Modeling and Analysis of Two-Stage Systems with Parallel Machines and Limited Repair Capacity

Proceedings of the 13th IFAC Symposium on Information Control Problems in Manufacturing Moscow, Russia, June 3-5, 2009

Modeling and Analysis of Two-Stage Systems with Parallel Machines and Limited Repair Capacity ⋆ M. Colledani ∗ S. B. Gershwin ∗∗ ∗

Mechanical Engineering Department, Politecnico di Milano, Milan, Italy, (e-mail: [email protected]). ∗∗ Department of Mechanical Engineering, Massachusetts Institute of Technology, Cambridge, MA, 02139, USA (e-mail: [email protected]) Abstract: A method to study the performance of two-stage systems with unreliable parallel machines is presented. Once failed, machines require the intervention of a repair operator. In contrast to previous works which consider one repairman dedicated to each machine, a limited number of repairmen is assumed to be available for each stage. The resulting two machine system is modeled as a general multiple up and multiple down two-stage fluid system, for which an exact analytical solution is available. Results show that while considering parallel stages integrated in a system the allocation of repairmen is all but trivial and a new behaviour of the production rate as a function of the number of machines at the stage is described. Furthermore, for non-identical machines, the method makes it possible to suitably set priorities among failed machines. Finally, the results suggest the definition of new cost-oriented design problems for the optimal allocation of the operators in parallel machine lines. Keywords: Limited Repair Capacity, Parallel Lines, Multiple up states, Two-Stage systems. 1. INTRODUCTION Production systems formed by stages in which parallel machines simultaneously perform processes on the incoming production flow are frequently found in companies to increase the capacity of specific stages. Indeed, especially while reconfiguring a system for production improvement needs, it may be more convenient to add new machines to the existing resources instead of replacing them with a new faster machine. As an example, in water bottling lines the first station dedicated to the blowing of PET bottles frequently represents an efficiency bottleneck for the system, due to the large set of operations it performs (heating of the preform, air pressure blowing and cooling). However, due to their complexity, the cost of high speed blowing machines is generally much higher then the cost of medium speed blowing machines. Therefore, production system designers, frequently prefer to position two or more blowing stations in parallel then adopting a unique fast machine in the line. Similar considerations drive the adoption of parallel resources in machining systems, like engine block production lines (Colledani et al. (2009)). In spite of the diffusion of this type of systems, they received relatively low attention in the literature and only few methods to estimate their performance are available. Methods to study parallel stages with identical machines can be found in Mitra (1988) concerning two-machine lines and in Burman (1995) and Patchong and Willaeys (2001) ⋆ The research is partially funded by the Roberto Rocca Project ”Interactions Among Quality and Productivity Performance Measures in Production Systems”. It is also partly funded by the SingaporeMIT Alliance (SMA).

978-3-902661-43-2/09/$20.00 © 2009 IFAC

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for multi-stage lines. Furthermore, parallel stages with non-identical machines, also named split/merge stages, have been considered in Tan (2001). In most of the cases, only approximate solutions are available (Burman (1995) and Patchong and Willaeys (2001)) or exact solution of a subset of this large class of systems is provided (Tan (2001)). Recently, in Tan and Gershwin (2007) the framework for modeling similar systems within the scheme of multiple up-multiple down machines was provided. In this paper we consider an additional feature of these systems that was never included in previous contributions. Indeed, all the existing methods assume unlimited repair capacity at each parallel machines stage. In other words, if one stage k is composed of mk = 10 machines in parallel and they are all simultaneously failed it is assumed that 10 operators are available for restoring the machines to operational conditions. However, in real systems, this situation may be unrealistic. In this paper, we assume that for a given stage k only a maximum number of ck operator is available for repair interventions. Thus, if more than ck machines are simultaneously down at stage k only ck of them may be repaired. This problem is known in the literature as the machine interference problem. Traditionally the machine interference problem has been addressed considering stand alone machines subject to random failures. In other words, machines are never considered to be integrated in a production line. For recent reviews on available machineinterference models refer to Armstrong and Haque (2007) and Altiok (1995). In Kuhn (2003) the machine interference problem in buffered production lines was presented

10.3182/20090603-3-RU-2001.0391

13th IFAC INCOM (INCOM'09) Moscow, Russia, June 3-5, 2009

Fig. 1. Two-stage system with parallel machines per stage.

