operations management 2

Chapter 14S

Maintenance

Teaching Notes:

Maintenance is critical with JIT 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 JIT 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, reduction in 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 minimize the total cost of keeping facilities and equipment in good working order.

2. Breakdown costs would include such items as repair of equipment (perhaps on an emergency basis), lost production, disrupted schedules, parts, possible damage to product, equipment, and facilities, possible injuries, late deliveries, and loss of morale.

3. Preventive maintenance may result from inspections that reveal the need for preventive maintenance, according to calendar (passage of time), or after a set number of operations.

4. Predictive maintenance is an attempt to predict when breakdowns are likely and schedule preventive maintenance just prior to that time. Generally, such predictions are made on the basis of historical breakdown and repair records. Consequently, it is essential to have accurate, up-to-date records on which to base the analysis.

5. Organizations use some combination of these approaches to deal with breakdowns: Standby or backup equipment; inventories of spare parts; reliance on operators to perform some (usually minor) repairs; and well-trained repair personnel.

6. The Pareto concept is that in any list of factors known to contribute to a certain result or outcome (accident, injury, equipment breakdown or whatever), 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 maintenance efforts (both preventative and remedial) will be most effective when they are directed towards those few factors (procedures, equipment, policies, or whatever) that are most significant, and give much less attention to minor factors.

7. The key points should involve these concepts:

a. The need for maintenance

b. Preventive maintenance

c. Breakdown maintenance

d. Predictive maintenance

8. As the Preventive Maintenance increases, the probability of breakdown and the associated repairs decrease. When a machine fails, the following related costs must be considered.

a. Cost of lost production capacity (when machines are down, production capacity is reduced accordingly).

b. Cost of employee idle time during machine down time.

c. Cost of producing defective units, scraps and/or reworks.

d. Cost of repairs (labor and spare parts).

e. Cost of reduced quality.

f. Cost of customer service.

g. Cost of employee safety.

9. When implementing a JIT system, there is very little, if any, work-in-process inventory. Therefore, the machines (operations) depend on each other for incoming materials and parts. When a breakdown occurs, not only the failing machine shuts down, but also many of the downstream work centers may shut down. Preventive maintenance will reduce the chance of a breakdown and increase the efficiency of the system. A Preventive Maintenance program is more important for a product layout than a job-shop layout. This importance is due to the fact that there is much more dependency among operations for a product layout—especially if there is no work-in-process inventory.

10. As the level of investment in Preventive Maintenance increases, the likelihood of machine/ equipment breakdowns decrease if the machine is well maintained. Properly maintained machines will produce less defects, which will result in higher quality products.

Solutions

1.	Number of recalibrations	Probability of occurrence	Expected number of recalibrations
	0	.15	0
	1	.25	.25
	2	.30	.60
	3	.20	.60
	4	.10	.40
		1.00	1.85

Expected cost of recalibration: 1.85/month x $500 = $925/month.

Service contract = $650. This is $275 less than a month.

2.	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 would have an expected cost of:

1.90 breakdown/month x $240/breakdown = $456/month

A service contract with a single fee and all repairs covered would cost $500

A service contract covering all but the first repair would cost:

Fee: Expected repair cost: P at least (1 breakdown) x repair cost = .90 ($240) =

$350 $216 $566

3. Using the formula:

Preventive cost Breakdown cost

= Probability

Equipment Probability Z Maintenance Interval

A201:

$300_ $2,300

= .1304

–1.12

20 –1.12(2) = 17.76 days

B400:

$200_ $3,500

= .0571

–1.58

30 –1.58(3) = 25.26 days

C850:

$530_ $4,800

= .1104

–1.22

40 –1.22(4) = 35.12 days

Enrichment Module: Maintenance Problems

The purpose of this enrichment module is to further demonstrate the various trade-offs 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 probabilities, expected values and finally, break even analysis.

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 $350 to repair and the associated average downtime cost is $300. ITL Inc. is considering to purchase a preventive maintenance (PM) service contract.

· The PM service contract will guarantee to cover all breakdowns after the first breakdown in any given month. In other words, after the first breakdown the PM firm will process the repairs free of charge.

· The monthly cost of this maintenance contract is $600.

It appears that ITC Inc. has two options:

1. Do not buy the service contract and continue implementing the current repair program.

2. Purchase the preventive maintenance service.

Determine the total cost associated with the two options listed above and make a recommendation.

2. A local cutlery manufacturing company has been analyzing the breakdown/failure records of a certain type of a cutting machine. This particular type of cutting machine had the following breakdown record.

# of breakdowns per day	Frequency
0	72
1	135
2	90
3	45
4	18
	360 days/year

Cost of repair for this type of machine is $80 per breakdown. The preventive maintenance program would cost the company $30/day and would repair the machine free of charge after the first breakdown.

a. Determine the total daily cost of repairs if the company does not implement the preventive maintenance program.

b. Determine the total cost associated with maintenance and repairs if the company adopts the preventive maintenance program.

3. Use the information given in Problem 1 and determine the maximum amount that ITL Inc. would be willing to pay for the preventive maintenance service contract (Breakeven point).

4. Use the information given in Problem 1 to solve this problem.

ITL Inc. has been presented with the opportunity to purchase another service maintenance contract. In a given month, this particular contract will cover the first repair free of charge. ITL Inc. will pay for any additional repairs after the first breakdown. This service contract will cost ITL $400.00.

a. Determine the total cost of repairs and maintenance associated with this service contract.

b. Compare the service contract in Problem 1 with the service contract in this problem. Which of the two service contracts should ITL choose, if any?

Solution to 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	0
1	10	.2778	.2778
2	9	.25	.50
3	4	.1111	.3333
4	8	.2222	.8889
5	2	.0556	.2778
	36	1.0	2.2775

· E[X] = 2.2775 Expected # of breakdowns per month

· The total cost of implementing the repair program without purchasing the maintenance contract is:

E[X] * [Cost of repairs]

(2.2775) * ($650) = $1,480.37

· The total cost associated with the maintenance service contract can be determined as follows:

X
# of breakdowns	Cumulative prob. of months with X breakdowns
0	.0833
1	.3611
2	.6111
3	.7222
4	.9444
5	1.0

· P (X ( 1) = 1–.0833

P (X ( 1) = .9167 (Probability of at least one breakdown)

· Total Cost = (.9167) (1) ($650) + $600

Total Cost = $1,195.85

· $1,195.85 < $1,480.37. Therefore, buy the maintenance service contract.

Solution to Problem 2

(x) – # of breakdowns/day	Frequency	P(x)	x * P(x)
0	72	.20	0
1	135	.375	.375
2	90	.25	.50
3	45	.125	.375
4	18	.05	.20
	360 days	1.0	1.45

· E [x] = 1.45 breakdowns/day

a. Total cost if the preventive maintenance program is not purchased:

Total daily repair cost = $80 * 1.45 = $116/day

b. Repair Cost = (.8) (1) (80) = $64

Preventive Maintenance cost = $30/month

· TC = Total Cost = Repair Cost + Preventive Maintenance Cost.

TC = $64 + $30 = $94

· Since 94 < 116, purchase the service contract.

Solution to Problem 3

· E [x] = Expected # of repairs/month

· RC = Repair cost/breakdown

· y = # of breakdowns not covered by the service contract

· P(Y > y) = Probability of at least y breakdowns

· PMC = Preventive Maintenance Service Contract cost/month

· TC = Total Cost

· E[x] * RC = (RC) (y) (P(Y > y) + RMC

· (2.2775) (650) = (650) (1) (.9167) + RMC

· 1,480.37 + 595.855 + RMC

RMC = 884.51

Solution to Problem 4

· E [x] = 2.2775 breakdown/month (from problem 1)

· Since the first breakdown is covered by the service contract

· Expected Number of breakdowns not covered by the service contract is = 2.2775 – 1 = (1.2775)

· TC = (1.2775) (650) + 400

· TC = 830 + 400 = 1,230.38.

Since the TC of the service contract in problem 1 is less than the TC of the service contract in this problem, ITL should choose the preventive maintenance contract in problem 1 (1,195.85 < 1,230.38).