Cost Benefit Analysis of Series Systems with Mixed Standby Components and k-stage Erlang Repair Time

In this paper, we compare the availability characteristics between three different series system configurations with mixed (cold and warm) standby components and k-stage Erlang repair times. Three configurations are studied under the assumption that the time-to-failure for each of the operative and warm standby components are assumed to be exponentially distributed with parameters λ andα , respectively. We present a recursive method, using the supplementary variable technique; we develop the explicit expressions for the steady-state availability. Under the cost/benefit criterion, comparisons are made based on assumed numerical values given to the distribution parameters, and to the cost of the operative and standby units.


Introduction
This type of problems is discussed by [2,3] have studied a two similar or dissimilar unit cold standby redundant system with preventive maintenance, inspection and two types of repairs. [4] carried out the cost-benefit analysis of a one-server two-unit system subject to different repair policies. They carried out the analysis under the assumption that a unit undergoes only one type of failure with no specific mention about inspection. [5,6] have studied the stochastic behaviour of some standby redundant systems. [9] considered the reliability and mean time to system failure (MTTF) analysis of a two-state complex with repairable system, consisting of two sub-systems A and B arranged in series, incorporating the concept of hardware and human failures. [7] considered reliability, availability and cost/benefit of four different series system configurations with mixed standby components. [11] studied a model representing a two-unit active and one-unit on standby human-machine system with general failed system repair time distribution. A two-unit warm standby system with constant failure rate and two types of repairmen and patience time were investigated by [8]. [1] studied the two unit standby system and obtained exact confidence limits for the steady-state availability of the system, when the failure rate of an operative unit is constant and the repair time of the failed unit is a two stage Erlang distribution. [10] studied the optimal system for series systems with mixed standby components.
The main contribution of this paper is three folds. First, we present a recursive method, using the supplementary variable technique and treating the supplementary variable as the remaining repair time, to develop the ( ) i Av ∞ , for configuration i, where 1, 2, 3 i = . The second purpose is to develop the explicit expressions for the steady-state availability, for the three configurations. Finally, we rank three configurations for the ( ) i Av ∞ based on assumed numerical values given to the system parameters.

Assumptions and Configurations Description
We consider a power plant of 10 MW satisfies the following assumptions: 1. The system comprises of operative components and mixed standby components "cold and warm".
2. The generators are available in components of both 10 and 5 MW.
3. Standby generators are always necessary in case of failure. 4. When the operative component fails, it is immediately replaced by a warm standby if it is available and the cold standby becomes warm standby component. 5. We assume that the switch is perfect and the switchover time from warm standby component to primary component, from cold standby component to warm standby component, from failure to repair, or from repair to cold standby component (or primary component if the system is short) is instantaneous. 6. When operative and warm standby components are repaired, they become as good as new.
7. The operative and warm standby components fail independently of the others and follow an exponential time to failure distribution with parameters λ and α respectively. 8. The time to repair of the components is independent and identically distributed random variable following k-stage Erlang distribution with probability density function ( ) b u and mean repair time 9. If one operative unit or warm standby unit is in repair, then arriving failed units have to wait in the queue until the server is available.
The above assumptions are common to all of the following three configurations: Configuration 1 is a serial system of one operative 10 MW component, one warm standby 10 MW component and one cold 10 MW component.
Configuration 2 is a serial system of two operative 5 MW components, one warm standby 5 MW component and one cold 5 MW component.
Configuration 3 is a serial system of one operative 10 MW component, two warm standby 10 MW components and one cold 10 MW component.

Cost-benefit factor
In this paper, we assume that the size-proportional costs for the operative components and warm standby components are given in Table 1. With this, we calculate the costs for each configuration ( ) 1, 2, 3 i i= which are shown in Table 2. Throughout this paper, we assume that i C be the cost of

Availability for Configuration 1
Relating the state of the system at time t and t dt + , we obtain: In steady-state, let us define and further define From (1)-(5), we obtain the following steady-state equations as follow: Now, from (6), we obtain: Further, we define ∫ Now, taking the Laplace-Stieltjes Transform (LST) on both sides of (7)-(9) and using (10) yield: We develop a recursive method to get the explicit ex- and, Again, setting s λ = in (11), it follows that: Setting s λ = in (12) yields: (17) Substituting (14) and (16) in (17), we have: Similarly, setting 0 s = in equation (12), we obtain: From (19) and (15), we have: Differentiating (13) with respect to s and setting s = 0 in the result, we obtain: Differentiating (11) with respect to s and then setting s = 0 in the result we have: Similarly, differentiating (12) with respect to s and setting s = 0 in the result, we find that: Substituting (23) in (21), we have: We can compute 3 P as follow: The explicit expression for the availability of configura- Using (15), (20), (25) and (26), we obtain the explicit expression for the

Availability for Configuration 2
Following the same procedures in the subsection 3.1, we can get the steady-state equations as follow: Now, from (28), we obtain: Taking the LST on both sides of (29)-(31) and using (32), we get that: and Again, setting 2 s λ = in (33), it follows that: Substituting (36) and (38) in (39), we have: Similarly, setting ( ) Differentiating (35) with respect to s and setting s = 0 in the result, we obtain: Differentiating (33) with respect to s and then setting s = 0 in the result yields: Likewise, differentiating (34) with respect to s and setting s = 0 in the result, we find that: Substituting (44) in (42), we have: where ( ) ( ) ( ) We can compute 4 P as follow. (49)

Availability for Configuration 3
We use the same procedure as above to obtain the steady-state equations as follow: Taking the LST on both sides of (51)-(54) and using (55), we obtain: Differentiating (59) with respect to s and setting s = 0 in the result, we obtain: Differentiating (56) with respect to s and then setting s = 0 in the result yields: Differentiating (57) with respect to s and then setting s = 0 in the result, it follows that: Likewise, differentiating (58) with respect to s and setting s = 0 in the result, we find that: Substituting (72) into (69), we have: Now, using the normalizing condition: We can compute 4 P as follow: The availability of configuration 3 ( )

Comparison between the three Configurations
The purpose of this section is to compare    Table 2. We compare i b , 1, 2,3 i = , in three cases as follow: