2. Problem formulation

In this section, we consider the stochastic process $\{\mathcal {X}_\pi (t): t \geq 0 \}$ where $\Pi$ is the set of possible supervisor assignment policies, $\pi \in \Pi$, and $\mathcal {X}_\pi (t)=x \in X = \{0,1, \ldots, N\}$ is the number of customers who have been served by a subordinate and are waiting for a supervisor at time $t$ under policy $\pi$. We assume that $\Pi$ consists of all Markovian stationary deterministic policies corresponding to the state space $X$ of the stochastic process $\{ \mathcal {X}_\pi (t)\}$. The policy $\pi \in \Pi$ specifies if each supervisor is serving the customers or working on her own responsibilities as a function of the current state $x \in X$ (i.e., the number of customers who are waiting for the supervisor). We note that $\{ \mathcal {X}_\pi (t)\}$ is a birth-and-death process with finite state space $X$ and there exists a finite scalar $q$ such that the transition rates $\{q_\pi (x,x^\prime )\}$ of $\{ \mathcal {X}_{\pi } (t)\}$ satisfy $\sum _{x^{\prime } \in X, x^{\prime } \neq x} q_\pi (x,x^\prime ) \leq q$ for all $x \in X$ and $\pi \in \Pi$. This indicates that $\{ \mathcal {X}_\pi (t)\}$ is uniformizable for all $\pi \in \Pi$. Let $\{Y_\pi (k)\}$ denote the corresponding discrete-time Markov chain, so that $\{Y_\pi (k)\}$ has the same state space ${X}$ as $\{ \mathcal {X}_\pi (t)\}$ and transition probabilities $p_{\pi }(x,x^\prime )=q_\pi (x,x^\prime )/q$ if $x^\prime \neq x$ and $p_{\pi }(x,x)=1-\sum _{x^\prime \in {X}, x^\prime \neq x}q_\pi (x,x^\prime )/q$ for all $x \in {X}$. We then translate the continuous-time optimization problem to a discrete-time Markov decision problem (see, e.g., [Reference Lippman20]). That is to say, we can generate sample paths of $\{ \mathcal {X}_\pi (t)\}$, where $\pi \in \Pi$, by generating a Poisson process $\{ K (t)\}$ with rate $q = N\mu _1+N\theta +M\mu _2 \lt \infty$ and at the times of events of $\{ K (t)\}$, the next state of $\{ \mathcal {X}_\pi (t)\}$ is generated using the transition probabilities of $\{Y_\pi (k)\}$.

Let $a \in A=\{0,1,\ldots,M\}$ denote the assignment of supervisors, where $a$ represents the number of supervisors who are working with the subordinates and $A$ is the action space. Let $A_{x}$ and $a_x$ denote the set of allowable actions in state $x \in X$. Note that $A_0 = \{0\}$, representing that supervisors can only work on their own responsibilities when there are no customers waiting for the second-stage service, and $A_N = \{1,\ldots,M\}$, representing that we have at least one supervisor serving the customers if the number of customers who are waiting for supervisors attains the maximum (since it would be unethical for supervisors not to serve customers when all the first-stage servers cannot serve any more customers). For $1 \leq x \leq N-1$, we have $A_{x} = \{0,1,\ldots,\min \{x,M\}\}$. Figure 1 illustrates the corresponding rate diagram when action $a_x \in A_x$ is selected in state $x \in X$.

Figure 1. State-transition diagram for the two-stage service system.

For the discrete-time Markov decision process problem with uniformization constant $q$, we have, for all $a_{x}\in A_{x}$, the following one-step transition probabilities:

(1) \begin{equation} p(x^\prime\,|\,x,a_{x})=\left\{\begin{array}{ll} \dfrac{(N-x)\mu_1}{q} & \text{for}\ x \in \{0,\ldots,N-1\}, x^\prime=x+1,\\ \dfrac{(x-a_{x})\theta+a_{x} \mu_2}{q} & \text{for}\ x \in \{1,\ldots,N\}, x^\prime=x-1,\\ 1-\dfrac{(N-x)\mu_1+(x-a_{x})\theta+a_{x} \mu_2}{q} & \text{for}\ x \in \{0,\ldots,N\}, x^\prime=x,\\ 0 & \text{otherwise}. \end{array}\right. \end{equation}

Furthermore, for all $x \in X$ and $a_{x} \in A_{x}$, we specify the immediate reward $r(x,a_{x})$ of choosing action $a_{x}$ in state $x$:

$$r(x,a_{x}) = \frac{(M-a_{x}) r_s \mu_s+a_{x} r_2 \mu_2-(x-a_{x})c\theta}{q}.$$

Note that due to the abandonments, this Markov decision process problem is unichain. Since $X$ is finite, $A_{x}$ is finite for each $x \in X$, and $r(x,a_{x})$ is bounded, there exists a stationary long-run average optimal policy (see [Reference Puterman23], Theorem 8.4.5).

For any policy $\pi \in \Pi$, let $g_{N,M}^\pi$ denote the gain (long-run average reward) of the continuous-time problem under policy $\pi$ for a system with $N$ subordinates and $M$ supervisors. Note that ${g_{N,M}^\pi }/{q}$ is the gain for the corresponding discrete-time problem. The objective is to identify the optimal policy $\pi ^* \in \Pi$ that attains the optimal gain $g_{N,M}^*$, that is, find $\pi ^*$ such that

$$g_{N,M}^{\pi^*}=g_{N,M}^*=\max_{\pi \in \Pi}g_{N,M}^\pi.$$

3. Optimal policy

In this section, we show that one of two policies is always optimal and characterize the conditions under which each policy is optimal. Note that a Markovian deterministic decision rule $d: X\rightarrow A$ specifies which action $d(x)\in A_{x}$ to choose in each state $x\in X$. Thus, a stationary policy $\pi$ can be defined using the corresponding decision rule $d$ which will be denoted as $\pi =d^{\infty }$.

Define $\pi ^{\mathcal {S}}=(d_{\mathcal {S}})^{\infty }$, where

$$d_{\mathcal{S}}(x)=\left\{\begin{array}{ll} 0 & \text{for}\ x=0,\ldots,N-1,\\ 1 & \text{for}\ x=N. \end{array}\right.$$

Similarly, define $\pi ^{\mathcal {C}}=(d_{\mathcal {C}})^{\infty }$ where $d_{\mathcal {C}}(x)=\min \{x,M\}$ for all $x\in \{0,\ldots,N\}$. Thus, $\pi ^{\mathcal {S}}$ gives priority to the $\mathcal {S}$upervisors’ own responsibilities and $\pi ^{\mathcal {C}}$ gives priority to the $\mathcal {C}$ustomers. The following theorem completely characterizes the optimal policy.

Proof of Theorem 1 and Remark 2. It follows from $1 \leq M \leq N$ that $N=1$ implies $M=1$, in which case $X=\{0,1\}$, $A_0=\{0\}$, and $A_1=\{1\}$. Thus, there is only one feasible policy when $N=1$ and $\pi ^{\mathcal {S}} = \pi ^{\mathcal {C}}$ are both optimal. Therefore, we assume $N \geq 2$ in the rest of the proof.

Without loss of generality, we assume that $q =1$ and use the value iteration algorithm for unichain Markov decision process problems (see p. 364 of [Reference Puterman23]).

To prove the optimality of $\pi ^{\mathcal {S}}$ and $\pi ^{\mathcal {C}}$ under different conditions, for all $x=0,\ldots,N$, we set

$$v_0(x)=(N-x-1)\times \frac{r_s\mu_s-r_2\mu_2}{\mu_1+\mu_2}$$

and compute $v_n(x)=\max _{a_{x} \in A_{x}}v_n^{a_{x}}(x)$ for $n\geq 1$, where for $x\in X$ and $a_{x}\in A_{x}$,

(2) \begin{align} v_n^{a_{x}}(x)& = (M-a_{x})r_s \mu_s+ a_{x} r_2\mu_2 -(x-a_{x}) c \theta+(N-x)\mu_1 v_{n-1}(x+1)\nonumber\\ & \quad +[(x-a_{x})\theta+a_{x}\mu_2] v_{n-1}(x-1)\nonumber\\ & \quad +[1-(N-x)\mu_1-(x-a_{x})\theta-a_{x}\mu_2] v_{n-1}(x). \end{align}

Note that since $A_0=\{0\}$, $v_n(0)=v_n^0(0)$ follows. For $a_{x}^{1},a_{x}^{2}\in A_{x}$ and $x\in \{1,\ldots,N\}$, define

(3) \begin{equation} \Delta_n^{a_{x}^{1},a_{x}^{2}}(x)=v_n^{a_{x}^{1}}(x)-v_n^{a_{x}^{2}}(x) =(a_{x}^{2}-a_{x}^{1})(r_s \mu_s -r_2 \mu_2-c \theta +(\theta -\mu_2)[v_{n-1}(x-1)-v_{n-1}(x)]). \end{equation}

We first prove part (i). We will show that $\Delta _n^{a_{x}^{1},a_{x}^{2}}(x)\geq 0$ for all $n\geq 1$, $x\in \{1,\ldots,N\}$ and $a_{x}^{1} \lt a_{x}^{2}\in A_{x}$, which implies that $v_n(x)=v_n^0(x)$ for all $x\in \{1,\ldots,N-1\}$ and $v_n(N)=v_n^1(N)$ ($v_n(N)=v_n^0(N)$ in Remark 2). First assume $\theta = \mu _2$. We then have:

$$\Delta_n^{a_{x}^{1},a_{x}^{2}}(x)=(a_{x}^{2}-a_{x}^{1})(r_s \mu_s -r_2 \mu_2-c \theta) \geq 0$$

for all $n\geq 1$ and $x\in \{1,\ldots,N\}$ as long as $c \leq {(r_s \mu _s -r_2 \mu _2)}/{\theta }=c_0$.

Next assume $\theta \neq \mu _2$. We use induction to prove that $\Delta _n^{a_{x}^{1},a_{x}^{2}}(s)\geq 0$ for all $n\geq 1$, $x\in \{1,\ldots,N\}$, and $a_{x}^{1} \lt a_{x}^{2}\in A_{x}$. For $n=1$ and $x \in \{1,\ldots,N\}$, (3) yields

(4) \begin{equation} \Delta_1^{a_{x}^{1},a_{x}^{2}}(x) =(a_{x}^{2}-a_{x}^{1})\left[\frac{(r_s\mu_s-r_2\mu_2)(\theta+\mu_1)}{\mu_1+\mu_2}-c\theta\right] \geq 0, \end{equation}

where the inequality follows since $c\leq c_0$. Now assume that $\Delta _k^{a_{x}^{1},a_{x}^{2}}(x) \geq 0$ for $k = 1, \ldots, n-1$, $x \in \{1,\ldots,N\}$ and $a_{x}^{1} \lt a_{x}^{2}$ (i.e., for $k = 1, \ldots, n-1$, $v_k(x)=v_k^0(x)$ for all $x\in \{0,\ldots,N-1\}$ and $v_k(N)=v_k^1(N)$ in Theorem 1; $v_k(N)=v_k^0(N)$ in Remark 2) as long as $c \leq c_0$. We will show that the same assertion holds for $k=n$. From the induction hypothesis, we have

$$v_{n-1}(x)=Mr_s \mu_s -xc\theta+(N-x)\mu_1 v_{n-2}(x+1)+x\theta v_{n-2}(x-1)+[1-(N-x)\mu_1-x\theta]v_{n-2}(x)$$

for $x \in \{0,\ldots, N-1\}$ ($x \in \{0,\ldots, N\}$ in Remark 2), and

\begin{align*} v_{n-1}(N)& =(M-1)r_s \mu_s +r_2\mu_2-(N-1)c\theta+[(N-1)\theta +\mu_2]v_{n-2}(N-1)\\ & \quad +[1-(N-1)\theta-\mu_2]v_{n-2}(N). \end{align*}

Furthermore, it follows from $\Delta _{n-1}^{a_{x}^{1},a_{x}^{2}}(x)\geq 0$ and (3) that for $x =1,\ldots,N$,

$$r_s \mu_s -r_2 \mu_2-c \theta +(\theta -\mu_2)[v_{n-2}(x-1)-v_{n-2}(x)]\geq 0,$$

which implies that

(5) \begin{equation} (\theta -\mu_2)[v_{n-2}(x-1)-v_{n-2}(x)] \geq{-}r_s \mu_s +r_2 \mu_2+c \theta. \end{equation}

Note that for $x \in \{1,\ldots,N-1\}$ ($x \in \{1,\ldots, N\}$ in Remark 2),

(6) \begin{align} v_{n-1}(x-1)-v_{n-1}(x) & =c \theta +(N-x)\mu_1[v_{n-2}(x)-v_{n-2}(x+1)]\nonumber\\ & \quad +(x-1)\theta [v_{n-2}(x-2)-v_{n-2}(x-1)]\nonumber\\ & \quad +[1-(N-x+1)\mu_1-x\theta][v_{n-2}(x-1)-v_{n-2}(x)] \end{align}

and

(7) \begin{align} v_{n-1}(N-1)-v_{n-1}(N) & =r_s \mu_s -r_2\mu_2+(N-1)\theta [v_{n-2}(N-2)-v_{n-2}(N-1)]\nonumber\\ & \quad +[1-(N-1)\theta-\mu_1-\mu_2][v_{n-2}(N-1)-v_{n-2}(N)]. \end{align}

Note that since $q = N\mu _1+N\theta +M\mu _2$ and we assumed, without loss of generality, that $q=1$, we have that $1-(N-x+1)\mu _1-x\theta$ and $1-(N-1)\theta -\mu _1-\mu _2$ are positive for $x \in \{1,\ldots,N\}$.

Observe that the multipliers $(N-x)\mu _1$ and $(x-1)\theta$ of $v_{n-2}(x) -v_{n-2}(x+1)$ and $v_{n-2}(x-2) -v_{n-2}(x-1)$ equal zero when $x=N$ and $x=1$, respectively. Equations (5), (6), and (7) yield

\begin{align*} (\theta -\mu_2)[v_{n-1}(x-1)-v_{n-1}(x)] & \geq c\theta(\theta -\mu_2) + (1-\theta-\mu_1){({-}r_s \mu_s +r_2 \mu_2+c \theta)} \\ & ={(1-\mu_1-\mu_2)\theta c}+{(\theta+\mu_1-1)(r_s \mu_s -r_2 \mu_2)} \end{align*}

for $x \in \{1,\ldots,N-1\}$ ($x \in \{1,\ldots, N\}$ in Remark 2), and

\begin{align*} (\theta -\mu_2)[v_{n-1}(N-1)-v_{n-1}(N)]& \geq (\theta -\mu_2)(r_s \mu_s -r_2\mu_2)+(1-\mu_1-\mu_2)({-r_s \mu_s +r_2 \mu_2+c \theta})\\ & ={(1-\mu_1-\mu_2)\theta c}+{(\theta+\mu_1-1)(r_s \mu_s -r_2 \mu_2)}. \end{align*}

Now Eq. (3) yields that for all $x=1,\ldots,N$ and $a_{x}^{1} \lt a_{x}^{2} \in A_{x}$,

(8) \begin{align} \Delta_n^{a_{x}^{1},a_{x}^{2}}(x) & \geq (a_{x}^{2}-a_{x}^{1})[r_s \mu_s -r_2 \mu_2-c \theta +(1-\mu_1-\mu_2)\theta c+(\theta+\mu_1-1)(r_s \mu_s -r_2 \mu_2)]\nonumber\\ & =(a_{x}^{2}-a_{x}^{1})[-(\mu_1+\mu_2)c\theta+(\theta+\mu_1)(r_s \mu_s -r_2 \mu_2)]\geq 0 \end{align}

as long as $c\leq c_0$.

