Equilibrium points and equilibrium sets of some $\textit{GI}/M/1$ queues

Abstract

Queues can be seen as a service facility where quality of service (QoS) is an important measure for the performance of the system. In many cases, the queue implements the optimal admission control (either discounted or average) policy in the presence of holding/congestion cost and revenue collected from admitted customers. In this paper, users offer an arrival rate at stationarity that depends on the QoS they experience. We study the interaction between arriving customers and such a queue under two different QoS measures—the asymptotic rate of the customers lost and the fraction of customers lost in the long run. In particular, we investigate the behaviour of equilibrium points and equilibrium sets associated with this interaction and their interpretations in terms of business cycles. We provide sufficient conditions for existence of equilibrium sets for M/M/1 queue. These conditions further help us to identify the relationship among system parameters for which equilibrium sets exist. Next, we consider $\textit{GI}/M/1$ queues and provide a sufficient condition for existence of multiple optimal revenue policies. We then specialize these results to study the equilibrium sets of (i) a D/M/1 queue and (ii) a queue where the arrival rate is locally continuous. The equilibrium behaviour in the latter case is more interesting as there may be multiple equilibrium points or sets. Motivated by such queues, we introduce a weaker version of monotonicity and investigate the existence of generalized equilibrium sets.

Appendices

Appendix A: Existence of control limits and properties of QoS

1.1 A.1 Proof of Lemma 2

Proof

Let $R_\lambda$ be the control limit when the arrival rate is $\lambda$ of an admission-controlled M/M/1 queue. Let $\varLambda _{R_\lambda }$ denote the rate at which customers are admitted into the system in the long run, $H_{R_\lambda }$ denote the average queue length and $g^{R_\lambda } = r\, \varLambda _{R_\lambda } - hH_{R_\lambda }$ denote the average utility (difference of reward and holding cost). Using the stationary distribution for a $M/M/1/R_\lambda$ queue,

$$\begin{aligned} \varLambda _{R_\lambda }= & {} \frac{\lambda (1-\rho ^{R_\lambda })}{1-\rho ^{R_\lambda +1}} ~~\text {and}\\ H_{R_\lambda }= & {} \frac{\rho (1-\rho ^{R_\lambda })}{(1-\rho )(1-\rho ^{R_\lambda +1})}-\frac{R_\lambda \rho ^{R_\lambda +1}}{(1-\rho ^{R_\lambda +1})}, \end{aligned}$$

when $\rho = \frac{\lambda }{\mu } < 1$. If there exists $R_\lambda$ such that, $g^{R_\lambda -1}$ is greater than $g^{R_\lambda }$, then we know that a finite control limit exists which is $R_\lambda$. Consider

$$\begin{aligned} g^{R_\lambda }-g^{R_\lambda -1}=r\,(\varLambda _{R_\lambda }-\varLambda _{R_\lambda -1})\,-\, h\,(H_{R_\lambda }-H_{R_\lambda -1}). \end{aligned}$$

(10)

After straightforward algebraic manipulations,

$$\begin{aligned} g^{R_\lambda }-g^{R_\lambda -1} = r\lambda \rho ^{R_\lambda -1}(1-\rho )^{2}-h\rho ^{R_\lambda }(-\rho -R_\lambda \rho +R_\lambda +\rho ^{R_\lambda +1}). \end{aligned}$$

(11)

Note that $\rho < 1$, and from the above equation, one can find a sufficiently large $R_\lambda$ for which $g^{R_\lambda }-g^{R_\lambda -1} < 0$. $\square$

1.2 A.2 Proof of Theorem 1

Proof

We already know that if multiple optimal control limit exist, they are consecutive integers. Let us assume that $R^*_\lambda$ and $R^*_\lambda +1$ are those optimal values. The QoS is given by $L(R^*_\lambda ) = \frac{(1-\rho )\rho ^{R^*_\lambda }}{1-\rho ^{{R^*_\lambda }+1}}$. For the first condition, let us solve the expected average reward of the queue as

$$\begin{aligned} g^{R_\lambda } = r\, \varLambda _{R_\lambda } - hH_{R_\lambda }, \end{aligned}$$