stage 2 cannot move and repair a failed machine at stage 1 and vice versa. This assumption is reasonable in case the two process stages are physically located in different areas of the production plant or in case the repair operations required in the two stages are so different as to call for specific expertise of the repairmen. In case of different machines per stage, a priority of intervention is set. In practice, when more machines are simultaneously down at one stage and there is interference, only the machines with highest priority are served. In case of identical machines, no priority is assigned but it is equally likely for each available operator to work on a failed machine.

for the first time. The main idea is that the existence of starvation and blocking phenomena when machines are integrated in a system affects the average time between two calls in row for the repairman. Symmetrically, the fact the the repairman is shared among different machines causes a delay in the repair intervention. Authors proposed two different approaches to approximately estimate the impact of the number of repairmen on the production rate of the system, named iterative and integrative procedures. However, only the case of operators shared among the machines in serial production lines was considered, thus parallel machines were not included.

Each machine Mk,i is assumed to process material with deterministic processing rate µk,i . The instantaneous processing rate of the entire stage is the sum of the processing rates of the operational machines. That is, mk X µk (t) = µk,i αk,i (t) (1)

In this paper an exact method for estimating the performance of two-stage lines with limited repair capacity is presented. Although such small systems are rarely found in real production plants, there are at least three reasons for focusing on similar systems. First, they are the simplest integrated production systems. Thus they are suitable to study the properties and the behaviour of uninvestigated system configurations. Second, given the generality of the proposed approach, entire portions of line can be represented by a unique production stage, enabling the analysis of more complex systems within a unique simplified model. Third, two-stage systems are traditionally the basis for the analysis of longer production lines, through decomposition techniques (Gershwin (1994)).

The asynchronous two-stage line behaviour is modeled through a continuous flow model. A continuous flow of material from outside is supposed to enter the system at the first stage, then moves to the buffer and visits the second stage before leaving the system. There is always available material at the input of the system (i.e. the first stage is never starved) and available space for material storage at the output of the system (i.e. the second stage is never blocked). The buffer has finite capacity N that is a real number.

2. PRODUCTION SYSTEM MODEL In this paper, linear two-stage systems with an intermediate finite capacity buffer B are considered. Stage k is composed of mk unreliable machines, k = 1, 2. The layout of the system is represented in Figure 1. At each stage, ck repair operators are assigned. Machine Mk,i with i = 1, .., mk is unreliable and subject to failures. Failures are random events. The time to fail is an independent exponentially distributed random variable with rate pk,i , 1 . Failures where the Mean Time to Failure M T T Fk,i = pk,i are assumed to be Operation Dependent Failures (ODF). Thus a machine can only fail while processing material. When machine Mk,i goes down and one of the ck operators is available, it is repaired. The time to repair is an independent exponentially distributed random variable with rate 1 rk,i , where the Mean Time to Repair M T T Rk,i = rk,i . Otherwise, if the ck operators are busy repairing other failed machines at the same stage, Mk,i must wait for an operator to be available before being repaired. Therefore, machines interfere with each other. We assume that the simultaneous intervention of more than one operator per failed machine is not possible. In other words, if machine Mk,i with i = 1, .., mk is down only one operator is allowed to attend the machine for the repairing intervention. Moreover, we assume that operators are not shared between different stages. A repairman allocated to repair tasks at

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i=1

where αk,i assumes value 1 or 0 respectively if Mk,i is up or down. The maximum processing rate at the stage is found when all machines are operational. It is the sum of all the processing rates of the machines composing the stage.

Performance Measures Given the system described in the previous section, the performance measures of interest in this analysis are: • The average production rate of the system P • The average level of buffer B, n Although the method makes it possible to calculate other measures related to the utilization of the repairmen, in this study we focus only on these two performance measures. 3. SINGLE STAGE (ISOLATED) SYSTEMS 3.1 Parallel Identical Machines In the case of identical parallel machines, each machine in stage k may fail with rate pk , may be repaired with rate rk and processes material at rate µk . Therefore, stage k has mk + 1 states. In state j = 0, .., mk , j machines are simultaneously operational: thus the entire stage processes material at rate jµk . Accordingly, the possible transitions for stage Mk are: • from state j to state j − 1 with rate jpk , for j = 1, .., mk • from state j to state j + 1 with rate qj rk , for j = 0, .., mk − 1. where qj is calculated considering the number of operators involved in stage k with the following equation:

13th IFAC INCOM (INCOM'09) Moscow, Russia, June 3-5, 2009

Fig. 2. State Transition Diagram for stage Mk with 3 parallel machines and 2 available operators. ½ ck , (mk − j) ≥ ck qj = (2) (mk − j), (mk − j) < ck Figure 2 represents the state transition diagram of stage Mk in case 3 identical machines and 2 operators are available. The transition matrix λk is given by: −2rk 2rk 0 0 pk −pk − 2rk 2rk 0 (3) λk = 0 2pk −2pk − rk rk 0 0 3pk −3pk where the states are ordered as [0, 1, 2, 3]. The processing rates are: µk = [0 µk 2µk 3µk ]. (4) It is worth mentioning that, as previously shown in [Burman, 1995], given this processing rate vector, the efficiency in isolation of the stage k when ck = mk can be calculated with the following formula: rk (5) ek = rk + pk In other words, the efficiency in isolation of the stage with mk parallel machines and unlimited repair capacity is equivalent to that of a single machine with failure and repair rates respectively pk and rk and processing rate equal to mk µk . In case of limited repair capacity the derivation of the efficiency in isolation of the stage k gets more complicated. The probability of the generic state j > 0 of the birthdeath process (Figure 2) can be expressed as a function of the state 0 using the following expressions:  j j   π ck ρk ,  j ≤ (mk − ck ) 0 j! πj = (6) ck ! mk j  k −ck   π0 cm ( )ρ , j > (m − c ) k k k k mk ! j where ρk = prkk . Therefore, π(0) can be calculated by using the normalization equation. After some manipulation, the efficiency in isolation of the stage is given by: m −c mkP −ck m j j Pk jρjk ck k k ck ! mk j ck ρk (j ) mk j! + mk mk ! ek,ck =

j=0 mkP −ck j=0

cjk ρjk j!

+

j=mk −ck +1 m −c m Pk ck k k ck ! mk j (j )ρk mk ! j=mk −ck +1

(7) It will be shown that, for a given number of operators ck < mk , this function is a decreasing function of mk . 3.2 Parallel Non-Identical Machines In the case of non-identical parallel machines, each machine in stage k may fail with rate pk,i , may be repaired

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Fig. 3. State Transition Diagram for stage Mk with 2 non identical parallel machines with rate rk,i and process material at rate µk,i . Therefore, stage k has Sk = 2mk states. Each state j is characterized by a vector of state indicators αk,i which assume values 1 if the machine i is up and 0 if machine i is down. Considering a stage with 2 non-identical machines, the possible states of the stage are [11, 10, 01, 00]. According to equation (1), the value of the processing rates in these state is µk = [µk,1 + µk,2 , µk,1 , µk,2 , 0]. Figure 3 represents the state transitions diagram of machine Mk with unlimited repair capacity. The transition rate matrix λk is given by (the index k is omitted): −p1 − p2 r2 λk = r1 0

p2 −p1 − r2 0 r1

p1 0 −p2 − r1 r2

0 p1 p2 −r1 − r2

(8)

In case of limited repair capacity, the priority matrix βk for stage k is considered in input. This binary square matrix has mk ×mk elements. β(i, i′ ) = 1 if machine i has priority over machine i′ , in case of interference. Otherwise, the element is set to 0. Elements on the diagonal are all 0. For consistency, the determinant of βk must always be 0. Once the priority matrix is defined, the transition rate matrix λ′k in case of limited repair capacity can be obtained as a function of λk . In particular, for those states j such that Sk P (1 − αk,i ) the total number of down machines I(j) = i=1

satisfies I(j) > ck the rows of the matrix λk must be changed according to the following equation: λ′k (j, j ′ ) = λk (j, j ′ )(1 − αk,i )Iiβ

(9)

 m Pk   β(i, i′ ) ≥ I(j) − ck  1,

(10)

where:

Iiβ =

i′ =1

m Pk   β(i, i′ ) < I(j) − ck  0, i′ =1

Moreover, the total number of machines attended by operators at state j is equal to q(j) = I(j) in case I(j) ≤ ck or q(j) = ck in case I(j) > ck . Example. In the previous case, if ck = 1 operator is available and priority goes to machine 1 at stage k, than β(1, 2) = 1 and all the other elements are set to 0. The value of I(j) for the four states is I(j) = [0, 1, 1, 2]. Therefore, I(j) > ck only for the last state denoted j = (00). According to equation (9) the last row of λk must be modified. The term (1−αk,i ) = 1 for both i = 1, 2. However, the term Iiβ = 1 only for i = 1. Therefore, the only possible repair operation happens with rate r1 . The new transitions matrix λk is:

13th IFAC INCOM (INCOM'09) Moscow, Russia, June 3-5, 2009

p2 −p1 − r2 0 r1

p1 0 −p2 − r1 0

0 p1 p2 −r1 − r2

P(c1)

(11)

Production rate

−p1 − p2 r2 λk = r1 0

4. SOLUTION FOR TWO-STAGE SYSTEMS

5. SYSTEM BEHAVIOUR 5.1 Impact of the Number of Available Operators We consider a two machine line in which m1 = 10 and m2 = 1. We set µ1 = 1/10 for each one of the identical machines at stage 1 and µ2 = 1. The behaviour of the system production rate as a function of the number of available operators at stage 1, c1 , is shown in Figure 4, for 4 different values of the failure rate p1 = [0.01, 0.015, 0.02, 0.025] corresponding to the following maximum values (case c1 = 10) for the efficiency in isolation emax = [0.909, 0.869, 0.833, 0.8]. 1 As it can be noticed the production rate increases as a function of the number of operators c1 . However, for different values of p1 , the number of operators required for having a significant impact on the system performance changes. For instance, if p1 = 0.01 the number of operators assigned to the stage M1 should be equal to 3, for the lines with p1 = [0.015, 0.02, 0.025] it should be respectively equal to [4, 5, 5]. Further increase in the number of operators does not involve any significant impact on the production rate of the system. This highlights the fact that having ck = mk may be a substantial loss of repair effort, not causing any positive impact on the system performance.

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p1=0.01 p1=0.015 p1=0.02 p1=0.025 1

2

3

4 5 6 7 8 n of operators at M 1

9

10

Fig. 4. Production rate as a function of the number of repair operators at M1 with 10 parallel machines at stage 1. p2 = 0.01, r1 = r2 = 0.1, N = 5, e2 = 0.909 P(m1): case of unlimited repair capacity 0.859 Production rate

The two-stage system with multiple parallel machines is analyzed by using the exact method proposed in Tan and Gershwin (2007). This method takes in input the transition rate matrices of the two stages namely λ1 = λ1 (i, i′ ) and λ2 = λ2 (j, j ′ ), the transition rate matrices ψ1 = ψ1 (i, i′ ) and ψ2 = ψ2 (j, j ′ ) representing the behaviour of the two machines in the boundary conditions, and the processing rate vectors µ1 = µ1 (i) and µ2 = µ2 (j). In case stage k is composed of identical parallel machines the transition rate matrix λk and the processing rate vector µk have the structure respectively specified in equations (3) and (4). In case stage k is composed of non-identical parallel machines, the transition rate matrix λk and the processing rate vector µk are instead obtained as described in section 3.2. The additional matrices ψ1 and ψ2 specify the behaviour of machines when respectively blocked or starved. In continuous flow models, two possible blocking (starvation) situations can be observed, named partial and complete blocking (starvation). Stage M1 is said to be completely blocked if the buffer is full and M2 is in state (j) characterized by µ2 (j) = 0. Stage M1 is said to be partially blocked if the buffer is full and M2 is in state (j) characterized by µ2 (j) > 0. In both cases, stage 1 is forced to process material at the rate of stage M2 in state (j), i.e. µ2 (j). Similar definitions hold also for the starvation of stage M2 .

0.86 0.84 0.82 0.8 0.78 0.76 0.74 0.72 0.7 0.68 0.66

0.858 0.857 0.856 0.855 0.854 0.853 2

4 6 8 10 n of identical machines at M 1

12

Fig. 5. Production rate as a function of the number of identical machines at stage m1 . p1 = p2 = 0.01, r1 = r2 = 0.1, N = 5, µ2 = 1, µ1 = 1/m1 5.2 Impact of the Number of Identical Machines Unlimited repair case The behaviour of a two machine system in which the number of identical machines at the first stage m1 is varied in the range [1, 12] is shown in Figure 5. The number of repair operators is set to c1 = m1 . In the case of unlimited repair capacity, the production rate is a monotonic, continuously increasing function which asymptotically tends to the maximum production rate. Notice that the efficiency in isolation of the machine is kept constant in the experiment, by varying the processing rate of each parallel machine. Thus the increment in the production rate of the system is only due to the reduction of the variance of the machine output. In fact, comparing the variance of the output for a stage composed of identical parallel machines (Burman (1995)) and the variance of the output of the single equivalent machine with efficiency in isolation calculated in equation (5) it can be noticed that the parallel configuration has a smaller variance of the output for every set of p and r. The variance of the output as a function of the number of machines at stage 1 calculated at time T = 100000 is reported in Figure 6. Limited repair case In case of limited repair capacity if the number of available operators is changed from 2 to 6 a counterintuitive behaviour is observed. In Figure 7 the production rate of the same line with c1 = [2, 3, 4, 5, 6] is shown. For each value of c1 the production rate curve is a concave function presenting a maximum. In other words, if the repair capacity is not unlimited, there is an optimal

