Machine Break Down per Week Answer 2022
VerifiedAdded on 2022/08/12
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Ans
. 6
Machine Break downs
per week Probability
Cumulative
Probability
Random
Number
Interval
0 0.10 0.10 00-09
1 0.10 0.20 10-19
2 0.20 0.40 20-39
3 0.25 0.65 40-64
4 0.30 0.95 65-94
5 0.05 1.00 95-99
a) Simulating the Machine breakdowns for twenty weeks
Weeks
Random
Numbers Machine Breakdowns
1 64 3
2 12 1
3 97 5
4 78 4
5 34 2
6 29 2
7 12 1
8 59 3
9 48 3
10 50 3
11 52 3
12 45 3
13 28 2
14 92 4
15 17 1
16 95 5
17 51 3
18 8 0
19 46 3
20 23 2
Total 53
b) Average Number of Machines that will Break Down per week
. 6
Machine Break downs
per week Probability
Cumulative
Probability
Random
Number
Interval
0 0.10 0.10 00-09
1 0.10 0.20 10-19
2 0.20 0.40 20-39
3 0.25 0.65 40-64
4 0.30 0.95 65-94
5 0.05 1.00 95-99
a) Simulating the Machine breakdowns for twenty weeks
Weeks
Random
Numbers Machine Breakdowns
1 64 3
2 12 1
3 97 5
4 78 4
5 34 2
6 29 2
7 12 1
8 59 3
9 48 3
10 50 3
11 52 3
12 45 3
13 28 2
14 92 4
15 17 1
16 95 5
17 51 3
18 8 0
19 46 3
20 23 2
Total 53
b) Average Number of Machines that will Break Down per week
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Total Number of Machines Breakdown for 20 Weeks 53
Number of Weeks 20
Average Number of Machines that will Break Down per week
(Georgescu, Ashhab and Nori, 2014) = 53/20
2.65
c) Repair Time (Hours) Probability
Cumulative
Probability
Random
Number
Interval
1 0.30 0.30 00-29
2 0.50 0.80 30-79
3 0.20 1.00 80-99
d) Simulating the Repair Time for twenty weeks
Weeks
Random
Numbers Repair Time
1 2 1
2 38 2
3 66 2
4 73 2
5 78 2
6 57 2
7 77 2
8 25 1
9 99 3
10 9 1
11 19 1
12 43 2
13 61 2
14 77 2
15 34 2
16 76 2
17 96 3
18 60 2
19 63 2
20 0 1
Total 37
Average Weekly Repair Time
Number of Weeks 20
Average Number of Machines that will Break Down per week
(Georgescu, Ashhab and Nori, 2014) = 53/20
2.65
c) Repair Time (Hours) Probability
Cumulative
Probability
Random
Number
Interval
1 0.30 0.30 00-29
2 0.50 0.80 30-79
3 0.20 1.00 80-99
d) Simulating the Repair Time for twenty weeks
Weeks
Random
Numbers Repair Time
1 2 1
2 38 2
3 66 2
4 73 2
5 78 2
6 57 2
7 77 2
8 25 1
9 99 3
10 9 1
11 19 1
12 43 2
13 61 2
14 77 2
15 34 2
16 76 2
17 96 3
18 60 2
19 63 2
20 0 1
Total 37
Average Weekly Repair Time
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Repair Time for 20 Weeks 37
Number of Weeks 20
Average Weekly Repair Time (In Hours) = 37/20
1.85
e) Simulating the Repair Time for twenty weeks - Random Numbers of Break downs
Weeks
Random
Numbers Repair Time
1 64 2
2 12 1
3 97 3
4 78 2
5 34 2
6 29 1
7 12 1
8 59 2
9 48 2
10 50 2
11 52 2
12 45 2
13 28 1
14 92 3
15 17 1
16 95 3
17 51 2
18 8 1
19 46 2
20 23 1
Total 36
Average Weekly Repair Time
Repair Time for 20 Weeks 36
Number ofWeeks 20
Average Weekly Repair Time (In Hours) = 36/20
1.80
Results will be different because of the changes in the random number interval
calculated on the basis of the probability.
f) Average Weekly Break down Costs
Number of Weeks 20
Average Weekly Repair Time (In Hours) = 37/20
1.85
e) Simulating the Repair Time for twenty weeks - Random Numbers of Break downs
Weeks
Random
Numbers Repair Time
1 64 2
2 12 1
3 97 3
4 78 2
5 34 2
6 29 1
7 12 1
8 59 2
9 48 2
10 50 2
11 52 2
12 45 2
13 28 1
14 92 3
15 17 1
16 95 3
17 51 2
18 8 1
19 46 2
20 23 1
Total 36
Average Weekly Repair Time
Repair Time for 20 Weeks 36
Number ofWeeks 20
Average Weekly Repair Time (In Hours) = 36/20
1.80
Results will be different because of the changes in the random number interval
calculated on the basis of the probability.
