Cost-Beneﬁt Analysis of a Standby Retrial System with an Unreliable Server and Switching Failure

: In many industries and plants, a stable power supply system with acceptable cost/beneﬁt is essential. This paper investigates the cost-effectiveness of an unreliable retrial system that includes standby generators and experiences the switchover failures of standby generators. Four different standby retrial conﬁgurations are included, and each conﬁguration includes various numbers of primary and standby generators. Upon arrival, a failed generator is repaired immediately if the server is available; otherwise, the failed generator will enter into orbit. The server is subject to breakdown even when the server is idle. The explicit expressions of the mean time-to-failure and steady-state availability for each conﬁguration are derived and compared. We also compare the cost/beneﬁt ratio among four conﬁgurations. The developed results can provide managers with decision reference for stable power supply system and cost reduction.


Introduction
A robust power supply system with acceptable cost/benefit is a fundamental part for high-tech fabrication plants (called fabs), which plays an important role in modern industries, such as packaging and testing, IC design, and wafer foundry. For fabs, a stable power supply system is essential for maintaining the competition. Liu et al. [1] applied a retrial system with standby switching failure to model the power supply system for a fab. In their work, the repair server is assumed to be reliable. They applied the supplementary variables techniques to obtain the explicit expressions of the steady-state availability. However, in real world application, the repair server may be malfunctioned, but can be repaired. To address this issue, in this paper, we extend the work of Liu et al. [1] to the case of a unreliable repair server. As we know, the more system features we take into consideration simultaneously, the more difficulty we have in modelling the system and deriving the equations. Hence, the investigated model has not been studied in the literature so far. In addition, we derive the explicit expressions of the mean time-to-failure and steady-state availability for each configuration. Four different standby retrial configurations are included, and each configuration includes various numbers of primary and standby generators. Except for obtaining the explicit expressions of steady-state availability for each configuration, we also obtain the explicit expressions of the mean time-to-failure for each configuration and make a comparison.
Due to the practical applicability of the retrial queues, more and more researchers have attracted attention to retrial systems. For existing works related to retrial systems, interested readers can refer to Falin and Templeton [2], Artalejo [3], Artalejo and Gomez-Corral [4], Yang and Chang [5], Chen [6], Chen and Wang [7], Yang and Tsao [8], Phung-Duc [9], and the references therein. Recently, Yen, Wang, and Wu [10] investigated the system availability and sensitivity analysis of a retrial queue with working breakdowns operating

System Description
By referring to Liu et al. [1], we consider that a system requires 24 megawatts (MW) power and assume that the power generating capacity of a generator is available in units of 24 MW, 12 MW, and 8 MW. In this system, the energy generator that is supplying electricity is called the primary generator, and the energy generator that is online but does not provide power is called the standby generator. Both the primary and standby energy generators may fail and can be repaired. The failure times of primary and standby generators obey an exponential distribution with rate λ and α (0 < α < λ), respectively. We assume that the switching device has a failure probability q. In the repair facility, only one server is responsible for repairing a failed generator and there is no waiting space in front of the server. Hence, a failed generator finding that the server is free is repaired immediately; otherwise, it will enter orbit. The retrial time obeys exponential distribution with the rate γ. The failed generators will continue to retry until it gets the required repair. The repair times of a failed generator obey exponentially distributed with rate µ. Additionally, the server may fail at any time even if it is idle. The server fails with an exponential breakdown rate of η. The repair time for the server obeys exponentially distributed with mean β −1 .
Four configurations are considered as follows: configuration 1 consists of one 24 MW primary generator and one 24 MW standby generator. Configuration 2 includes two 12 MW primary generator s and one 12 MW standby generator. Configuration 3 contains two 12 MW primary generators and two 12 MW standby generators. Configuration 4 contains three 8 MW primary generators and two 8 MW standby generators.

Systematic Methodology
We first draw the transition-rate diagram for each configuration. According to the diagram, we develop the differential-difference equations. Finally, we utilize the matrixanalytic method to obtain the explicit expressions of MTTF and Av. We define the following probabilities throughout the paper.
where W(t) represents the number of failed generators in orbit at time t, and J(t) denotes the states of the server at time t, and 0, the server is free, 1, the server is busy, 2, the free server is under repair, 3, the busy server is under repair.

Configuration 1
For the reliability case, the state-transition-rate diagram of configuration 1 is provided in Figure 1. It is assumed that the system is characterized as a failure as soon as the remaining electricity generation capacity is less than 24 MW. So, if the system cannot be resumed, the states (0, 2), (1, 1), (2,2), and (3, 1) are absorbing states. Based on the Figure 1, the differential-difference equations can be written as the following matrix form: where To evaluate MTTF 1 , we define A 1 as the transpose matrix of Q 1 omitting the row and column for the absorbing states. First, 1 represents a column vector with all elements equal to 1, and the initial condition is P(0) = [P 0,0 (0), P 0,1 (0), P 1,0 (0), P 2,0 (0), P 2,1 (0), P 3,0 (0)] T = [1, 0, 0, 0, 0, 0] T . Then, we can obtain where MTTF 1 can be expressed as Due to the complexity of the explicit expression for MTTF 1 , this formula is difficult to display here. However, by using a suitable computer program, it can be evaluated numerically.  To evaluate MTTF1, we define A1 as the transpose matrix of Q1 omitting the row and column for the absorbing states. First, 1 represents a column vector with all elements equal to 1, and the initial condition is Then, we can obtain where 10 11 To discuss the availability of this configuration, we need the process below to obtain the steady-state availability. In steady-state, we let the derivatives of the state probabilities be zero. Then, we have By partitioning the probability vector as (4) can be rewritten as where and 0 is a zero matrix with the appropriate dimension. By solving Equation (5) and after some routine substitutions, we have Hence, we can calculate π 1 (1) by solving Equation (9) and the normalizing condition Thus, the steady-state availability of configuration, , can be computed once the steady-state probabilities are obtained. Figure 2 presents the state-transition-rate diagram of configuration 2. Similarly, the transition from (2, 2) to (0, 2), the transition from (0, 2) to (1, 1), the transition from state (1, 1) to (0, 1) and (3,1), and the transition from (3, 1) to (1, 1) are ignored when investigating the system MTTF. The associated probability vector for this configuration is defined as:

Configuration 2
We can write the differential-difference equations of configuration 2 as the following matrix form: where We define A 2 as the transpose matrix of Q 2 omitting the row and column for the absorbing state. Let represent the initial condition. We can obtain where For this configuration, MTTF 2 can be expressed as As mentioned earlier, the explicit expression of MTTF 2 is not shown here due to its complexity. We compare the four configurations in Section 4 based on the calculated numerical results. We utilize the same process in the prior subsection to get the steady-state availability. In steady-state, we let the derivatives of the state probabilities be zero, then Solving the above equation and using the normalizing condition ∑ 2 we can obtain the steady-state probabilities. Once the steady-state probabilities are obtained, the availability of configuration 2, For this configuration, MTTF2 can be expressed as As mentioned earlier, the explicit expression of MTTF2 is not shown here due to its complexity. We compare the four configurations in Section 4 based on the calculated numerical results. We utilize the same process in the prior subsection to get the steady-state availability. In steady-state, we let the derivatives of the state probabilities be zero, then from state (1, 2) to states (0, 2) and (3,2), and transition from state (3,2) to (1,2) should be deleted. Let P 3 (0) = [P 0,0 (0), P 0,1 (0), P 0,2 (0), P 1,0 (0), P 1,1 (0), P 2,0 (0), P 2,1 (0), P 2,2 (0), P 3,0 (0), P 3,1 (0)] T = [1, 0, 0, 0, 0, 0, 0, 0, 0, 0] T represent the initial condition. We can write the differential-difference equations of this configuration as the following matrix form: where We define A 3 as the transpose matrix of Q 3 omitting the row and column for the absorbing state. After the transpose operation, we have We can obtain the mean time-to-failure for configuration 3 as follows. For the availability case of this configuration, we utilize the same process in the prior subsection to get the steady-state availability. In steady-state, we let the derivatives of the state probabilities be zero, then we have where Similarly, solving Equation (17) recursively with the normalizing condition ∑ 3 j=0 P 0,j (∞) + ∑ 2 j=0 P 1,j (∞) + ∑ 3 j=0 P 2,j (∞) + ∑ 2 j=0 P 3,j (∞) = 1 can obtain the steady-state probabilities. Once the steady-state probabilities are obtained, then the availability of this configuration, We can obtain the mean time-to-failure for configuration 3 as follows.

Comparison of MTTF and Av
The cases listed below are provided in this section to compare four configurations according to their MTTF i and Av i (i = 1,2,3,4).

Scope of λ Results
Scope of q Results
Considering the same cases as in the previous subsection, we compare th / ratio for each configuration i (i = 1,2,3,4), i.e., / and / . Th results depicted in Figures 5-10 indicate that when λ, , or η increase, / an / increase for any configuration, but / and / decrease as μ, or β increase for any configuration. Figure 5 displays that configuration 2 has minimum / for different ranges of λ . The best configuration of / ratio 3 / 3 if 0.01 < λ < 0.2731. Otherwise, the best configuration of / r tio is 1 / 1 . The worst configuration of / ratio is 2 / 2 .
(a) / (b) /          From Figures 6-10, the optimum configuration based on the / is dependent on μ, , γ, η, and β. With the / ratio, the worst configuration is configuration 4 for different ranges of μ, , γ, η, and β. 2 / 2 has the minimum / ratio if μ > 1.8495, < 0.4514, γ > 1.6048 or η < 0.0780. Otherwise, the best configuration of the / is 1 / 1 . In addition, they also show that the optimum configuration based on the / will not change as μ, , γ, η, and β vary. 2 / 2 has the minimum / ratio. The worst configuration of / ratio is 2 / 2 for all ranges of β. The worst configuration is 2 / 2 if γ > 2.1352, < 0.2352, η > 0.1245 or μ > 1.8469. Otherwise, the worst configuration is configuration 4. From Figures 6-10, the optimum configuration based on the cost/Av is dependent on µ, q, γ, η, and β. With the cost/Av ratio, the worst configuration is configuration 4 for different ranges of µ, q, γ, η, and β. C 2 /Av 2 has the minimum cost/availability ratio if µ > 1.8495, q < 0.4514, γ > 1.6048 or η < 0.0780. Otherwise, the best configuration of the cost/availability is C 1 /Av 1 . In addition, they also show that the optimum configuration based on the cost/MTTF will not change as µ, q, γ, η, and β vary. C 2 /MTTF 2 has the minimum cost/MTTF ratio. The worst configuration of cost/bene f it ratio is C 2 /MTTF 2 for all ranges of β. The worst configuration is C 2 /MTTF 2 if γ > 2.1352, q < 0.2352, η > 0.1245 or µ > 1.8469. Otherwise, the worst configuration is configuration 4.

Conclusions
This work evaluated the cost-benefit of four standby unreliable retrial configurations with standby switching failure. The explicit and computationally tractable expressions for MTTF i and Av i (i = 1, 2, 3, 4) were derived for each configuration. A Matlab computer program is utilized to carry out the proposed approach. We ranked four configurations based on MTTF, Av and the cost/benefit ratio. According to our numerical results, the system with configuration 1 displayed the highest performance in most cases. Therefore, the procedure proposed in this paper can provide managers with a valuable tool to select the configuration with greatest benefit in terms of MTTF or availability. For future research, we may consider the general repair times of failed generators and the server. Moreover, we may also design the optimal management system.