To study losses propagation and to show how these dynamics affect OEE in a complex system [25] this chapter presents some examples taken from an OEE study of a real manufacturing

system carried out by the authors through a process simulation analysis [19].

scraps

Using Overall Equipment Effectiveness for Manufacturing System Design

**1 2**

Case 1)

)()( 2211 11 == *sQsQ*

rework

**Figure 12.** Different cases of quality rate calculation

the design phase and the production.

station for each activity of the process.

process during the design (figure 13).

**Figure 13.** Two approaches for quality efficiency

**5. The simulation model**

**1 2 C**

Case 3)

http://dx.doi.org/10.5772/56089

))(( 2121 11 == *ssQQ*

**Production**

scraps

73

**Figure 11.** Rel OEE depending on buffer size in system affected by variability due to speed losses

### *4.5.4. Quality losses*

In this paragraph we analyze how quality losses propagate in the system and if it is possible to assess the effect of quality control on OEE and OTE.

First of all we have to consider that quality rate for a station is usually calculated considering only the time spent for the manufacturing of products that have been rejected in the same station. This traditional approach focuses on stations that cause defects but doesn't allow to point out completely the effect of the machine defectiveness on the system. In order to do so, the total time wasted by a station due to quality losses should include even the time spent for manufacturing of good products that will be rejected for defectiveness caused by other stations. In this sense quality losses depends on where scraps are identified and rejected. For example, scraps in the last station should be considered loss of time for the upstream station to estimate the real impact of the loss on the system and to estimate the theoretical production capacity needed in the upstream station. In conclusion the authors propose to calculate quality rate for a station considering as quality loss all time spent to manufacture products that will not complete the whole process successfully.

From a theoretical point of view we could consider the following case for calculation of quality rate of a station that depends on types of rejection (scraps or rework) and on quality controls positioning. If we consider two stations with an assigned defectiveness *Sj* and each station reworks its scraps with a rework cycle time equal to theoretical cycle time, quality rate could be formulate as shown in case 1 in figure 12. Each station will have quality losses (time spent to rework products) due its own defectiveness. If we consider two stations with an assigned defectiveness *Sj* and a quality control station at downstream each station, quality rate could be formulate as shown in case 2 in figure 12. The station 1, that is the upstream station, will have quality losses (time spent to work products that will be discarded) due to its own and station 2 defectiveness. If we consider two stations with an assigned defectiveness *Sj* and quality control station is only at the end of the line, quality rate quality rate could be formulate as shown in case 3 in figure 12. In this case both stations will have quality losses due to the propagation of defectiveness in the line. Case 2 and 3 point out that quality losses could be not simple to evaluate if we consider a long process both in design and management of system. In particular in the quality rate of station 1 we consider time lost for reject in the station 2.

**Figure 12.** Different cases of quality rate calculation

50.00%

1 21 41 61 81 101 121 141

In this paragraph we analyze how quality losses propagate in the system and if it is possible

First of all we have to consider that quality rate for a station is usually calculated considering only the time spent for the manufacturing of products that have been rejected in the same station. This traditional approach focuses on stations that cause defects but doesn't allow to point out completely the effect of the machine defectiveness on the system. In order to do so, the total time wasted by a station due to quality losses should include even the time spent for manufacturing of good products that will be rejected for defectiveness caused by other stations. In this sense quality losses depends on where scraps are identified and rejected. For example, scraps in the last station should be considered loss of time for the upstream station to estimate the real impact of the loss on the system and to estimate the theoretical production capacity needed in the upstream station. In conclusion the authors propose to calculate quality rate for a station considering as quality loss all time spent to manufacture products that will not

From a theoretical point of view we could consider the following case for calculation of quality rate of a station that depends on types of rejection (scraps or rework) and on quality controls positioning. If we consider two stations with an assigned defectiveness *Sj* and each station reworks its scraps with a rework cycle time equal to theoretical cycle time, quality rate could be formulate as shown in case 1 in figure 12. Each station will have quality losses (time spent to rework products) due its own defectiveness. If we consider two stations with an assigned

be formulate as shown in case 2 in figure 12. The station 1, that is the upstream station, will

and a quality control station at downstream each station, quality rate could

**Figure 11.** Rel OEE depending on buffer size in system affected by variability due to speed losses

to assess the effect of quality control on OEE and OTE.

complete the whole process successfully.

MIN MAX

**Buffer Size (j)**

60.00%

70.00%

80.00%

**Rel.P(j)**

72 Operations Management

*4.5.4. Quality losses*

defectiveness *Sj*

90.00%

100.00%

110.00%

Finally, it is important to highlight the different role that the quality efficiency plays during the design phase and the production.

When the system is producing, Operations Manager focuses his attention on the causes of the delectability with the aim to reduce it. When it is to design the production system, Operations Manager focuses on the expected quality efficiency of each station, on the location of quality control, on the process (rework or scraps) to identify the correct number of equipments or station for each activity of the process.

In this sense, the analysis is vertical during the production phase, but it follows the whole process during the design (figure 13).

**Figure 13.** Two approaches for quality efficiency

### **5. The simulation model**

To study losses propagation and to show how these dynamics affect OEE in a complex system [25] this chapter presents some examples taken from an OEE study of a real manufacturing system carried out by the authors through a process simulation analysis [19].

Simulation is run for each kind of time losses (Availability, Performance and Quality), to clearly show how each equipment ineffectiveness may compromise the performance of the whole system.

Considering equipment isolated from the system the OEE for the single machine is equal to its availability; in particular, relating to previous data, machines have an OEE equal to 0,67 and 0,5 respectively for insulating and packaging. The case points out how the losses due to

Using Overall Equipment Effectiveness for Manufacturing System Design

http://dx.doi.org/10.5772/56089

75

A simulation has been run to study the effect of buffer capacity in this case. Capacity of buffer downstream of insulating has been changed from 0 to 30 coils for different simulations. The results of simulations are shown in figure 15a. The OEE for both machines is equal to 0,33 with no buffer capacity. This results is the composition of availability of insulating and packaging (0,67 x 0,5) as expected. The OEEs increase in function of buffer dimension that avoids the propagation of major stoppage and availability losses propagation. Also the OTE is equal to 0,33 that is, according to formulation in (1) and as previously explained, equal to OEE of the

Insulating and packaging increase rapidly OEEs since a structural limits of buffer capacity of 15 coils; from this value OEEs of two stations converge on value of 0,5. The upstream insulating station, that has an availability greater than packaging, has to adapt itself to real cycle time of

It's important to point out that in performance monitoring of manufacturing plant the propagation of the previous losses is often gathered as performance losses (reduced speed or minor stoppage) in absence of specific data collection relating to major stoppage due to absence of material flow. So, if we consider also all other efficiency looses ignored in this sample, we can understand how much could be difficult to identify the real impact of this kind of efficiency losses monitoring the real system. Moreover simulation supports in system design in order to dimension buffer capacity (e.g. in this case structural limit for OEE is reached for 16 coils). Moreover through simulation it is possible to point out that the positive effect of buffer is

reduced with an higher cycle time of machine as shown in figure 15b.

**Figure 15.** OEE in function of buffer dimension (a) and cycle time (b)

major stoppage spread to other station in function of buffer capacity dimension.

last station but assessed in the system.

packaging that is the bottleneck station.

The simulation model is about a manufacturing plant for production of electrical cable. In particular we focuses on production of unipolar electrical cable that takes place by a flow-shop process. In the floor plant the production equipment is grouped in production areas arranged according to their functions (process layout). The different production areas are located along the line of product flow (product layout). Buffers are present amongst the production areas to stock the product in process. This particular plant allows to analyze deeply the problem of OEE-OTE investigation due to its complexity.

In terms of layout the production system was realized as a job shop system, although the flow of material from a station to another was continuous and typical of flow shop process. As stated in (§2) the reason lies on due to the huge size of the products that passes from a station to another. For this reason the buffer amid station, although present, couldn't contain huge amount of material.

The process implemented in the simulation model is shown in figure 14. Entities are unit quantity of cable that have different mass amongst stations. Parameters that are data input in the model are equipment speed, defectiveness, equipment failure rate and mean time to repair. Each parameters is described by a statistical distribution in order to simulate random condi‐ tion. In particular equipment speed has been simulated with a triangular distribution in order to simulate performance losses due to speed reduction.

The model evaluates OTE and OEE for each station as usually measured in manufacturing plant. The model has been validated through a plan of tests and its results of OEE has been compared with results obtained from an analytic evaluation.

**Figure 14.** ASME representation of manufacturing process

### **5.1. Example of availability losses propagation**

In accordance with the proposed method (§ 3.5) we show how availability losses propagate in the system and to assess the effect of buffer capacity on OEE through the simulation. We focuses on the insulating and packaging working stations. Technical data about availability of equipment are: mean time between failure for insulating is 20000 sec while for packaging is 30000 sec; mean time between repair for insulating is 10000 sec while for packaging is 30000 sec. The cycle time of the working stations are the same equal to 2800 sec for coil. The quality rates are set to 1. Idling, minor stoppages and reduced speed are not considered and set to 0.

Considering equipment isolated from the system the OEE for the single machine is equal to its availability; in particular, relating to previous data, machines have an OEE equal to 0,67 and 0,5 respectively for insulating and packaging. The case points out how the losses due to major stoppage spread to other station in function of buffer capacity dimension.

Simulation is run for each kind of time losses (Availability, Performance and Quality), to clearly show how each equipment ineffectiveness may compromise the performance of the

The simulation model is about a manufacturing plant for production of electrical cable. In particular we focuses on production of unipolar electrical cable that takes place by a flow-shop process. In the floor plant the production equipment is grouped in production areas arranged according to their functions (process layout). The different production areas are located along the line of product flow (product layout). Buffers are present amongst the production areas to stock the product in process. This particular plant allows to analyze deeply the problem of

In terms of layout the production system was realized as a job shop system, although the flow of material from a station to another was continuous and typical of flow shop process. As stated in (§2) the reason lies on due to the huge size of the products that passes from a station to another. For this reason the buffer amid station, although present, couldn't contain huge

The process implemented in the simulation model is shown in figure 14. Entities are unit quantity of cable that have different mass amongst stations. Parameters that are data input in the model are equipment speed, defectiveness, equipment failure rate and mean time to repair. Each parameters is described by a statistical distribution in order to simulate random condi‐ tion. In particular equipment speed has been simulated with a triangular distribution in order

The model evaluates OTE and OEE for each station as usually measured in manufacturing plant. The model has been validated through a plan of tests and its results of OEE has been

Roughing Drawing Bunching Insulating Packaging

In accordance with the proposed method (§ 3.5) we show how availability losses propagate in the system and to assess the effect of buffer capacity on OEE through the simulation. We focuses on the insulating and packaging working stations. Technical data about availability of equipment are: mean time between failure for insulating is 20000 sec while for packaging is 30000 sec; mean time between repair for insulating is 10000 sec while for packaging is 30000 sec. The cycle time of the working stations are the same equal to 2800 sec for coil. The quality rates are set to 1. Idling, minor stoppages and reduced speed are not considered and set to 0.

whole system.

74 Operations Management

amount of material.

OEE-OTE investigation due to its complexity.

to simulate performance losses due to speed reduction.

**Figure 14.** ASME representation of manufacturing process

**5.1. Example of availability losses propagation**

compared with results obtained from an analytic evaluation.

A simulation has been run to study the effect of buffer capacity in this case. Capacity of buffer downstream of insulating has been changed from 0 to 30 coils for different simulations. The results of simulations are shown in figure 15a. The OEE for both machines is equal to 0,33 with no buffer capacity. This results is the composition of availability of insulating and packaging (0,67 x 0,5) as expected. The OEEs increase in function of buffer dimension that avoids the propagation of major stoppage and availability losses propagation. Also the OTE is equal to 0,33 that is, according to formulation in (1) and as previously explained, equal to OEE of the last station but assessed in the system.

Insulating and packaging increase rapidly OEEs since a structural limits of buffer capacity of 15 coils; from this value OEEs of two stations converge on value of 0,5. The upstream insulating station, that has an availability greater than packaging, has to adapt itself to real cycle time of packaging that is the bottleneck station.

It's important to point out that in performance monitoring of manufacturing plant the propagation of the previous losses is often gathered as performance losses (reduced speed or minor stoppage) in absence of specific data collection relating to major stoppage due to absence of material flow. So, if we consider also all other efficiency looses ignored in this sample, we can understand how much could be difficult to identify the real impact of this kind of efficiency losses monitoring the real system. Moreover simulation supports in system design in order to dimension buffer capacity (e.g. in this case structural limit for OEE is reached for 16 coils). Moreover through simulation it is possible to point out that the positive effect of buffer is reduced with an higher cycle time of machine as shown in figure 15b.

**Figure 15.** OEE in function of buffer dimension (a) and cycle time (b)

### **5.2. Minor stoppages and speed reduction**

We run the simulation also for the case study (§ 4). The simulation shown how two stations, with the same theoretical cycle time (200 sec/coil) affected by a triangular distribution with a performance rate of 52% as single machine, have: 48% of performance rate with a capacity buffer of 1 coil and 50% of performance rate with a capacity buffer of 2 coils. But if we consider two stations with the same theoretic cycle time but affects by different triangular distributions so that theoretic performance rates differ, simulation shows how the performance rates of two stations converge towards the lowest one as expected (19), (20).

cases are shown in table 6 in which the proposal method has compared with the traditional one. The proposal method allowed to identify the correct efficiency, for example to dimension the drawing station, because it considers time wasted to manufacture products rejected in bunching station. The difference between values of Q2 and OTE is explained by the value of P2=0,95 that is due to the propagation of quality losses for the upstream station in performance losses for the downstream station. Moreover about positioning of quality control the case 2 has to be prefer because the simulation shows a positive effect on the OTE if the bunching

Case 2) 0,952 0,95 0,952 0,95 0,95 0,952

Case 3) 0,952 0,952 0,952 -- 0,952 0,952

**Table 6.** Comparison of quality rate calculation and evaluation of impact of quality control positioning on quality rates

The evaluation of Overall Equipment Effectiveness (OEE) and Overall Throughput Effective‐ ness (OTE) can be critical for the correct estimation of workstations number needed to realize the desired throughput (production system design), as also for the analysis and the continuous

The use of OEE as performance improvement tool has been widely described in the literature. But it has been less approached in system design for a correct evaluation of the system efficiency (OTE), in order to study losses propagation, overlapping of efficiency losses and

In this chapter, starting by the available literature on time losses, we identified a simplified set of relevant time-losses that need to be considered during the design phase. Then, through the simulation, we shown how OEE of single machine and the value of OTE of the whole system are interconnected and mutually influencing each other, due to the propagation of availability,

For each category of time losses we described the effects of efficiency losses propagation from a station to the system, for a correct estimation and analysis of OEE and OTE during manu‐ facturing system design. We also shown how to avoid losses propagation through adequate technical solutions which can be defined during system design as the buffer sizing, the

The simulation model shown in this chapter was based on a real production system and it used real data to study the losses propagation in a manufacturing plant for production of electrical

improvement of the system performance (during the system management).

**Proposal method Traditional method** Q1 Q2 OTE Q1 Q2 OTE

Using Overall Equipment Effectiveness for Manufacturing System Design

http://dx.doi.org/10.5772/56089

77

station is the system bottleneck (as it happens in the real system).

and on OTE

**6. Conclusions**

effective actions for losses reduction.

performance and quality losses throughout the system.

equipment configuration and the positioning of control stations.

Through the same simulation model we considered also the second type of performance losses propagation, due to the propagation of reduced speed caused by unbalanced line. Figure 16 shown the effect of unbalanced cycle time of stations relating to insulating and packaging. The station have the same P as single machine equal to 67% but different theoretical cycle time. In particular insulating, the upstream station, is faster than packag‐ ing. Availability and quality rate of stations is set to 1. The buffer capacity is set to 1 coil. A simulation has been run to study the effect of unbalancing station. Theoretical cycle time of insulating has been changed since theoretical cycle time of packaging that is fixed in mean. The simulation points out that insulating has to adapt itself to cycle time of packaging that is the bottleneck station. This results in the model as a lower value for performance rate of insulating station. The same happens often in real systems where the result is influenced by all the efficiency losses at the same time. The effect disappears gradually with a better balancing of two stations as in figure 16.

**Figure 16.** Performance rate of insulating and packaging in function of insulating cycle time

### **5.3. Quality losses**

In relation to the model, this sample focuses on the drawing and bunching working stations that have defectiveness set to 5%, the same cycle times and no other efficiency losses. The quality control has been changed simulating case 2 and 3. The results of simulation for the two cases are shown in table 6 in which the proposal method has compared with the traditional one. The proposal method allowed to identify the correct efficiency, for example to dimension the drawing station, because it considers time wasted to manufacture products rejected in bunching station. The difference between values of Q2 and OTE is explained by the value of P2=0,95 that is due to the propagation of quality losses for the upstream station in performance losses for the downstream station. Moreover about positioning of quality control the case 2 has to be prefer because the simulation shows a positive effect on the OTE if the bunching station is the system bottleneck (as it happens in the real system).

**Table 6.** Comparison of quality rate calculation and evaluation of impact of quality control positioning on quality rates and on OTE

### **6. Conclusions**

**5.2. Minor stoppages and speed reduction**

76 Operations Management

stations converge towards the lowest one as expected (19), (20).

a better balancing of two stations as in figure 16.

**performance rate**

**5.3. Quality losses**

**Figure 16.** Performance rate of insulating and packaging in function of insulating cycle time

In relation to the model, this sample focuses on the drawing and bunching working stations that have defectiveness set to 5%, the same cycle times and no other efficiency losses. The quality control has been changed simulating case 2 and 3. The results of simulation for the two

We run the simulation also for the case study (§ 4). The simulation shown how two stations, with the same theoretical cycle time (200 sec/coil) affected by a triangular distribution with a performance rate of 52% as single machine, have: 48% of performance rate with a capacity buffer of 1 coil and 50% of performance rate with a capacity buffer of 2 coils. But if we consider two stations with the same theoretic cycle time but affects by different triangular distributions so that theoretic performance rates differ, simulation shows how the performance rates of two

Through the same simulation model we considered also the second type of performance losses propagation, due to the propagation of reduced speed caused by unbalanced line. Figure 16 shown the effect of unbalanced cycle time of stations relating to insulating and packaging. The station have the same P as single machine equal to 67% but different theoretical cycle time. In particular insulating, the upstream station, is faster than packag‐ ing. Availability and quality rate of stations is set to 1. The buffer capacity is set to 1 coil. A simulation has been run to study the effect of unbalancing station. Theoretical cycle time of insulating has been changed since theoretical cycle time of packaging that is fixed in mean. The simulation points out that insulating has to adapt itself to cycle time of packaging that is the bottleneck station. This results in the model as a lower value for performance rate of insulating station. The same happens often in real systems where the result is influenced by all the efficiency losses at the same time. The effect disappears gradually with

> The evaluation of Overall Equipment Effectiveness (OEE) and Overall Throughput Effective‐ ness (OTE) can be critical for the correct estimation of workstations number needed to realize the desired throughput (production system design), as also for the analysis and the continuous improvement of the system performance (during the system management).

> The use of OEE as performance improvement tool has been widely described in the literature. But it has been less approached in system design for a correct evaluation of the system efficiency (OTE), in order to study losses propagation, overlapping of efficiency losses and effective actions for losses reduction.

> In this chapter, starting by the available literature on time losses, we identified a simplified set of relevant time-losses that need to be considered during the design phase. Then, through the simulation, we shown how OEE of single machine and the value of OTE of the whole system are interconnected and mutually influencing each other, due to the propagation of availability, performance and quality losses throughout the system.

> For each category of time losses we described the effects of efficiency losses propagation from a station to the system, for a correct estimation and analysis of OEE and OTE during manu‐ facturing system design. We also shown how to avoid losses propagation through adequate technical solutions which can be defined during system design as the buffer sizing, the equipment configuration and the positioning of control stations.

> The simulation model shown in this chapter was based on a real production system and it used real data to study the losses propagation in a manufacturing plant for production of electrical

cable. The validation of the model ensures the meaningful of the approach and of the identified set of possible solutions and hints.

[7] Dixon, N. A. V. J.R., The new performance challenge. Measuring operations for

Using Overall Equipment Effectiveness for Manufacturing System Design

http://dx.doi.org/10.5772/56089

79

[8] S. D., «Can CIM improve overall factory effetivenes,» in *Pan Pacific Microelectronic*

[11] A. V. A., Semiconductor Manufacturing Productivity- Overall Equipment Effective‐

[12] Rooda, D. R. A. J. J.E., «Equipment effectiveness: OEE revisited,» *IEEE Transactions on*

[13] Gamberini, G. L. R. B. R., «Alternative approaches for OEE evaluation: some guide‐ lines directing the choice,» in *XVII Summer School Francesco Turco*, Venice, (2012). [14] Grando, T. F. A., «Modelling Plant Capacity and Productivity,» *Production Planning*

[15] Spada, C. V. C., «The Impact of Cultural Issues and Interpersonal Behavior on Sus‐ tainable Excellence and Competitiveness: An Analysis of the Italian Context,» *Contri‐*

[16] Badiger, G. R, & Proposal, A. , «A. evaluation of OEE and impact of six big losses on equipment earning capacity,» *International Journal Process Management & Benchmark‐*

[17] Jacobs, A. N. C. B, & Operations, R. F. and supply chain management, McGraw-Hill,

[18] Oechsner, R. From overall equipment efficiency(OEE) to overall Fab effectiveness

[19] Introna, D. S. B. C. V, & Flow-shop, V. process oee calculation and improvement us‐

[20] R. MA, «Factory Level Metrics: Basis for Productivity Improvement,» in *Proceedings of the International Conference on Modeling and Analysis of Semiconductor*, Tempe, Arizo‐

[21] Scott, P. R. D., «Can overall factory effetiveness prolong Moore's Law?,» *Solid State*

[22] B. D., «Buffer size design linked to reliability performance: A simulative study,»

[23] Introna, G. A. V. V., «Increasing Availability of Production Flow lines through Opti‐ mal Buffer Sizing: a Simulative Study,» in *The 23rd European Modeling & Simulation*

*Computers & Industrial Engineering,* vol. 56, p. 1633-1641, 2009.

*Symposium (Simulation in Industry)*, Rome, (2011).

(OFE),» *Materials Science in Semiconductor Processing,* (2003). , 5 , 333-339.

[9] Waters, W. D. D.J., Operations Management, Kogan Page Publishers, (1999).

[10] Chase, A. N. J. F. R.B., Operations Management, McGraw-Hill, (2008).

world-class competition, Dow Jones Irwin, (1990).

ness (OEE) guidebook, SEMATECH, 1995.

*and Control,* n. 3, (2005). , 16 , 209-322.

*ing,* (2008). , 235 - 247 .

A cura di, (2010).

na, USA, 2002.

*Technology,* (1998). , 41 , 75-82.

*butions to Management Science,* (2008). , 95 - 113 .

ing simulation analysis,» in *MITIP*, Florence, (2007).

*Semiconductor Manufacturing,* n. 1, (2005). , 18 , 189-196.

*Symposium*, Kauai, HI, 1999.

By analyzing and each time losses we also shown how the choices taken during the design of the production system to increase the OTE (e.g. buffer size, maintenance configuration, etc.) affect the successive management of the operations.

### **Acknowledgements**

The realization of this chapter would not have been possible without the support of a person whose cooperated with the chair of Operations Management of University of Rome "Tor Vergata" in the last years, producing valuable research. The authors wish to express their gratitude to Dr. Bruna Di Silvio without whose knowledge, diligence and assistance this work would not have been successful.

### **Author details**

Vittorio Cesarotti1 , Alessio Giuiusa1,2 and Vito Introna1

1 University of Rome "Tor Vergata", Italy

2 Area Manager Inbound Operations at Amazon.com

### **References**

cable. The validation of the model ensures the meaningful of the approach and of the identified

By analyzing and each time losses we also shown how the choices taken during the design of the production system to increase the OTE (e.g. buffer size, maintenance configuration, etc.)

The realization of this chapter would not have been possible without the support of a person whose cooperated with the chair of Operations Management of University of Rome "Tor Vergata" in the last years, producing valuable research. The authors wish to express their gratitude to Dr. Bruna Di Silvio without whose knowledge, diligence and assistance this work

[1] H. H. S., «Manufacturing productivity improvement using effectivenes metrics and

[2] B. I., «Effective measurement and successful elements of company productivity: the basis of competitiveness and world prosperity,» *International Journal of Production*

[3] Jeong, P. D. K.Y., «Operational efficiency and effectiveness measurement,» *Interna‐ tional Journal of Operations and Production Management,* n. 1404-1416, (2001). , 21

[4] N. S., Introduction to TPM- Total Productive Maintenance, Productivity Press, 1988.

[5] S. R.J., World Class Manufacturing. The lesson of simplicity Applied, The Free Press,

[6] Womack, J. D. J.P., Lean Thinking, Simon & Schuster, (1996).

, Alessio Giuiusa1,2 and Vito Introna1

set of possible solutions and hints.

**Acknowledgements**

78 Operations Management

would not have been successful.

1 University of Rome "Tor Vergata", Italy

simulation analysis,» 2002.

*Economics,* vol. 52, pp. 203-213, 1997.

2 Area Manager Inbound Operations at Amazon.com

**Author details**

Vittorio Cesarotti1

**References**

1987.

affect the successive management of the operations.

[24] Connor, P. D. T. O. Practical Reliability Engineering (Fourth Ed.), New York: John Wiley & Sons, (2002).

**Chapter 4**

**Reliability and Maintainability in Operations**

The study of component and process reliability is the basis of many efficiency evaluations in Operations Management discipline. For example, in the calculation of the Overall Equipment Effectiveness (OEE) introduced by Nakajima [1], it is necessary to estimate a crucial parameter called availability. This is strictly related to reliability. Still as an example, consider how, in the study of service level, it is important to know the availability of machines, which again depends

Reliability is defined as the probability that a component (or an entire system) will perform its function for a specified period of time, when operating in its design environment. The elements necessary for the definition of reliability are, therefore, an unambiguous criterion for judging whether something is working or not and the exact definition of environmental conditions and usage. Then, reliability can be defined as the time dependent probability of correct operation if we assume that a component is used for its intended function in its design environment and if we clearly define what we mean with "failure". For this definition, any discussion on the

A broader definition of reliability is that "reliability is the science to predict, analyze, prevent and mitigate failures over time." It is a science, with its theoretical basis and principles. It also has sub-disciplines, all related - in some way - to the study and knowledge of faults. Reliability is closely related to mathematics, and especially to statistics, physics, chemistry, mechanics and electronics. In the end, given that the human element is almost always part of the systems,

In addition to the prediction of system durability, reliability also tries to give answers to other questions. Indeed, we can try to derive from reliability also the availability performance of a

> © 2013 De Carlo; licensee InTech. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use,

© 2013 De Carlo; licensee InTech. This is a paper distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

distribution, and reproduction in any medium, provided the original work is properly cited.

reliability basics starts with the coverage of the key concepts of probability.

Additional information is available at the end of the chapter

**Management**

Filippo De Carlo

**1. Introduction**

http://dx.doi.org/10.5772/54161

on their reliability and maintainability.

it often has to do with psychology and psychiatry.

## **Reliability and Maintainability in Operations Management**

Filippo De Carlo

[24] Connor, P. D. T. O. Practical Reliability Engineering (Fourth Ed.), New York: John

[25] Kane, J. O. Simulating production performance: cross case analysis and policy impli‐ cations,» *Industrial Management & Data Systems,* n. 4, (2004). , 104 , 309-321.

[26] Gondhinathan, B. A, & Proposal, R. , «A. evaluation of OEE and impact of six big losses on equipment earning capacity,» *International Journal Process Management &*

Wiley & Sons, (2002).

80 Operations Management

*Benchmarking,* n. 3, (2008). , 2 , 235-247.

Additional information is available at the end of the chapter

http://dx.doi.org/10.5772/54161

### **1. Introduction**

The study of component and process reliability is the basis of many efficiency evaluations in Operations Management discipline. For example, in the calculation of the Overall Equipment Effectiveness (OEE) introduced by Nakajima [1], it is necessary to estimate a crucial parameter called availability. This is strictly related to reliability. Still as an example, consider how, in the study of service level, it is important to know the availability of machines, which again depends on their reliability and maintainability.

Reliability is defined as the probability that a component (or an entire system) will perform its function for a specified period of time, when operating in its design environment. The elements necessary for the definition of reliability are, therefore, an unambiguous criterion for judging whether something is working or not and the exact definition of environmental conditions and usage. Then, reliability can be defined as the time dependent probability of correct operation if we assume that a component is used for its intended function in its design environment and if we clearly define what we mean with "failure". For this definition, any discussion on the reliability basics starts with the coverage of the key concepts of probability.

A broader definition of reliability is that "reliability is the science to predict, analyze, prevent and mitigate failures over time." It is a science, with its theoretical basis and principles. It also has sub-disciplines, all related - in some way - to the study and knowledge of faults. Reliability is closely related to mathematics, and especially to statistics, physics, chemistry, mechanics and electronics. In the end, given that the human element is almost always part of the systems, it often has to do with psychology and psychiatry.

In addition to the prediction of system durability, reliability also tries to give answers to other questions. Indeed, we can try to derive from reliability also the availability performance of a

© 2013 De Carlo; licensee InTech. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. © 2013 De Carlo; licensee InTech. This is a paper distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/3.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

system. In fact, availability depends on the time between two consecutive failures and on how long it takes to restore the system. Reliability study can be also used to understand how faults can be avoided. You can try to prevent potential failures, acting on the design, materials and maintenance.

faults. **Dynamic reliability**, instead, assumes that some failures, so-called primary failures, promote the emergence of secondary and tertiary faults, with a cascading effect. In this text

In the traditional paradigm of static reliability, individual components have a binary status: either working or failed. Systems, in turn, are composed by an integer number *n* of compo‐ nents, all mutually independent. Depending on how the components are configured in creating the system and according to the operation or failure of individual components, the

Let's consider a generic *X* system consisting of *n*elements. The static reliability modeling implies that the operating status of the *i* - *th* component is represented by the state function *Xi*

> 1 *if the i* - *th component works* 0 *if the i* - *th component fails*

> > 1 *if the system works* 0 *if the system fails*

The most common configuration of the components is the series system. A series system works if and only if all components work. Therefore, the status of a series system is given by the state

> *Xi* = min *i*∈{1,2,…,*n*}

System configurations are often represented graphically with Reliability Block Diagrams (RBDs) where each component is represented by a block and the connections between them express the configuration of the system. The operation of the system depends on the ability to cross the diagram from left to right only by passing through the elements in operation. Figure

The state of operation of the system is modeled by the state function Φ(*X* )

Φ(*X* )=∏ *i*=1 *n*

where the symbol ∏ indicates the product of the arguments.

1 contains the RBD of a four components series system.

**Figure 1.** Reliability block diagram for a four components (1,2,3,4) series system.

Φ(*X* )={

(1)

83

(2)

*Xi* (3)

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

we will only deal with static models of reliability.

*Xi* ={

system either works or does not work.

defined as:

function:

Reliability involves almost all aspects related to the possession of a property: cost management, customer satisfaction, the proper management of resources, passing through the ability to sell products or services, safety and quality of the product.

This chapter presents a discussion of reliability theory, supported by practical examples of interest in operations management. Basic elements of probability theory, as the sample space, random events and Bayes' theorem should be revised for a deeper understanding.

### **2. Reliability basics**

The period of regular operation of an equipment ends when any chemical-physical phenom‐ enon, said fault, occurred in one or more of its parts, determines a variation of its nominal performances. This makes the behavior of the device unacceptable. The equipment passes from the state of operation to that of non-functioning.

In Table 1 faults are classified according to their origin. For each failure mode an extended description is given.

**Table 1.** Main causes of failure. The table shows the main cases of failure with a detailed description

To study reliability you need to transform reality into a model, which allows the analysis by applying laws and analyzing its behavior [2]. Reliability models can be divided into static and dynamic ones. **Static models** assume that a failure does not result in the occurrence of other faults. **Dynamic reliability**, instead, assumes that some failures, so-called primary failures, promote the emergence of secondary and tertiary faults, with a cascading effect. In this text we will only deal with static models of reliability.

system. In fact, availability depends on the time between two consecutive failures and on how long it takes to restore the system. Reliability study can be also used to understand how faults can be avoided. You can try to prevent potential failures, acting on the design, materials and

Reliability involves almost all aspects related to the possession of a property: cost management, customer satisfaction, the proper management of resources, passing through the ability to sell

This chapter presents a discussion of reliability theory, supported by practical examples of interest in operations management. Basic elements of probability theory, as the sample space,

The period of regular operation of an equipment ends when any chemical-physical phenom‐ enon, said fault, occurred in one or more of its parts, determines a variation of its nominal performances. This makes the behavior of the device unacceptable. The equipment passes from

In Table 1 faults are classified according to their origin. For each failure mode an extended

Temperature Operational variable that depends mainly on the specific characteristics of the material (thermal inertia), as well as the spatial and temporal distribution of heat sources.

Wear State of physical degradation of the component; it manifests itself as a result of aging phenomena

Corrosion Phenomenon that depends on the characteristics of the environment in which the component is

To study reliability you need to transform reality into a model, which allows the analysis by applying laws and analyzing its behavior [2]. Reliability models can be divided into static and dynamic ones. **Static models** assume that a failure does not result in the occurrence of other

that make the component no longer suitable.

**Table 1.** Main causes of failure. The table shows the main cases of failure with a detailed description

Function of the temporal and spatial distribution of the load conditions and of the response of the material. The structural characteristics of the component play an important role, and should be assessed in the broadest form as possible, incorporating also possible design errors, embodiments,

that accompany the normal activities (friction between the materials, exposure to harmful agents,

operating. These conditions can lead to material degradation or chemical and physical processes

random events and Bayes' theorem should be revised for a deeper understanding.

products or services, safety and quality of the product.

the state of operation to that of non-functioning.

material defects, etc..

etc..)

maintenance.

82 Operations Management

**2. Reliability basics**

description is given.

Stress, shock, fatigue

**Failure cause Description**

In the traditional paradigm of static reliability, individual components have a binary status: either working or failed. Systems, in turn, are composed by an integer number *n* of compo‐ nents, all mutually independent. Depending on how the components are configured in creating the system and according to the operation or failure of individual components, the system either works or does not work.

Let's consider a generic *X* system consisting of *n*elements. The static reliability modeling implies that the operating status of the *i* - *th* component is represented by the state function *Xi* defined as:

$$\mathbf{X}\_i = \begin{cases} 1 & \text{if the } i \text{ - } th \text{ component works} \\ 0 & \text{if the } i \text{ - } th \text{ component fails} \end{cases} \tag{1}$$

The state of operation of the system is modeled by the state function Φ(*X* )

$$\Phi(\mathbf{X}) = \begin{cases} 1 & \text{if } \text{the system works} \\ 0 & \text{if the system fails} \end{cases} \tag{2}$$

The most common configuration of the components is the series system. A series system works if and only if all components work. Therefore, the status of a series system is given by the state function:

$$\text{sp(X \)}=\prod\_{i=1}^{n} X\_i = \min\_{i \in \{1,2,\ldots,n\}} X\_i \tag{3}$$

where the symbol ∏ indicates the product of the arguments.

System configurations are often represented graphically with Reliability Block Diagrams (RBDs) where each component is represented by a block and the connections between them express the configuration of the system. The operation of the system depends on the ability to cross the diagram from left to right only by passing through the elements in operation. Figure 1 contains the RBD of a four components series system.

**Figure 1.** Reliability block diagram for a four components (1,2,3,4) series system.

The second most common configuration of the components is the parallel system. A parallel system works if and only if at least one component is working. A parallel system does not work if and only if all components do not work. So, if Φ - (*X* ) is the function that represents the state of not functioning of the system and *X i* indicates the non-functioning of the *i* - *th* element, you can write:

$$\stackrel{\circ}{\Phi}(X) = \prod\_{i=1}^{n} X\_i \tag{4}$$

**Figure 3.** Series-parallel system. The picture shows the RBD of a system due to the series-parallel model of 9 elementa‐

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

85

**Figure 4.** Calculation of the state function of a series-parallel. Referring to the configuration of Figure 3, the state function of the system is calculated by first making the state functions of the parallel of{1,2}, of {3,4, 5} and of

{6,7, 8 , 9}. Then we evaluate the state function of the series of the three groups just obtained.

ry units.

Accordingly, the state of a parallel system is given by the state function:

$$\mathfrak{O}(X) = 1 - \prod\_{i=1}^{n} \{1 - X\_i\} = \coprod\_{i=1}^{n} X\_i = \max\_{i \in \{1, 2, \dots, n\}} X\_i \tag{5}$$

where the symbol ∐ indicates the complement of the product of the complements of the arguments. Figure 2 contains a RBD for a system of four components arranged in parallel.

**Figure 2.** Parallel system. The image represents the RBD of a system of four elements (1,2,3,4) arranged in a reliability parallel configuration.

Another common configuration of the components is the series-parallel systems. In these systems, components are configured using combinations in series and parallel configurations. An example of such a system is shown in Figure 3.

State functions for series-parallel systems are obtained by decomposition of the system. With this approach, the system is broken down into subsystems or configurations that are in series or in parallel. The state functions of the subsystems are then combined appropriately, de‐ pending on how they are configured. A schematic example is shown in Figure 4.

The second most common configuration of the components is the parallel system. A parallel system works if and only if at least one component is working. A parallel system does not

(*X* )=∏ *i*=1 *n X* -

Φ -

(1 - *X i*

) =∐ *i*=1 *n*

where the symbol ∐ indicates the complement of the product of the complements of the arguments. Figure 2 contains a RBD for a system of four components arranged in parallel.

**Figure 2.** Parallel system. The image represents the RBD of a system of four elements (1,2,3,4) arranged in a reliability

Another common configuration of the components is the series-parallel systems. In these systems, components are configured using combinations in series and parallel configurations.

State functions for series-parallel systems are obtained by decomposition of the system. With this approach, the system is broken down into subsystems or configurations that are in series or in parallel. The state functions of the subsystems are then combined appropriately, de‐

pending on how they are configured. A schematic example is shown in Figure 4.

*Xi* = max *i*∈{1,2,…,*n*}

Accordingly, the state of a parallel system is given by the state function:

*i*=1 *n*

Φ(*X* )=1 - ∏

(*X* ) is the function that represents the

*i* indicates the non-functioning of the *i* - *th* element,

*i* (4)

*Xi* (5)

work if and only if all components do not work. So, if Φ

state of not functioning of the system and *X*

you can write:

84 Operations Management

parallel configuration.

An example of such a system is shown in Figure 3.

**Figure 3.** Series-parallel system. The picture shows the RBD of a system due to the series-parallel model of 9 elementa‐ ry units.

**Figure 4.** Calculation of the state function of a series-parallel. Referring to the configuration of Figure 3, the state function of the system is calculated by first making the state functions of the parallel of{1,2}, of {3,4, 5} and of {6,7, 8 , 9}. Then we evaluate the state function of the series of the three groups just obtained.

A particular component configuration, widely recognized and used, is the **parallel** *k***out of***n*. A system *k*out of *n*works if and only if at least *k*of the *n*components works. Note that a series system can be seen as a system *n*out of *n*and a parallel system is a system 1 out of*n*. The state function of a system *k*out of *n*is given by the following algebraic system:

$$\Phi(X) = \begin{cases} 1 & \text{if } \sum\_{i=1}^{n} X\_i \ge k \\ 0 & \text{otherwise} \end{cases} \tag{6}$$

of the system. This configuration is, therefore, constructed with the creation of a series subsystem for each path using only the minimum components of that set. Then, these subsys‐ tems are connected in parallel. An example of an equivalent system is shown in Figure 7.

**Figure 6.** Minimal Cut Set. The system of the left contains the minimal cut set, indicated by the dashed lines, shown in the right part. Each of them represents a minimum subset of the components of the system such that the failure of all

**Figure 7.** Equivalent configurations with MPS. You build a series subsystem for each MPS. Then such subsystems are

The second equivalent configuration, is based on the logical principle that the failure of all the components of any MCS implies the fault of the system. This configuration is built with the creation of a parallel subsystem for each MCS using only the components of that group. Then,

After examining the components and the status of the system, the next step in the static modeling of reliability is that of considering the probability of operation of the component and

1 2

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

87

1 3 5

4 3 2

4 5

5

2

4

connected in parallel.

of the system.

3

components in the subset does not imply the operation of the system.

these subsystems are connected in series (see Figure 8).

1

The RBD for a system *k*out of *n*has an appearance identical to the RBD schema of a parallel system of *n*components with the addition of a label "*k* out of *n*". For other more complex system configurations, such as the bridge configuration (see Figure 5), we may use more intricate techniques such as the minimal path set and the minimal cut set, to construct the system state function.

A Minimal Path Set - MPS is a subset of the components of the system such that the operation of all the components in the subset implies the operation of the system. The set is minimal because the removal of any element from the subset eliminates this property. An example is shown in Figure 5.

**Figure 5.** Minimal Path Set. The system on the left contains the minimal path set indicated by the arrows and shown in the right part. Each of them represents a minimal subset of the components of the system such that the operation of all the components in the subset implies the operation of the system.

A Minimal Cut Set - MCS is a subset of the components of the system such that the failure of all components in the subset does not imply the operation of the system. Still, the set is called min‐ imal because the removal of any component from the subset clears this property (see Figure 6).

MCS and MPS can be used to build equivalent configurations of more complex systems, not referable to the simple series-parallel model. The first equivalent configuration is based on the consideration that the operation of all the components, in at least a MPS, entails the operation

Reliability and Maintainability in Operations Management http://dx.doi.org/10.5772/54161 87

A particular component configuration, widely recognized and used, is the **parallel** *k***out of***n*. A system *k*out of *n*works if and only if at least *k*of the *n*components works. Note that a series system can be seen as a system *n*out of *n*and a parallel system is a system 1 out of*n*. The state

> *i*=1 *n Xi* ≥*k*

The RBD for a system *k*out of *n*has an appearance identical to the RBD schema of a parallel system of *n*components with the addition of a label "*k* out of *n*". For other more complex system configurations, such as the bridge configuration (see Figure 5), we may use more intricate techniques such as the minimal path set and the minimal cut set, to construct the system state

A Minimal Path Set - MPS is a subset of the components of the system such that the operation of all the components in the subset implies the operation of the system. The set is minimal because the removal of any element from the subset eliminates this property. An example is

**Figure 5.** Minimal Path Set. The system on the left contains the minimal path set indicated by the arrows and shown in the right part. Each of them represents a minimal subset of the components of the system such that the operation of

A Minimal Cut Set - MCS is a subset of the components of the system such that the failure of all components in the subset does not imply the operation of the system. Still, the set is called min‐ imal because the removal of any component from the subset clears this property (see Figure 6).

MCS and MPS can be used to build equivalent configurations of more complex systems, not referable to the simple series-parallel model. The first equivalent configuration is based on the consideration that the operation of all the components, in at least a MPS, entails the operation

all the components in the subset implies the operation of the system.

(6)

0 *otherwise*

function of a system *k*out of *n*is given by the following algebraic system:

function.

86 Operations Management

shown in Figure 5.

Φ(*X* )={1 *if* ∑

**Figure 6.** Minimal Cut Set. The system of the left contains the minimal cut set, indicated by the dashed lines, shown in the right part. Each of them represents a minimum subset of the components of the system such that the failure of all components in the subset does not imply the operation of the system.

of the system. This configuration is, therefore, constructed with the creation of a series subsystem for each path using only the minimum components of that set. Then, these subsys‐ tems are connected in parallel. An example of an equivalent system is shown in Figure 7.

**Figure 7.** Equivalent configurations with MPS. You build a series subsystem for each MPS. Then such subsystems are connected in parallel.

The second equivalent configuration, is based on the logical principle that the failure of all the components of any MCS implies the fault of the system. This configuration is built with the creation of a parallel subsystem for each MCS using only the components of that group. Then, these subsystems are connected in series (see Figure 8).

After examining the components and the status of the system, the next step in the static modeling of reliability is that of considering the probability of operation of the component and of the system.

**Figure 8.** Equivalent configurations with MCS. You build a subsystem in parallel for each MCS. Then the subsystems are connected in series.

The reliability *Ri* of the *i* - *th* component is defined by:

$$R\_i = P\left(X\_i = 1\right) \tag{7}$$

*R* =1 - ∏ *i*=1 *n* (1 - *Ri*

> *i*=1 *n*

follows:

*R* =*P*(⋃ *i*=1 *n*

(*Xi* =1)) =1 - *P*(⋂

system with *n* elements is given by:

their co-product: 1 - (1 - 0.85)4 =0.9995.

system.

*i*=1 *n*

(*Xi* =0)) =1 - ∏

) =∐ *i*=1 *n*

In fact, from the definition of system reliability and by the properties of event probabilities, it

*P*(*Xi* =0) = =1 - ∏

In many parallel systems, components are identical. In this case, the reliability of a parallel

**Figure 10.** A parallel system consisting of 4 elements with the same reliability of 0.85. The system reliability s given by

For a series-parallel system, system reliability is determined using the same approach of decomposition used to construct the state function for such systems. Consider, for instance, the system drawn in Figure 11, consisting of 9 elements with reliability *R*1 =*R*2 =0.9; *R*3 =*R*4 =*R*5 =0.8 and *R*6 =*R*7 =*R*8 =*R*9 =0.7. Let's calculate the overall reliability of the

*R* =1 - (1 - *Ri*

*i*=1 *n*

1 - *P*(*Xi* =1) =1 - ∏

*Ri* (10)

Reliability and Maintainability in Operations Management

*i*=1 *n* (1 - *Ri*

)*n* (12)

) =∐ *i*=1 *n*

http://dx.doi.org/10.5772/54161

*Ri* (11)

89

while the **reliability of the system***R* is defined as in equation 8:

$$R = P\{\uplus(X) = 1\}\tag{8}$$

The methodology used to calculate the reliability of the system depends on the configuration of the system itself. For a series system, the reliability of the system is given by the product of the individual reliability (law of Lusser, defined by German engineer Robert Lusser in the 50s):

$$R = \prod\_{i=1}^{n} R\_i \qquad \text{since} \qquad R = P\left(\bigcap\_{i=1}^{n} \left(X\_i = 1\right)\right) = \prod\_{i=1}^{n} P\left(X\_i = 1\right) = \prod\_{i=1}^{n} R\_i \tag{9}$$

For an example, see Figure 9.

**Figure 9.** serial system consisting of 4 elements with reliability equal to 0.98, 0.99, 0.995 and 0.975. The reliability of the whole system is given by their product: *R* = 0.98 · 0.99 · 0.995 · 0.975 = 0.941

For a parallel system, reliability is:

Reliability and Maintainability in Operations Management http://dx.doi.org/10.5772/54161 89

)*n* (12)

$$R = 1 \ -\prod\_{i=1}^{n} \left(1 \ -R\_i\right) = \coprod\_{i=1}^{n} R\_i \tag{10}$$

In fact, from the definition of system reliability and by the properties of event probabilities, it follows:

$$R = P\left(\bigcup\_{i=1}^{n} \left(X\_i = 1\right)\right) = 1 - P\left(\bigcap\_{i=1}^{n} \left(X\_i = 0\right)\right) = 1 - \prod\_{i=1}^{n} P\left(X\_i = 0\right) = = 1 - \prod\_{i=1}^{n} \left[1 - P\left(X\_i = 1\right)\right] = 1 - \prod\_{i=1}^{n} \left(1 - R\_i\right) = \prod\_{i=1}^{n} R\_i \tag{11}$$

In many parallel systems, components are identical. In this case, the reliability of a parallel system with *n* elements is given by:

*R* =1 - (1 - *Ri*

The reliability *Ri*

are connected in series.

88 Operations Management

of the *i* - *th* component is defined by:

The methodology used to calculate the reliability of the system depends on the configuration of the system itself. For a series system, the reliability of the system is given by the product of the individual reliability (law of Lusser, defined by German engineer Robert Lusser in the 50s):

**Figure 8.** Equivalent configurations with MCS. You build a subsystem in parallel for each MCS. Then the subsystems

(*Xi* =1)) =∏

**Figure 9.** serial system consisting of 4 elements with reliability equal to 0.98, 0.99, 0.995 and 0.975. The reliability of

*i*=1 *n*

*P*(*Xi* =1) =∏

*i*=1 *n*

*Ri* (9)

*i*=1 *n*

while the **reliability of the system***R* is defined as in equation 8:

*Ri* since *R* =*P*(⋂

the whole system is given by their product: *R* = 0.98 · 0.99 · 0.995 · 0.975 = 0.941

*R* =∏ *i*=1 *n*

For an example, see Figure 9.

For a parallel system, reliability is:

*Ri* =*P*(*Xi* =1) (7)

*R* =*P*(Φ(*X* )=1) (8)

**Figure 10.** A parallel system consisting of 4 elements with the same reliability of 0.85. The system reliability s given by their co-product: 1 - (1 - 0.85)4 =0.9995.

For a series-parallel system, system reliability is determined using the same approach of decomposition used to construct the state function for such systems. Consider, for instance, the system drawn in Figure 11, consisting of 9 elements with reliability *R*1 =*R*2 =0.9; *R*3 =*R*4 =*R*5 =0.8 and *R*6 =*R*7 =*R*8 =*R*9 =0.7. Let's calculate the overall reliability of the system.

**Figure 11.** The system consists of three groups of blocks arranged in series. Each block is, in turn, formed by elements in parallel. First we must calculate *R*1,2 =1 - (1 - 0.8)2 =0.99. So it is possible to estimated *R*3,4,5 =1 - (1 - 0.8)3 =0.992. Then we must calculate the reliability of the last parallel block *R*6,7,8,9 =1 - (1 - 0.7)4 =0.9919. Finally, we proceed to the series of the three blocks: *R* = *R*1,2 ∙ *R*3,4,5 ∙ *R*6,7,8,9 =0.974.

To calculate the overall reliability, for all other types of systems which can't be brought back to a series-parallel scheme, it must be adopted a more intensive calculation approach [3] that is normally done with the aid of special software.

Reliability functions of the system can also be used to calculate measures of **reliability importance**.

These measurements are used to assess which components of a system offer the greatest opportunity to improve the overall reliability. The most widely recognized definition of reliability importance *I ' i* of the components is the **reliability marginal gain**, in terms of overall system rise of functionality, obtained by a marginal increase of the component reliability:

$$\begin{array}{c} I \stackrel{\cdot}{.i} = \frac{\partial \mathcal{R}}{\partial \mathcal{R}\_i} \end{array} \tag{13}$$

*Ii* = ∏ *j*=1 *j*≠*i*

If the system is arranged in parallel, the reliability importance becomes as follows:

*n*

(1 - *R j*

It follows that the most important component in a parallel system is the more reliable. With the same data as the previous example, this time having a parallel arrangement, we can verify Eq. 16 for the first item: *I*1 =*R*(11) - *R*(01) = 1 - (1 - 1) ·(1 - 0.8)∙(1 - 0.7) - 1 - (1 - 0)· (1 - 0.8)∙(1 - 0.7)

For the calculation of the reliability importance of components belonging to complex systems, which are not attributable to the series-parallel simple scheme, reliability of different systems must be counted. For this reason the calculation is often done using automated algorithms.

Suppose you have studied the reliability of a component, and found that it is 80% for a mission duration of 3 hours. Knowing that we have 5 identical items simultaneously active, we might be interested in knowing what the overall reliability of the group would be. In other words, we want to know what is the probability of having a certain number of items functioning at

Consider a set of *m*identical and independent systems in a same instant, each having a reliability*R*. The group may represent a set of systems in use, independent and identical, or could represent a set of devices under test, independent and identical. A discrete random variable of great interest reliability is*N* , the number of functioning items. Under the assump‐ tions specified, *N* is a binomial random variable, which expresses the probability of a Bernoulli process. The corresponding probabilistic model is, therefore, the one that describes the extraction of balls from an urn filled with a known number of red and green balls. Suppose that the percentage *R*of green balls is coincident with the reliability after 3 hours. After each extraction from the urn, the ball is put back in the container. Extraction is repeated *m*times, and we look for the probability of finding *n*green. The sequence of random variables thus obtained is a Bernoulli process of which each extraction is a test. Since the probability of obtaining *N* successes in *m*extractions from an urn, with restitution of the ball, follows the

binomial distribution B(*m*, *R*)B, the probability mass function of *N* is the well-known:

the end of the 3 hours of mission. This issue is best known as fleet reliability.

*Ii* = ∏ *j*=1 *j*≠*i*

=1 - 0 - 1 + (1 - 0.8)∙(1 - 0.7)=(1 - 0.8)∙(1 - 0.7).

**3. Fleet reliability**

*n*

Thus, the most important component (in terms of reliability) in a series system is the less reliable. For example, consider three elements of reliability *R*1 =0.9, *R*2 =0.8 e *R*3 =0.7. It is therefore: *I*1 =0.8∙0.7=0.56, *I*2 =0.9∙0.7=0.63 and *I*3 =0.9 · 0.8=0.72 which is the higher value.

*R j* (15)

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

91

) (16)

For other system configurations, an alternative approach facilitates the calculation of reliability importance of the components. Let *R*(1*i* )be the reliability of the system modified so that *Ri* =1 and *R*(0*i* )be the reliability of the system modified with*Ri* =0, always keeping unchanged the other components. In this context, the reliability importance *Ii* is given by:

$$I\_i = R\begin{pmatrix} 1\_i \end{pmatrix} - R\begin{pmatrix} 0\_i \end{pmatrix} \tag{14}$$

In a series system, this formulation is equivalent to writing:

$$I\_i = \prod\_{j=1 \atop j \neq i}^{n} \mathbb{R}\_{\,j} \tag{15}$$

Thus, the most important component (in terms of reliability) in a series system is the less reliable. For example, consider three elements of reliability *R*1 =0.9, *R*2 =0.8 e *R*3 =0.7. It is therefore: *I*1 =0.8∙0.7=0.56, *I*2 =0.9∙0.7=0.63 and *I*3 =0.9 · 0.8=0.72 which is the higher value.

If the system is arranged in parallel, the reliability importance becomes as follows:

$$I\_i = \prod\_{j=1 \atop j \neq i}^{n} \{1 - R\_{\,j}\} \tag{16}$$

It follows that the most important component in a parallel system is the more reliable. With the same data as the previous example, this time having a parallel arrangement, we can verify Eq. 16 for the first item: *I*1 =*R*(11) - *R*(01) = 1 - (1 - 1) ·(1 - 0.8)∙(1 - 0.7) - 1 - (1 - 0)· (1 - 0.8)∙(1 - 0.7) =1 - 0 - 1 + (1 - 0.8)∙(1 - 0.7)=(1 - 0.8)∙(1 - 0.7).

For the calculation of the reliability importance of components belonging to complex systems, which are not attributable to the series-parallel simple scheme, reliability of different systems must be counted. For this reason the calculation is often done using automated algorithms.

### **3. Fleet reliability**

**Figure 11.** The system consists of three groups of blocks arranged in series. Each block is, in turn, formed by elements in parallel. First we must calculate *R*1,2 =1 - (1 - 0.8)2 =0.99. So it is possible to estimated *R*3,4,5 =1 - (1 - 0.8)3 =0.992. Then we must calculate the reliability of the last parallel block *R*6,7,8,9 =1 - (1 - 0.7)4 =0.9919. Finally, we proceed to the

To calculate the overall reliability, for all other types of systems which can't be brought back to a series-parallel scheme, it must be adopted a more intensive calculation approach [3] that

Reliability functions of the system can also be used to calculate measures of **reliability**

These measurements are used to assess which components of a system offer the greatest opportunity to improve the overall reliability. The most widely recognized definition of

system rise of functionality, obtained by a marginal increase of the component reliability:

For other system configurations, an alternative approach facilitates the calculation of reliability

) - *R*(0*i*

)be the reliability of the system modified with*Ri* =0, always keeping unchanged the

*I* ' *i* = ∂ *R* ∂ *Ri*

*Ii* =*R*(1*i*

other components. In this context, the reliability importance *Ii*

In a series system, this formulation is equivalent to writing:

*i* of the components is the **reliability marginal gain**, in terms of overall

)be the reliability of the system modified so that *Ri* =1

is given by:

) (14)

(13)

series of the three blocks: *R* = *R*1,2 ∙ *R*3,4,5 ∙ *R*6,7,8,9 =0.974.

**importance**.

90 Operations Management

and *R*(0*i*

reliability importance *I '*

importance of the components. Let *R*(1*i*

is normally done with the aid of special software.

Suppose you have studied the reliability of a component, and found that it is 80% for a mission duration of 3 hours. Knowing that we have 5 identical items simultaneously active, we might be interested in knowing what the overall reliability of the group would be. In other words, we want to know what is the probability of having a certain number of items functioning at the end of the 3 hours of mission. This issue is best known as fleet reliability.

Consider a set of *m*identical and independent systems in a same instant, each having a reliability*R*. The group may represent a set of systems in use, independent and identical, or could represent a set of devices under test, independent and identical. A discrete random variable of great interest reliability is*N* , the number of functioning items. Under the assump‐ tions specified, *N* is a binomial random variable, which expresses the probability of a Bernoulli process. The corresponding probabilistic model is, therefore, the one that describes the extraction of balls from an urn filled with a known number of red and green balls. Suppose that the percentage *R*of green balls is coincident with the reliability after 3 hours. After each extraction from the urn, the ball is put back in the container. Extraction is repeated *m*times, and we look for the probability of finding *n*green. The sequence of random variables thus obtained is a Bernoulli process of which each extraction is a test. Since the probability of obtaining *N* successes in *m*extractions from an urn, with restitution of the ball, follows the binomial distribution B(*m*, *R*)B, the probability mass function of *N* is the well-known:

$$P\{\mathbf{N}=n\} = \frac{m\,\mathrm{l}}{n\,\mathrm{l}\,\mathrm{(m\,-\,n)}\,\mathrm{l}}\,\mathrm{R}\,\,^{n}\{\mathbf{1}\,-\,\mathrm{R}\}^{m\,\mathrm{-}n} \tag{17}$$

failure rate is the ratio between the instantaneous probability of failure in a neighborhood of *t*- conditioned to the fact that the element is healthy in *t*- and the amplitude of the same

The hazard function *λ*(*t*) [5] coincides with the intensity function *z*(*t*) of a Poisson process. The

*P*(*t* ≤ *T* < *t* + ∆ *t* | *T* ≥ *t*)

Thanks to Bayes' theorem, it can be shown that the relationship between the hazard function,

Thanks to the previous equation, with some simple mathematical manipulations, we obtain

*f* (*t*)=*λ*(*t*)∙*e*

The most popular conceptual model of the hazard function is the **bathtub curve**. According to this model, the failure rate of the device is relatively high and descending in the first part of the device life, due to the potential manufacturing defects, called **early failures**. They manifest themselves in the first phase of operation of the system and their causes are often linked to structural deficiencies, design or installation defects. In terms of reliability, a system

Later, at the end of the life of the device, the failure rate increases due to wear phenomena. They are caused by alterations of the component for material and structural aging. The beginning of the period of wear is identified by an increase in the frequency of failures which continues as time goes by. The **wear-out failures** occur around the average age of operating;

the only way to avoid this type of failure is to replace the population in advance.

*λ*(*t*)= *f* (*t*)

*R*(*t*)=*e* -*∫* 0 *t*

∆ *t* (22)

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

93

*R*(*t*) (23)

*λ*(*u*)∙*du* (24)

0 *t*

*λ*(*u*)∙*du* (26)

*λ*(*u*)*du* (25)

*R*(*t*) *dR*(*t*)= - *λ*(*t*)*dt* →ln *R*(*t*) - ln *R*(0) = - *∫*

*λ*(*t*)= *lim Δt*→0

density of probability of failure and reliability is the following:

neighborhood.

hazard function is given by:

the following relation:

*R*(*t*)= *f* (*t*)

*λ*(*t*) = 1

In fact, since ln *R*(0) =ln 1 =0, we have:

*dt* = - 1

*λ*(*t*) ∙ *dR*(*t*)

From equation 24 derive the other two fundamental relations:

*F* (*t*)=1 - *e*

*dt* → 1

that manifests infantile failures improves over the course of time.

*λ*(*t*) ∙ *dF* (*t*)

The expected value of *N* is given by: *E*(*N* )=*μN* =*m*∙*R*and the standard deviation is: *σN* = *m*∙*R* ∙(1 - *R*).

Let's consider, for example, a corporate fleet consisting of 100 independent and identical systems. All systems have the same mission, independent from the other missions. Each system has a reliability of mission equal to 90%. We want to calculate the average number of missions completed and also what is the probability that at least 95% of systems would complete their mission. This involves analyzing the distribution of the binomial random variable character‐ ized by *R* = 0.90and*m* = 100. The expected value is given by *E*(*N* )=*μN* =100∙0.9=90.

The probability that at least 95% of the systems complete their mission can be calculated as the sum of the probabilities that complete their mission 95, 96, 97, 98, 99 and 100 elements of the fleet:

$$P\{N \ge n\} = \sum\_{n=99}^{100} \left[ \frac{m!}{n! \left\{ (m-n)! \right\}!} R \, ^n \{1 - R\}^{m-n} \right] = 0,058 \tag{18}$$

### **4. Time dependent reliability models**

When reliability is expressed as a function of time, the continuous random variable, not negative, of interest is *T* , the instant of failure of the device. Let *f* (*t*) be the probability density function of *T* , and let *F* (*t*) be the cumulative distribution function of *T* . *F* (*t*) is also known as failure function or unreliability function [4].

In the context of reliability, two additional functions are often used: the **reliability** and the hazard function. Let's define **Reliability** *R*(*t*)as the survival function:

$$R(t) = P(T \ge t) = 1 - F(t) \tag{19}$$

The **Mean Time To Failure - MTTF** is defined as the expected value of the failure time:

$$MTTF = E\left\{T\right\} = \left\{\_0^\infty t \,\,\bullet \, f\left(t\right) \,\bullet \, dt\right.\tag{20}$$

Integrating by parts, we can prove the equivalent expression:

$$MTTF = E\left(T\right) = \left\| \begin{smallmatrix} T \\ \end{smallmatrix} R\left(t\right) \bullet dt\right\|\tag{21}$$

### **5. Hazard function**

Another very important function is the **hazard function**, denoted by *λ*(*t*), defined as the trend of the instantaneous failure rate at time *t* of an element that has survived up to that time *t*. The failure rate is the ratio between the instantaneous probability of failure in a neighborhood of *t*- conditioned to the fact that the element is healthy in *t*- and the amplitude of the same neighborhood.

The hazard function *λ*(*t*) [5] coincides with the intensity function *z*(*t*) of a Poisson process. The hazard function is given by:

$$\lambda\left(t\right) = \lim\_{\Delta t \to 0} \frac{P\left(t \le T \le t \star \Delta t \mid T \ge t\right)}{\Delta t} \tag{22}$$

Thanks to Bayes' theorem, it can be shown that the relationship between the hazard function, density of probability of failure and reliability is the following:

$$
\lambda\left(t\right) = \frac{f\left(t\right)}{R\left(t\right)}\tag{23}
$$

Thanks to the previous equation, with some simple mathematical manipulations, we obtain the following relation:

$$R(t) = e^{\int\_{t}^{t} \lambda(u) \bullet du} \tag{24}$$

In fact, since ln *R*(0) =ln 1 =0, we have:

*P*(*N* =*n*)= *m* !

*P*(*N* ≥*n*)= ∑

**4. Time dependent reliability models**

failure function or unreliability function [4].

**5. Hazard function**

*n*=95

hazard function. Let's define **Reliability** *R*(*t*)as the survival function:

*MTTF* = *E*(*T* )=*∫*

*MTTF* = *E*(*T* )=*∫*

Integrating by parts, we can prove the equivalent expression:

100 *m* !

*σN* = *m*∙*R* ∙(1 - *R*).

92 Operations Management

fleet:

The expected value of *N* is given by: *E*(*N* )=*μN* =*m*∙*R*and the standard deviation is:

Let's consider, for example, a corporate fleet consisting of 100 independent and identical systems. All systems have the same mission, independent from the other missions. Each system has a reliability of mission equal to 90%. We want to calculate the average number of missions completed and also what is the probability that at least 95% of systems would complete their mission. This involves analyzing the distribution of the binomial random variable character‐

The probability that at least 95% of the systems complete their mission can be calculated as the sum of the probabilities that complete their mission 95, 96, 97, 98, 99 and 100 elements of the

When reliability is expressed as a function of time, the continuous random variable, not negative, of interest is *T* , the instant of failure of the device. Let *f* (*t*) be the probability density function of *T* , and let *F* (*t*) be the cumulative distribution function of *T* . *F* (*t*) is also known as

In the context of reliability, two additional functions are often used: the **reliability** and the

The **Mean Time To Failure - MTTF** is defined as the expected value of the failure time:

0

0

Another very important function is the **hazard function**, denoted by *λ*(*t*), defined as the trend of the instantaneous failure rate at time *t* of an element that has survived up to that time *t*. The

ized by *R* = 0.90and*m* = 100. The expected value is given by *E*(*N* )=*μN* =100∙0.9=90.

*n* ! (*m* - *n*)! *R n*(1 - *R*)*m*-*n* (17)

*n* ! (*m* - *n*)! *R n*(1 - *R*)*m*-*n* =0,058 (18)

*R*(*t*)=*P*(*T* ≥*t*)=1 - *F* (*t*) (19)

∞*t* ∙ *f* (*t*)∙*dt* (20)

∞*R*(*t*)∙*dt* (21)

$$\text{R}\,\text{R}\,(t) = \frac{f(t)}{\Lambda(t)} = \frac{1}{\Lambda(t)} \bullet \frac{d\mathbb{P}\,(t)}{dt} = \text{--}\,\frac{1}{\Lambda(t)} \bullet \frac{d\mathbb{R}\,(t)}{dt} \to \frac{1}{\mathbb{R}\,(t)}\\d\mathbb{R}\,(t) = \text{-}\,\lambda(t)dt \to \text{ln}\,\mathbb{R}\,(t)\,\mathbb{I} - \ln\mathbb{I}\\ \text{R}\,(0) \mathbb{I} = \text{-}\,\mathbf{j}\_0^t \,\lambda(u)du \tag{25}$$

From equation 24 derive the other two fundamental relations:

$$F(t) = 1 \cdot e^{\frac{\int\_{\lambda}^{\lambda} \lambda \, (u) \bullet du}{\int\_{\lambda}^{\lambda} \lambda \, du}} f(t) = \lambda(t) \bullet e^{\frac{\int\_{\lambda}^{\lambda} \lambda \, (u) \bullet du}{\int\_{\lambda}^{\lambda} \lambda \, du}} \tag{26}$$

The most popular conceptual model of the hazard function is the **bathtub curve**. According to this model, the failure rate of the device is relatively high and descending in the first part of the device life, due to the potential manufacturing defects, called **early failures**. They manifest themselves in the first phase of operation of the system and their causes are often linked to structural deficiencies, design or installation defects. In terms of reliability, a system that manifests infantile failures improves over the course of time.

Later, at the end of the life of the device, the failure rate increases due to wear phenomena. They are caused by alterations of the component for material and structural aging. The beginning of the period of wear is identified by an increase in the frequency of failures which continues as time goes by. The **wear-out failures** occur around the average age of operating; the only way to avoid this type of failure is to replace the population in advance.

Between the period of early failures and of wear-out, the failure rate is about constant: failures are due to random events and are called **random failures**. They occur in non-nominal operating conditions, which put a strain on the components, resulting in the inevitable changes and the consequent loss of operational capabilities. This type of failure occurs during the useful life of the system and corresponds to unpredictable situations. The central period with constant failure rate is called **useful life**. The juxtaposition of the three periods in a graph which represents the trend of the failure rate of the system, gives rise to a curve whose characteristic shape recalls the section of a bathtub, as shown in Figure 12.

*R*(*t* + *t*0 |*t*0) =*P*(*T* >*t* + *t*0 |*T* >*t*0) (27)

*P*(*T* > *t*0) (28)

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

95

*R*(*t*0) (29)

∞*R*(*t* + *t*0 |*t*0)∙*dt* (30)

∞*R*(*t*)∙*dt* (31)

*λ*(*u*)∙*du* =1 (32)

*P*(*T* >*t* + *t*0 |*T* >*t*0) = *P*(*T* > *t*0 | *T* > *t* + *t*0) ∙ *P*(*T* > *t* + *t*0)

*R*(*t* + *t*0 |*t*0) = *R*(*t* + *t*0)

residual life of a device that has already survived a time *t*0:

running (burn-in) to avoid errors in the field.

*MTTF*(*t*0) = *E*(*T* - *t*0 |*T* >*t*0) =*∫*

And, given that *P*(*T* >*t*0 |*T* >*t* + *t*0) =1, we obtain the final expression, which determines the

The **residual Mean Time To Failure** – **residual MTTF** measures the expected value of the

For an IFR device, the residual reliability and the residual MTTF, decrease progressively as the device accumulates hours of operation. This behavior explains the use of preventive actions to avoid failures. For a DFR device, both the residual reliability and the residual MTTF increase while the device accumulates hours of operation. This behavior motivates the use of an intense

The **Mean Time To Failure** –**MTTF**, measures the expected value of the life of a device and coincides with the residual time to failure, where *t*0 =0. In this case we have the following

The **characteristic life** of a device is the time *tC* corresponding to a reliability *R*(*tC*) equal to

Let us consider a CFR device with a constant failure rate *λ*. The time-to-failure is an exponential random variable. In fact, the probability density function of a failure, is typical of an expo‐

0 *tC* 0

*MTTF* =*MTTF* (0)=*E*(*T* |*T* >0)=*∫*

, that is the time for which the area under the hazard function is unitary:

*R*(*tC*) =*e*-1 =0,368 →*R*(*tC*) = *∫*

0

Applying Bayes' theorem we have:

residual reliability:

relationship:

nential distribution:

1 *e*

**Figure 12.** Bathtub curve. The hazard function shape allows us to identify three areas: the initial period of the early failures, the middle time of the useful life and the final area of wear-out.

### The most common mathematical classifications of the hazard curve are the so called **Constant Failure Rate - CFR**, **Increasing Failure Rate - IFR** and **Decreasing Failure Rate - DFR**.

The CFR model is based on the assumption that the failure rate does not change over time. Mathematically, this model is the most simple and is based on the principle that the faults are purely random events. The IFR model is based on the assumption that the failure rate grows up over time. The model assumes that faults become more likely over time because of wear, as is frequently found in mechanical components. The DFR model is based on the assumption that the failure rate decreases over time. This model assumes that failures become less likely as time goes by, as it occurs in some electronic components.

Since the failure rate may change over time, one can define a reliability parameter that behaves as if there was a kind of counter that accumulates hours of operation. The **residual reliabili‐ ty** function *R*(*t* + *t*0 |*t*0), in fact, measures the reliability of a given device which has already survived a determined time *t*0. The function is defined as follows:

Reliability and Maintainability in Operations Management http://dx.doi.org/10.5772/54161 95

$$R\left(t + t\_0 \mid t\_0\right) = P\left(T > t + t\_0 \mid T > t\_0\right) \tag{27}$$

Applying Bayes' theorem we have:

Between the period of early failures and of wear-out, the failure rate is about constant: failures are due to random events and are called **random failures**. They occur in non-nominal operating conditions, which put a strain on the components, resulting in the inevitable changes and the consequent loss of operational capabilities. This type of failure occurs during the useful life of the system and corresponds to unpredictable situations. The central period with constant failure rate is called **useful life**. The juxtaposition of the three periods in a graph which represents the trend of the failure rate of the system, gives rise to a curve whose characteristic

**Figure 12.** Bathtub curve. The hazard function shape allows us to identify three areas: the initial period of the early

The most common mathematical classifications of the hazard curve are the so called **Constant Failure Rate - CFR**, **Increasing Failure Rate - IFR** and **Decreasing Failure Rate - DFR**.

The CFR model is based on the assumption that the failure rate does not change over time. Mathematically, this model is the most simple and is based on the principle that the faults are purely random events. The IFR model is based on the assumption that the failure rate grows up over time. The model assumes that faults become more likely over time because of wear, as is frequently found in mechanical components. The DFR model is based on the assumption that the failure rate decreases over time. This model assumes that failures become less likely

Since the failure rate may change over time, one can define a reliability parameter that behaves as if there was a kind of counter that accumulates hours of operation. The **residual reliabili‐ ty** function *R*(*t* + *t*0 |*t*0), in fact, measures the reliability of a given device which has already

shape recalls the section of a bathtub, as shown in Figure 12.

94 Operations Management

failures, the middle time of the useful life and the final area of wear-out.

as time goes by, as it occurs in some electronic components.

survived a determined time *t*0. The function is defined as follows:

$$P\left(T > t + t\_0 \mid T > t\_0\right) = \frac{P\left(T > t\_0 \mid T > t + t\_0\right) \bullet P\left(T > t + t\_0\right)}{P\left(T > t\_0\right)}\tag{28}$$

And, given that *P*(*T* >*t*0 |*T* >*t* + *t*0) =1, we obtain the final expression, which determines the residual reliability:

$$R\left(t+t\_0 \mid t\_0\right) = \frac{R\left(t+t\_0\right)}{R\left(t\_0\right)}\tag{29}$$

The **residual Mean Time To Failure** – **residual MTTF** measures the expected value of the residual life of a device that has already survived a time *t*0:

$$MTTF\left(t\_0\right) = E\left(T \cdot t\_0 \mid T > t\_0\right) = \left\| \Box^\* R\left(t + t\_0 \mid t\_0\right) \bullet dt\right\|\tag{30}$$

For an IFR device, the residual reliability and the residual MTTF, decrease progressively as the device accumulates hours of operation. This behavior explains the use of preventive actions to avoid failures. For a DFR device, both the residual reliability and the residual MTTF increase while the device accumulates hours of operation. This behavior motivates the use of an intense running (burn-in) to avoid errors in the field.

The **Mean Time To Failure** –**MTTF**, measures the expected value of the life of a device and coincides with the residual time to failure, where *t*0 =0. In this case we have the following relationship:

$$MTTF = MTTF\left(0\right) = E\left(T \mid T > 0\right) = \left\|\_{0}^{\infty} R\left(t\right) \bullet dt\right\|\tag{31}$$

The **characteristic life** of a device is the time *tC* corresponding to a reliability *R*(*tC*) equal to 1 *e* , that is the time for which the area under the hazard function is unitary:

$$R\left(t\_{\mathcal{C}}\right) = e^{-1} = 0,\\ \text{368} \quad -R\left(t\_{\mathcal{C}}\right) = \stackrel{t\_{\mathcal{C}}}{\int\_{0}^{t\_{\mathcal{C}}}} \lambda\left(u\right) \bullet du = 1\tag{32}$$

Let us consider a CFR device with a constant failure rate *λ*. The time-to-failure is an exponential random variable. In fact, the probability density function of a failure, is typical of an expo‐ nential distribution:

$$f\begin{pmatrix} t \end{pmatrix} = \lambda\begin{pmatrix} t \end{pmatrix} \bullet \ e^{\int\_{0}^{t} \lambda\begin{pmatrix} u \end{pmatrix} \bullet du} = \lambda e^{-\lambda \bullet t} \tag{33}$$

The corresponding cumulative distribution function *F* (*t*)is:

$$F\left(t\right) = \stackrel{t}{\int} f\left(z\right) dz = \stackrel{t}{\int} \lambda e^{-\lambda \bullet z} dz = 1 - e^{-\lambda \bullet t} \tag{34}$$

The reliability function *R*(*t*)is the survival function:

$$R\left(t\right) = 1 - F\left(t\right) = e^{-\lambda\_1 \bullet t} \tag{35}$$

*P*(*t* <*T* <*t* + *dt* |*T* >*t*)= *P*(*T* > *t* | *t* < *T* < *t* + *dt*) ∙ *P*(*t* < *T* < *t* + *dt*)

with λ =1.

**Figure 13.** Probability density function and cumulative distribution of an exponential function. In the figure is seen

*P*(*T* > *t*) = *f* (*t*)*dt*

As can be seen, this probability does not depend on *t*, i.e. it is not function of the life time already elapsed. It is as if the component does not have a memory of its own history and it is

The use of the constant failure rate model, facilitates the calculation of the characteristic life of

Therefore, the characteristic life, in addition to be calculated as the time value *tC* for which the

*λ e*-*λ*∙*t*| ∞

0 = - 0 *λ* + 1 *λ* = 1

reliability is 0.368, can more easily be evaluated as the reciprocal of the failure rate.

∞*e*-*λ*∙*t* ∙*dt* =- 1

In the CFR model, then, the MTTF and the characteristic life coincide and are equal to 1

The definition of MTTF, in the CFR model, can be integrated by parts and give:

0

*e* -*λ*∙*t* = *λe* -*λ*∙*t*

*dt*

Since *P*(*T* >*t* |*t* <*T* <*t* + *dt*)=1, being a certainty, it follows:

and of *f* (*t*)=λ ∙*e* -λ∙*t*

the trend of *f* (*t*)=λ ∙*e* -λ∙*t*

*P*(*t* <*T* <*t* + *dt* |*T* >*t*)= *P*(*t* < *T* < *t* + *dt*)

for this reason that the exponential distribution is called **memoryless**.

*R*(*tC*) =*e*

∞*R*(*t*)∙*dt* =*∫*

*MTTF* =*∫* 0

a device. In fact for a CFR item, *tC*is the reciprocal of the failure rate. In fact:

*P*(*T* > *t*) (40)

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

97

*e* -*λ*∙*t* =*λ* ∙*dt* (41)

*λ* (42)

*λ* (43)

*λ* .

For CFR items, the residual reliability and the residual MTTF both remain constant when the device accumulates hours of operation. In fact, from the definition of residual reliability, ∀*t*0∈ 0, *∞* , we have:

$$R\left(t + t\_0 \mid t\_0\right) = \frac{R\left(t + t\_0\right)}{R\left(t\_0\right)} = \frac{e^{-\lambda \bullet \left(t + t\_0\right)}}{e^{-\lambda \bullet t\_0}} = e^{-\lambda \bullet \left(t + t\_0\right) + \lambda \bullet t\_0} = e^{-\lambda \bullet t} = R\left(t\right) \tag{36}$$

Similarly, for the residual MTTF, is true the invariance in time:

$$MTTF\left(t\_0\right) = \left\| \begin{pmatrix} t \ \left(t + t\_0 \mid t\_0\right) \bullet dt = \emptyset \end{pmatrix} \bullet \begin{pmatrix} t \ \left(\bullet \ \left.dt\right\rangle \end{pmatrix} \begin{pmatrix} 0 \ \left.\begin{pmatrix} 0 \ \left.\begin{pmatrix} 0 \ \end{pmatrix}\right\} \end{pmatrix} \end{pmatrix} \tag{37}$$

This behavior implies that the actions of prevention and running are useless for CFR devices. Figure 13 shows the trend of the function *f* (*t*)=*λ* ∙*e*-*λ*∙*t* and of the cumulative distribution function *F* (*t*)=1 - *e*-*λ*∙*t* for a constant failure rate *λ* =1. In this case, since *λ* =1, the probability density function and the reliability function, overlap: *f* (*t*)=*R*(*t*)=*e*-*t* .

The probability of having a fault, not yet occurred at time *t*, in the next *dt*, can be written as follows:

$$P\{t \le T \le t + dt \mid T > t\}\tag{38}$$

Recalling the Bayes' theorem, in which we consider the probability of an hypothesis H, being known the evidence E:

$$P(H \mid E) = \frac{P(E \mid H) \bullet P(H)}{P(E)} \tag{39}$$

we can replace the evidence E with the fact that the fault has not yet taken place, from which we obtain *P*(*E*)→*P*(*T* >*t*). We also exchange the hypothesis H with the occurrence of the fault in the neighborhood of *t*, obtaining *P*(*H* )→ *P*(*t* <*T* <*t* + *dt*). So we get:

**Figure 13.** Probability density function and cumulative distribution of an exponential function. In the figure is seen the trend of *f* (*t*)=λ ∙*e* -λ∙*t* and of *f* (*t*)=λ ∙*e* -λ∙*t* with λ =1.

$$P\left\{t \le T < t + dt \mid T > t\right\} = \frac{P\left\{T > t \mid t < T < t + dt\right\} \bullet P\left\{t < T < t + dt\right\}}{P\left\{T > t\right\}}\tag{40}$$

Since *P*(*T* >*t* |*t* <*T* <*t* + *dt*)=1, being a certainty, it follows:

*f* (*t*)=*λ*(*t*)∙*e*

*f* (*z*)*dz* = *∫* -∞ *t λe*-*λ*∙*z*

The corresponding cumulative distribution function *F* (*t*)is:

*F* (*t*)= *∫* -∞ *t*

The reliability function *R*(*t*)is the survival function:

*R*(*t* + *t*0 |*t*0) = *R*(*t* + *t*0)

0

Figure 13 shows the trend of the function *f* (*t*)=*λ* ∙*e*-*λ*∙*t*

*MTTF*(*t*0) =*∫*

*R*(*t*0) = *e*

∞*R*(*t* + *t*0 |*t*0)∙*dt* =*∫*

density function and the reliability function, overlap: *f* (*t*)=*R*(*t*)=*e*-*t*

in the neighborhood of *t*, obtaining *P*(*H* )→ *P*(*t* <*T* <*t* + *dt*). So we get:

Similarly, for the residual MTTF, is true the invariance in time:

*e* -*λ*∙*t* 0 =*e*

∀*t*0∈ 0, *∞* , we have:

96 Operations Management

follows:

known the evidence E:

For CFR items, the residual reliability and the residual MTTF both remain constant when the device accumulates hours of operation. In fact, from the definition of residual reliability,

0

This behavior implies that the actions of prevention and running are useless for CFR devices.

function *F* (*t*)=1 - *e*-*λ*∙*t* for a constant failure rate *λ* =1. In this case, since *λ* =1, the probability

The probability of having a fault, not yet occurred at time *t*, in the next *dt*, can be written as

Recalling the Bayes' theorem, in which we consider the probability of an hypothesis H, being

we can replace the evidence E with the fact that the fault has not yet taken place, from which we obtain *P*(*E*)→*P*(*T* >*t*). We also exchange the hypothesis H with the occurrence of the fault

*P*(*H* |*E*)= *P*(*E* | *H* ) ∙ *P*(*H* )

=*λe*-*λ*∙*t* (33)

*dz* =1 - *e*-*λ*∙*t* (34)

∞*R*(*t*)∙*dt* ∀*t*0∈ 0, ∞ (37)

.