From (8), we have $\Delta _n^{a_{x}^{1},a_{x}^{2}}(x) \geq 0$ for all $n \geq 1$, $x \in \{1,\ldots,N\}$, and $a_{x}^{1} \lt a_{x}^{2}$ when $c \leq c_0$. Therefore, $v_n(x)=v_n^0(x)$ for $x=0,\ldots,N-1$ ($x=0,\ldots,N$ in Remark 2) and $v_n(N)=v_n^1(N)$ for all $n\geq 1$ when $c \leq c_0$. Since we have a finite state space $X$ and ${A}_{x}$ is finite for all $x$, $r(x,a_{x})$ is bounded and the model is unichain, there exists a stationary long-run average optimal policy (see [Reference Puterman23], Theorem 8.4.5). Note that from (1) and $q = N\mu _1+N\theta +M\mu _2=1$, regardless of the action $a_{x}$ chosen in each state $x$, we have $p(x\,|\,x,a_{x}) =1-{[(N-x)\mu _1+(x-a_{x})\theta +a_{x} \mu _2]}/{q}$ $= x\mu _1+(N-x+a_{x})\theta +(M-a_{x})\mu _2 \gt 0$ for $\forall x \in X$ and $a_{x} \in A_{x}$, which indicates that the transition matrix for any feasible stationary policy is aperiodic. Therefore, since the stationary policies are unichain and every optimal policy has an aperiodic transition matrix, it follows from Theorems 8.5.4 and 8.5.6 of Puterman [Reference Puterman23] that for any $\epsilon \gt 0$, value iteration will stop after a finite number of iterations with an $\epsilon$-optimal policy. Furthermore, since $\epsilon$ is arbitrary and the state and action spaces are finite, an $\epsilon$-optimal policy (for $\epsilon$ small enough) is indeed an optimal policy.

For part (ii), it follows from the proof of part (i) that $\Delta _n^{a_{x}^{1},a_{x}^{2}}(x)\leq 0$ for all $n\geq 1$, $x\in \{1,\ldots,N\}$, and $a_{x}^{1} \lt a_{x}^{2}\in A_{x}$, when $c\geq c_0$ and $\theta = \mu _2$. When $\theta \neq \mu _2$, we again use induction to prove that $\Delta _n^{a_{x}^{1},a_{x}^{2}}(x)\leq 0$ for all $n\geq 1$, $x \in \{1,\ldots,N\}$, and $a_{x}^{1} \lt a_{x}^{2}\in A_{x}$. From (4), we know that $\Delta _1^{a_{x}^{1},a_{x}^{2}}(x) \leq 0$ for $x\in \{ 1,\ldots,N\}$ and $a_{x}^{1} \lt a_{x}^{2} \in A_{x}$. Assume that $\Delta _k^{a_{x}^{1},a_{x}^{2}}(x) \leq 0$ for $k = 1, \ldots, n-1$, $x \in \{1,\ldots,N\}$, and $a_{x}^{1} \lt a_{x}^{2}\in A_{x}$. Therefore, we have $v_{n-1}(x)=v_{n-1}^{\min \{x,M\}}(x)$ for $x \in \{0,\ldots, N\}$.

Note that from $\Delta _{n-1}^{a_{x}^{1},a_{x}^{2}}(x)\leq 0$ and (3), for $x =1,\ldots,N$, we have

(9) \begin{equation} (\theta -\mu_2)[v_{n-2}(x-1)-v_{n-2}(x)] \leq{-}r_s \mu_s +r_2 \mu_2+c \theta. \end{equation}

Furthermore, for $x \in \{1,\ldots,N\}$, (2) yields

\begin{align*} & v_{n-1}(x-1)-v_{n-1}(x)\\ & \quad = v_{n-1}^{\min\{x-1,M\}}(x-1)-v_{n-1}^{\min\{x,M\}}(x)\\ & \quad =(M-\min\{x-1,M\})r_s \mu_s +\min\{x-1,M\}r_2\mu_2 -(x-1-\min\{x-1,M\})c\theta\\ & \qquad +(N-x+1)\mu_1 v_{n-2}(x) +[(x-1-\min\{x-1,M\})\theta+\min\{x-1,M\}\mu_2] v_{n-2}(x-2)\\ & \qquad +[1-(N-x+1)\mu_1-(x-1-\min\{x-1,M\})\theta-\min\{x-1,M\}\mu_2]v_{n-2}(x-1)\\ & \qquad -(M-\min\{x,M\})r_s \mu_s -\min\{x,M\}r_2\mu_2 +(x-\min\{x,M\})c\theta\\ & \qquad -(N-x)\mu_1 v_{n-2}(x+1)-[(x-\min\{x,M\})\theta+\min\{x,M\}\mu_2] v_{n-2}(x-1)\\ & \qquad -[1-(N-x)\mu_1-(x-\min\{x,M\})\theta-\min\{x,M\}\mu_2]v_{n-2}(x)\\ & \quad =c\theta+(r_s\mu_s-r_2\mu_2-c\theta)\times {1}_{\{x\leq M\}}+(N-x)\mu_1[v_{n-2}(x)-v_{n-2}(x+1)]\\ & \qquad +[(x-1-\min\{x-1,M\})\theta+\min\{x-1,M\}\mu_2]\times[v_{n-2}(x-2)-v_{n-2}(x-1)]\\ & \qquad +[1-(N-x+1)\mu_1-(x-\min\{x,M\})\theta-\min\{x,M\}\mu_2]\times[v_{n-2}(x-1)-v_{n-2}(x)], \end{align*}

where ${1}_{\{x\leq M\}}$ is an indicator function defined as

$${1}_{\{x\leq M\}}= \left\{\begin{array}{ll} 1 & \text{when}\ x\leq M, \\ 0 & \text{when}\ x \gt M. \end{array}\right.$$

Since $q =N\mu _1+N\theta +M\mu _2$ and we assumed, without loss of generality, that $q=1$, we have that $1-(N-x+1)\mu _1-(x-\min \{x,M\})\theta -\min \{x,M\}\mu _2$ is positive for $x \in \{1,\ldots, N\}$. Therefore, for $x \in \{1,\dots,M\}$, (9) yields

\begin{align*} (\theta -\mu_2)[v_{n-1}(x-1)-v_{n-1}(x)] & \leq (\theta -\mu_2)(r_s \mu_s -r_2 \mu_2)+(N-x)\mu_1\times({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & \quad +(x-1)\mu_2\times({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & \quad +[1-(N-x+1)\mu_1-x\mu_2]\times({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & =(\theta -\mu_2)(r_s \mu_s -r_2 \mu_2)+(1-\mu_1-\mu_2)({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & =(1-\mu_1-\mu_2)c\theta - (1-\mu_1-\theta)(r_s \mu_s -r_2 \mu_2), \end{align*}

and for $x \in \{M+1,\dots,N\}$, we have

\begin{align*} (\theta -\mu_2)[v_{n-1}(x-1)-v_{n-1}(x)] & \leq (\theta -\mu_2)c\theta+(N-x)\mu_1\times({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & \quad +[(x-1-M)\theta+M\mu_2]\times({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & \quad +[1-(N-x+1)\mu_1-(x-M)\theta-M\mu_2]\times({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & = (\theta -\mu_2)c\theta+(1-\mu_1-\theta)({-}r_s \mu_s +r_2 \mu_2+c \theta)\\ & =(1-\mu_1-\mu_2)c\theta - (1-\mu_1-\theta)(r_s \mu_s -r_2 \mu_2). \end{align*}

Now for all $x=1,\ldots,N$ and $a_{x}^{1} \lt a_{x}^{2} \in A_s$, Eq. (3) yields that

(10) \begin{align} \Delta_n^{a_{x}^{1},a_{x}^{2}}(x) & \leq (a_{x}^{2}-a_{x}^{1})[r_s \mu_s -r_2 \mu_2-c \theta +(1-\mu_1-\mu_2)c\theta -(1-\mu_1-\theta)(r_s \mu_s -r_2 \mu_2)]\nonumber\\ & =(a_{x}^{2}-a_{x}^{1})[-(\mu_1+\mu_2)c\theta+(\theta+\mu_1)(r_s \mu_s -r_2 \mu_2)]\leq 0 \end{align}

as long as $c\geq c_0$.