(12)

where $\varLambda _{R_\lambda }$ is the rate at which customers are admitted into the system in the long run and $H_{R_\lambda }$ denotes average queue length. The above equation is nothing but the difference between reward and holding cost. Using the stationary distribution for an $M/M/1/R_\lambda$ queue it is easy to see that

$$\begin{aligned} \begin{aligned} \varLambda _{R^*_\lambda }&= \lambda \bigg (\frac{1-\rho ^{R^*_\lambda }}{1-\rho ^{R^*_\lambda +1}}\bigg ), \\ H_{R^*_\lambda }&= \frac{\rho (1-\rho ^{R^*_\lambda })}{(1-\rho )(1-\rho ^{R^*_\lambda +1})}-\frac{R^*_\lambda \rho ^{R^*_\lambda +1}}{1-\rho ^{R^*_\lambda +1}}. \end{aligned} \end{aligned}$$

(13)

The firm’s profit remains the same for these multiple optimal control limits $R_\lambda ^*$ and $R_\lambda ^* + 1$, i.e.

$$\begin{aligned} g^{R^*_\lambda } - g^{R^*_\lambda +1} = r\,(\varLambda _{R^*_\lambda }-\varLambda _{R^*_\lambda +1})\,-\,h\,(H_{R^*_\lambda +1}-H_{R^*_\lambda }) = 0. \end{aligned}$$

Using (13) in the above, straightforward algebraic manipulations yield the desired result.

The second condition is straightforward. Using the value of the QoS L in the affine utility function $f(L(\lambda ))$ we obtain

$$\begin{aligned} m - e\,\frac{(1-\rho )\,\rho ^{R^*_\lambda }}{1-\rho ^{{R^*_\lambda +1}}}< \lambda < m - e\,\frac{(1-\rho )\,\rho ^{R^*_\lambda +1}}{1-\rho ^{{R^*_\lambda }+2}}. \end{aligned}$$

(14)

This completes the proof. $\square$

1.3 A.3 Proof of Theorem 2

Proof

Let $\{\tau _i\}_{i\ge 1}$ be the sequence of inter-arrival times and y be the expected inter-arrival time. Also, let $\{\sigma _i\}_{i\ge 0}$ be the sequence of times which denotes successive arrivals and $W_t$ denote the state of the natural process (number in the system) at time t.

Let $X_i$, $Y_i$ be the random variables denoting the state and action at ith decision epoch, $k(X_i,Y_i)$ be the lump sum portion of the reward and $c(W_t,X_i,Y_i)$ be the rate at which the continuous portion of the reward is received between the decision epoch i and $i+1$ for $i \in \{0,1,2,\ldots \}.$ From [3, Chapter 11], the average reward is as follows:

$$\begin{aligned} g^{{R_\lambda }}(s)= & {} \liminf \limits _{n\rightarrow \infty } \frac{E_{s}^{R_\lambda } \bigg \{ \sum _{i=0}^{n}\bigg [ k(X_i,Y_i) + \int \limits _{\sigma _i}^{\sigma _{i+1}}c(W_t,X_i,Y_i)\mathrm{d}t \bigg ] \bigg \}}{E_s^{R_\lambda } \bigg \{ \sum _{i=0}^{n}\tau _{i} \bigg \}} \\= & {} \liminf \limits _{n\rightarrow \infty } \frac{ \text {Expected total reward up to } n\text {th}\,\text {decision epoch}}{\text {Expected total time until the }n\text {th}\,\text {decision epoch}}. \end{aligned}$$

Substitute $r=1$ and $h=0$ and note that this is a finite state Markov chain. We have

$$\begin{aligned} g^{R{_\lambda }}(s)= & {} \lim \limits _{n\rightarrow \infty } \frac{E_{s}^{R_\lambda } \bigg \{ \sum _{i=0}^{n} k(X_i,Y_i) \bigg \}}{E_s^{R_\lambda } \bigg \{ \sum _{i=0}^{n}\tau _{i} \bigg \} } \\= & {} \lim \limits _{n\rightarrow \infty } \frac{\text {Expected number of customers admitted up to } n\text {th}\,\text { decision epoch}}{\text {Expected number of arrivals up to } n\text {th}\,\text {decision epoch} \times E(\tau _1)} \\= & {} \lim \limits _{n\rightarrow \infty } \frac{\text {Expected number of customers admitted up to } n\text {th}\,\text {decision epoch}}{ n \times y}. \end{aligned}$$

Thus,

$$\begin{aligned} g^{{R_\lambda }}(s) \times y= & {} \lim \limits _{n\rightarrow \infty } \frac{\text {Expected number of customers admitted up to } n\text {th}\,\text {decision epoch}}{n} \\= & {} \text {Fraction of customers admitted in the long run.} \end{aligned}$$

