Pooling in tandem queueing networks with non-collaborative servers

Abstract

This paper considers pooling several adjacent stations in a tandem network of single-server stations with finite buffers. When stations are pooled, we assume that the tasks at those stations are pooled but the servers are not. More specifically, each server at the pooled station picks a job from the incoming buffer of the pooled station and conducts all tasks required for that job at the pooled station before that job is placed in the outgoing buffer. For such a system, we provide sufficient conditions on the buffer capacities and service times under which pooling increases the system throughput by means of sample-path comparisons. Our numerical results suggest that pooling in a tandem line generally improves the system throughput—substantially in many cases. Finally, our analytical and numerical results suggest that pooling servers in addition to tasks results in even larger throughput when service rates are additive and the two systems have the same total number of storage spaces.

Appendix

In this appendix, we provide proofs of our theoretical results, lemmas that are used in some of our proofs, and other supplementary material. We use Lemmas 1 and 2 to prove Proposition 1. The proof of Lemma 1 is trivial and hence is omitted.

Lemma 1

If $a_i$ and $b_i$ are some real numbers for $i=1,\ldots ,n$, where n is a positive integer, then we have

$$\begin{aligned} \max _{i=1,\ldots ,n}\left\{ a_i\right\} -\max _{i=1,\ldots ,n}\left\{ b_i\right\} \ge \min _{i=1,\ldots ,n}\left\{ a_i-b_i\right\} . \end{aligned}$$

Lemma 2

Let $\{a_i\}_{i=1}^n$ be a sequence of real numbers, where n is a positive integer. Then, for $J\in \{2,\ldots ,n\}$, $k\in \{1,\ldots ,J-1\}$, and $\ell _j\in \{1,\ldots ,n\}$, for all $j\in \{1,\ldots ,k\}$, we have

$$\begin{aligned} \Phi _J^{(n)}\left\{ a_i:i\in \{1,\ldots ,n\}\right\} \le \Phi _{J-k}^{(n-k)}\left\{ a_i:i\in \{1,\ldots ,n\}{\setminus }\{\ell _1,\ldots ,\ell _k\}\right\} . \end{aligned}$$

Proof of Lemma 2

Let $m\le k$ be the number of elements in $\{a_{\ell _1},\ldots ,a_{\ell _k}\}$ that are greater than $\Phi _J^{(n)}\left\{ a_i:i\in \{1,\ldots ,n\}\right\}$. Then,

$$\begin{aligned} \Phi _J^{(n)}\left\{ a_i:i\in \{1,\ldots ,n\}\right\}= & {} \Phi _{J-m}^{(n-k)}\left\{ a_i:i\in \{1,\ldots ,n\}{\setminus }\{\ell _1,\ldots ,\ell _k\}\right\} \\\le & {} \Phi _{J-k}^{(n-k)}\left\{ a_i:i\in \{1,\ldots ,n\}{\setminus }\{\ell _1,\ldots ,\ell _k\}\right\} , \end{aligned}$$

where the equality holds because when m elements that are larger than $\Phi _J^{(n)}\{a_i:i\in \{1,\ldots ,n\}\}$ are taken out from the set, then the Jth largest element becomes the $(J-m)$th largest in the new set. $\square$

Proof of Proposition 1

For $j\in \{1,\ldots ,K-1,M,\ldots ,N\}$ and $i\ge 1$, let $\Delta _j(i)=D_j(i)-D_j^{[K,M]}(i)$. Also, let $\Delta _j(i)=D_M(i)-A_{j-K+1}(i-M+j+1)$ for $j\in \{K,\ldots ,M-1\}$ and $i\ge 1$. For convenience, assume that $\Delta _j(i)=0$ when $j\notin \{1,\ldots ,N\}$ or $i\le 0$. Consider now the following inequalities for $i\ge 1$:

$$\begin{aligned} \Delta _j\left( i\right)\ge & {} \min \left\{ \Delta _{j-1}\left( i\right) ,\Delta _{j}\left( i-1\right) ,\Delta _{j+1}\left( i-b_{j+1}-1\right) \right\} ,\nonumber \\&\forall j\in \{1,\ldots ,K-2,M+1,\ldots ,N\}; \end{aligned}$$

(8)

$$\begin{aligned} \Delta _{K-1}\left( i\right)\ge & {} \min \left\{ \Delta _{K-2}\left( i\right) ,\Delta _{K-1}\left( i-1\right) ,\Delta _{M}\left( i-P^{[K,M]}-M+K-1\right) \right\} ;\quad \end{aligned}$$

(9)

$$\begin{aligned} \Delta _K\left( i\right)\ge & {} \min \left\{ \Delta _{K-1}\left( i\right) ,\Delta _{M}\left( i-M+K-1\right) ,\Delta _{K}\left( i-1\right) \right\} ; \end{aligned}$$

(10)

$$\begin{aligned} \Delta _j\left( i\right)\ge & {} \Delta _{j-1}\left( i\right) , \forall j\in \{K+1,\ldots ,M-1\}; \end{aligned}$$

(11)

$$\begin{aligned} \Delta _M\left( i\right)\ge & {} \min \left\{ \Delta _{M-1}\left( i\right) ,\Delta _{M+1}\left( i-Q^{[K,M]}-1\right) \right\} . \end{aligned}$$

(12)

It is easy to see that the inequalities (8) through (12) imply that $\Delta _j(i) \ge 0$ for all $i\ge 1$ and $j\in \{1,\ldots ,N\}$. Then, it remains to show that the inequalities (8) through (12) are true.

We first provide a recursive formula that the departure times $D_j(i)$ must satisfy. For convenience, we assume that $D_j(i)=X_j(i) =0$ if $j\notin \{1,\ldots ,N\}$ or $i\le 0$. Then, for all $i\ge 1,$ we have

$$\begin{aligned} D_j\left( i\right) =\max \left\{ D_{j-1}\left( i\right) +X_j\left( i\right) ,D_j\left( i-1\right) +X_j\left( i\right) ,D_{j+1}\left( i-b_{j+1}-1\right) \right\} ,\forall j\in \{1,\ldots ,N\}.\nonumber \\ \end{aligned}$$

(13)

Now, using condition (i), Lemma 1, and Eqs. (2) and (13) gives inequality (8). Similarly, using condition (i), Lemma 1, and Eqs. (3) and (13), we obtain

$$\begin{aligned} \Delta _{K-1}\left( i\right)\ge & {} \min \left\{ \Delta _{K-2}\left( i\right) ,\Delta _{K-1}\left( i-1\right) , D_K\left( i-b_K-1\right) \right. \\&\left. -D_M^{[K,M]}\left( i-P^{[K,M]}-M+K-1\right) \right\} . \end{aligned}$$

Then, using Eq. (13) and the condition that $b_j=0$ for $j\in \{K+1,\ldots ,M\}$ iteratively yields $D_K(i-b_K-1)\ge D_M(i-b_K-M+K-1)$ for all $i\ge 1$. The condition that $P^{[K,M]}\ge b_K$ now yields $D_K(i-b_K-1)\ge D_M(i-P^{[K,M]}-M+K-1)$ for all $i\ge 1$, which completes the proof of inequality (9).

Next, we prove inequality (10). Since $A_1(i)\ge A_j(i)$ for all $i\ge 1$ and $j\in \{1,\ldots ,M-K\}$, Eq. (6) gives

$$\begin{aligned} A_1(i+1)= & {} \max \Big \{D_{K-1}^{[K,M]}(i+M-K)+X^{[K,M]}(i+M-K),\\&D_M^{[K,M]}(i-1)+X^{[K,M]}(i+M-K),A_1(i)\Big \}, \end{aligned}$$

for all $i\ge 1$. Then, it is easy to obtain that

$$\begin{aligned} \Delta _K\left( i\right)= & {} \min \Bigg \{ D_M\left( i\right) -D^{[K,M]}_{K-1}\left( i\right) -X^{[K,M]}\left( i\right) ,D_M\left( i\right) -D^{[K,M]}_{M}\left( i-M+K-1\right) \nonumber \\&\qquad -X^{[K,M]}\left( i\right) ,D_M\left( i\right) -A_1\left( i-M+K\right) \Bigg \}. \end{aligned}$$

(14)

It now follows from condition (ii) and the fact that $D_M(i)\ge D_{K-1}(i)+\sum _{j=K}^MX_j(i)$ for all $i\ge 1$ that the first term of the minimum operator in Eq. (14) is greater than or equal to $\Delta _{K-1} (i)$. Similarly, note that $D_M(i)\ge D_{K}(i)+\sum _{j=K+1}^MX_j(i)\ge D_{K}(i-1)+\sum _{j=K}^MX_j(i)$, for all $i\ge 1$, so that condition (ii) implies that the second term of the minimum operator in Eq. (14) is greater than or equal to $D_{K}(i-1)-D_{M}^{[K,M]}( i-M+K-1)$, for all $i\ge 1$. Moreover, using Eq. (13) and the condition that $b_j=0$ for $j\in \{K+1,\ldots ,M\}$ iteratively, one can obtain that $D_{K}(i-1)\ge D_{M}( i-M+K-1)$ and hence that the second term of the minimum operator in Eq. (14) is greater than or equal to $\Delta _{M}(i-M+K-1)$. Noting that $D_M(i)\ge D_M(i-1)$ for all $i\ge 1$ yields inequality (10).

We next prove inequality (11). Using Lemma 2 with $k=j-1$ and Eq. (6), we have

$$\begin{aligned} A_j\left( i+1\right) \le \max \left\{ A_{j-1}\left( i\right) ,\ldots ,A_{M-K}\left( i\right) \right\} =A_{j-1}\left( i\right) , \end{aligned}$$

for $j\in \{2,\ldots ,M-K\}$ and $i\ge 1$. Then, inequality (11) is immediate. Finally, we show that inequality (12) is true. Equation (5) implies that $D_M^{[K,M]}(i)\le \max \{A_{M-K}(i),D_{M+1}^{[K,M]}(i-Q^{[K,M]}-1)\}$, and hence that

$$\begin{aligned} \Delta _M\left( i\right) \ge \min \left\{ D_{M}\left( i\right) -A_{M-K}(i),D_{M}\left( i\right) -D_{M+1}^{[K,M]}\left( i-Q^{[K,M]}-1\right) \right\} ,\qquad \end{aligned}$$

(15)

for all $i\ge 1$. Note that the first term of the minimum operator in inequality (15) is equal to $\Delta _{M-1}(i)$, for all $i\ge 1$. Moreover, using Eq. (13) and the condition that $Q^{[K,M]}\ge b_{M+1}$, we obtain that $D_M(i)\ge D_{M+1}(i-Q^{[K,M]}-1)$, for all $i\ge 1$, which immediately yields that the second term of the minimum operator in inequality (15) is greater than or equal to $\Delta _{M+1}(i-Q^{[K,M]}-1)$ for all $i\ge 1$, and the proof is complete. $\square$

To prove Proposition 2, we need the following lemma, whose proof is immediate.

Lemma 3

Let $\mathcal {Y}=\{\mathbf {Y}(i)\}_{i\ge 1}$ and $\mathcal {Z}=\{\mathbf {Z}(i)\}_{i\ge 1}$ be two stochastic processes. If $\mathcal {Y}\le _{st}\mathcal {Z}$, then $\phi (\mathcal {Y})\le _{st}\phi (\mathcal {Z})$ for every non-decreasing functional $\phi :\mathbb {R}^{\infty }\rightarrow \mathbb {R}^{\infty }$.

Proof of Proposition 2

Let $\phi :\mathbb {R}_+^{\infty }\rightarrow \mathbb {R}_+^{\infty }$ be defined by $\{\mathbf {D}^{[K,M]}(i)\}_{i\ge 1}=\phi (\{\mathbf {X}^{[K,M]}(i)\}_{i\ge 1})$, see Eqs. (2), (3), (5), and (6). Define also $\tilde{\mathbf {X}}^{[K,M]}(i)=\big (\mathbf {X}_{1,K-1},\sum _{k=K}^MX_k(i),\mathbf {X}_{M+1,N}(i)\big )$, for all $i\ge 1$, and $\{\tilde{\mathbf {D}}^{[K,M]}(i)\}_{i\ge 1}=\phi \big (\big \{\tilde{\mathbf {X}}^{[K,M]}(i)\big \}_{i\ge 1}\big )$. Then, Proposition 1 yields that $\{\tilde{\mathbf {D}}^{[K,M]}(i)\}_{i\ge 1}$ $\le \{\mathbf {D}(i)\}_{i\ge 1}$. It is clear that $\phi$ is a non-decreasing functional. Hence, by Lemma 3 and inequality (7), we have $\{\mathbf {D}^{[K,M]}(i)\}_{i\ge 1}\le _{st}\{\tilde{\mathbf {D}}^{[K,M]}(i)\}_{i\ge 1}$, and the result follows. $\square$

We defer the proofs of Proposition 3 and Corollary 1 as they are based on Proposition 7. We need the following lemma to prove Proposition 4.

Lemma 4

If $a_i$ and $b_i$ are some positive real numbers for $i=1,\ldots ,n$, where n is a positive integer, then we have

$$\begin{aligned} \min _{i=1,\ldots ,n}\left\{ \frac{a_i}{b_i}\right\} \le \frac{\sum _{i=1}^na_i}{\sum _{i=1}^nb_i}. \end{aligned}$$

(16)

Proof of Lemma 4

Let $J\in {\{1,\ldots ,n\}}$ be the argument that achieves the minimum in (16), so that $a_Jb_i\le a_ib_J$ for all $i=1,\ldots ,n$. Then, we have

$$\begin{aligned} b_J\sum _{i=1}^na_i-a_J\sum _{i=1}^nb_i=\sum _{i=1}^n(a_i b_J-a_J b_i)\ge 0. \end{aligned}$$

$\square$

Proof of Proposition 4

Let $T_{\infty }^{[K,M]}$ be the throughput of the tandem line where stations K through M are parallel pooled and all buffers in the system are replaced by infinite capacity buffers. Then, due to the monotonicity of the throughput of a tandem line in the buffer sizes (see, for example, page 186 in Buzacott and Shanthikumar [8]), we have $T^{[K,M]}\le T_{\infty }^{[K,M]}$. We will next show that $T_{\infty }^{[K,M]}\le T^{[1,N]}$, which will complete the proof.

Under the assumptions on service times and Assumptions 1 and 2, $T_{\infty }^{[K,M]}$ exists and satisfies

$$\begin{aligned} T_{\infty }^{[K,M]}= & {} \min \left\{ \min _{j\in \{1,\ldots ,K-1,M+1,\ldots ,N\}}\left\{ \frac{1}{E[X_j(1)]}\right\} ,\sum \limits _{\ell =K}^{M} \theta _\ell /\sum \limits _{j=K}^M\theta _jE[X_j(1)]\right\} \\\le & {} \frac{\sum _{\ell =1}^{N} \theta _\ell }{\sum _{j=1}^N\theta _jE[X_j(1)]}=T^{[1,N]}, \end{aligned}$$

where the inequality follows from Lemma 4. $\square$

Proof of Proposition 5

Because the service times are i.i.d. and the buffers are infinite, the throughput of the original line is given by $T=1/\mathrm {E}[X_J(1)]$. Similarly, the throughput of the pooled line where stations K through M are pooled will be determined by the bottleneck station, i.e.,

$$\begin{aligned} T^{[K,M]}=\min \Bigg \{\min _{j\in \{1,\ldots ,N\}{\setminus }\{K,\ldots ,M\}}\left\{ \frac{1}{\mathrm {E}[X_j(1)]}\right\} , \frac{M-K+1}{\sum _{j=K}^M \mathrm {E}\left[ X_{j}(1)\right] }\Bigg \}, \end{aligned}$$

under Assumption 1 and the condition that servers K through M are identical. Hence, if $K\le J\le M$ and $\mathrm {E}[X_j(1)]/\mathrm {E}[X_J(1)]\rightarrow 0$ for all $j\in \{1,\ldots ,N\}{\setminus }\{J\}$,

$$\begin{aligned} \frac{T^{[K,M]}}{T}\rightarrow M-K+1. \end{aligned}$$

$\square$

Example 2

Suppose that we pool stations 1 and 2 in a tandem line of three stations where $b_1=\infty$ and $b_2=b_3=0$. Suppose also that the service times at the pooled station satisfy $X_{\ell }^{[1,2]}(i)=X_1(i)+X_2(i)$ for $\ell =1,2$ and $i\ge 1$, $P^{[1,2]}=\infty$, and $Q^{[1,2]}=0$. For the original line, consider a sample path where $(X_0(1),X_0(2))=(0,1)$, $(X_1(1),X_1(2))=(1,1)$, $(X_2(1),X_2(2))=(3,1)$, and $(X_3(1),X_3(2))=(1,3)$ minutes. For the pooled line, suppose that $(X_0^{[1,2]}(1),X_0^{[1,2]}(2))=(0,1)$ and $(X_3^{[1,2]}(1),X_3^{[1,2]}(2))=(1,2)$ minutes. Note that this example satisfies all conditions of Proposition 1 and the condition that $X_0^{[K,M]}(i)\le X_0(i)$ for all $i\ge 1$, and hence $D_3^{[1,2]}(i)\le D_3(i)$ for $i=1,2$. However, in the pooled line, the first job to arrive at the system departs as the second job from the system. This results in a longer sojourn time for this job by pooling. In particular, the sojourn time of the first job arriving to the original line is five minutes, whereas its sojourn time in the pooled line is six minutes.

Proof of Proposition 6

For all $t\ge 0$, let $B_j^{[K,M]}(t)$ be the total number of departures from station $j\in \{1,\ldots ,K-1,M,\ldots ,N\}$ by time t in the pooled system and $B_j(t)$ be the total number of departures from station $j\in \{1,\ldots ,N\}$ by time t in the unpooled system. Let also $B_0^{[K,M]}(t)=B_0(t)$ be the total number of arrivals by time $t\ge 0$ and $D_0^{[K,M]}(i)=D_0(i)$ be the arrival time of job $i\ge 1$ at each system. For notational convenience, assume that $D_K^{[K,M]}(i)=D_M^{[K,M]}(i)$ and $B_K^{[K,M]} (t) = B_M^{[K,M]} (t)$, for all $i\ge 1$ and $t \ge 0$. Then, for all $t\ge 0$, we have

$$\begin{aligned} H(t)= & {} \sum \limits _{j=0}^{N-1}h_{j+1}\sum \limits _{i=1}^{B_j(t)} \bigg (\min \{t,D_{j+1}(i)\}-D_j(i)\bigg ) \text{ and }\\ H^{[K,M]}(t)= & {} \sum \limits _{j\in \{0,\ldots ,K-2\}\bigcup \{M,\ldots ,N-1\}}h_{j+1} \sum \limits _{i=1}^{B_j^{[K,M]}(t)}\left( \min \{t,D_{j+1}^{[K,M]}(i)\}-D_j^{[K,M]}(i)\right) \\&+\, h^{[K,M]}\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}\left( \min \{t,D_K^{[K,M]}(i)\}-D_{K-1}^{[K,M]}(i)\right) . \end{aligned}$$

Consequently, for all $t\ge 0$, we obtain

$$\begin{aligned}&H(t)-H^{[K,M]}(t)\nonumber \\&\quad =\sum \limits _{j=K-1}^{N-1}h_{j+1}\sum \limits _{i=1}^{B_j(t)} \bigg (\min \{t,D_{j+1}(i)\}-D_j(i)\bigg )\nonumber \\&\qquad -\sum \limits _{j=M}^{N-1}h_{j+1}\sum \limits _{i=1}^{B_j^{[K,M]}(t)} \left( \min \{t,D_{j+1}^{[K,M]}(i)\}-D_j^{[K,M]}(i)\right) \nonumber \\&\qquad -\, h^{[K,M]}\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}\left( \min \{t,D_K^{[K,M]}(i)\}-D_{K-1}^{[K,M]}(i)\right) \nonumber \\&\qquad +\sum \limits _{j=0}^{K-2}h_{j+1}\Bigg (\sum \limits _{i=1}^{B_j(t)} \bigg (\min \{t,D_{j+1}(i)\}-D_j(i)\bigg )\nonumber \\&\qquad -\sum \limits _{i=1}^{B_j^{[K,M]}(t)} \bigg (\min \{t,D^{[K,M]}_{j+1}(i)\}-D^{[K,M]}_j(i)\bigg )\Bigg ). \end{aligned}$$

(17)

We start by dealing with the sum of the first three terms of Eq. (17). First, note that for all $\ell ,m\in \{0,\ldots ,N-1\}$, $\ell \le m$, and $t\ge 0$, we have

$$\begin{aligned}&\sum \limits _{j=l}^{m}\sum \limits _{i=1}^{B_j(t)}\bigg (\min \{t,D_{j+1}(i)\}-D_j(i)\bigg )\nonumber \\&\quad =\sum \limits _{j=\ell }^{m}\sum \limits _{i=1}^{B_{j+1}(t)}D_{j+1}(i)+\sum \limits _{j=\ell }^{m}\sum \limits _{i=B_{j+1}(t)+1}^{B_j(t)}t -\sum \limits _{j=\ell }^{m}\sum \limits _{i=1}^{B_j(t)}D_j(i)\nonumber \\&\quad =\sum \limits _{i=1}^{B_{m+1}(t)}D_{m+1}(i)+\sum \limits _{i=B_{m+1}(t)+1}^{B_{\ell }(t)}t-\sum \limits _{i=1}^{B_{\ell }(t)}D_{\ell }(i)\nonumber \\&\quad =\sum \limits _{i=1}^{B_{\ell }(t)}\bigg (\min \{t,D_{m+1}(i)\}-D_{\ell }(i)\bigg ). \end{aligned}$$

(18)