Equation (10) shows that $\Delta _n^{a_{x}^{1},a_{x}^{2}}(x) \leq 0$ for all $n \geq 1, x \in \{1,\ldots,N\}$, and $a_{x}^{1} \lt a_{x}^{2} \in A_{x}$ when $c \geq c_0$. Therefore, we have $v_n(x)=v_n^{\min \{x,M\}}(x)$ for all $x \in \{0,\ldots,N\}$. The remaining proof of part (ii) regarding the $\epsilon$-optimality of the policy generated from value iteration is identical to the corresponding arguments in part (i).

The threshold $c_0$ increases in $r_s$, $\mu _s$ and decreases in $r_2$, $\mu _2$. That is to say, the threshold on the abandonment cost where the supervisors switch from focusing on their own responsibilities to focusing on the customers increases with the rewards and processing rate of the supervisor on their own responsibilities, and decreases when the rewards or processing rate of the supervisors on the customers increase. This is because larger $r_s$, $\mu _s$ and smaller $r_2$, $\mu _2$ all imply relatively greater rewards when the supervisors are working on their own responsibilities. The fact that $c_0$ does not depend on $M,N$ reflects the linearity of the rewards and lack of switching times and costs, as well as the fact that each supervisor's choices on whether to work with a subordinate or not has limited immediate impact on other supervisors and subordinates.

Moreover, when $r_s\mu _s \gt r_2\mu _2$, i.e., $\pi ^{\mathcal {C}}$ is ineffective from the perspective of immediate revenue, then $c_0$ decreases when $\theta$ increases. This means that when the supervisors earn greater rewards per unit time working on their own responsibilities, then as the abandonment rate $\theta$ increases, the supervisors switch from prioritizing their own responsibilities to prioritizing customers earlier (for lower abandonment costs). The condition $\mu _2 \gt \theta$ determines whether the rate of supervisors finishing the second-stage is larger than the abandonment rate. Therefore, if $\mu _2 \gt \theta$, then $\pi ^{\mathcal {C}}$ is effective in reducing future abandonments. If $\mu _2 \lt \theta$, then $c_0$ decreases in $\mu _1$. In this case, $\pi ^{\mathcal {C}}$ is ineffective in both increasing immediate revenue and reducing future abandonments, and if $\pi ^{\mathcal {C}}$ is optimal for a particular $\mu _1$, then Policy $\pi ^{\mathcal {C}}$ will remain optimal for a larger $\mu _1$. However, if $\mu _2 \gt \theta$, then $c_0$ increases in $\mu _1$. In this case, $\pi ^{\mathcal {C}}$ is of mixed effectiveness in improving immediate revenue and reducing future abandonments, and if $\pi ^{\mathcal {S}}$ is optimal for a particular $\mu _1$, then Policy $\pi ^{\mathcal {S}}$ will remain optimal for a larger $\mu _1$.

When $r_s\mu _s \lt r_2\mu _2$, i.e., $c_0$ is negative, then $c_0$ increases in $\theta$. When $c_0$ is negative, $\pi ^{\mathcal {C}}$ is always optimal if there is a cost when a customer leaves the system without the second-stage service. However, when there is a reward for each customer leaving the system with the completion of the first-stage service only, an increase in the abandonment rate can lead supervisors to switch to serving the customers earlier (for lower abandonment rewards). If $\mu _2 \gt \theta$, then $c_0$ decreases in $\mu _1$ and if $\mu _2 \lt \theta$, then $c_0$ increases in $\mu _1$. This is because if $r_s\mu _s \lt r_2\mu _2$ and $\mu _2 \gt \theta$, then Policy $\pi ^{\mathcal {C}}$ is effective in both increasing immediate revenues and reducing future abandonments, which leads to the conclusion that when Policy $\pi ^{\mathcal {C}}$ is optimal for a specific $\mu _1$, it will remain optimal for larger $\mu _1$. Conversely, if $r_s\mu _s \lt r_2\mu _2$ and $\mu _2 \lt \theta$, then Policy $\pi ^{\mathcal {C}}$ is effective in increasing immediate revenues but ineffective in reducing future abandonments. In this case, if $\pi ^{\mathcal {S}}$ is optimal for a particular $\mu _1$, then Policy $\pi ^{\mathcal {S}}$ will remain optimal for a larger $\mu _1$.

The next corollary specifies the optimal policy when the abandonment rate $\theta$ is small or large.

Proof. Theorem 1 introduces optimal policies with a threshold $c_0$ on $c$. We have

$$c_0=\frac{(r_s \mu_s -r_2 \mu_2)(\theta+\mu_1)}{(\mu_1+\mu_2)\theta}\nonumber=\frac{r_s \mu_s -r_2 \mu_2}{\mu_1+\mu_2}+\frac{(r_s \mu_s -r_2 \mu_2)\mu_1}{(\mu_1+\mu_2)\theta},$$

which leads to:

$$\lim_{\theta \rightarrow 0} c_0 = \left\{\begin{array}{ll} +\infty & \text{when}\ r_s \mu_s -r_2 \mu_2 \gt 0, \\ 0 & \text{when}\ r_s \mu_s -r_2 \mu_2 =0,\\ -\infty & \text{when}\ r_s \mu_s -r_2 \mu_2 \lt 0, \end{array}\right.\quad \lim_{\theta \rightarrow \infty} c_0 =\frac{r_s \mu_s -r_2 \mu_2}{\mu_1+\mu_2}.$$

Corollary 5 indicates that when the abandonment rate approaches 0, the optimal policy (whether a supervisor prioritizes her own responsibilities or serving customers) will maximize the immediate reward associated with the action. On the other hand, when the abandonment rate approaches infinity, the optimal policy still depends on how the abandonment cost $c$ compares with a threshold. To better understand the value of the threshold, consider the case where $N \gt M =1$ as an example. When $\pi ^{\mathcal {S}}$ is adopted, in the limit all customers will abandon and the long-run average reward of the system approaches $r_s\mu _s-Nc\mu _1$. When $\pi ^{\mathcal {C}}$ is adopted, in the limit the system behaves as a birth-death process with states 0, 1, birth rate $N\mu _1$, death rate $\mu _2$, and the long-run average reward approaches ${(\mu _2r_s\mu _s+N\mu _1[r_2\mu _2-(N-1)c\mu _1])}/{(\mu _2+N\mu _1)}$. The comparison of the long-run average rewards of the two systems leads to a threshold of ${(r_s \mu _s -r_2 \mu _2)}/{(\mu _1+\mu _2)}$ for the parameter $c$.

The next proposition shows the closed-form expressions of the gains for policies $\pi ^{\mathcal {S}}$ and $\pi ^{\mathcal {C}}$.

Proposition 6. For $1 \leq M \leq N$, the gains of $\pi ^{\mathcal {S}}$ and $\pi ^{\mathcal {C}}$ are

(11) \begin{equation} g_{N,M}^{\pi^{\mathcal{S}}} = \frac{\begin{array}{c} \sum_{j=0}^{N-1}{N \choose j}\theta^{N-j}\mu_1^j[\mu_2+(N-1)\theta](Mr_s \mu_s-jc\theta)\\ +N\theta\mu_1^N[(M-1)r_s \mu_s+r_2 \mu_2-(N-1)c\theta] \end{array}}{(\theta+\mu_1)^N [{\mu_2+(N-1)\theta}]+\mu_1^N ({\theta -\mu_2})}\end{equation}

and

(12) \begin{align} g_{N,M}^{\pi^{\mathcal{C}}} & = \frac{1}{\sum_{k=0}^M {N \choose k}(\frac{\mu_1}{\mu_2})^k+\sum_{k=M+1}^{N} \frac{\prod_{i=1}^k[(N+1-i)\mu_1]}{M!\mu_2^M(\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta])}}\nonumber\\ & \quad \times \left\{ \sum_{k=0}^M {N \choose k}\left(\frac{\mu_1}{\mu_2}\right)^k[(M-k)r_s\mu_s+kr_2\mu_2]\right.\nonumber\\ & \quad \left.+\sum_{k=M+1}^{N} \frac{\prod_{i=1}^k[(N+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]} [Mr_2\mu_2-(k-M)c\theta]\right\}, \end{align}

respectively (with the convention that the summation over an empty set is 0).

Proof. The long-run average rewards $g_{N,M}^{\pi ^{\mathcal {C}}}$ and $g_{N,M}^{\pi ^{\mathcal {S}}}$ can be computed using the birth-death structure of the underlying Markov chains under $\pi ^{\mathcal {C}}$ and $\pi ^{\mathcal {S}}$, respectively. Specifically, the closed-form expression of the gain (long-run average reward) for any specific policy can be uniquely determined by

(13) \begin{equation} g_{N,M}^\pi = \sum_{x=0}^N \eta_x^\pi r(x,d_{\pi}({x}))q, \end{equation}

where $\eta _x^\pi$ is the limiting probability of $\{\mathcal {X}_\pi (t)\}$ being in state $x$ under policy $\pi$.

Let $\pmb {\eta }^\pi$ denote the limiting probability vector under policy $\pi$. The limiting probabilities can be obtained by solving the set of equations $\eta _x^\pi [\sum _{k \neq x} q_\pi (x,k)] =\sum _{k \neq x} \eta _k^\pi q_\pi (k,x)$ for all $x \in X$, along with the equation $\sum _{k=0}^N \eta _k^\pi =1$.

Therefore, for policy $\pi ^{\mathcal {S}}$, we have

(14) \begin{equation} \eta^{\pi^{\mathcal{S}}}_x = \left\{\begin{array}{ll} \dfrac{{N \choose x}\theta^{N-x}\mu_1^x[\mu_2+(N-1)\theta]}{\sum_{j=0}^{N-1}{N \choose j}\theta^{N-j}\mu_1^j[\mu_2+(N-1)\theta]+N\theta\mu_1^N} & \text{for}\ x=0,\ldots,N-1, \\ \dfrac{N\theta\mu_1^N}{\sum_{j=0}^{N-1}{N \choose j}\theta^{N-j}\mu_1^j[\mu_2+(N-1)\theta]+N\theta\mu_1^N} & \text{for}\ x=N, \end{array}\right.\end{equation}

and for policy $\pi ^{\mathcal {C}}$, we have

(15) \begin{equation} \eta^{\pi^{\mathcal{C}}}_x = \left\{\begin{array}{ll} \dfrac{{N \choose x}(\dfrac{\mu_1}{\mu_2})^x}{\sum_{k=0}^M {N \choose k}(\dfrac{\mu_1}{\mu_2})^k+\sum_{k=M+1}^{N} \dfrac{\prod_{i=1}^k[(N+1-i)\mu_1]}{M!\mu_2^M(\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta])}} & \text{for}\ x=0,\ldots,M, \\ \dfrac{\dfrac{\prod_{i=1}^x[(N+1-i)\mu_1]}{M!\mu_2^M(\Pi_{l=M+1}^x[M\mu_2+(l-M)\theta])}}{\sum_{k=0}^M {N \choose k}(\dfrac{\mu_1}{\mu_2})^k+\sum_{k=M+1}^{N} \dfrac{\prod_{i=1}^k[(N+1-i)\mu_1]}{M!\mu_2^M(\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta])}} & \text{for}\ x=M+1,\ldots,N. \end{array}\right.\end{equation}

By plugging Eqs. (14) ((15)) and the corresponding rewards $r(x,d_{\pi ^{\mathcal {S}}}({x}))$ ($r(x,d_{\pi ^{\mathcal {C}}}({x}))$) into (13), we can obtain the gains of $\pi ^{\mathcal {S}}$ ($\pi ^{\mathcal {C}}$).

Until now, we have assumed that customers will not abandon while they are receiving (first- or second-stage) service. This is motivated by service applications where it is unlikely that customers will abandon when in service. However, there are situations where abandonments may occur during the service (e.g., in healthcare applications).

The next corollary extends Theorem 1 to the case where abandonments can also occur during the first- and second-stage service. In particular, the corollary show that the structure of the optimal threshold policy remains the same when abandonments can also occur during service. Let $\theta _1$ ($\theta _2$) denote the abandonment rate during the first-stage (second-stage) service and $c_1$ ($c_2$) denote the corresponding abandonment cost. Note that $\theta _1$, $\theta _2$ or $c_1$, $c_2$ do not necessarily equal $\theta$ or $c$. Then, one of $\pi ^{\mathcal {C}}$ and $\pi ^{\mathcal {S}}$ is always optimal, but the threshold on the value of $c$ is different.

The proof of Corollary 7 follows similar techniques as the proof of Theorem 1 by setting

$$v_0(x)=(N-x-1)\times \frac{r_s\mu_s-r_2\mu_2+c_2\theta_2-c_1\theta_1}{\mu_1+\mu_2+\theta_2}.$$

Alternatively, the new threshold can be obtained in an intuitive way as follows. Note that when abandonments can also occur during the first- and second-stage service, the birth rates in Figure 1 remain the same, while the death rate in state $x \in \{1,\ldots,N\}$ is now $(x-a_x)\theta + a_x(\mu _2+\theta _2)$ due to the abandonments that may take place during the second-stage service. Similarly, the immediate reward of the second-stage service is now $r_2 \times {\mu _2}/{(\mu _2+\theta _2)}-c_2\times {\theta _2}/{(\mu _2+\theta _2)}$. Moreover, for state $x \in \{1,\ldots,N\}$, the costs from the first-stage abandonments are $(N-x)c_1\theta _1$ per unit time. That is to say, the immediate reward $r(x,a_x)$ of choosing action $a_x$ in state $x$ now is

\begin{align*} r(x,a_x) & = \frac{(M-a_x) r_s \mu_s+a_x (r_2 \times \frac{\mu_2}{\mu_2+\theta_2}-c_2\times \frac{\theta_2}{\mu_2+\theta_2})(\mu_2+\theta_2) -(x-a_x)c\theta - (N-x)c_1\theta_1}{q} \\ & =\frac{(M-a_x) r_s \mu_s+a_x (\frac{r_2\mu_2}{\mu_2+\theta_2} + \frac{c_1\theta_1}{\mu_2+\theta_2}-\frac{c_2\theta_2}{\mu_2+\theta_2} )(\mu_2+\theta_2)-(x-a_x)(c-\frac{c_1\theta_1}{\theta})\theta - Nc_1\theta_1}{q}. \end{align*}

By ignoring the ${Nc_1\theta _1}/{q}$ term as it is constant in $x$, replacing the $r_2$ term in $c_0$ by ${r_2\mu _2}/{(\mu _2+\theta _2)} + {c_1\theta _1}/{(\mu _2+\theta _2)}-{c_2\theta _2}/{(\mu _2+\theta _2)}$, replacing the $c$ term by $c-{c_1\theta _1}/{\theta }$, replacing the $\mu _2$ term by $\mu _2+\theta _2$, and replacing the $(\mu _1+\mu _2)$ term by $(\mu _1+\mu _2+\theta _2)$ in (8) and (10), the structure of the optimal policy remains unchanged, and the new threshold $c_0^{\prime }$ should satisfy

$$c_0^\prime = \frac{c_1\theta_1(\mu_2+\theta_2-\theta)+(r_s \mu_s -r_2 \mu_2+c_2\theta_2)(\theta+\mu_1)}{(\mu_1+\mu_2+\theta_2)\theta},$$

as in Corollary 7. We note that $c_0^\prime$ increases in $c_2$. This is because when the second-stage abandonment cost increases, the actual reward of a supervisor serving a customer ($r_2 \times {\mu _2}/{(\mu _2+\theta _2)}-c_2\times {\theta _2}/{(\mu _2+\theta _2)}$) decreases. Therefore, the supervisors will only switch to serve the customers for larger abandonment costs while they are waiting for the second-stage service.

Moreover, $c_0^\prime$ is constant in $c_1$ or $\theta _1$ when $\mu _2+\theta _2 = \theta$; $c_0^\prime$ increases in $c_1$ or $\theta _1$ when $\mu _2+\theta _2 \gt \theta$; and $c_0^\prime$ decreases in $c_1$ or $\theta _1$ when $\mu _2+\theta _2 \lt \theta$. This is because when $\mu _2+\theta _2 =\theta$, the death rate in state $x \in \{1,\ldots,N\}$ is $a_x(\mu _2+\theta _2)+(x-a_x)\theta =x\theta$. Thus, the death rate in state $x$ is the same regardless of the chosen action $a_x$, which leads to the threshold $c_0^\prime$ remaining the same. However, when $\mu _2+\theta _2 \gt \theta$, larger $a_x$ results in higher death rates. Since the $c_1$, $\theta _1$ terms in the immediate reward $r(x,a_x)$ equal $-(N-x)c_1\theta _1$, the effects of $c_1$, $\theta _1$ are less for smaller $x$, leading to an increment in the threshold $c_0^\prime$. On the contrary, when $\mu _2+\theta _2 \lt \theta$, larger $a_x$ results in lower death rates, and hence, the supervisors will switch to serve the customers for smaller abandonment costs $c$ while they are waiting for the second-stage service.

Meanwhile, we note that $c_0^\prime$ increases in $\theta _2$ when $c_1\theta _1+c_2(\mu _1+\mu _2) \gt r_s\mu _s-r_2\mu _2$; decreases in $\theta _2$ when $c_1\theta _1+c_2(\mu _1+\mu _2) \lt r_s\mu _s-r_2\mu _2$; and is constant in $\theta _2$ when $c_1\theta _1+c_2(\mu _1+\mu _2) = r_s\mu _s-r_2\mu _2$. Thus, when the abandonment costs $c_1, c_2$ and rate $\theta _1$ during service are large (small) relative to the benefit $r_s\mu _s-r_2\mu _2$ of supervisors focusing on their own responsibilities, the supervisors will switch later (earlier) from their own responsibilities to serving the customers as the abandonment rate $\theta _2$ increases.

Note that when $c_1=c_2=c$ and $\theta _1=\theta _2=\theta$, the immediate reward $r(x,a_x)$ of choosing action $a_x$ in state $x$ is

(16) \begin{align} r(x,a_x) & = \frac{(M-a_x) r_s \mu_s+a_x (r_2 \times \frac{\mu_2}{\mu_2+\theta}-c\times \frac{\theta}{\mu_2+\theta})(\mu_2+\theta) -(x-a_x)c\theta - (N-x)c\theta}{q} \nonumber\\ & =\frac{M r_s \mu_s+a_x{(r_2\mu_2-r_s \mu_s)}- Nc\theta}{q}. \end{align}

Since ${M r_s \mu _s}/{q}$ and ${Nc\theta }/{q}$ in (16) are constant in $x$, the optimal policy depends solely on the comparison of $r_2\mu _2$ and $r_s\mu _s$ in this case.

Note that in this case, $\pi ^{\mathcal {S}}$ is optimal when $r_s \mu _s \geq r_2 \mu _2 +c\theta$ and $\pi ^{\mathcal {C}}$ is optimal otherwise. This is because the immediate reward $r(x,a_x)$ of choosing action $a_x$ in state $x$ now is

\begin{align*} r(x,a_x) & = \frac{(M-a_x) r_s \mu_s+a_x (r_2 \mu_2+c\theta) - Nc\theta}{q} \\ & = \frac{Mr_s \mu_s+a_x{(r_2\mu_2+{c\theta}-r_s\mu_s)}- Nc\theta}{q}, \end{align*}

which does not depend on $x$. Therefore, the optimal policy depends on the comparison of $r_2\mu _2+{c\theta }$ and $r_s\mu _s$ in this case.

Note that in this case, the immediate reward $r(x,a_x)$ of choosing action $a_x$ in state $x$ is

\begin{align*} r(x,a_x) & = \frac{(M-a_x) r_s \mu_s+a_x (\frac{r_2 \mu_2}{\mu_2+\theta}-\frac{c\theta}{\mu_2+\theta})(\mu_2+\theta)-(x-a_x)c\theta }{q}\\ & = \frac{M r_s \mu_s+a_x ({r_2 \mu_2}-r_s \mu_s)-xc\theta }{q}, \end{align*}

which depends on the state $x$. However, when $r_s\mu _s-r_2\mu _2 \gt 0$, the threshold in Remark 10 satisfies ${[(r_s \mu _s - r_2 \mu _2)(\theta +\mu _1)]}/{\theta \mu _2} \gt c_0={[(r_s \mu _s -r_2 \mu _2)(\theta +\mu _1)]}/{(\mu _1+\mu _2)\theta } \gt 0$. This implies that when the supervisors earn greater rewards per unit time working on their own responsibilities, if abandonments can also occur during the second-stage service and $c_2=c \gt 0$, the expected rewards for supervisors serving the customers decrease. Therefore, the supervisors will switch to serve the customers for larger abandonment costs. However, when $r_s\mu _s-r_2\mu _2 \lt 0$, we have ${[(r_s \mu _s - r_2 \mu _2)(\theta +\mu _1)]}/{\theta \mu _2} \lt c_0 \lt 0$. That is to say, when the supervisors earn greater rewards per unit time serving the customers, if abandonments can also occur during the second-stage service and there is a reward for customers leaving the system without second-stage service (i.e., $-c$ as $c \lt 0$), there are added benefits for supervisors to serve the customers and hence they will switch to serving the customers earlier.

4. Benefits of pooling the subordinates of several supervisors

In this section, we investigate the effects of pooling supervisors (and their subordinates) on the system performance. When there are multiple supervisors, each of whom has her own subordinates, a natural question arises: should each supervisor work with her subordinates only, or should all supervisors work with all the subordinates? Consider our example of government services in Section 1. One possibility is that the waiting people form several queues and each official is responsible for one queue in addition to her other responsibilities. Alternatively, all officials can be jointly responsible for serving all the waiting people (in addition to their own responsibilities).

In particular, we consider two cases. In case 1, there are $KM$ subordinates and $M$ supervisors (a pooled system with $K$ subordinates per supervisor); in case 2, there are $M$ systems each with $K \geq 1$ subordinates and one supervisor ($M$ dedicated systems). Note that since $c_0$ does not depend on the number of subordinates or supervisors, the optimality condition is the same for both cases. We then have Proposition 12. Before we elaborate on Proposition 12, we first prove Lemma 11.

Proof. Observe that

$${KM \choose k }k = {KM \choose k-1}[KM-(k-1)],$$

and

\begin{align*} {K \choose j}j! (M-k+j)& \leq K(K-1)^{j-1}(M-k+j)\\ & =[KM-(k-j) -(k-j)(K-1)](K-1)^{j-1} \end{align*}

for $1 \leq k \leq KM$ and $1 \leq j \leq K$. When $1 \leq k \leq KM$, we have

\begin{align*} & {KM \choose k}k-\sum_{j=1}^{\min\{K,k\}} {KM \choose k-j}{K \choose j}j!(M-k+j)\\ & \quad \geq {KM \choose k-1}[KM-(k-1)]-\sum_{j=1}^{\min\{K,k\}} {KM \choose k-j}[KM-(k-j)-(k-j)(K-1)](K-1)^{j-1}\nonumber\\ & \quad = {KM \choose k-1}(k-1)(K-1)-\sum_{j=2}^{\min\{K,k\}} {KM \choose k-j}[KM-(k-j)-(k-j)(K-1)](K-1)^{j-1}\nonumber\\ & \quad = {KM \choose k-2}[KM-(k-2)](K-1)-\sum_{j=2}^{\min\{K,k\}} {KM \choose k-j}[KM-(k-j)-(k-j)(K-1)](K-1)^{j-1}\\ & \quad \geq{KM \choose k-2}(k-2)(K-1)^2-\sum_{j=3}^{\min\{K,k\}} {KM \choose k-j}[KM-(k-j)-(k-j)(K-1)](K-1)^{j-1}\\ & \qquad \vdots \\ & \quad \geq {KM \choose k-\min\{K,k\}}(k-\min\{K,k\})(K-1)^{\min\{K,k\}} \geq 0. \end{align*}

Proof. (i) Using the closed-form expression of the gain for policy $\pi ^{\mathcal {S}}$ in (11), with some algebra, we have

(17) \begin{align} g_{KM,M}^{\pi^{\mathcal{S}}} - Mg_{K,1}^{\pi^{\mathcal{S}}} &=\frac{1}{\begin{matrix}(\mu_1+\theta)\{[(K-1)\theta+\mu_2](\theta+\mu_1)^K+\mu_1^K(\theta-\mu_2)\} \\ \{[(KM-1)\theta+\mu_2](\theta+\mu_1)^{KM}+\mu_1^{KM}(\theta-\mu_2)\}\end{matrix}}\nonumber\\ & \qquad \times KM\theta[(\mu_1+\theta)(r_s\mu_s-r_2\mu_2) - c\theta(\mu_1+\mu_2)]\nonumber\\ & \qquad \times\{\mu_1^K(\mu_1+\theta)^{KM}[(KM-1)\theta+\mu_2]-\mu_1^{KM}(\mu_1+\theta)^{K}[(K-1)\theta+\mu_2]\}. \end{align}

We now proceed to show that each term here is non-negative. Since $c\leq c_0$, $(\mu _1+\theta )(r_s\mu _s-r_2\mu _2) - c\theta (\mu _1+\mu _2)\geq 0$. Furthermore,

$$\mu_1^K(\mu_1+\theta)^{KM}[(KM-1)\theta+\mu_2]-\mu_1^{KM}(\mu_1+\theta)^{K}[(K-1)\theta+\mu_2] \geq \mu_1^{KM}(\mu_1+\theta)^{K}K(M-1)\theta \geq 0.$$

Similarly,

$$[(K-1)\theta+\mu_2](\theta+\mu_1)^K+\mu_1^K(\theta-\mu_2)\geq K\theta\mu_1^K \gt 0$$

and

$$[(KM-1)\theta+\mu_2](\theta+\mu_1)^{KM}+\mu_1^{KM}(\theta-\mu_2)\geq KM\theta\mu_1^{KM} \gt 0.$$

Thus, $g_{KM,M}^{\pi ^{\mathcal {S}}} - Mg_{K,1}^{\pi ^{\mathcal {S}}}\geq 0$ with the equality holding when $c=c_0$ or $M=1$. We now proceed to obtain the limit of ${(g_{KM,M}^{\pi ^{\mathcal {S}}} - Mg_{K,1}^{\pi ^{\mathcal {S}}})}/{M}$ when $M$ goes to infinity. Based on (17), we have:

\begin{align*} & \lim_{M \rightarrow \infty} \frac{g^{\pi^{\mathcal{S}}}_{KM,M}- Mg^{\pi^{\mathcal{S}}}_{K,1}}{M}\\ & \quad = \frac{K\theta\mu_1^K[(\mu_1+\theta)(r_s\mu_s-r_2\mu_2)- c\theta(\mu_1+\mu_2)]}{(\mu_1+\theta)\{[(K-1)\theta+\mu_2](\theta+\mu_1)^K+\mu_1^K(\theta-\mu_2)\}}\\ & \qquad \times \lim_{M \rightarrow \infty}\frac{(\mu_1+\theta)^{KM}[(KM-1)\theta+\mu_2]-\mu_1^{K(M-1)}(\mu_1+\theta)^{K}[(K-1)\theta+\mu_2]}{[(KM-1)\theta+\mu_2](\theta+\mu_1)^{KM}+\mu_1^{KM}(\theta-\mu_2)}. \end{align*}

The result now follows from the fact that

\begin{align*} \lim_{M \rightarrow \infty} \frac{(\mu_1+\theta)^{KM}[(KM-1)\theta+\mu_2]-\mu_1^{K(M-1)}(\mu_1+\theta)^{K}[(K-1)\theta+\mu_2]}{[(KM-1)\theta+\mu_2](\theta+\mu_1)^{KM}+\mu_1^{KM}(\theta-\mu_2)}=1. \end{align*}

(ii) Using the closed-form expression of the gain for policy $\pi ^{\mathcal {C}}$ in (12), with some algebra, we have

(18) \begin{align} g_{KM,M}^{\pi^{\mathcal{C}}} - Mg_{K,1}^{\pi^{\mathcal{C}}}& =\frac{[c\theta(\mu_1+\mu_2)-(\mu_1+\theta)(r_s\mu_s-r_2\mu_2)]}{(\mu_1+\theta)[1+T(K)]}\nonumber\\ & \quad \times \frac{\Gamma}{\sum_{k=0}^M {KM \choose k}(\frac{\mu_1}{\mu_2})^k+\sum_{k=M+1}^{KM} \frac{\prod_{i=1}^k[(KM+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}}, \end{align}

where

\begin{align*} \Gamma & =M\sum_{k=M+1}^{KM} \frac{\prod_{i=1}^k[(KM+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}+\sum_{k=1}^M {KM \choose k}\left(\frac{\mu_1}{\mu_2}\right)^kk\\ & \quad -\left(\sum_{k=1}^{M} {KM \choose {M-k}}\left(\frac{\mu_1}{\mu_2}\right)^{M-k}k \right)\times T(K). \end{align*}

We now show that the expression (18) is non-negative. Note that the term $c\theta (\mu _1+\mu _2)-(r_s\mu _s-r_2\mu _2)(\mu _1+\theta )\geq 0$ since $c \geq c_0$. We will show that $\Gamma \geq 0$. Define

$$\alpha_k=\sum_{j=\max\{0,k-K\}}^{k-1}\frac{ {KM \choose j}{K \choose k-j}(k-j)!(M-j)}{\mu_2^j \Pi_{l=1}^{{k-j}}[\mu_2+(l-1)\theta]}$$

for $1 \leq k \leq M$, and

$$\beta_{k}=\sum_{j=\max\{0,k-K\}}^{M-1} \frac{{KM \choose j}{K \choose k-j}(k-j)!(M-j)}{\mu_2^j \Pi_{l=1}^{{k-j}}[\mu_2+(l-1)\theta]}$$

for $M+1 \leq k \leq M+K-1$. Note that by expanding $(\sum _{k=1}^{M} {KM \choose {M-k}}(\frac {\mu _1}{\mu _2})^{M-k}k )\times T(K)$ and ordering the terms by ascending exponent of $\mu _1$, we have

\begin{align*} \left(\sum_{k=1}^{M} {KM \choose {M-k}}\left(\frac{\mu_1}{\mu_2}\right)^{M-k}k \right)\times T(K) & =\left(\sum_{j=0}^{M-1} {KM \choose j}\left(\frac{\mu_1}{\mu_2}\right)^j(M-j) \right)\\ & \quad \times \left(\sum_{k=1}^{K} \frac{\prod_{i=1}^k[(K+1-i)\mu_1]}{\Pi_{l=1}^k[\mu_2+(l-1)\theta]}\right)\\ & =\sum_{k=1}^{M+K-1}\mu_1^k \sum_{j=\max\{0,k-K\}}^{\min\{k-1,M-1\}}\frac{ {KM \choose j}{K \choose k-j}(k-j)!(M-j)}{\mu_2^j \Pi_{l=1}^{{k-j}}[\mu_2+(l-1)\theta]}\\ & =\sum_{k=1}^{M}\mu_1^k \alpha_k + \sum_{k=M+1}^{M+K-1}\mu_1^k\beta_k. \end{align*}

By grouping the terms in $\Gamma$ based on the exponent of $\mu _1$, we obtain $\Gamma =\Gamma _1+\Gamma _2+\Gamma _3$, where

\begin{align*} \Gamma_1& =\sum_{k=1}^M\mu_1^k\left[{KM \choose k}\frac{k}{\mu_2^k}-\alpha_k\right],\\ \Gamma_2& =\sum_{k=M+1}^{M+K-1}\mu_1^k\left[\frac{M\prod_{i=1}^k(KM+1-i)}{M!\mu_2^M \Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}-\beta_{k}\right],\\ \Gamma_3& =\sum_{k=M+K}^{KM}\mu_1^k\frac{M\prod_{i=1}^k(KM+1-i)}{M!\mu_2^M \Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]} \end{align*}

(recall that the summation over an empty set equals zero).

Note that $\Gamma _3$ is positive if $M,K \geq 2$ and $\Gamma _3$ is zero if $M =1$ or $K=1$. Moreover, when $M=1$, $\Gamma _1 = 0$ and $\Gamma _2 = \sum _{k=M+1}^{M+N-1} \mu _1^k \times 0 = 0$. Thus, $\Gamma =0$ when $M=1$ and it suffices to show that $\Gamma _1$ and $\Gamma _2$ are non-negative, which we will prove by showing that each term of the respective summation is non-negative. For $\Gamma _1$, when $1 \leq k \leq M$, we have

(19) \begin{align} {KM \choose k}\frac{k}{\mu_2^k}-{\alpha_k} & \geq \frac{1}{\mu_2^k}\left[ {KM \choose k}k-\sum_{j=\max\{0,k-K\}}^{k-1} {KM \choose j}{K \choose k-j}(k-j)!(M-j)\right]\nonumber\\ & =\frac{1}{\mu_2^k}\left[{KM \choose k}k-\sum_{j=1}^{\min\{K,k\}} {KM \choose k-j}{K \choose j}j!(M-k+j)\right] \geq 0, \end{align}

where the last inequality follows from Lemma 11. Similarly, note that

$$\beta_{k} \leq \sum_{j=\max\{0,k-K\}}^{M-1}\frac{ {KM \choose j}{K \choose k-j}(k-j)!(M-j)}{\mu_2^{M} \Pi_{l=1}^{{k-M}}(\mu_2+l\theta)},\quad \text{for}\ M+1 \leq k \leq M+K-1.$$

Therefore, for $M+1 \leq k \leq M+K-1$,

(20) \begin{align} & \frac{M\prod_{i=1}^{k}(KM+1-i)}{M!\mu_2^M \Pi_{l=M+1}^{k}[M\mu_2+(l-M)\theta]}-\beta_{k}\nonumber\\ & \quad\geq \frac{M{KM \choose k}k!}{M!\mu_2^M M^{k-M} \prod_{l=1}^{k-M} (\mu_2+l\theta)}-\frac{\sum_{j=\max\{0,k-K\}}^{M-1}{KM \choose j}{K \choose k-j}(k-j)!(M-j)}{\mu_2^M \prod_{l=1}^{k-M} (\mu_2+l\theta)}\nonumber\\ & \quad=\frac{1}{\mu_2^M\prod_{l=1}^{k-M} (\mu_2+l\theta)}\left[\frac{{KM \choose k}k!}{(M-1)!M^{k-M}}-\sum_{j=\max\{0,k-K\}}^{M-1}{KM \choose j}{K \choose k-j}(k-j)!(M-j)\right]\nonumber\\ & \quad\geq\frac{1}{\mu_2^M \prod_{l=1}^{k-M} (\mu_2+l\theta)} \left[{KM \choose k}k-\sum_{j=\max\{0,k-K\}}^{M-1}{KM \choose j}{K \choose k-j}(k-j)!(M-j)\right]\nonumber\\ & \quad=\frac{1}{\mu_2^M \prod_{l=1}^{k-M} (\mu_2+l\theta)} \left[{KM \choose k}k-\sum_{j=k-M+1}^{\min\{K,k\}}{KM \choose k-j}{K \choose j}j!(M-k+j)\right]\nonumber\\ & \quad\geq\frac{1}{\mu_2^M \prod_{l=1}^{k-M} (\mu_2+l\theta)} \left[{KM \choose k}k-\sum_{j=1}^{\min\{K,k\}}{KM \choose k-j}{K \choose j}j!(M-k+j)\right] \geq 0, \end{align}

where the last inequality follows from Lemma 11. It follows from (19) and (20) that $\Gamma _1$ and $\Gamma _2$ are non-negative, which implies that $\Gamma$ is positive when $M \geq 2$ and $K \geq 2$. When $M \geq 2$ and $K=1$, since $\Gamma _2 = \Gamma _3 = 0$, we have

$$\Gamma =\Gamma_1=\sum_{k=1}^M\mu_1^k\left[{M \choose k}\frac{k}{\mu_2^k}-\alpha_k\right]= \sum_{k=1}^M(\frac{\mu_1}{\mu_2})^k\left[{M \choose k}{k}-{ {M \choose k-1}(M-k+1)}\right] = 0.$$

Therefore, we have $g_{KM,M}^{\pi ^{\mathcal {C}}} - Mg_{K,1}^{\pi ^{\mathcal {C}}} \geq 0$ with equality holding only when $c=c_0$ or $M=1$ or $K=1$ ($\Gamma =0$ in the last two cases).

We now proceed to obtain the lower and upper bounds of ${(g_{KM,M}^{\pi ^{\mathcal {C}}} - Mg_{K,1}^{\pi ^{\mathcal {C}}})}/{M}$ when $M$ goes to infinity. Note that

\begin{align*} & {\frac{\Gamma}{\sum_{k=0}^M {KM \choose k}\left(\frac{\mu_1}{\mu_2}\right)^k+\sum_{k=M+1}^{KM} \frac{\prod_{i=1}^k[(KM+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}}}\times \frac{1}{M}\nonumber\\ & \quad= \frac{\begin{array}{c} M\sum_{k=M+1}^{KM} \dfrac{\prod_{i=1}^k[(KM+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}+\sum_{k=1}^M {KM \choose k}\left(\dfrac{\mu_1}{\mu_2}\right)^k k-T(K) \\ \times \left[\sum_{k=0}^{M-1} {KM \choose k}\left(\dfrac{\mu_1}{\mu_2}\right)^k(M-k)\right]\end{array}}{M\sum_{k=0}^M {KM \choose k}\left(\frac{\mu_1}{\mu_2}\right)^k+M\sum_{k=M+1}^{KM} \frac{\prod_{i=1}^k[(KM+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}}\nonumber\\ & \quad=1-\frac{[1+T(K)]\sum_{k=0}^M {KM \choose k}\left(\frac{\mu_1}{\mu_2}\right)^k\times (M-k)}{M\sum_{k=0}^M {KM \choose k}\left(\frac{\mu_1}{\mu_2}\right)^k+M\sum_{k=M+1}^{KM} \frac{\prod_{i=1}^k[(KM+1-i)\mu_1]}{M!\mu_2^M\Pi_{l=M+1}^k[M\mu_2+(l-M)\theta]}} \lt 1. \nonumber \end{align*}

Since ${g_{KM,M}^{\pi ^{\mathcal {C}}} - Mg_{K,1}^{\pi ^{\mathcal {C}}}} \geq 0$ , it now follows from (18) that for all $M \geq 1$,

$$0\leq \frac{g_{KM,M}^{\pi^{\mathcal{C}}} - Mg_{K,1}^{\pi^{\mathcal{C}}}}{M} \leq \frac{c\theta(\mu_1+\mu_2)-(\mu_1+\theta)(r_s\mu_s-r_2\mu_2)}{(\mu_1+\theta)[1+T(K)]}.$$

Proposition 12 shows that pooling supervisors and their subordinates improves the performance of the system in terms of the long-run average reward. However, the improvement per pooled supervisor is bounded. We utilize numerical examples to illustrate the comparison of dedicated and pooled systems and to quantify the incremental benefit per pooled supervisor of pooling systems as more supervisors are pooled. In our numerical examples, we have: $\mu _1 = 4$, $r_1 = 5$, $\mu _2 = 6$, $r_2 = 8$, $\mu _s = 11$, $r_s= 6$, $\theta = 2$. Consider the dedicated and pooled systems where there are $M$ supervisors, each of whom has $K=4$ subordinates. Note that the threshold $c_0$ is $\frac {27}{5} = 5.4$. Figure 2 shows the value of pooling $M$ supervisors as a function of $M$ when $c =2$ (where $\pi ^{\mathcal {S}}$ is optimal) and when $c =10$ (where $\pi ^{\mathcal {C}}$ is optimal). The incremental value of pooling more than 5 (10) supervisors is small when $c =2$ ($c =10$).

Figure 2. The incremental value of pooling $M$ supervisors as a function of $M$.