One can compute $g^{{R_\lambda }}$ by substituting $r=1$ and $h=0$ in [14, Equation 13] after making the changes given in Sect. 5.2. This is same as the system of equations in (8). Let the solution of this system of equations be ${\tilde{g}}^{R_\lambda }$. Then, the fraction of customers lost in long run is

$$\begin{aligned} L(\lambda ,R_\lambda )= (1- {\tilde{g}}^{{R_\lambda }}y) \end{aligned}$$

and the rate of customers lost is

$$\begin{aligned} L_1(\lambda ,R_\lambda ) = \lambda (1- {\tilde{g}}^{{R_\lambda }}y). \end{aligned}$$

$\square$

1.4 A.4 Proof of Theorem 3

Proof

Let $R^*_{\lambda _0}$ be the unique control limit at $\lambda _0$. Then,

$$\begin{aligned} v_\alpha ^{R^*_{\lambda _0}}(s) > v_\alpha ^{R_{\lambda _0}}(s), ~~\text {where } R_{\lambda _0}\ne R^*_{\lambda _0}~~ \forall ~s. \end{aligned}$$

For the linear system of equations in (6), the coefficient matrix w.r.t. the variables $v_\alpha ^{R_\lambda }(0),\ldots ,v_\alpha ^{R_{\lambda }}(R_{\lambda })$ is same as the matrix $(I-M_{R^*_{\lambda _{0}}})$ from [3, Theorem 11.3.1], where $M_{R^*_{\lambda _{0}}}$ is the transition probability matrix corresponding the Markov Deterministic policy $R^*_{\lambda _0}$. The solution of this system of equations is the value vector corresponding to the policy $R^*_{\lambda _0}$ which exists and is unique. This implies that the matrix $(I-M_{R^*_{\lambda _0}})$ is invertible. Due to the continuity of matrix inverse, the system of equations in (6) will have a solution $\phi _{i}(\lambda ), i=0,\ldots ,R_\lambda$, which is continuous in a small enough neighbourhood (a, b) of $\lambda _0$ (which can also be obtained using the implicit function theorem [26, 27] on the above linear system), given below:

$$\begin{aligned} v_\alpha (0)= & {} \phi _0(\lambda ), \\ v_\alpha (1)= & {} \phi _1(\lambda ), \\&\vdots&\\ v_\alpha (R_\lambda )= & {} \phi _{R_\lambda }(\lambda ). \end{aligned}$$

So, for a small enough interval (a, b) around $\lambda _0$ we have

$$\begin{aligned} v_\alpha ^{R^*_{\lambda _0}}(s) > v_\alpha ^{R_{\lambda }}(s)~~ \forall ~~ s,~~ \forall ~~ \lambda \in (a,b). \end{aligned}$$

Therefore, in the interval (a, b) the optimal control limit will be $R^*_{\lambda _0}$. $\square$

1.5 A.5 Proof of Theorem 4

Proof

From Theorem 3, the optimal control limit at $\lambda _2$ cannot be unique as it will give rise to an open interval around $\lambda _2$ overlapping with $(\lambda _1,\lambda _2)$ on the left (and $(\lambda _2,\lambda _3)$ on the right) leading to a contradiction on uniqueness of $R^*_\lambda$ (and ${\tilde{R}}^{*}_{\lambda }$). So, multiple optima exists at the common point $\lambda _2$ of these open intervals. From Corollary 3 in [14], the value $v_\alpha ^{R^*_\lambda }(s)$ is unimodal with respect to $R^*_\lambda$ with possible ties at multiple optimal control limits which necessarily occur consecutively. Also,

$$\begin{aligned} v_\alpha ^{R^*_{\lambda }}(s) \ge v_\alpha ^{{\tilde{R}}^{*}_{\lambda }}(s)~~ \text {for}~~ \lambda < \lambda _2. \end{aligned}$$

Taking the limit $\lambda \uparrow \lambda _2$ on both sides and using the fact that continuous and unique functions have a unique limit, we get

$$\begin{aligned} \lim _{\lambda \uparrow \lambda _2}v_{\alpha }^{R^*_{\lambda }}(s)\ge & {} \lim _{\lambda \uparrow \lambda _2}v_{\alpha }^{{\tilde{R}}^{*}_{\lambda }}(s), \nonumber \\ v_{\alpha }^{R^*_{\lambda _2}}(s)\ge & {} v_{\alpha }^{{\tilde{R}}^{*}_{\lambda _2}}(s). \end{aligned}$$