Similarly, for all $\ell ,m\in \{0,\ldots ,K-1\}\cup \{M,\ldots ,N-1\}$, $\ell \le m$, and $t\ge 0$, we can obtain

$$\begin{aligned}&\sum \limits _{\begin{array}{c} j\in \{\ell ,\ldots ,m\}{\setminus }\\ \{K,\ldots ,M-1\} \end{array}}\sum \limits _{i=1}^{B_j^{[K,M]}(t)}\left( \min \{t,D_{j+1}^{[K,M]}(i)\}-D_j^{[K,M]}(i)\right) \nonumber \\&\qquad =\sum \limits _{i=1}^{B_\ell ^{[K,M]}(t)}\left( \min \{t,D_{m+1}^{[K,M]}(i)\}-D_\ell ^{[K,M]}(i)\right) . \end{aligned}$$

(19)

Now, by conditions (iii), (iv), and (v), and Eqs. (18) and (19), the sum of the first three terms of Eq. (17) is greater than or equal to

$$\begin{aligned} h_K\left\{ \sum \limits _{i=1}^{B_{K-1}(t)}\bigg (\min \{t,D_N(i)\}-D_{K-1}(i)\bigg )-\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}\left( \min \{t,D_N^{[K,M]}(i)\}-D_{K-1}^{[K,M]}(i)\right) \right\} . \nonumber \\ \end{aligned}$$

(20)

Next, suppose that condition (ii)(a) holds. Then, the fourth term of Eq. (17) reduces to

$$\begin{aligned} h_K\sum \limits _{i=1}^{B_0(t)}\left( \min \{t,D_{K-1}(i)\}-\min \{t,D^{[K,M]}_{K-1}(i)\}\right) , \end{aligned}$$

by Eqs. (18) and (19). Then, using Eq. (20), we have

$$\begin{aligned}&H(t)-H^{[K,M]}(t)\ge h_K\left\{ \sum \limits _{i=1}^{B_{K-1}(t)}\bigg (\min \{t,D_N(i)\}-D_{K-1}(i)\bigg )\right. \\&\qquad -\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}\left( \min \{t,D_N^{[K,M]}(i)\}-D_{K-1}^{[K,M]}(i)\right) +\sum \limits _{i=1}^{B_0(t)} \min \{t,D_{K-1}(i)\}\\&\left. \qquad -\sum \limits _{i=1}^{B_0(t)} \min \{t,D^{[K,M]}_{K-1}(i)\}\right\} \\&\quad =h_K\left\{ \sum \limits _{i=1}^{B_{K-1}(t)}\min \{t,D_N(i)\}-\sum \limits _{i=1}^{B_{K-1}(t)}D_{K-1}(i) -\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}\min \{t,D_N^{[K,M]}(i)\}\right. \\&\qquad +\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}D_{K-1}^{[K,M]}(i) +\sum \limits _{i=1}^{B_{K-1}(t)}D_{K-1}(i)+\sum \limits _{i=B_{K-1}(t)+1}^{B_0(t)}t\\&\left. \qquad -\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}D^{[K,M]}_{K-1}(i)-\sum \limits _{i=B_{K-1}^{[K,M]}(t)+1}^{B_0(t)}t\right\} \\&\quad =h_K\left\{ \sum \limits _{i=1}^{B_{K-1}(t)}\min \{t,D_N(i)\} -\sum \limits _{i=1}^{B_{K-1}^{[K,M]}(t)}\min \{t,D_N^{[K,M]}(i)\}+\sum \limits _{i=B_{K-1}(t)+1}^{B_{0}(t)}t\right. \\&\qquad \left. -\sum \limits _{i=B_{K-1}^{[K,M]}(t)+1}^{B_{0}(t)}t\right\} \\&\quad =h_K\sum \limits _{i=1}^{B_{0}(t)}\bigg (\min \{t,D_N(i)\}-\min \{t,D_N^{[K,M]}(i)\}\bigg ), \end{aligned}$$

which is nonnegative by condition (i).

Finally, suppose that condition (ii)(b) holds, in which case we have $D_j(i)=D_j^{[K,M]}(i)$ and $B_j(t)=B_j^{[K,M]}(t)$ for $j\in \{0,\ldots ,K-1\}$, $i\ge 1$, and $t\ge 0$. Then, the fourth term of Eq. (17) becomes zero, and using Eq. (20) and condition (i), we have

$$\begin{aligned} H(t)-H^{[K,M]}(t)\ge h_K\sum \limits _{i=1}^{B_{K-1}(t)} \left( \min \{t,D_N(i)\}-\min \{t,D_N^{[K,M]}(i)\}\right) \ge 0. \end{aligned}$$

$\square$

Proof of Proposition 7

Under Assumptions 2 and 3, we have

$$\begin{aligned} T^{(1,N)}=\lim _{n\rightarrow \infty }\frac{n}{\sum _{i=1}^nX^{(1,N)}(i)}= \lim _{n\rightarrow \infty }\frac{n\sum _{\ell =1}^{N}\theta _\ell }{\sum _{i=1}^n\sum _{j=1}^N\theta _jX_j(i)}, \end{aligned}$$

and, under Assumptions 1 and 2, we have

$$\begin{aligned} T^{[1,N]}=\sum \limits _{\ell =1}^{N}\lim _{n\rightarrow \infty }\frac{n}{\sum _{i=1}^nX^{[1,N]}_{\ell }(i)}=\lim _{n\rightarrow \infty }\frac{n\sum _{\ell =1}^{N} \theta _\ell }{\sum _{i=1}^n\sum _{j=1}^N\theta _jX_j(i)}. \end{aligned}$$

These limits exist and are equal by the strong law of large numbers because $\left\{ \sum _{j=1}^N\theta _jX_j(i)\right\} _{i\ge 1}$ is an i.i.d. sequence of random variables with finite mean, which completes the proof. $\square$

Proof of Proposition 3

Because $\{\mathbf {X}(i)\}_{i\ge 1}$ is a sequence of i.i.d. random vectors with finite component means and $\theta _j\in [0,\infty )$ for all $j\in \{1,\ldots ,N\}$, $\{\sum _{j=1}^N\theta _jX_j(i)\}_{i\ge 1}$ is a sequence of i.i.d. random variables with finite mean. Hence, Proposition 7 yields that $T^{[1,N]}=T^{(1,N)}$ if Assumptions 1, 2, and 3 hold. Combining this with the fact that $T^{(1,N)}\ge T$ under Assumption 3 by Theorem 1 in Argon and Andradóttir [4] completes the proof. $\square$

Proof of Corollary 1

Let $t_{m,n}$ denote the throughput of the tandem line that is obtained by removing stations 1 through $m-1$ and stations $n+1$ through N in the original line, where $1\le m\le n\le N$. If in the original line $b_{K}=b_{M+1}=\infty$, then its throughput will exist and be equal to $\min \{t_{1,K-1},t_{K,M},t_{M+1,N}\}$ (see, for example, Muth [19]) under the assumption that the service times are i.i.d. with finite mean. Moreover, since the throughput of a tandem line decreases with a decrease in the buffer sizes (see, for example, page 186 in Buzacott and Shanthikumar [8]), we have

$$\begin{aligned} T\le \min \{t_{1,K-1},t_{K,M},t_{M+1,N}\} \end{aligned}$$

(21)

as $b_K$ and $b_{M+1}$ are not necessarily infinite in the original line.

Now, let $t^{[K,M]}$ be the throughput of the system that consists of only the pooled station with an infinite supply of jobs in front of the pooled station and infinite room following it. If the buffers before and after the pooled station are infinite, then we have $T^{[K,M]}=\min \{t_{1,K-1},t^{[K,M]},t_{M+1,N}\}$. Using Proposition 3, which implies that $t^{[K,M]}\ge t_{K,M}$, and inequality (21), we have $T\le T^{[K,M]}$. (Note that we here use the fact that Proposition 3 is still valid assuming that there is a stochastic arrival stream at the first station and $b_1=\infty$. See Sect. 3 for this result.) $\square$

Proof of Corollary 2

Let $t^{(K,M)}$ be the throughput of the system that consists of only the pooled station under cooperative pooling with an infinite supply of jobs in front of the pooled station and infinite room following it. If the buffers before and after the pooled stations are infinite, then we have $T^{[K,M]}=\min \{t_{1,K-1},t^{[K,M]},t_{M+1,N}\}$ and $T^{(K,M)}=\min \{t_{1,K-1},t^{(K,M)},t_{M+1,N}\}$ under the given assumption on service times. (See the proof of Corollary 1 for definitions of $t_{1,K-1}$ and $t_{M+1,N}$.) Now, using Proposition 7, we have $t^{[K,M]}=t^{(K,M)}$, which implies that $T^{(K,M)}=T^{[K,M]}$. (Note that we here use the fact that Proposition 7 is still valid assuming that there is a stochastic arrival stream at the first station and $b_1=\infty$. See Sect. 3 for this result.) $\square$

Proof of Proposition 8

Each one of the four systems can be modeled as a birth-death process. We start with System 0; the others can be derived from this birth–death model by simple substitution. Let the system state be the number of jobs that have finished service at station 1 but not at station 2. Then, the state space will be given by $\mathcal {S}=\{0,1,\ldots ,L_1+L_2+B_0\}$. Let $\lambda (i)$ be the birth rate in state i for $i=0,1,\ldots ,L_1+L_2+B_0-1$ and $\theta (i)$ be the death rate in state i for $i=1,2,\ldots ,L_1+L_2+B_0$. We have:

$$\begin{aligned} \lambda (i)= & {} \left\{ \begin{array}{ll}L_1\mu _1,&{}\text{ for } i=0,\ldots ,L_2+B_0\text{, }\\ (L_1+L_2+B_0-i)\mu _1,&{}\text{ for } i=L_2+B_0+1,\ldots ,L_1+L_2+B_0-1\text{; }\end{array}\right. \\ \theta (i)= & {} \left\{ \begin{array}{ll}i\mu _2,&{}\text{ for } i=1,\ldots ,L_2-1\text{, }\\ L_2\mu _2,&{}\text{ for } i=L_2,\ldots ,L_1+L_2+B_0\text{. }\end{array}\right. \end{aligned}$$

Next, we let $\pi (i)$ be the limiting probability of being in state $i\in \mathcal {S}$. Note that the limiting distribution for this birth–death process exists (because the state space is finite) and is given by $\pi (i)=f(i)\alpha ^i\pi (0)$ for $i\in \mathcal {S}$, where $\alpha =L_1\mu _1/(L_2\mu _2)$,

$$\begin{aligned} f(i)= & {} \left\{ \begin{array}{ll}\frac{L_2^i}{i!},&{}\text{ for } i=0,\ldots ,L_2-1\text{, }\\ \frac{L_2^{L_2}}{L_2!},&{}\text{ for } i=L_2,\ldots ,L_2+B_0+1\text{, }\\ \frac{L_2^{L_2}L_1^{L_2+B_0-i}L_1!}{L_2!(L_1+L_2+B_0-i)!},&{}\text{ for } i=L_2+B_0+2,\ldots ,L_1+L_2+B_0,\end{array}\right. \\ \end{aligned}$$

and

$$\begin{aligned} \pi (0)=\left( \sum _{i=0}^{L_1+L_2+B_0}f(i)\alpha ^i\right) ^{-1}. \end{aligned}$$

Then, the steady-state throughput of System 0 is given by $T_0=\pi (0)\sum _{i=1}^{L_1+L_2+B_0}\theta (i)f(i)\alpha ^i$.

To obtain the steady-state throughput for System j (for $j=1,2,3$), replace $B_0$ with $B_j$ in the above expressions for System 0. Furthermore, for System j, where $j=1,2$, replace $L_j$ and $\mu _j$ with 1 and $L_j\mu _j$, respectively. Finally, for System 3, replace $L_i$ and $\mu _i$ with 1 and $L_i\mu _i$, respectively, for $i=1,2$. The steady-state throughputs are then given as follows:

$$\begin{aligned} T_0= & {} L_1\mu _1\left( \frac{\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^i}{i!}+\frac{L_2^{L_2}\alpha ^{L_2}}{L_2!}\sum _{i=0}^{B_0-1}\alpha ^i+\frac{L_2^{L_2}L_1!\alpha ^{L_1+L_2+B_0-1}}{L_1^{L_1}L_2!}\sum _{i=0}^{L_1-1}\frac{\alpha ^{-i}L_1^i}{i!}}{\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^i}{i!}+\frac{L_2^{L_2}\alpha ^{L_2}}{L_2!}\sum _{i=0}^{B_0-1}\alpha ^i+\frac{L_2^{L_2}L_1!\alpha ^{L_1+L_2+B_0}}{L_1^{L_1}L_2!}\sum _{i=0}^{L_1}\frac{\alpha ^{-i}L_1^i}{i!}}\right) ,\nonumber \\ \end{aligned}$$

(22)

$$\begin{aligned} T_1= & {} L_1\mu _1\left( \frac{\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^i}{i!}+\frac{L_2^{L_2}\alpha ^{L_2}}{L_2!}\sum _{i=0}^{B_1}\alpha ^i}{\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^i}{i!}+\frac{L_2^{L_2}\alpha ^{L_2}}{L_2!}\sum _{i=0}^{B_1+1}\alpha ^i}\right) , \end{aligned}$$

(23)

$$\begin{aligned} T_2= & {} L_1\mu _1\left( \frac{\sum _{i=0}^{B_2}\alpha ^i+\frac{L_1!\alpha ^{L_1+B_2}}{L_1^{L_1}}\sum _{i=0}^{L_1-1}\frac{\alpha ^{-i}L_1^i}{i!}}{\sum _{i=0}^{B_2}\alpha ^i+\frac{L_1!\alpha ^{L_1+B_2+1}}{L_1^{L_1}}\sum _{i=0}^{L_1}\frac{\alpha ^{-i}L_1^i}{i!}}\right) , \end{aligned}$$

(24)

$$\begin{aligned} T_3= & {} L_1\mu _1\left( \frac{\sum _{i=0}^{B_3+1}\alpha ^i}{\sum _{i=0}^{B_3+2}\alpha ^i}\right) . \end{aligned}$$

(25)

We next perform a pairwise comparison of the steady-state throughputs of these four systems.

System 0 versus System 1: From Eqs. (22) and (23), we find that $T_0\le T_1$ if and only if

$$\begin{aligned} \left( \sum _{i=0}^{B_1}\alpha ^i-\sum _{i=0}^{B_0}\alpha ^i-\frac{L_1!\alpha ^{L_1+B_0-1}}{L_1^{L_1}}\sum _{i=0}^{L_1-2}\frac{\alpha ^{-i}L_1^i}{i!}\right) \left( \sum _{i=0}^{L_2}\frac{L_2^i\alpha ^{i}}{i!}-\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^{i+1}}{i!}\right) \ge 0. \end{aligned}$$

(26)

The term in the second parentheses above reduces to

$$\begin{aligned} 1+\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^{i+1}}{(i+1)!}(L_2-1-i), \end{aligned}$$

which is greater than zero. Hence, $T_0\le T_1$ if and only if the term in the first parentheses in (26) is nonnegative.

We first consider the case where $B_1=B_0+L_1-1$, so that System 1 has the same number of spaces for jobs as System 0. For this case, the term in the first parentheses in (26) reduces to

$$\begin{aligned} \frac{L_1!\alpha ^{B_0+1}}{L_1^{L_1}}\sum _{i=0}^{L_1-2}\alpha ^{L_1-2-i}\left( \frac{L_1^{L_1}}{L_1!}-\frac{L_1^i}{i!}\right) , \end{aligned}$$

which is greater than zero because $L_1^{L_1-i}i!>L_1!$ for all $i=0,1,\ldots ,L_1-2$. Thus, when $B_1=B_0+L_1-1$, we have $T_0< T_1$.

We next consider the case where $B_1=B_0$, so that Systems 0 and 1 have the same number of buffer spaces excluding the spaces for servers. For this case, the term in the first parentheses in (26) reduces to $-L_1!\sum _{i=0}^{L_1-2}\alpha ^{L_1+B_0-1-i}L_1^{i-L_1}/i!<0$. Thus, when $B_1=B_0$, we have $T_0> T_1$.

System 0 versus System 2: From Eqs. (22) and (24), we find that $T_0\le T_2$ if and only if

$$\begin{aligned}&\left( \frac{L_2^{L_2}\alpha ^{B_0+L_2-B_2-1}}{L_2!}\sum _{i=0}^{B_2}\alpha ^i-\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^i}{i!}-\frac{L_2^{L_2}\alpha ^{L_2}}{L_2!}\sum _{i=0}^{B_0-1}\alpha ^i\right) \nonumber \\&\qquad \left( \frac{L_1^{L_1}\alpha ^{1-L_1}}{L_1!}-(1-\alpha )\sum _{i=0}^{L_1-1}\frac{\alpha ^{-i}L_1^i}{i!}\right) \ge 0. \end{aligned}$$

(27)

We can show that the term in the second parentheses above is positive as follows:

$$\begin{aligned} \frac{L_1^{L_1}\alpha ^{1-L_1}}{L_1!}-(1-\alpha )\sum _{i=0}^{L_1-1}\frac{\alpha ^{-i}L_1^i}{i!}= & {} \sum _{i=0}^{L_1}\frac{\alpha ^{1-i}L_1^i}{i!}-\sum _{i=0}^{L_1-1}\frac{\alpha ^{-i}L_1^i}{i!}\\= & {} \alpha +\sum _{i=0}^{L_1-1}\frac{L_1^i\alpha ^{-i}}{(i+1)!}\left( L_1-1-i\right) > 0. \end{aligned}$$

Hence, $T_0\le T_2$ if and only if the term in the first parentheses in (27) is nonnegative.

We first consider the case where $B_2=B_0+L_2-1$, so that System 2 has the same number of spaces for jobs as System 0. For this case, the term in the first parentheses in (27) reduces to

$$\begin{aligned} \sum _{i=0}^{L_2-2}\left( \frac{L_2^{L_2}}{L_2!}-\frac{L_2^i}{i!}\right) \alpha ^i, \end{aligned}$$

(28)

which is positive because $L_2^{L_2-i}i!>L_2!$ for all $i=0,1,\ldots ,L_2-2$. Thus, when $B_2=B_0+L_2-1$, we have $T_0< T_2$.

We next consider the case where $B_2=B_0$, so that Systems 0 and 2 have the same number of buffer spaces excluding the spaces for servers. For this case, the term in the first parentheses in (27) reduces to $-\sum _{i=0}^{L_2-2}L_2^i\alpha ^i/i!<0$. Thus, when $B_2=B_0$, we have $T_0> T_2$.

System 1 versus System 3: From Eqs. (23) and (25), we find that $T_1\le T_3$ if and only if

$$\begin{aligned} \sum _{i=0}^{L_2-1}\frac{L_2^{i}\alpha ^{i}}{i!}+\frac{L_2^{L_2}\alpha ^{L_2}}{L_2!}\left( \sum _{i=0}^{B_1}\alpha ^{i}-\sum _{i=0}^{B_3+1}\alpha ^{i+B_1-B_3-1}\right) \le 0. \end{aligned}$$

(29)

We first consider the case where $B_3=B_1+L_2-1$, so that System 3 has the same number of spaces for jobs as System 1. For this case, the left-hand side of (29) reduces to (28) multiplied by negative one, which is less than zero. Thus, when $B_3=B_1+L_2-1$, we have $T_1< T_3$.

We next consider the case where $B_3=B_1$, so that Systems 1 and 3 have the same number of buffer spaces excluding the spaces for servers. For this case, the left-hand side of (29) reduces to $\sum _{i=0}^{L_2-2}\alpha ^{i}L_2^i/i!$, which is positive. Hence, in this case, we have $T_1> T_3$.

System 2 versus System 3: From Eqs. (24) and (25), we find that $T_2\le T_3$ if and only if

$$\begin{aligned} \sum _{i=0}^{B_2}\alpha ^{i+B_3+1-B_2}+\frac{L_1!\alpha ^{L_1-1}}{L_1^{L_1}}\sum _{i=0}^{L_1-1}\frac{L_1^i\alpha ^{-i}}{i!}-\sum _{i=0}^{B_3+1}\alpha ^i\le 0. \end{aligned}$$

(30)

We first consider the case where $B_3=B_2+L_1-1$, so that System 3 has the same number of spaces for jobs as System 2. For this case, the left-hand side of (30) reduces to $\sum _{i=0}^{L_1-2}(L_1!/(L_1^{L_1-i}i!)-1)\alpha ^{L_1-1-i}$, which is less than zero because $L_1^{L_1-i}i!>L_1!$ for all $i=0,1,\ldots ,L_1-2$. Thus, when $B_3=B_2+L_1-1$, we have $T_2< T_3$.

We next consider the case where $B_3=B_2$, so that Systems 2 and 3 have the same number of buffer spaces excluding the spaces for servers. For this case, the left-hand side of (30) reduces to $L_1!\sum _{i=0}^{L_1-2}L_1^{i-L_1}\alpha ^{L_1-1-i}/i!>0$. Hence, in this case, we have $T_2> T_3$.

In order to complete the proof, we need to show that $T_0$ is a non-decreasing function of $B_0$, which implies that $T_j$ is a non-decreasing function of $B_j$ for $j=1,2,3$. (Our literature search failed to find the exact monotonicity result in published work. For a similar monotonicity result in the case of tandem lines with two or more stations each having a single server, see [18].) From Eq. (22), we find that $T_0$ with $B_0+1$ buffers is greater than equal to $T_0$ with $B_0$ buffers if and only if

$$\begin{aligned} \left( \sum _{i=0}^{L_1-1}\frac{\alpha ^{L_1-i}L_1^i}{i!}-\sum _{i=0}^{L_1-2}\frac{\alpha ^{L_1-1-i}L_1^i}{i!}\right) \left( \sum _{i=0}^{L_2}\frac{L_2^i\alpha ^{i}}{i!}-\sum _{i=0}^{L_2-1}\frac{L_2^i\alpha ^{i+1}}{i!}\right) \ge 0. \end{aligned}$$

(31)

We know from the argument following (26) that the term in the second parentheses above is positive. Furthermore, the term in the first parentheses in (31) reduces to

$$\begin{aligned} \alpha ^{L_1}+\sum _{i=1}^{L_1-1}\frac{\alpha ^{L_1-i}L_1^{i-1}}{i!}(L_1-i), \end{aligned}$$

which is greater than zero. This concludes the proof. $\square$