Overflow in systems with two servers: the negative consequences

Abstract

We investigate a system with two types of customers wherein each customer type has a dedicated server. When the corresponding dedicated server is busy, we allow customer overflow to the non-dedicated server, and the service time of each server is customer-dependent. The objective of this study is to assess the negative consequences of overflow. On the basis of the analytical stationary distribution of the proposed two-server model, we first identify the conditions under which overflow leads to improvement of throughputs. Second, we obtain customers’ overflow rates and ratios. For a symmetric system under heavy traffic, the wrong assignment ratio comes close to 50%. Third, we analyze the probability that a customer is served by a non-dedicated server. The probability that both servers are serving non-dedicated customers approaches 25% in a symmetric system under heavy traffic. Finally, we determine various overflow conditions while including the overflow costs.

Appendices

Appendix

A Proof of Theorem 1

Proof

The balance equations for X(t) are

$$\begin{aligned} (\lambda _1+\lambda _2)\pi (0,0)&=\mu _1[\pi (1,0)+\pi (0,1)]+\mu _2[\pi (0,2)+\pi (2,0)],\\ (\lambda _1+\lambda _2+\mu _1)\pi (1,0)&=\lambda _1\pi (0,0)+\mu _1\pi (1,1)+\mu _2\pi (1,2),\\ (\mu _1+\mu _2)\pi (1,2)&=\lambda _1\pi (0,2)+\lambda _2\pi (1,0),\\ (\lambda _1+\lambda _2+\mu _2)\pi (0,2)&=\lambda _2\pi (0,0)+\mu _1\pi (1,2)+\mu _2\pi (2,2),\\ 2\mu _1\pi (1,1)&=\lambda _1[\pi (0,1)+\pi (1,0)],\\ (\lambda _1+\lambda _2+\mu _1)\pi (0,1)&=\mu _1\pi (1,1)+\mu _2\pi (2,1),\\ (\lambda _1+\lambda _2+\mu _2)\pi (2,0)&=\mu _1\pi (2,1)+\mu _2\pi (2,2),\\ (\mu _1+\mu _2)\pi (2,1)&=\lambda _1\pi (2,0)+\lambda _2\pi (0,1),\\ 2\mu _2\pi (2,2)&=\lambda _2[\pi (2,0)+\pi (0,2)], \end{aligned}$$

The normalization requirement for the probability measure is $\sum _{i = 0}^{2} \sum _{j = 0}^{2} \pi (i,j) = 1$. It can be readily verified that Equations (1)-(9) satisfy the balance equations and the normalization requirement, which completes the proof. $\square$

B Proof of Proposition 1

Proof

Note that $P_{loss}<\min \left\{ \frac{\rho _1}{1+\rho _1}, \frac{\rho _2}{1+\rho _2}\right\}$ holds iff $\frac{(\rho _1+\rho _2)^2}{(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)}<\frac{\rho _1}{1+\rho _1}$ and $\frac{(\rho _1+\rho _2)^2}{(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)}<\frac{\rho _2}{1+\rho _2}$. Using the non-negativity of $\rho _1$ and $\rho _2$, these two inequalities are simplified to $\rho _2^2<\rho _1^2+2\rho _1$ and $\rho _1^2<\rho _2^2+2\rho _2$. Again using the non-negativity of $\rho _1$ and $\rho _2$, these two inequalities are summarized as $\rho _{\max }<\sqrt{\rho _{\min }^2+2\rho _{\min }}$ where $\rho _{\max }=\max \{\rho _1,\rho _2\}$ and $\rho _{\min }=\min \{\rho _1,\rho _2\}$. This completes the proof. $\square$

C Proof of Corollary 1

Proof

For the first statement, substituting the expression for $P_{loss}$ into the relative loss reduction yields $(\lambda _1 + \lambda _2) \frac{(\rho _1 +\rho _2)^2}{(\rho _1 + \rho _2)^2 + 2(\rho _1 + \rho _2 + 1)} < \lambda _1 \frac{\rho _1}{1+\rho _1} +\lambda _2 \frac{\rho _2}{1+\rho _2}$. When the occupation rates are the same, i.e., $\rho _1 = \rho _2 = \rho$, this inequality becomes $(\lambda _1 + \lambda _2) \frac{4 \rho ^2}{4 \rho ^2 + 2(2 \rho + 1)} < (\lambda _1 + \lambda _2) \frac{\rho }{1 + \rho }$. The simplification of this inequality leads to $4 \rho ^2 (1 + \rho ) < 4 \rho ^3 + 2 \rho (2 \rho + 1)$. It can be readily verified that this inequality holds for all $\rho > 0$. This indicates that the relative loss reduction requirement always holds when the intensity rates $\rho _1$ and $\rho _2$ are the same.

For the second statement, when the arrival rates are the same, i.e., $\lambda _1 = \lambda _2$, the relative loss reduction becomes $2 P_{loss} < \frac{\rho _1}{1 + \rho _1} + \frac{\rho _2}{1 + \rho _2}$. Substituting the expression for $P_{loss}$ into the inequality we obtain $2 \frac{(\rho _1 +\rho _2)^2}{(\rho _1 + \rho _2)^2 + 2(\rho _1 + \rho _2 + 1)} < \frac{\rho _1}{1 + \rho _1} + \frac{\rho _2}{1 + \rho _2}$. This inequality can be simplified to $\rho _1^3 + \rho _2^3 < \rho _1^2 \rho _2 + \rho _2^2 \rho _1 + 4 \rho _1 \rho _2 + 2 \rho _1 + 2 \rho _2$. Notice that the reversed derivation of the above inequalities also hold, hence, when $\rho _1^3 + \rho _2^3 < \rho _1^2 \rho _2 + \rho _2^2 \rho _1 + 4 \rho _1 \rho _2 + 2 \rho _1 + 2 \rho _2$ and $\lambda _1 = \lambda _2$, the relative loss reduction requirement holds.

For the third statement, substituting the expression for $P_{loss}$ into the relative loss reduction requirement yields $(\lambda _1 + \lambda _2) \frac{(\rho _1 +\rho _2)^2}{(\rho _1 + \rho _2)^2 + 2(\rho _1 + \rho _2 + 1)} < \lambda _1 \frac{\rho _1}{1+\rho _1} +\lambda _2 \frac{\rho _2}{1+\rho _2}$. When the service rates are the same, this inequality becomes $(\lambda _1 + \lambda _2) \frac{ \frac{1}{\mu ^2}(\lambda _1 +\lambda _2)^2}{\frac{1}{\mu ^2}(\lambda _1 + \lambda _2)^2 + 2 \frac{1}{\mu } (\lambda _1 + \lambda _2 + \mu )} < \lambda _1 \frac{\lambda _1}{\mu +\lambda _1} +\lambda _2 \frac{\lambda _2}{\mu +\lambda _2}$. The simplification of this inequality leads $\frac{(\lambda _1 + \lambda _2)^3}{(\lambda _1 + \lambda _2)^2 + 2 \mu (\lambda _1 + \lambda _2 + \mu )} < \frac{\lambda _1^2 (\mu + \lambda _2) + \lambda _2^2 (\mu + \lambda _1)}{(\mu + \lambda _1)(\mu +\lambda _2)}$. Moreover, we have $\big (\lambda _1^2 (\mu + \lambda _2) + \lambda _2^2 (\mu + \lambda _1)\big )\big ((\lambda _1 + \lambda _2)^2 + 2 \mu (\lambda _1 + \lambda _2 + \mu )\big ) - (\lambda _1 + \lambda _2)^3(\mu + \lambda _1)(\mu +\lambda _2) = \mu ^2(\lambda _1^2 + \lambda _2^2)(\lambda _1 + \lambda _2 + 2\mu ) > 0$, this indicates that the relative loss reduction requirement always holds when the service rates $\mu _1$ and $\mu _2$ are the same. $\square$