(15)

Also,

$$\begin{aligned} v_{\alpha }^{{\tilde{R}}^{*}_{\lambda }}(s) \ge v_\alpha ^{{R}^{*}_{\lambda }}(s)~~ \text {for}~~ \lambda > \lambda _2. \end{aligned}$$

Taking $\lambda \downarrow \lambda _2$, we get

$$\begin{aligned} \lim _{\lambda \downarrow \lambda _2}v_\alpha ^{{\tilde{R}}^{*}_{\lambda }}(s),\ge & {} \lim _{\lambda \downarrow \lambda _2}v_\alpha ^{R^*_{\lambda }}(s), \nonumber \\ v_\alpha ^{{\tilde{R}}^{*}_{\lambda _2}}(s)\ge & {} v_\alpha ^{R^{*}_{\lambda _2}}(s). \end{aligned}$$

(16)

From (15) and (16), we get

$$\begin{aligned} v_\alpha ^{{\tilde{R}}^{*}_{\lambda _2}}(s)= & {} v_\alpha ^{R^*_{\lambda _2}}(s) ~~\forall ~~s. \end{aligned}$$

Suppose ${\hat{R}}^{*}_{\lambda _2}$ is optimal at $\lambda _2$ but $R^{*}_{\lambda _2}$ is not optimal at $\lambda _2$. Then,

$$\begin{aligned} v_\alpha ^{{\hat{R}}^*_{\lambda _2}}(s) > v_\alpha ^{{R}^*_{\lambda _2}}(s) ~~\forall ~~s. \end{aligned}$$

We also know that $v_\alpha ^{{R}^*_{\lambda _2}}(s)$ is a continuous function of $\lambda _2$. Therefore, we can always find an $\epsilon$ small enough such that

$$\begin{aligned} v_\alpha ^{{\hat{R}}^*_{(\lambda _2 - \epsilon )}}(s) > v_\alpha ^{{R}^*_{(\lambda _2-\epsilon )}}(s) ~~\forall ~~s. \end{aligned}$$

(17)

However, (17) contradicts the definition of $R^{*}_{\lambda }$ as the unique optimal control limit in $(\lambda _1,\lambda _2)$. Hence,

$$\begin{aligned} v_\alpha ^{{\hat{R}}^*_{\lambda _2}}(s) = v_\alpha ^{{R}^*_{\lambda _2}}(s) = v_\alpha ^{{\tilde{R}}^{*}_{\lambda _2}}(s) ~~\forall ~~s. \end{aligned}$$

(18)

Since ${\hat{R}}^*_{\lambda }$ is the optimal control limit and (18) holds, it can be concluded that $R^*_\lambda$, ${\tilde{R}}^{*}_{\lambda }$ and ${\hat{R}}^*_{\lambda }$ are optimal at $\lambda _2$, and others also if they exist. $\square$

1.6 A.6 Proof of Theorem 5

Proof

If there exist multiple optima for a given $\lambda$, then

$$\begin{aligned} v_{\alpha }^{R_{\lambda }}(s) = v_{\alpha }^{R_{\lambda }+1}(s)~~~~~\forall ~ s\in S. \end{aligned}$$

The value vector can be uniquely determined by the first $R_{\lambda }+1$ states, as the remaining $v_{\alpha }^{R_{\lambda }}(s)$’s for $s>R_{\lambda }$ can be evaluated recursively using the first $R_{\lambda }+1$ components of the value vector. Therefore, it is enough to check that the above equation holds for all $s\le R_{\lambda }$.

As given by Van Nunen and Puterman in [8],

$$\begin{aligned} v_{\alpha }^{R_{\lambda }}(s)= & {} a_{s-1}v_{\alpha }^{R_{\lambda }}(s-1) - b_{s}, \\ v_{\alpha }^{R_{\lambda }+1}(s)= & {} a_{s-1}v_{\alpha }^{R_{\lambda }+1}(s-1) -b_{s}. \end{aligned}$$

Now, $v_{\alpha }^{R_{\lambda }}(s)=v_{\alpha }^{R_{\lambda }+1}(s)$ gives $v_{\alpha }^{R_{\lambda }}(s-1)=v_{\alpha }^{R_{\lambda }+1}(s-1)$ and doing this recursively ultimately leads to $v_{\alpha }^{R_{\lambda }}(0)=v_{\alpha }^{R_{\lambda }+1}(0)$. So, this condition is a necessary one. We can see that it is also a sufficient condition. $\square$

Appendix B: Methods for calculating $\lambda$ when $R^*_\lambda = 1,2$ in a discounted cost criteria M/M/1 queue

After showing the existence of multiple control limits, we want to identify the point $\lambda$ where multiple optimal control limits exist. The following three methods involve finding $\varDelta \lambda$, if we start from some arbitrary $\lambda _0$. We only consider the case when $R_{\lambda }=1,2$. For higher value of control limits, these methods becomes very cumbersome.

1.1 B.1 Method 1

The first method involves equating $v_\alpha ^{R_\lambda }(0)$ and $v_\alpha ^{R_{\lambda }+1}(0)$ as a function of $\varDelta \lambda$. In particular, for $R_{\lambda }=1$ and $R_{\lambda }=2$, the value at state 0 is given as follows:

$$\begin{aligned} v_{\alpha }^{1}(0)= & {} \frac{r(1-p_0) - c(1)}{1-q_1 -p_0}. \\ v_{\alpha }^{2}(0)= & {} \frac{r(1-p_1 -p_0^2) - c(1)(1-p_0-p_1) - p_0c(2)}{(1-q_1)(1-p_0-p_1)-{p_0}{q_2}}. \end{aligned}$$

After adding each of the roots obtained from solving $v_{\alpha }^{1}(0) = v_{\alpha }^{2}(0)$ to $\lambda _0$, one of these value is the $\lambda$ where the multiple optimal control exists. However, we need to check and find which one is the correct root by finding the control limit at the each of them. Also, some roots might be invalid. The following example illustrate the method for a M/M/1 queue with given set of costs and parameters $r,h,\mu ,\alpha$. The computations were done in Mathematica 9 on Paaspoli server (OS: Debian, AMD OPTERON 6212, 2.6 GHz processors and 32GB RAM).

• 1. ${h = 1}$, ${r=0.4}$, ${{\mu } = 5.5}$, ${{\alpha }=0.54}$ and ${{\lambda _0} = 10}$

  The simplified version of the final equation to be solved is $1.45185\varDelta \lambda ^4+61.9742\varDelta \lambda ^3+955.654\varDelta \lambda ^2+6309.78\varDelta \lambda +14988.1$. Roots for this equation are

  • (a) $\varDelta \lambda _1 =-10.000000$

  • (b) $\varDelta \lambda _2 =-6.106282$

  • (c) $\varDelta \lambda _3 =-10.540051$

  • (d) $\varDelta \lambda _4 =-16.040032$

  Root (b) is the point which after adding to $\lambda _0$ gives the point of multiple optima, i.e. $\lambda = 3.893718$.

1.2 B.2 Method 2

This method involves solving an optimization problem which minimizes the square of the difference between the value corresponding to control limits $R_\lambda =1,2$ at state zero. The optimization problem is formulated as follows:

$$\begin{aligned} \begin{array}{lll} M2 &{}&{}\min \limits _{\varDelta \lambda } ~(v_\alpha ^{1,0}(\lambda _0 + \varDelta \lambda )- v_\alpha ^{2,0}(\lambda _0 + \varDelta \lambda ))^2\\ &{}\hbox {subject to} &{} \\ &{}v_\alpha ^{1,0}(\lambda _0 + \varDelta \lambda ) &{}=\frac{r(1-p_0)-c(1)}{(1-p_0-p_1)} \\ &{}v_\alpha ^{2,0}(\lambda _0 + \varDelta \lambda ) &{}=\frac{r(1-p_0^2-p_1)-(1-p_0-p_1)c(1)-p_0c(2)}{(1-q_1)(1-p_1-p_0) -p_0q_2} \\ &{}\varDelta \lambda \quad \text {unrestricted}. \end{array} \end{aligned}$$

(19)

The above constrained optimization problem is a non convex minimization problem. It can be solved easily using solvers like SNOPT, KNITRO, etc. This approach has an edge over the earlier one as it gives an if and only if condition for obtaining the $\lambda$ where multiple optimal control limits exist.

Theorem 7

For an M/M/1 queue with discounted criteria, multiple optimal control limits 1 and 2 will exist if and only if M2 attains its global minimum.

