3.6 Numerical experience

To clarify how the proposed algorithm works, the following example is given. Consider an integrating problem with parameters as follows: number of parts q = 300 units, the setup time between batches s = 30 min, unit processing time of part t = 20 min, the length of the preventive maintenance time tPM = 60 min = 1/μ (constant), the shape parameter of Weibull distribution β = 1.69, and scale parameter α = 2857.14, constant repair rate μ = 1/60, common due date d = 10000.00, the unit inventory holding cost of finished parts c1 = US\$ 0.20 per unit per minute, the unit inventory holding cost of in-process parts c2 = US\$ 0.10 per unit per minute, the unit cost of PM cPM = US\$ 30.00, unit setup cost cs = US\$ 3.00, probability of defect part on in-control state p1 = 0.00, probability of defect part on out-of-control state p2 = 0.30, unit rework cost per unit part for nonconforming parts cw = US \$100.00, and unit corrective maintenance cost cr = US\$ 120.00.

The computational steps to solve the problem are the followings.

Step-1. Yields α = 2,857.14.

Step-2. Yields 30 + 20 x 300 = 6,030.00 ≤10,000.00 is met, then the problem is feasible.

Step-3.Yields g = 3 and Nk = 125, k = 1, 2, 3.

Step-4. Yields R = 1, 2, 3.

Step-5 to Step-21, for R = 1, yield the best solution TC[1]\* = 201,313.00.

1st looping

Step-5 to Step-21, for R = 2, yield the best solution TC[2]\* = 201,124.80. 2nd looping

Step-5 to Step-21, for R = 3, yield the best solution TC[3]\* = 201,158.80.

Step-23. Yields a set of the best solutions for R = 1, 2, 3 as a set of {TC[1]\*,TC[2]\*, TC[3]\*} = {201,313.00, 201,124.80, 201,158.80}.

Step-24. Yields a minimal TC = min {201,313.00, 201,124.80, 201,158.80} =

201,124.80 that occurs in two production runs with two PMs.

Step-25. Yield the complete solution as shown in Table 1 and Figure 6.

## 3.7 Conclusion

This chapter proposes a model of integrating batch scheduling and maintenance scheduling by criterion of minimization in holding cost, setup cost, PM cost, rework cost, and restoration cost (CM). The criterion of scheduling used is the minimization of the total actual flow time. The first preventive maintenance (PM) scheduling policy (from the direction of due date) is made precisely on due date. The second PM and so on would be done before the time of first deterioration according to Weibull ROCOF cumulative function.

production schedule, estimate the number of nonconforming parts and total restoration (CM). Next, compute estimated rework cost and estimated restoration cost, and then compute total cost. This step is done for two batches and so on until an increase on total cost is found. Write the best total cost for one production run and one PM. This process is carried out for two production runs with two PMs until the best total cost is found for two production runs with two PMs. Continue the process until g production runs with g PMs. The best solution of algorithm is a minimization

L½ � ikk Q½ � ikk B½ � ikk BPM[k] CPM[k] TC[2]\* L½ � <sup>11</sup> <sup>1</sup> 42.46 9150.77 BPM[1] = 10000.00 CPM[1] = 10060.00 201124.80

Integrated Batch Production and Maintenance Scheduling to Minimize Total Production…

L½ � <sup>101</sup> <sup>1</sup> 15.46 3937.69 BPM[2] = 3847.69 CPM[2] = 3907.69

L½ � <sup>21</sup> <sup>1</sup> 39.46 8331.54 L½ � <sup>31</sup> <sup>1</sup> 36.46 7572.31 L½ � <sup>41</sup> <sup>1</sup> 33.46 6873.08 L½ � <sup>51</sup> <sup>1</sup> 30.46 6233.85 L½ � <sup>61</sup> <sup>1</sup> 27.46 5654.62 L½ � <sup>71</sup> <sup>1</sup> 24.46 5135.39 L½ � <sup>81</sup> <sup>1</sup> 21.46 4676.15 L½ � <sup>91</sup> <sup>1</sup> 18.46 4276.92

DOI: http://dx.doi.org/10.5772/intechopen.85004

L½ � <sup>12</sup> <sup>2</sup> 6.46 3718.46 L½ � <sup>22</sup> <sup>2</sup> 3.46 3619.23 L½ � <sup>32</sup> <sup>2</sup> 0.46 3580.00

Best schedule for the example.

Table 1.

Figure 6.

41

The model makes a trade-off in the following two things. An increase in the number of batch (length of production run) up to a certain limit will minimize the

of all of the best total cost for k = 1, 2, …, g.

Gantt chart of the best solution for the example.

In the model developed, searching for a solution begins with problem resolving without involving restoration cost (CM) and rework cost for nonconforming part. It begins with one batch in one production run and one PM. After finding a


Integrated Batch Production and Maintenance Scheduling to Minimize Total Production… DOI: http://dx.doi.org/10.5772/intechopen.85004

Table 1. Best schedule for the example.



Min {TC[R], R = 1, 2, …, g}. Go to Step-25.

\$100.00, and unit corrective maintenance cost cr = US\$ 120.00.

Step-3.Yields g = 3 and Nk = 125, k = 1, 2, 3.

TC[3]\*} = {201,313.00, 201,124.80, 201,158.80}.

The computational steps to solve the problem are the followings.

Step-5 to Step-21, for R = 1, yield the best solution TC[1]\* = 201,313.00.

Step-5 to Step-21, for R = 2, yield the best solution TC[2]\* = 201,124.80.

Step-5 to Step-21, for R = 3, yield the best solution TC[3]\* = 201,158.80.

Step-25. Yield the complete solution as shown in Table 1 and Figure 6.

201,124.80 that occurs in two production runs with two PMs.

Step-2. Yields 30 + 20 x 300 = 6,030.00 ≤10,000.00 is met, then the problem is

Step-23. Yields a set of the best solutions for R = 1, 2, 3 as a set of {TC[1]\*,TC[2]\*,

This chapter proposes a model of integrating batch scheduling and maintenance scheduling by criterion of minimization in holding cost, setup cost, PM cost, rework cost, and restoration cost (CM). The criterion of scheduling used is the minimization of the total actual flow time. The first preventive maintenance (PM) scheduling policy (from the direction of due date) is made precisely on due date. The second PM and so on would be done before the time of first deterioration according to

In the model developed, searching for a solution begins with problem resolving without involving restoration cost (CM) and rework cost for nonconforming part. It

begins with one batch in one production run and one PM. After finding a

Step-24. Yields a minimal TC = min {201,313.00, 201,124.80, 201,158.80} =

To clarify how the proposed algorithm works, the following example is given. Consider an integrating problem with parameters as follows: number of parts q = 300 units, the setup time between batches s = 30 min, unit processing time of part t = 20 min, the length of the preventive maintenance time tPM = 60 min = 1/μ (constant), the shape parameter of Weibull distribution β = 1.69, and scale parameter α = 2857.14, constant repair rate μ = 1/60, common due date d = 10000.00, the unit inventory holding cost of finished parts c1 = US\$ 0.20 per unit per minute, the unit inventory holding cost of in-process parts c2 = US\$ 0.10 per unit per minute, the unit cost of PM cPM = US\$ 30.00, unit setup cost cs = US\$ 3.00, probability of defect part on in-control state p1 = 0.00, probability of defect part on out-of-control state p2 = 0.30, unit rework cost per unit part for nonconforming parts cw = US

Step 22. Observe whether R = g

3.6 Numerical experience

Industrial Engineering

Step-1. Yields α = 2,857.14.

Step-4. Yields R = 1, 2, 3.

Weibull ROCOF cumulative function.

feasible.

1st looping

2nd looping

3.7 Conclusion

40


Step-25. Write all values of decision variables.

Step 23. Write {TC[R], R = 1, 2, …, g}. Step-24. The minimal solution is

Figure 6. Gantt chart of the best solution for the example.

production schedule, estimate the number of nonconforming parts and total restoration (CM). Next, compute estimated rework cost and estimated restoration cost, and then compute total cost. This step is done for two batches and so on until an increase on total cost is found. Write the best total cost for one production run and one PM. This process is carried out for two production runs with two PMs until the best total cost is found for two production runs with two PMs. Continue the process until g production runs with g PMs. The best solution of algorithm is a minimization of all of the best total cost for k = 1, 2, …, g.

The model makes a trade-off in the following two things. An increase in the number of batch (length of production run) up to a certain limit will minimize the

### Industrial Engineering

total inventory holding cost. Meanwhile, an increase in the length of production run will imply on an increase in the number of nonconforming parts and in a number of restoration (CM).

References

33:185-195

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DOI: http://dx.doi.org/10.5772/intechopen.85004

Integrated Batch Production and Maintenance Scheduling to Minimize Total Production…

[8] Jiang RY, Murthy DNP. Maintenance Decision Models for Management. Beijing: Science Press; 2008

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[4] Zahedi Z, Ari Samadhi TMA, Suprayogi S, Halim AH. Integrated batch production and maintenance scheduling for multiple items processed on a deteriorating machine to minimize total production and maintenance costs with due date constraint. International Journal of Industrial Engineering Computations. 2016;7(2):229-240

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[6] Zahedi Z, Salim A, Yusriski R, Haris H. Optimization of integrated batch production and maintenance scheduling on flow shop with two machines. International Journal of Industrial Engineering Computations (IJIEC). 2019;10(2):225-238. DOI: 10.5267/j.

[7] Daellenbach HG, McNickle DC. Management Science, Decision Making Through Systems Thinking. New York,

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For readers who want to learn more about the integration of batch production and machine maintenance scheduling, they can read Zahedi et al. [4–6].