D Proof of Lemma 1

Proof

The absolute loss reduction $P_{loss}<\min \left\{ \frac{\rho _1}{1+\rho _1}, \frac{\rho _2}{1+\rho _2}\right\}$ indicates $P_{loss}<\frac{\rho _1}{1+\rho _1}$ and $P_{loss}< \frac{\rho _2}{1+\rho _2}$. Multiplying the first inequality by $\lambda _1$, the second inequality by $\lambda _2$ and adding these two inequalities yields $(\lambda _1 + \lambda _2) P_{loss} < \lambda _1 \frac{\rho _1}{1 + \rho _1} + \lambda _2 \frac{\rho _2}{1 + \rho _2}$, which is exactly the relative loss reduction. $\square$

E Proof of Proposition 3

Proof

For the first statement, it can be readily verified that $r_1= \frac{(\rho _1+\rho _2)^2+2\rho _1}{2(\rho _1+\rho _2+1)^2}<\frac{1}{2}$, $r_2= \frac{(\rho _1+\rho _2)^2+2\rho _2}{2(\rho _1+\rho _2+1)^2}<\frac{1}{2}$ and $r = \frac{(\lambda _1+\lambda _2)(\rho _1+\rho _2)^2+2(\lambda _1\rho _1+\lambda _2\rho _2)}{2(\lambda _1+\lambda _2)(\rho _1+\rho _2+1)^2}<\frac{1}{2}$. In addition, when $\lambda _1=\lambda _2$ and $\rho _1=\rho _2=\rho$, we have $\lim _{\rho \rightarrow \infty } r_1= \lim _{\rho \rightarrow \infty } r_2 = \lim _{\rho \rightarrow \infty } = \frac{1}{2}$.

The second statement can be obtained by verifying the following first-order partial derivatives $\frac{\partial { r_1}}{\partial {\rho _1}} = \frac{2\rho _2+1}{(\rho _1+\rho _2+1)^3}$, $\frac{\partial { r_1 }}{ \partial {\rho _2} } =\frac{\rho _2-\rho _1}{(\rho _1+\rho _2+1)^3}$, $\frac{\partial { r_2}}{\partial {\rho _1}} = \frac{\rho _1 - \rho _2 }{(\rho _1+\rho _2+1)^3}$, and $\frac{\partial {r_2}}{ \partial {\rho _2} } =\frac{2\rho _1+1}{(\rho _1+\rho _2+1)^3}$.

For the third statement, when $\lambda _1 =\lambda _2$, the total wrong assignment ratio reduces to $r = \frac{(\rho _1+\rho _2)^2+(\rho _1+\rho _2)}{2(\rho _1+\rho _2+1)^2} = \frac{(\rho _1+\rho _2+1)^2-(\rho _1+\rho _2+1)}{2(\rho _1+\rho _2+1)^2} = \frac{1}{2} - \frac{1}{2(\rho _1+\rho _2+1)}$. Therefore, we conclude that r is monotonically increasing in $\rho _1+\rho _2$, which completes the proof. $\square$

F Proof of Proposition 4

Proof

For the first statement, we have

$$\begin{aligned} \pi (2,1)=&\frac{\rho _1\rho _2}{(\rho _1+\rho _2+1)[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]}\left( \rho _1+\rho _2-\frac{\rho _1+\alpha \rho _2}{\rho _1+\alpha \rho _2+\alpha +1}\right) \\ \le&\frac{\rho _1\rho _2}{(\rho _1+\rho _2+1)[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]}\left( \rho _1+\rho _2 \right) \\ \le&\frac{(\rho _1+ \rho _2)^3}{4 (\rho _1+\rho _2+1)[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]} \\ \le&\frac{1}{4}. \end{aligned}$$

For the second statement, when $\rho _1=\rho _2=\rho$, applying Equation (9) we have $\pi (2,1)=\frac{\rho ^3}{2(\rho +1)(2\rho ^2+2\rho +1)}.$ For any positive $\rho$, we have $\frac{\mathrm {d}\pi (2,1)}{\mathrm {d}\rho }=\frac{\rho ^2(4\rho ^2+6\rho +3)}{2(\rho +1)^2(2\rho ^2+2\rho +1)^2}> 0$ and $\lim _{\rho \rightarrow \infty }\pi (2,1)=\lim _{\rho \rightarrow \infty }{\frac{\rho ^3}{2(\rho +1)(2\rho ^2+2\rho +1)}}=\frac{1}{4}$. Therefore, we conclude that when the occupation rates are the same, the serious malpositioning probability $\pi (2,1)$ is increasing in $\rho$ and approaches $\frac{1}{4}$.