Proof

Suppose we have multiple optimal control limits 1 and 2 for a given $\lambda _m$ (say $\lambda _0 + \varDelta \lambda _m$). Then, at $\lambda _0 + \varDelta \lambda _m$, the value vectors for control limit 1 and 2 are equal, i.e.

$$\begin{aligned} v_\alpha ^{1,s}(\lambda _0 + \varDelta \lambda _m) = v_\alpha ^{2,s}(\lambda _0 + \varDelta \lambda _m)~~ \forall ~s. \end{aligned}$$

In particular, we have $v_\alpha ^{1,0}(\lambda _0 + \varDelta \lambda _m) = v_\alpha ^{2,0}(\lambda _0 + \varDelta \lambda _m)$. Hence, we conclude that at the point of multiple optima $\lambda _m$, the objective value of M2 is 0, and it attains its global minima. Let us consider the converse case where we are given that the global optimal solution of M2 with objective value 0 is attained at $\varDelta \lambda _m$. Then, we have that for $\varDelta \lambda _m$,

$$\begin{aligned} v_\alpha ^{1,0}(\lambda _0 + \varDelta \lambda _m) = v_\alpha ^{2,0}(\lambda _0 + \varDelta \lambda _m). \end{aligned}$$

This is a sufficient condition for multiple optimal control limit to exist at $\lambda _0 + \varDelta \lambda _m$ as given in Theorem 5.

Hence, the solution set of the above minimization problem characterizes the set of points at which multiple optimal control limit exists. $\square$

The following example illustrate this constrained optimization method. The optimization problem was modelled in AMPL and solved using the SNOPT solver on OPTIMUS server (OS Linux, Intel Quad core Xeon E5506 2.13 GHz and 64GB RAM).

• 1. Costs and parameters ${h = 1}$, ${r=0.4}$, ${{\mu } = 5.5}$, ${{\alpha }=0.54}$ and ${{\lambda _0} = 5}$ of an M/M/1 queue when used in M2 gives objective value $= 1.349202e^{-12}$ and $\varDelta \lambda = -1.10629$, giving $\lambda = 3.89371$.

1.3 B.3 Method 3

If we substitute the value of $v_\alpha ^{1,0}(\lambda _0 + \varDelta \lambda )$ and $v_\alpha ^{2,0}(\lambda _0 + \varDelta \lambda )$ in the objective function of (19), then we get the unconstrained version of M2 given as follows:

$$\begin{aligned}&M3 \quad \quad \min \limits _{\varDelta \lambda } ~\left( \frac{r(1-p_0)-c(1)}{(1-p_0-p_1)} - \frac{r(1-p_0^2-p_1)-(1-p_0-p_1)c(1)-p_0c(2)}{(1-q_1)(1-p_1-p_0) -p_0q_2}\right) ^2&\nonumber \\&\quad \varDelta \lambda \quad \text {unrestricted}.&\end{aligned}$$

(20)

This can be solved by equating the gradient of the objective function in M3 to 0 as shown below:

$$\begin{aligned} 2(v_\alpha ^1(\varDelta \lambda )-v_\alpha ^2(\varDelta \lambda ))\dfrac{d(v_\alpha ^1(\varDelta \lambda )-v_\alpha ^2(\varDelta \lambda ))}{\mathrm{d}\lambda } = 0. \end{aligned}$$

(21)

The solution of the unconstrained optimization problem can be root of either of the terms in (21). However, we are only interested in the root of the first term which gives global minima. The roots of second term might lead to some unwanted solution with non-zero objective value.

The following example illustrate this method using AMPL as modelling language. The optimization problem was solved using the SNOPT solver on OPTIMUS server (OS Linux, Intel Quad core Xeon E5506 2.13 GHz and 64GB RAM).

• Costs and parameters ${h = 1}$, ${r=0.4}$, ${{\mu } = 5.5}$, ${{\alpha }=0.54}$ and ${{\lambda _0} = 10}$ of an M/M/1 queue when used in M3 gives objective value $=3.722129e^{-10}$ and $\varDelta \lambda = -6.10629$, giving $\lambda = 3.89371$.

