin{aligned} h_\theta ^{(k+1)}(0) = {\left\{ \begin{array}{ll} 0 &{} \quad 1 \le k < \theta ; \\ -1 &{} \quad k = \theta . \end{array}\right. } \end{aligned}$$

(125)

$\square$