For the third statement, we first denote ${\bar{\rho }} = \rho _1+\rho _2$. When $\alpha =1$ and $\rho _1+\rho _2$ is fixed, i.e., ${\bar{\rho }}$ is fixed, deploying Equations (1) and (9), we obtain $\pi (2,1)=\phi ({\bar{\rho }})\rho _1\rho _2$, where $\phi ({\bar{\rho }})$ is a function of ${\bar{\rho }}$ and consequently is a constant for fixed ${\bar{\rho }}$. Notice that $\rho _1\rho _2$ is decreasing in $|\rho _2-\rho _1|$ when $\rho _1+\rho _2$ is fixed. Hence, when $\rho _1+\rho _2$ is fixed, the malpositioning probability $\pi (2,1)$ is decreasing in $|\rho _2-\rho _1 |$.

When $\rho _2$ is fixed, the serious malpositioning probability $\pi (2,1)$ is a function of $\rho _1$. It can be readily verified that $\frac{\partial \pi (2,1)}{\partial \rho _1} = \frac{\rho _2(\rho _1 + \rho _2)^2(\rho _2^2 - \rho _1^2 + 4 \rho _2 + 6) + 4 \rho _2(2\rho _1 +\rho _2)}{(\rho _1 + \rho _2 + 2)^2[(\rho _1 + \rho _2)^2 + 2(\rho _1 + \rho _2 +1)]^2}$, the denominator of this partial derivative is positive, hence we focus on the sign of the nominator. Since $\rho _2$ is fixed, we consider the residual of the nominator of this partial derivative as a function of $\rho _1$, by replacing $\rho _1$ by x, we obtain g(x) where $g(x) =(x+\rho _2)^2(\rho _2^2-x^2+4\rho _2+6)+4(2x+\rho _2)$, the derivatives of g(x) regarding to x are $g'(x)=-2(x+\rho _2)(2x^2+x\rho _2-\rho _2^2-4\rho _2-6)+8$, $g''(x)=-4(3x^2+3x\rho _2-2\rho _2-3)$ and $g'''(x)=-12(2x+\rho _2)$. Notice that for positive x, we have $g'''(x)<0$, $g''(0)=4(2\rho _2+3)>0$ and $\lim _{x\rightarrow \infty } g''(x)=-\infty$. Hence, there exists a unique positive number $x_1$ such that $g''(x_1)=0$. Besides, we have $g''(x)>0$ if $0<x<x_1$ and $g''(x)<0$ if $x>x_1$. Moreover, we have $g'(0)=2\rho _2(\rho _2^2+4\rho _2+6)+8>0,\quad \lim _{x\rightarrow \infty } g'(x)=-\infty$. Hence we conclude that there exists a unique positive number $x_2(>x_1)$ such that $g'(x_2)=0$. Obviously, we have $g'(x)>0$ if $0<x<x_2$ and $g'(x)<0$ if $x>x_2$. Similarly, we can conclude that there exists a unique positive number $x_3(>x_2)$ such that $g(x_3)=0$. In addition, we have $g(x)>0$ if $0<x<x_3$ and $g(x)<0$ if $x>x_3$. Moreover, we know that $\frac{\partial \pi (2,1)}{\partial \rho _1}=\frac{\rho _2g(\rho _1)}{(\rho _1+\rho _2+2)^2{[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]^2}}$. Therefore, the serious malpositioning probability $\pi (2,1)$ is increasing in $\rho _1$ if $\rho _1<x_3$ and decreasing in $\rho _1$ if $\rho _1>x_3$. Observe that $g(\rho _2)=4\rho _2(4\rho _2^2+6\rho _2+3)>0$, which implies $\rho _2<x_3$. This completes the proof. $\square$

G Proof of Proposition 5

Proof

The first statement follows by noting that $\rho _1\rho _2= \frac{(\rho _1+\rho _2)^2-(\rho _1-\rho _2)^2}{4}$. For the second statement, when $\rho _1=\rho _2=\rho$, we have $\pi (0,1)+\pi (2,0) = \frac{\rho ^2}{(\rho +1)(2\rho ^2+2\rho +1)}$. Therefore, it can be readily verified that $\pi (0,1)+\pi (2,0)$ is unimodal and there exists a specific $\rho$ that maximizes $\pi (0,1)+\pi (2,0)$. $\square$

H Proof of Proposition 6

Proof

The first statement follows by noting that $\pi (1,1)+\pi (2,2)<1$ and $\lim _{\rho _1 \rightarrow \infty } [\pi (1,1)+\pi (2,2)]=1$ for $\rho _2\in (0,\infty )$. The second statement can be verified by checking the sign property of the first-order derivative of $\pi (1,1)+\pi (2,2)$ as follows, when $\rho _1=\rho _2=\rho$, we have $\lim _{\rho \rightarrow \infty } [\pi (1,1)+\pi (2,2)]= \lim _{\rho \rightarrow \infty } \frac{\rho ^2}{2\rho ^2+2\rho +1 }=\frac{1}{2}$ and $\frac{\mathrm {d}(\pi (1,1)+\pi (2,2))}{\mathrm {d} \rho }= \frac{2\rho (\rho +1)}{(2\rho ^2+2\rho +1)^2}>0$. The third statement follows by noting that $\rho _1^2+\rho _2^2 = \frac{1}{2}\left[ (\rho _1+\rho _2)^2 + (\rho _1 - \rho _2)^2 \right]$. For the last statement, we have

$$\begin{aligned} \frac{\partial (\pi (1,1)+\pi (2,2))}{\partial \rho _1} =&\frac{2\rho _1[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)] -2(\rho _1^2+\rho _2^2)(\rho _1+\rho _2+1)}{[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]^2} \\ =&\frac{2(\rho _2+1)(\rho _1^2+2\rho _1-\rho _2^2)}{[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]^2}. \end{aligned}$$

It can be verified that $\frac{\partial (\pi (1,1)+\pi (2,2))}{\partial \rho _1}\le 0$ for $0<\rho _1\le \sqrt{\rho _2^2+1}-1$ and $\frac{\partial (\pi (1,1)+\pi (2,2))}{\partial \rho _1}\ge 0$ for $\rho _1\ge \sqrt{\rho _2^2+1}-1$, which completes the proof. $\square$

I Proof of Proposition 7

Proof

Since the light malpositioning probability is a part of the total malpositioning probability, the first statement follows the first statement of Proposition 6. We now verify the second statement, when $\rho _1=\rho _2=\rho$, we have

$$\begin{aligned} \lim _{\rho \rightarrow \infty } P_{mal}&=\lim _{\rho \rightarrow \infty } \frac{3\rho ^3+4\rho ^2}{2(\rho +1)(2\rho ^2+2\rho +1)}= \frac{3}{4}, \\ \frac{\mathrm {d} P_{mal} }{\mathrm {d} \rho }&= \frac{4\rho ^4+18\rho ^3+21\rho ^2+8\rho }{2(\rho +1)^2(2\rho ^2+2\rho +1)^2}>0. \end{aligned}$$

$\square$

For the last statement, when $\alpha =1$, it can be readily verified that

$$\begin{aligned} P_{mal} = \frac{(\rho _1+\rho _2)^3(\rho _1+\rho _2+3)+(\rho _1^2+\rho _2^2)[(\rho _1+\rho _2)^2+5(\rho _1+\rho _2)+8]}{2(\rho _1+\rho _2+1)(\rho _1+\rho _2+2)[(\rho _1+\rho _2)^2+2(\rho _1+\rho _2+1)]}. \end{aligned}$$

Because $\rho _1^2 + \rho _2^2 = \frac{1}{2}[(\rho _1 - \rho _2)^2 + (\rho _1 + \rho _2)^2]$, we conclude that the total malpositioning probability $P_{mal}^{total}$ is increasing in $|\rho _1-\rho _2 |$ when $\rho _1+\rho _2$ is fixed and $\alpha =1$, which completes the proof.