Appendix C: Generation of the arrival rates from a discrete distribution

Consider the following discrete distribution whose mean is required to be locally continuous:

$$\begin{aligned} G(.)={\left\{ \begin{array}{ll} u_{_{1,k}}\,\,\,\,\,\,\,\,\,\,\, &{} \text {when arrival rate is } 1,\\ u_{_k}\,\,\,\,\,\,\,\,\,\,\, &{} \text {when arrival rate is } k,\\ u_{_{k+1}}\,\,\,\,\,\,\,\,\,\,\, &{} \text {when arrival rate is } k+1. \end{array}\right. } \end{aligned}$$

Mean arrival rate is $\lambda =u_{_{1,k}}+k\times u_{_{k}}+(k+1)\times u_{_{k+1}}$. We will fix $u_{_{1,k}}$ in order to generate mean arrival rate for a suitable interval. Let us take the support as $\{1,\, k,\, k+1\}$; the arrival rate is $\lambda \in (\underline{m}_{k},{\overline{m}}_{k}]$, where ${\overline{m}}_{k}$ and $\underline{m}_{k}$ can be derived as follows:

$$\begin{aligned} \lambda= & {} u_{_{1,k}}+k\times u_{_{k}}+(k+1)u_{_{k+1}} \\= & {} u_{_{1,k}}+k\times u_{_{k}}+(k+1)(1-u_{_{1,k}}-u_{_{k}}) \\= & {} (k+1)-k\times u_{_{1,k}}-u_{_{k}}. \end{aligned}$$

Now, $u_{_{k}}\in [0,\,1-u_{_{1,k}}]$. Thus, ${\overline{m}}_{k}=(k+1)-k\times u_{_{1,k}}$ and $\underline{m}_{k}=(k+1)-k\times u_{_{1,k}}-(1-u_{_{1,k}}).$.

To generate arrival rate whose mean is $\lambda \in (\underline{m}_{k+1},{\overline{m}}_{k+1}]$, we change the support of the distribution to $\{1,\, k+1,\, k+2\}$ and so on. While changing the support we are fixing $u_{_{1,k+1}} \ge u_{_{1,k}}$, where ${\overline{m}}_{k+1}=(k+2)-(k+1)\times u_{_{1,k+1}}$ and $\underline{m}_{k+1}=(k+2)-(k+1)\times u_{_{1,k+1}}-(1-u_{_{1,k+1}})$.

Example 11

In this example, we will illustrate the above method to generate arrival rates whose mean lies in $(3.7,\,6.25]$.

• 1. We take the support as $\{1,\,4,\,5\}$ and fix $u_{_{1,4}}=0.1$. Then, $\lambda =0.1+4u_{_{4}}+5(0.9-u_{_{4}})$, i.e. $\lambda =4.6-u_{_{4}}$. Thus, with support $\{1,\,4,\,5\}$, $u_k \in [0, 0.9)$ and $u_{5} = 1 -u_4$ we can generate a set of arrival rates with above mass functions such that the arrival rate $\lambda$ is continuous over the interval $(3.7,\,4.6]$.

• 2. The support now is $\{1,\,5,\,6\}$ and we fix $u_{_{1,5}}=0.15$. So, $\lambda =0.15+4u_{_{5}}+5(0.85-u_{_{4}})$, i.e. $\lambda =5.25-u_{_{5}}$. Thus, with support $\{1,\,5,\,6\}$ we can generate $\lambda \in (4.4,\,5.25]$ but we have already generated $\lambda$ until 4.6 from the previous support; so, we generate a family of arrivals whose rates continuously change over $\lambda \in (4.6,\,5.25]$.

• 3. We take the support as $\{1,\,6,\,7\}$ and we take $u_{_{1,6}}=0.2$. Then, $\lambda =0.2+6u_{_{6}}+7(0.8-u_{_{6}})$, i.e. $\lambda =5.8-u_{_{6}}$. Thus, with support $\{1,\,6,\,7\}$ we can generate $\lambda \in (5,\,5.8]$ but we have already generated $\lambda$ until 5.25 from the previous support; so, we now have arrivals whose rate is locally continuous over $\lambda \in (5.25,\,5.8]$.

• 4. We take $\{1,\,7,\,8\}$ as support and $u_{_{1,7}}=0.25$ so that $\lambda =0.25+7u_{_{7}}+8(0.75-u_{_{7}})$, i.e. $\lambda =6.25-u_{_{7}}$. Thus, with support $\{1,\,7,\,8\}$ we can generate $\lambda \in (5.5,\,6.25]$ but we have already generated $\lambda$ until 5.25 from the previous support; so, we have the arrivals with the desired property over $\lambda \in (5.8,\,6.25]$.