f) Average Weekly Break down Costs
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Cost per hour
€
50.00
Average Weekly Repair time (in hours) 1.85
Average Machine Breakdowns (numbers) 2.65
Average Weekly Break down Costs
= 50 X 1.85
X 2.65
€
245.13
g) ANALYSIS OF ADOPTION OF PREVENTIVE MAINTENANCE PROGRAM
Machine Break downs
per week Probability
Cumulative
Probability
Random
Number
Interval
0 0.20 0.20 00-19
1 0.30 0.50 20-49
2 0.20 0.70 50-69
3 0.15 0.85 70-84
4 0.10 0.95 85-94
5 0.05 1.00 95-99
Simulating the Machine breakdowns for twenty weeks
Weeks
Random
Numbers Machine Breakdowns
1 64 2
2 12 0
3 97 5
4 78 3
5 34 1
6 29 1
7 12 0
8 59 2
9 48 1
10 50 2
11 52 2
12 45 1
13 28 1
14 92 4
€
50.00
Average Weekly Repair time (in hours) 1.85
Average Machine Breakdowns (numbers) 2.65
Average Weekly Break down Costs
= 50 X 1.85
X 2.65
€
245.13
g) ANALYSIS OF ADOPTION OF PREVENTIVE MAINTENANCE PROGRAM
Machine Break downs
per week Probability
Cumulative
Probability
Random
Number
Interval
0 0.20 0.20 00-19
1 0.30 0.50 20-49
2 0.20 0.70 50-69
3 0.15 0.85 70-84
4 0.10 0.95 85-94
5 0.05 1.00 95-99
Simulating the Machine breakdowns for twenty weeks
Weeks
Random
Numbers Machine Breakdowns
1 64 2
2 12 0
3 97 5
4 78 3
5 34 1
6 29 1
7 12 0
8 59 2
9 48 1
10 50 2
11 52 2
12 45 1
13 28 1
14 92 4
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15 17 0
16 95 5
17 51 2
18 8 0
19 46 1
20 23 1
Total 34
Average Number of Machines that will Break Down per week
Total Number of Machines Breakdown for 20 Weeks 34
Number of Weeks 20
Average Number of Machines that will Break Down per week = 34/20
1.70
Average Weekly Break down Costs
Cost per hour
€
50.00
Average Weekly Repair time (in hours) 1.85
Average Machine Breakdowns (numbers) 1.70
Average Weekly Break down Costs (A)
= 50 X 1.85
X 1.70
€
157.25
Weekly Cost of Preventive Maintenance Program (B)
€
150.00
Average Weekly Break down Costs without Preventive
€
245.13
Maintenance Program( C )
Saving in Average Weekly Break Down Cost (D) = ( C ) - (A)
€
87.88
Loss in adoption of Preventive Maintenance Program (B) - (D)
€
62.12
(Rubinstein, 2016)
Decision - Company should not initiate the adoption of preventive maintenance
program as it will incur extra weekly cost of Euro 62.12
16 95 5
17 51 2
18 8 0
19 46 1
20 23 1
Total 34
Average Number of Machines that will Break Down per week
Total Number of Machines Breakdown for 20 Weeks 34
Number of Weeks 20
Average Number of Machines that will Break Down per week = 34/20
1.70
Average Weekly Break down Costs
Cost per hour
€
50.00
Average Weekly Repair time (in hours) 1.85
Average Machine Breakdowns (numbers) 1.70
Average Weekly Break down Costs (A)
= 50 X 1.85
X 1.70
€
157.25
Weekly Cost of Preventive Maintenance Program (B)
€
150.00
Average Weekly Break down Costs without Preventive
€
245.13
Maintenance Program( C )
Saving in Average Weekly Break Down Cost (D) = ( C ) - (A)
€
87.88
Loss in adoption of Preventive Maintenance Program (B) - (D)
€
62.12
(Rubinstein, 2016)
Decision - Company should not initiate the adoption of preventive maintenance
program as it will incur extra weekly cost of Euro 62.12
![Document Page](https://desklib.com/media/document/docfile/pages/machine-break-down-per-week/2024/09/28/b8a54175-0bb2-4293-b4ee-ec91b2ed141e-page-6.webp)
REFERENCES
Georgescu, I.M., Ashhab, S. and Nori, F., (2014), “Quantum simulation”, Reviews of Modern
Physics, 86(1), p.153.
Rubinstein, R.Y., (2016), “Simulation and the Monte Carlo method “ John Wiley & Sons, 10 (2),
p.66-75
Georgescu, I.M., Ashhab, S. and Nori, F., (2014), “Quantum simulation”, Reviews of Modern
Physics, 86(1), p.153.
Rubinstein, R.Y., (2016), “Simulation and the Monte Carlo method “ John Wiley & Sons, 10 (2),
p.66-75
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