13th IFAC INCOM (INCOM'09) Moscow, Russia, June 3-5, 2009

e1,c : case of limited repair capacity Efficiency in isolation of M1

Variance of the output

VAR(m1): case of unlimited repair capacity 160000 140000 120000 100000 80000 60000 40000 20000 0 0

2

4 6 8 10 n of identical machines at M 1

12

Fig. 6. Variance of the output of stage 1 in isolation as a function of the number of identical machines m1 .

1

0.91 0.905 0.9 0.895

c1=2 c1=3 0.89 c1=4 c1=5 c1=6 0.885 0

µk,ck =

0.858 0.856

12

Fig. 8. Efficiency in isolation of M1 on the number of machines m1 . p1 = 0.01, r1 = 0.1, µ1 = 1/m1 .

0.86 Production rate

4 6 8 10 n of identical machines at M 1

obtained as:

P(m1): case of limited repair capacity

rk rk +pk µk

(12)

ek,ck

Therefore µk,ck > µk .

0.854 0.852 0.85 0.848 c1=2 c =3 0.846 c1=4 1 0.844 c1=5 c1=6 0.842 2

2

4 6 8 10 n of identical machines at M 1

12

Fig. 7. Production rate as a function of the number of machines at stage m1 . p1 = p2 = 0.01, r1 = r2 = 0.1, N = 5, µ2 = 1, µ1 = 1/m1 . number of identical machines that should be included in the production stage. It is interesting to notice that the point of maximum production rate is not found when the number of operators c1 equals the number of parallel machines present at the stage m1 . For c1 = [2, 3, 4, 5, 6] the optimal number of parallel machines is respectively equal to mopt = [4, 8, 12, 17, 18]. The motivation for this 1 behaviour is the following. In Figure 8 the efficiency in isolation of the stage 1 as a function of the number of identical machines, for different number of available operators c1 = [2, 3, 4, 5, 6] is reported. As already highlighted in Section 3.1, the efficiency in isolation is decreasing in m1 . Therefore there are two opposed phenomena that simultaneously affect the production rate. While increasing m1 the efficiency in isolation decreases with negative impact on the line production rate, especially for larger values of m2 . At the same time, while increasing m1 the variance of the output decreases with positive impact, especially for small values of m1 . Therefore, where the second positive effect is more important than the first, the system production rate increases. When the first effect dominates the second the production rate starts decreasing. This fact has two relevant consequences. Consider the case in which one can select among a more expensive faster machine and a set of identical less expensive mk slower machines, with the same p and r. We are looking at the processing rate of the slow machines in order to have the same efficiency in isolation of the stage. If ck < mk , then the speed of the slow machines must be greater than the speed of the fast machine divided by mk . In fact, the processing rate of each single machine can be

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This phenomenon has very interesting economic consequences. Normally, multiple parallel machines are adopted when using a single machine with the same overall processing rate is more expensive. In other words, more slow machines are less expensive than few fast machines. However, the previous graph shows that if money is invested in locating more machines to stage M1 without increasing the number of available operators c1 the production rate of the whole system may also decrease. This result may be extremely useful in the phase of design of the production system as well as in the phase of allocation and management of the repair crew. 5.3 Impact of Buffers on the Optimal Repairmen Allocation In the following the problem of allocating a given number of operators in the system is tackled. Indeed, when distributing the workforce in the line, it is of relevance to know in advance what will be the impact of the decision on the productivity performance of the system. In this experiment, we consider a two-stage line characterized by the following number of machines per stage: m1 = 5, m2 = 5. Failure and repair rates are p1 = 0.01, p2 = 0.02 and r1 = 0.1, r1 = 0.09. A total number of 6 operators is considered to be available in the line. Considering that allocating to each stage more than 5 operators does not make sense and that a minimum number of operators of 1 per stage should be guaranteed, this leads to 5 possible allocations of operators. The results when N = 1 and N = 12 are reported in Table 1. As it can be noticed, the Table 1. Performance measure as a function of the repairmen allocation for N = 1, 12. N =1 P n N = 12 P n

5−1 0.66918 0.8169 5−1 0.6806 11.051

4−2 0.78185 0.7295 4−2 0.8017 10.328

3−3 0.79414 0.72132 3−3 0.81535 10.239

2−4 0.79387 0.7184 2−4 0.81573 10.146

1−5 0.7729 0.663 1−5 0.80124 8.785

optimal allocation of the operations changes for different buffer capacities. When the buffer is small, a balanced allocation is preferable, while when the buffer is larger

13th IFAC INCOM (INCOM'09) Moscow, Russia, June 3-5, 2009

the second stage fails more frequently and an unbalanced allocation which accommodates the lowest efficiency stage down times is preferable. This result suggests that the design of buffers and the repair crew allocation should be performed jointly for obtaining improved system configurations. 5.4 Priority Assignment in Non-identical Machines In this experiment the impact of different priority policies assigned to parallel non-identical machine stages is investigated. For this purpose, we consider a system with 3 machines at stage 1 and 1 machine at stage 2. The system data are reported in Table 2. In this paragraph we refer to a priority as 1 − 2 − 3 meaning that M1,1 as highest priority and M1,3 has lowest priority. When c1 = 3 there is no need to assign priorities. When c1 = 2 three possible independent priorities can be assigned [1 − 2 − 3], [2 − 3 − 1], [1 − 3 − 2]. Basically, only the lowest priority machine needs to be specified. In case c1 = 1 there are 6 independent priorities. The production rate of the 10 configurations is reported in table 3. As it can Table 2. Data of the analyzed test case. N =5 pk,i rk,i µk,i Pk AsVarRatek

M1,1 0.02 0.05 1.3 0.92 9.89

M1,2 0.15 0.9 1.05 0.9 0.257

M1,3 0.005 0.009 1.25 0.8 51.8

M2,1 0.005 0.1 2.5 2.38 5.398

Priority 1−3−2 2−1−3 2−3−1 1−2−3

P 2.2287 2.1686 2.2238 2.2147 1.9915

c1 1 1 1 1 1

Priority 1−3−2 2−1−3 2−3−1 3−1−2 3−2−1

The paper models and analyzes the impact of limited repair capacity in parallel lines. The most relevant results can be summarized in the following principles: • Assigning one repairman to each machine may be a waste of resources. There are cases in which allocating fewer operators does not change the production rate of the system significantly. • When having unlimited repair capacity, increasing the number of parallel machines is always beneficial, since the variance of the stage output is reduced. When having limited repair capacity, increasing the number of parallel machines at one stage may cause a degradation of the system performance. • For non-identical machines the priority of intervention should be assigned according to the HRR and LAVR policies. Since this area is almost new, several research issues are still open. Primarily, the extension of this approach to the case of repairmen shared among different stages will be addressed. Furthermore, the general decomposition approach that is currently under development will enable to study multi-stage lines with the same characteristics. Finally, advanced design techniques to jointly configure buffers, repair crew and number of machines will be addressed. REFERENCES

Table 3. Production rate, different priority. c1 3 2 2 2 1

6. CONCLUSION

P 1.7198 2.1304 1.9309 1.607 1.7128

be noticed, for c1 = 1 the production rate differs among the best and the worst allocation of a 32%, highlighting the importance of this analysis. Furthermore, the loss in reducing the number of operators from 3 to 2 is irrelevant (0.2%), if the priority is correctly assigned. From this analysis, but also from several other test cases we analyzed (not reported due to space limitations), important insights on the best priority policy can be derived. Contrary to what can be expected, priority should not be assigned according to the lowest efficiency in isolation, to the lowest production rate in isolation nor to the highest processing rate of machines. These policies proved to be ineffective. According to our results, two policies performed best in terms of production rate. These are the Highest Repair Rate (HRR) and the Lowest Asymptotic Variance Rate (LAVR) policies. In other words, machines that can be repaired faster should be repaired first as well as machines with low variance rate. By assigning priority according to these policies, we always obtained the highest production rate, with slight difference among these two policies. Although further analysis is required to scientifically support this statement, these preliminary results provide practical outcomes to be applied by industrialists.

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