*P*(*E*) (39)

*P*(*t* <*T* <*t* + *dt* |*T* >*t*) (38)

and of the cumulative distribution

*R*(*t*)=1 - *F* (*t*)=*e*-*λ*∙*t* (35)

$$P\left(t < T < t + dt \mid T > t\right) = \frac{P\left(t < T < t + dt\right)}{P(T > t)} = \frac{f\left(t\right)dt}{e^{-\lambda \bullet t}} = \frac{\lambda e^{-\lambda \bullet t}dt}{e^{-\lambda \bullet t}} = \lambda \bullet dt \tag{41}$$

As can be seen, this probability does not depend on *t*, i.e. it is not function of the life time already elapsed. It is as if the component does not have a memory of its own history and it is for this reason that the exponential distribution is called **memoryless**.

The use of the constant failure rate model, facilitates the calculation of the characteristic life of a device. In fact for a CFR item, *tC*is the reciprocal of the failure rate. In fact:

$$R\left(t\_{\mathcal{C}}\right) = e^{\cdot \lambda \cdot t\_{\mathcal{C}}} = e^{-1} \to t\_{\mathcal{C}} = \frac{1}{\lambda} \tag{42}$$

Therefore, the characteristic life, in addition to be calculated as the time value *tC* for which the reliability is 0.368, can more easily be evaluated as the reciprocal of the failure rate.

The definition of MTTF, in the CFR model, can be integrated by parts and give:

$$MTTF = \left\| \stackrel{\circ \circ}{\ast} \mathbb{R}(t) \bullet dt = \left\| \stackrel{\circ \circ}{\ast} e^{\cdot \cdot \lambda \bullet t} \bullet dt = \text{--} \frac{1}{\overline{\lambda}} e^{-\lambda \bullet t} \right\|\_{0}^{\circ \circ} = \text{--} \frac{0}{\overline{\lambda}} + \frac{1}{\overline{\lambda}} = \frac{1}{\overline{\lambda}} \tag{43}$$

In the CFR model, then, the MTTF and the characteristic life coincide and are equal to 1 *λ* . Let us consider, for example, a component with constant failure rate equal to *λ* =0.0002failures per hour. We want to calculate the MTTF of the component and its reliability after 10000 hours of operation. We'll calculate, then, what is the probability that the component survives other 10000 hours. Assuming, finally, that it has worked without failure for the first 6000 hours, we'll calculate the expected value of the remaining life of the component.

From equation 43 we have:

$$MTTF = \frac{1}{\lambda} = \frac{1}{0.0002\left[\frac{\text{fürhren}}{\text{h}}\right]} = 5000\left[\text{fh}\right] \tag{44}$$

In a system of CFR elements arranged in series, then, the failure rate of the system is equal to the sum of failure rates of the components. The MTTF can thus be calculated using the simple

> *λs* = 1 ∑ *i*=1 *n λi*

For example, let me show the following example. A system consists of a pump and a filter, used to separate two parts of a mixture: the concentrate and the squeezing. Knowing that the failure rate of the pump is constant and is *λP* =1,5∙10-4 failures per hour and that the failure

To begin, we compare the physical arrangement with the reliability one, as represented in the

*λi* =*λP* + *λF* =1.8∙10-4 failures

*λs* = 1

As a year of continuous operation is 24 · 365=8,760 hours, the reliability after one year is:

, let's try to assess the failure rate of the system,

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

h (51)

1.8 ∙ 10-4 =5,555 h (52)

(50)

99

*MTTF* = 1

the MTTF and the reliability after one year of continuous operation.

**Figure 14.** physical and reliability modeling of a pump and a filter producing orange juice.

As can be seen, it is a simple series, for which we can write:

*λs* = ∑ *i*=1 *n*

·8760 =0.2066

*MTTF* is the reciprocal of the failure rate and can be written:

*MTTF* = 1

rate of the filter is also CFR and is *λF* =3∙10-5

relation:

following figure:

*RS* =*e* -*λs*∙*t* =*e*-1.8∙10-4

For the law of the reliability *R*(*t*)=*e*-*λ*∙*t* , you get the reliability at 10000 hours:

$$R\,\text{(10000)} = e^{-0.0002 \cdot 10000} = 0.135\tag{45}$$

The probability that the component survives other 10000 hours, is calculated with the residual reliability. Knowing that this, in the model CFR, is independent from time, we have:

$$R\left(t + t\_0 \mid t\_0\right) = R\left(t\right) \to R\left(20000 \mid 100000\right) = R\left(10000\right) = 0.135\tag{46}$$

Suppose now that it has worked without failure for 6000 hours. The expected value of the residual life of the component is calculated using the residual MTTF, that is invariant. In fact:

$$\text{MTTF}\left(t\_0\right) = \left\|\ulcorner\text{R}\left(t + t\_0 \mid t\_0\right) \bullet dt \to \text{MTTF}\left(6000\right) = \left\|\ulcorner\text{R}\left(t + 6000 \mid 6000\right) \bullet dt = \left\|\ulcorner\text{R}\left(t\right) \bullet dt = \text{MTTF} = 5000\right\|\text{h}\mathbf{I} \tag{47}$$

### **6. CFR in series**

Let us consider *n* different elements, each with its own constant failure rate *λi* and reliability *Ri* =*e* -*λi* ∙*t* , arranged in series and let us evaluate the overall reliability *RS* . From equation 9 we have:

$$R\_S = \prod\_{i=1}^{n} R\_i = \prod\_{i=1}^{n} \mathbf{e}^{\cdot \cdot \lambda\_i \bullet t} = \mathbf{e}^{\cdot \cdot \sum\_{i=1}^{n} \lambda\_i \bullet t} \tag{48}$$

Since the reliability of the overall system will take the form of the type *RS* =*e* -*λs*∙*t* , we can conclude that:

$$R\_S = \mathbf{e}^{\cdot \cdot \sum\_{i=1}^n \lambda\_i \bullet t} = e^{\cdot \cdot \lambda\_{s\*} t} \to \lambda\_s = \sum\_{i=1}^n \lambda\_i \tag{49}$$

In a system of CFR elements arranged in series, then, the failure rate of the system is equal to the sum of failure rates of the components. The MTTF can thus be calculated using the simple relation:

Let us consider, for example, a component with constant failure rate equal to *λ* =0.0002failures per hour. We want to calculate the MTTF of the component and its reliability after 10000 hours of operation. We'll calculate, then, what is the probability that the component survives other 10000 hours. Assuming, finally, that it has worked without failure for the first 6000 hours, we'll

The probability that the component survives other 10000 hours, is calculated with the residual

Suppose now that it has worked without failure for 6000 hours. The expected value of the residual life of the component is calculated using the residual MTTF, that is invariant. In fact:

reliability. Knowing that this, in the model CFR, is independent from time, we have:

0

Let us consider *n* different elements, each with its own constant failure rate *λi*

Since the reliability of the overall system will take the form of the type *RS* =*e*

*RS* =∏ *i*=1 *n Ri* =∏ *i*=1 *n* e -*λi* ∙*t* =e -∑ *i*=1 *n λi* ∙*t*

*RS* =e -∑ *i*=1 *n λi* ∙*t* =*e* -*λs*∙*t* =5000 h (44)

, you get the reliability at 10000 hours:

*R*(*t* + *t*0 |*t*0) =*R*(*t*)→*R*(20000|10000)=*R*(10000)=0.135 (46)

∞*R*(*t* + 6000|6000)∙*dt* =*∫*

, arranged in series and let us evaluate the overall reliability *RS* . From equation 9 we

→*λs* = ∑ *i*=1 *n*

*R*(10000)=*e*-0.0002∙10000 =0.135 (45)

0

∞*R*(*t*)∙*dt* =*MTTF* =5000 h (47)

and reliability

*λi* (49)

(48)

, we can

calculate the expected value of the remaining life of the component.

*λ* = 1 0.0002 failures h

*MTTF* = 1

From equation 43 we have:

98 Operations Management

*MTTF* (*t*0) =*∫*

*Ri* =*e* -*λi* ∙*t*

have:

conclude that:

0

**6. CFR in series**

For the law of the reliability *R*(*t*)=*e*-*λ*∙*t*

∞*R*(*t* + *t*0 |*t*0)∙*dt* →*MTTF* (6000)=*∫*

$$MTTF = \frac{1}{\lambda\_\*} = \frac{1}{\sum\_{i=1}^k \lambda\_i} \tag{50}$$

For example, let me show the following example. A system consists of a pump and a filter, used to separate two parts of a mixture: the concentrate and the squeezing. Knowing that the failure rate of the pump is constant and is *λP* =1,5∙10-4 failures per hour and that the failure rate of the filter is also CFR and is *λF* =3∙10-5 , let's try to assess the failure rate of the system, the MTTF and the reliability after one year of continuous operation.

To begin, we compare the physical arrangement with the reliability one, as represented in the following figure:

**Figure 14.** physical and reliability modeling of a pump and a filter producing orange juice.

As can be seen, it is a simple series, for which we can write:

$$
\lambda\_s = \sum\_{i=1}^{n} \lambda\_i = \lambda\_P + \lambda\_F = 1.8 \bullet 10^{-4} \left[ \frac{\text{failures}}{\text{h}} \right] \tag{51}
$$

*MTTF* is the reciprocal of the failure rate and can be written:

$$MTTF = \frac{1}{\lambda\_{\ast}} = \frac{1}{1.8 \cdot 10^{-4}} = 5,555 \text{[h]} \tag{52}$$

As a year of continuous operation is 24 · 365=8,760 hours, the reliability after one year is: *RS* =*e* -*λs*∙*t* =*e*-1.8∙10-4 ·8760 =0.2066

### **7. CFR in parallel**

If two components arranged in parallel are similar and have constant failure rate λ, the reliability of the system *RP* can be calculated with equation 10, wherein *RC* is the reliability of the component *RC* =*e*-*λt* :

$$R\_P = 1 \cdot \prod\_{i=1}^{2} \left(1 \cdot R\_i\right) = 1 \cdot \left(1 \cdot R\_1\right)^2 = 2R\_\odot \text{ - } R\_\odot ^2 = 2 \text{e}^{\cdot \lambda \bullet t} \text{ - e}^{\cdot 2 \lambda \bullet t} \tag{53}$$

The calculation of the MTTF leads to *MTTF* = 3 2*λ* . In fact we have:

$$\text{MTTF} = \text{f}\_0^\infty \text{R}(t) \bullet dt = \text{f}\_0^\infty \text{2e}^{-\lambda \text{rt}} \text{ - e}^{\cdot 2\lambda \text{ } \bullet t} \bullet dt = \text{ - } \frac{2}{\lambda} \text{e}^{-\lambda t} + \frac{1}{2\lambda} \text{e}^{-2\lambda t} \Big|\_{0}^{\prime\prime} = \frac{2}{\lambda} (0 \text{ - 1}) + \frac{1}{2\lambda} (0 \text{ - 1}) = \frac{3}{2\lambda} \tag{54}$$

Therefore, the MTTF increases compared to the single component CFR. The failure rate of the parallel system *λP*, reciprocal of the MTTF, is:

$$
\lambda\_P = \frac{1}{MTTF} = \frac{2}{3}\lambda \tag{55}
$$

*λP* = 2

The MTTF is the reciprocal of the failure rate and is:

3 *λ* = 2

*MTTF* = 1

*RP* =*e*

to withstand the load of the system. The reliability is:

rate *λ* =9 · 10-6

follows:

= 3 !

*R*2 *out of* 3 = ∑

*j*=2 3 ( 3 *j* )*R j*

2 ! (3 - 2)! *e*-2*λt* ∙(1 - *e*-*λt*)3-2 +

*Rk* out of *n* =*P*(*k* ≤ *j* ≤*n*)= ∑

operation. Let's get the reliability after one year of operation.

=( 3

3 !

∙(1 - *R*)3- *j*

nent 2. For simplicity, it is assumed that S is not affected by faults.

=3∙*e*-2*λt* ∙(1 - *e*-*λt*)3-2 + 1∙*e*-*λt* ∙(1 - *e*-*λt*)3-3

*λp* = 1

As a year of continuous operation is 24 · 365=8,760 hours, the reliability after one year is:

It is interesting to calculate the reliability of a system of identical elements arranged in a parallel configuration *k* out of *n*. The system is partially redundant since a group of *k* elements is able

> *j*=*k n* ( *n j* )*R j*

Let us consider, for example, three electric generators, arranged in parallel and with failure

We'll have: *n* =3, *k* =2. So, after a year of operation (*t* =8760 *h* ), reliability can be calculated as

2 )e-2*λt*(1 - e-*λt*)3-2 + ( 3

=0.963

second component comes into operation only when the first fails. Otherwise, it is idle.

A particular arrangement of components is that of the so-called parallel with stand-by: the

**Figure 16.** RBD diagram of a parallel system with stand-by. When component 1 fails, the switch S activates compo‐

3 ! (3 - 3)! *e*-*λt* ∙(1 - *e*-*λt*)3-3

. In order for the system to be active, it is sufficient that only two items are in

=

3 )e-*λt*(1 - e-*λt*)3-3

=

=*e*-6∙10-6

3 9∙10-6 =6∙10-6 guasti

h (56)

http://dx.doi.org/10.5772/54161

101

Reliability and Maintainability in Operations Management

6 ∙ 10-6 =166,666 h (57)

·8,760 =0.9488 (58)

∙(1 - *R*)*n*- *j* (59)

As you can see, the failure rate is not halved, but was reduced by one third.

For example, let us consider a safety system which consists of two batteries and each one is able to compensate for the lack of electric power of the grid. The two generators are equal and have a constant failure rate *λB* =9∙10-6 failures per hour. We'd like to calculate the failure rate of the system, the MTTF and reliability after one year of continuous operation.

As in the previous case, we start with a reliability block diagram of the problem, as visible in Figure 15.

**Figure 15.** Physical and reliability modeling of an energy supply system.

It is a parallel arrangement, for which the following equation is applicable:

Reliability and Maintainability in Operations Management http://dx.doi.org/10.5772/54161 101

$$
\lambda\_P = \frac{2}{3}\lambda = \frac{2}{3}\Theta \bullet 10^{-6} = 6 \bullet 10^6 \left[\frac{\text{guasti}}{\text{h}}\right] \tag{56}
$$

The MTTF is the reciprocal of the failure rate and is:

**7. CFR in parallel**

100 Operations Management

the component *RC* =*e*-*λt*

*MTTF* =*∫* 0

Figure 15.

:

The calculation of the MTTF leads to *MTTF* = 3

∞2e-*λ*∙*t* - e-2*λ*∙*t* ∙*dt* = - 2

0

parallel system *λP*, reciprocal of the MTTF, is:

*RP* =1 - ∏ *i*=1 2 (1 - *Ri*

∞*R*(*t*)∙*dt* =*∫*

If two components arranged in parallel are similar and have constant failure rate λ, the reliability of the system *RP* can be calculated with equation 10, wherein *RC* is the reliability of

*λ e*-*λt* +

Therefore, the MTTF increases compared to the single component CFR. The failure rate of the

*MTTF* = 2

For example, let us consider a safety system which consists of two batteries and each one is able to compensate for the lack of electric power of the grid. The two generators are equal and have a constant failure rate *λB* =9∙10-6 failures per hour. We'd like to calculate the failure rate

As in the previous case, we start with a reliability block diagram of the problem, as visible in

*λP* = 1

of the system, the MTTF and reliability after one year of continuous operation.

**Figure 15.** Physical and reliability modeling of an energy supply system.

It is a parallel arrangement, for which the following equation is applicable:

As you can see, the failure rate is not halved, but was reduced by one third.

2*λ* . In fact we have:

0 = 2

*λ* (0 - 1) +

3 *λ* (55)

1 2*λ e*-2*λt*| ∞

2 =2e-*λ*∙*t* - e-2*λ*∙*t* (53)

1

2*λ* (0 - 1)= 3

2*λ* (54)

) =1 - (1 - *R*1)2 =2*RC* - *RC*

$$MTTF = \frac{1}{\lambda\_p} = \frac{1}{6 \bullet 10^6} = 166,666 \,\text{[h]}\tag{57}$$

As a year of continuous operation is 24 · 365=8,760 hours, the reliability after one year is:

$$R\_p = e^{-\lambda\_p} \mathbf{r}^t = e^{-6 \bullet 10^{-6} \cdot 8.760} = 0.9488 \tag{58}$$

It is interesting to calculate the reliability of a system of identical elements arranged in a parallel configuration *k* out of *n*. The system is partially redundant since a group of *k* elements is able to withstand the load of the system. The reliability is:

$$R\_{k\text{ out of }n} = P\{k \le j \le n\} = \sum\_{j=k}^{n} {n\choose j} R^{\nkern-1.1em]} \bullet \{1\ -R\}^{n\text{-}j} \tag{59}$$

Let us consider, for example, three electric generators, arranged in parallel and with failure rate *λ* =9 · 10-6 . In order for the system to be active, it is sufficient that only two items are in operation. Let's get the reliability after one year of operation.

We'll have: *n* =3, *k* =2. So, after a year of operation (*t* =8760 *h* ), reliability can be calculated as follows:

$$\begin{split} &R\_{2\text{ out of }\frac{3}{3}} = \sum\_{j=2}^{3} \binom{3}{j} \text{R}^{j} \bullet \{1 \text{ } - \text{R}\}^{3 - j} = \binom{3}{2} \text{e}^{-2\lambda t} \{1 \text{ } - \text{e}^{\cdot \lambda t}\}^{3 - 2} + \binom{3}{3} \text{e}^{-\lambda t} \{1 \text{ } - \text{e}^{-\lambda t}\}^{3 - 3} = \\ &= \frac{3!}{2!(3 \cdot 2)!} \text{e}^{-2\lambda t} \bullet \{1 \text{ } - \text{e}^{-\lambda t}\}^{3 - 2} + \frac{3!}{3!(3 \cdot 3)!} \text{e}^{-\lambda t} \bullet \{1 \text{ } - \text{e}^{-\lambda t}\}^{3 - 3} = \\ &= 3 \bullet \text{e}^{-2\lambda t} \bullet \{1 \text{ } - \text{e}^{-\lambda t}\}^{3 - 2} + 1 \bullet \text{e}^{-\lambda t} \bullet \{1 \text{ } - \text{e}^{-\lambda t}\}^{3 - 3} = 0.963 \end{split}$$

A particular arrangement of components is that of the so-called parallel with stand-by: the second component comes into operation only when the first fails. Otherwise, it is idle.

**Figure 16.** RBD diagram of a parallel system with stand-by. When component 1 fails, the switch S activates compo‐ nent 2. For simplicity, it is assumed that S is not affected by faults.

If the components are similar, then *λ*1 =*λ*2. It's possible to demonstrate that for the stand-by parallel system we have:

$$MTTF = \frac{2}{\lambda} \tag{60}$$

**•** diagnose the condition;

create two scenarios:

**9. Availability**

maintenance of condition.

**•** estimate the Remaining Useful Life – RUL;

parts, the maximization of the availability and so on.

used in the initial phase of design of the systems themselves.

a fault, but also for preventive or corrective maintenance.

**•** decide whether to maintain or to continue to operate normally.

that identify ranges of values for which maintenance action must arise [9].

CBM schedule is modeled with algorithms aiming at high effectiveness, in terms of cost minimization, being subject to constraints such as, for example, the maximum time for the maintenance action, the periods of high production rate, the timing of supply of the pieces

In support of the prognosis, it is now widespread the use of diagrams that do understand, even graphically, when the sensor outputs reach alarm levels. They also set out the alert thresholds

Starting from a state of degradation, detected by a measurement at the time *tk* , we calculate the likelihood that the system will still be functioning within the next instant of inspection *tk* +1. The choice to act with a preventive maintenance is based on the comparison of the expected value of the cost of unavailability, with the costs associated with the repair. Therefore, you

**•** continue to operate: if we are in the area of not alarming values. It is also possible that being in the area of preventive maintenance, we opt for a postponement of maintenance because it has already been established replacement intervention within a short interval of time **•** stop the task: if we are in the area of values above the threshold established for preventive

The modeling of repairable systems is commonly used to evaluate the performance of one or more repairable systems and of the related maintenance policies. The information can also be

In the traditional paradigm of modeling, a repairable system can only be in one of two states: working (up) or inoperative (down). Note that a system may not be functioning not only for

Availability may be generically be defined as the percentage of time that a repairable system is in an operating condition. However, in the literature, there are four specific measures of repairable system availability. We consider only the **limit availability**, defined with the limit

The limit availability just seen is also called **intrinsic availability**, to distinguish it from the **technical availability**, which also includes the logistics cycle times incidental to maintenance

*A*(*t*) (61)

Reliability and Maintainability in Operations Management

http://dx.doi.org/10.5772/54161

103

of the probability *A*(*t*) that the system is working at time *t*, when *t* tends to infinity.

*A*=lim *t*→∞

Thus, in parallel with stand-by, the MTTF is doubled.=

### **8. Repairable systems**

The devices for which it is possible to perform some operations that allow to reactivate the functionality, deserve special attention. A repairable system [6] is a system that, after the failure, can be restored to a functional condition from any action of maintenance, including replacement of the entire system. Maintenance actions performed on a repairable system can be classified into two groups: **Corrective Maintenance - CM** and **Preventive Maintenance - PM**. Corrective maintenance is performed in response to system errors and might correspond to a specific activity of both repair of replacement. Preventive maintenance actions, however, are not performed in response to the failure of the system to repair, but are intended to delay or prevent system failures. Note that the preventive activities are not necessarily cheaper or faster than the corrective actions.

As corrective actions, preventive activities may correspond to both repair and replacement activities. Finally, note that the actions of operational maintenance (servicing) such as, for example, put gas in a vehicle, are not considered PM [7].

Preventative maintenance can be divided into two subcategories: **scheduled** and **on-condi‐ tion**. Scheduled maintenance (hard-time maintenance) consists of routine maintenance operations, scheduled on the basis of precise measures of elapsed operating time.

**Condition-Based Maintenance - CBM** [8] (also known as predictive maintenance) is one of the most widely used tools for monitoring of industrial plants and for the management of maintenance policies. The main aim of this approach is to optimize maintenance by reduc‐ ing costs and increasing availability. In CBM it is necessary to identify, if it exists, a measur‐ able parameter, which expresses, with accuracy, the conditions of degradation of the system. What is needed, therefore, is a physical system of sensors and transducers capable of moni‐ toring the parameter and, thereby, the reliability performance of the plant. The choice of the monitored parameter is crucial, as is its time evolution that lets you know when mainte‐ nance action must be undertaken, whether corrective or preventive.

To adopt a CBM policy requires investment in instrumentation and prediction and control systems: you must run a thorough feasibility study to see if the cost of implementing the apparatus are truly sustainable in the system by reducing maintenance costs.

The CBM approach consists of the following steps:

**•** group the data from the sensors;

**•** diagnose the condition;

If the components are similar, then *λ*1 =*λ*2. It's possible to demonstrate that for the stand-by

The devices for which it is possible to perform some operations that allow to reactivate the functionality, deserve special attention. A repairable system [6] is a system that, after the failure, can be restored to a functional condition from any action of maintenance, including replacement of the entire system. Maintenance actions performed on a repairable system can be classified into two groups: **Corrective Maintenance - CM** and **Preventive Maintenance - PM**. Corrective maintenance is performed in response to system errors and might correspond to a specific activity of both repair of replacement. Preventive maintenance actions, however, are not performed in response to the failure of the system to repair, but are intended to delay or prevent system failures. Note that the preventive activities are not necessarily cheaper or

As corrective actions, preventive activities may correspond to both repair and replacement activities. Finally, note that the actions of operational maintenance (servicing) such as, for

Preventative maintenance can be divided into two subcategories: **scheduled** and **on-condi‐ tion**. Scheduled maintenance (hard-time maintenance) consists of routine maintenance

**Condition-Based Maintenance - CBM** [8] (also known as predictive maintenance) is one of the most widely used tools for monitoring of industrial plants and for the management of maintenance policies. The main aim of this approach is to optimize maintenance by reduc‐ ing costs and increasing availability. In CBM it is necessary to identify, if it exists, a measur‐ able parameter, which expresses, with accuracy, the conditions of degradation of the system. What is needed, therefore, is a physical system of sensors and transducers capable of moni‐ toring the parameter and, thereby, the reliability performance of the plant. The choice of the monitored parameter is crucial, as is its time evolution that lets you know when mainte‐

To adopt a CBM policy requires investment in instrumentation and prediction and control systems: you must run a thorough feasibility study to see if the cost of implementing the

operations, scheduled on the basis of precise measures of elapsed operating time.

nance action must be undertaken, whether corrective or preventive.

The CBM approach consists of the following steps:

**•** group the data from the sensors;

apparatus are truly sustainable in the system by reducing maintenance costs.

*λ* (60)

*MTTF* = 2

Thus, in parallel with stand-by, the MTTF is doubled.=

example, put gas in a vehicle, are not considered PM [7].

parallel system we have:

102 Operations Management

**8. Repairable systems**

faster than the corrective actions.

CBM schedule is modeled with algorithms aiming at high effectiveness, in terms of cost minimization, being subject to constraints such as, for example, the maximum time for the maintenance action, the periods of high production rate, the timing of supply of the pieces parts, the maximization of the availability and so on.

In support of the prognosis, it is now widespread the use of diagrams that do understand, even graphically, when the sensor outputs reach alarm levels. They also set out the alert thresholds that identify ranges of values for which maintenance action must arise [9].

Starting from a state of degradation, detected by a measurement at the time *tk* , we calculate the likelihood that the system will still be functioning within the next instant of inspection *tk* +1. The choice to act with a preventive maintenance is based on the comparison of the expected value of the cost of unavailability, with the costs associated with the repair. Therefore, you create two scenarios: