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Chapter 14S - Maintenance
CHAPTER 14S
MAINTENANCE
Teaching Notes:
Maintenance is critical to Lean and TQM implementation. Poor maintenance leads to direct repair
expenses and other expenses resulting from lost production. The production line disruptions are
particularly important in lean environments. An effective maintenance program ensures that machines
are functioning and performing the operations so that quality of the product is not compromised. An
effective maintenance program can lead to improved capacity; reduced defects, scrap, and rework;
smaller inventories; higher productivity; and lower product costs.
Answers to Discussion and Review Questions
1. The goal of a maintenance program is to keep the production system in good working order at
minimal cost.
2. Breakdown costs would include:
a. Cost of lost production capacity (when machines are down, production capacity is reduced
3. Preventive maintenance may result from planned inspections that reveal the need for
4. Predictive maintenance is an attempt to determine when to perform preventive maintenance
activities based on historical records and analysis of technical data to predict when
5. Organizations use some combination of these approaches to deal with breakdowns: Standby or
6. a. Preventive maintenance: The Pareto phenomenon tells us that in any list of factors known
to contribute to a certain result or outcome (e.g., accident, injury, equipment breakdown,
etc.), a relatively few factors will account for a disproportionate share of the result (e.g.,
90% of accidents are caused by carelessness). The importance of this is that preventive
14S-2
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
b. Breakdown maintenance: Similarly, the largest percentage of breakdown maintenance
costs will be spent on those breakdowns that account for the largest percentage of
breakdown and repair costs.
7. The key points of this supplement that relate to maintenance of an automobile should involve
these concepts:
8. As preventive maintenance increases, the probability of breakdown and associated repairs
9. When implementing a lean system, there is very little, if any, work-in-process inventory.
Therefore, the machines (operations) depend on each other for incoming materials and parts.
10. As the level of investment in preventive maintenance increases, the likelihood of machine/
equipment breakdowns decreases if the machine is well maintained. Properly maintained
machines will produce fewer defects, which will result in higher quality products.
Chapter 14S - Maintenance
Solutions
1. Given:
The probability that equipment in a hospital lab will need recalibration is given in the table
below. A service firm will provide maintenance and provide any necessary calibrations for
$650 per month. Recalibration costs $500 per time.
Number of
Recalibrations
0
1
2
3
4
Probability of
Occurrence
.15
.25
.30
.20
.10
Determine which approach, recalibration as needed or the service contract, would cost less:
Step 1:
Determine the expected number of recalibrations without the service contract. For each
number of recalibrations, multiply the number of recalibrations by the probability of
occurrence and sum:
Number of
Recalibrations
Probability of
Occurrence
Expected Number of
Recalibrations
0
.15
.00
1
.25
.25
2
.30
.60
3
.20
.60
4
.10
.40
1.00
1.85
Step 2:
Determine the expected cost of recalibration to the cost of the service contract.
14S-4
2. Given:
The frequency of breakdown of a machine is given in the table below. Repairs cost an average
of $240. A service firm will provide preventive maintenance under two options: #1 is
$500/month and covers all necessary repairs, and #2 is $350/month and covers any repairs
after the first one.
Number of
Breakdowns/Month
0
1
2
3
4
Frequency of
Occurrence
.10
.30
.30
.20
.10
Pay for All Repairs Approach:
Determine expected cost of the repair approach. For each number of breakdowns, multiply the
number of breakdowns by the probability of occurrence and sum:
Number of
breakdowns
Probability of
occurrence
Expected number of
breakdowns
0
.10
0
1
.30
.30
2
.30
.60
3
.20
.60
4
.10
.40
1.00
1.90
Repair Approach Expected Cost: 1.90 breakdowns/month x $240/breakdown = $456/month
Option 1 Cost:
$500/month
Option 2 Cost:
Determine expected cost of Option #2: PM contract costing $350 and covering all repairs after
the first one:
When there are 1 or more repairs, 1 of the repairs is not covered by the service contract (this
will occur when we have 1, 2, 3, or 4 breakdowns). We need to apply the probabilities to
calculate expected repair cost:
Expected Repair Cost:
1 breakdown: (1 breakdown * .30 * $240/breakdown) = $ 72
2 breakdowns: (1 breakdown * .30 * $240/breakdown) = $ 72
Chapter 14S - Maintenance
3. Given:
Determine the optimum preventive maintenance frequency for each of the pieces of equipment
below if breakdown time is normally distributed:
Equipment
Average Time
(days) between
Breakdowns
Standard
Deviation
Preventive
Maintenance Cost
Breakdown
Cost
A201
20
2
$300
$2,300
B400
30
3
$200
$3,500
C850
40
4
$530
$4,800
Step 1:
Compute the ratio of preventive cost to breakdown cost for each piece of equipment (round to
a maximum of four decimals). 𝑃𝑟𝑒𝑣𝑒𝑛𝑡𝑖𝑣𝑒 𝑐𝑜𝑠𝑡
𝐵𝑟𝑒𝑎𝑘𝑑𝑜𝑤𝑛 𝑐𝑜𝑠𝑡
Step 2:
Find the z value corresponding to the ratio from Step 1 and use this value of z to compute the
maintenance interval. Round z and maintenance interval to two decimals.
Chapter 14S - Maintenance
Enrichment Module: Maintenance Problems
The purpose of this enrichment module is to demonstrate further the various tradeoffs in maintenance
management by using realistic problems. In addition, these problems are designed to add richness and
variety to the material and problems covered in this supplement. These problems reinforce the
concepts students should have learned earlier, either in a basic statistics course, or in earlier chapters
of this book. These concepts include basic probabilities, frequency tables, converting frequencies to
1. ITL Inc. has been keeping track of breakdowns per month of their automated assembly line
over the past three years.
# of Breakdowns
# of Months This Occurred
0
3
1
10
2
9
3
4
4
8
5
2
Each failure (breakdown) of the assembly line will cost the company a repair cost of $650.
ITL Inc. is considering purchasing a preventive maintenance (PM) service contract.
The PM service contract will guarantee to cover all repair costs after the first breakdown
in any given month.
The monthly cost of this PM service contract is $600.
It appears that ITL Inc. has two options:
1. Do not buy the PM service contract and continue using the current repair program.
2. Purchase the PM service contract.
Determine the total cost associated with the two options listed above and make a
14S-8
Problem 1
Option 1:
x
F(x)
P(x)
x*P(x)
# of breakdowns
# of months this
occurred
Probability of # of
breakdowns/months
0
3
.0833
.0000
1
10
.2778
.2778
2
9
.2500
.5000
3
4
.1111
.3333
4
8
.2222
.8888
5
2
.0556
.2780
36
1.0000
2.2779
E[x] = 2.2779 (Expected # of breakdowns per month)
2.2779 * $650 = $1,480.64
Option 2:
Determine expected cost of Option #2: PM service contract costing $600 covering all repairs after
the first one:
When there are 1 or more repairs, 1 of the repairs is not covered by the PM service contract (this
will occur when we have 1, 2, 3, 4, or 5 breakdowns). We need to apply the probabilities to
calculate expected repair cost:
Expected Repair Cost:
1 breakdown: (1 breakdown * .2778 * $650/breakdown) = $180.57
2 breakdowns: (1 breakdown * .2500 * $650/breakdown) = $162.50
3 breakdowns: (1 breakdown * .1111 * $650/breakdown) = $ 72.22
4 breakdowns: (1 breakdown * .2222 * $650/breakdown) = $144.43
5 breakdowns: (1 breakdown * .0556 * $650/breakdown) = $ 36.14
Expected Repair Cost = $595.86
Alternative Method of Computing Expected Repair Cost:
P (X ≥ 1) = 1 – P(X = 0) = 1 - .0833 = .9167
Expected Repair Cost = .9167 * $650 = $595.86
Total Cost of Option 2 = Expected Repair Cost + PM Service Contract Cost
= $595.86 + $600 = $1,195.86/month
Chapter 14S - Maintenance
14S-9
Solution to Problem 2
x
# of
Breakdowns/Day
Frequency
P(x)
x * P(x)
0
72
.200
.000
1
135
.375
.375
2
90
.250
.500
3
45
.125
.375
4
18
.050
.200
360 days
1.000
1.450
E [x] = 1.450 breakdowns/day
a. Total cost if the preventive maintenance (PM) program is not purchased:
Expected Daily Repair Cost = 1.450 breakdowns/day * $80/breakdown = $116/day
b. Total Cost: Expected Repair Cost + PM Program Cost
When the there are 1 or more repairs, 1 of the repairs is not covered by the PM program (this
will occur when we have 1, 2, 3, or 4 breakdowns). We need to apply the probabilities to
calculate expected repair cost:
Alternative Method of Computing Expected Repair Cost:
P (X ≥ 1) = 1 – P(X = 0) = 1 - .200 = .800
Expected Repair Cost = .800 * $80 = $64
14S-10
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
McGraw-Hill Education.
