How to
calculate return on investment for maintenance
improvement projects
by Terry Wireman,
GenesisSolutions
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In an effort to become more competitive, companies have
found that maintenance represents from 15 to 40% of the
total product cost and dollars saved in maintenance are
a cost avoidance. In larger companies, reducing
maintenance expenditures by $1 million contributes as
much to profits as increasing sales by $3 million.
Improving maintenance and decreasing unnecessary
maintenance expenditures by $1 million is considerably
easier and more likely to occur than obtaining $3
million in new sales.
In this article,
guidelines are presented for calculating possible
savings that may be achieved by investing in improved
maintenance policies and practices, including a
computerized maintenance management system.
The objective is to
present a different method for examining the effect of
maintenance on a company’s costs. The material is
divided into sections to allow various parts to be used
where applicable and omitted where not.
Standard cost
justification
Standard cost
justification is composed of four main parts:
-
Maintenance labor
costs.
-
Maintenance materials
costs.
-
Project cost savings.
-
Downtime/availability
costs.
Maintenance labor costs –
Maintenance productivity in most American companies
averages between 25 and 35%, which is equivalent to less
than 3-hr/8-hr shift of hands-on activities. Most lost
productivity can be attributed to:
-
Waiting on parts.
-
Waiting on information,
drawings, instructions, etc.
-
Waiting for equipment
to be shut down.
-
Waiting on rental
equipment to arrive.
-
Waiting on other crafts
to finish their part of the job.
-
Running from emergency
to emergency.
While 100% productivity
is an unrealistic goal for any maintenance organization,
a more realistic percentage of 60% is achievable.
The productivity of
maintenance technicians can be improved by concentrating
on basic management techniques, such as:
-
Planning jobs in
advance.
-
Scheduling jobs and
coordinating schedules with operations.
-
Arranging for parts to
be ready.
-
Coordinating
availability of tools, rental equipment, etc.
-
Reducing
emergency work below the 50% level by preventative
maintenance
.
With computer
assistance, planning time per job is reduced, resulting
in more jobs planned and coordinated. This results in
more time for preventative maintenance activities which,
in turn, helps to reduce the amount of emergency and
breakdown activities. This results in fewer schedule
changes and increases productivity by reducing travel
and waiting times. Successful users of computerized
maintenance management systems have indicated an
increase in productivity of 28%.
A procedure for
calculating potential savings in maintenance labor costs
is shown in Table I.
Inventory and stores savings –
Maintenance material costs are related to the frequency
and size of the repairs made to the company’s
equipment. The total number of parts, in addition to
the stores’ policies, purchasing policies and overall
inventory management practices contribute to the overall
maintenance materials costs. Since little attention is
paid to maintenance materials in some companies,
inventories may be higher than necessary by 20 to 30%.
TABLE I – Procedure for
calculating savings in
maintenance labor costs
1.
Time wasted by personnel looking for spare
equipment parts, % _____
No
inventory system, 15-25%
Manual inventory system, 10-20%
Work
order system and inventory system, 5-15%
Computerized inventory and manual work order
system, 0-5%
2.
Time spent looking for information about a work
order, % _____
Manual work order system, 5-15%
No
work order system, 10-20%
3.
Time wasted by starting wrong priority work
order, % _____
Manual work order system, 0-5%
No
work order system, 5-10%
4.
Time wasted by equipment not being ready to work
on
(still in production),
% _____
Manual work order system, 0-5%
No
work order system, 10-15%
5.
Total wasted time,
%
_____
(add
lines 1, 2, 3, and 4)
6.
Total number of
craftsmen
_____
7.
Multiply line 6 by 2080,
hr
_____
(normal hours worked by an employee for a
year)
8.
Total number of hours for a craftsman,
hr _____
(multiply lines 5 and 7)
9.
Average labor rate, including benefits for a
craftsman, $/hr _____
10.
Potential savings, $ (multiply lines 8 and
9) _____
11.
Total savings,
$
_____
(multiply line 10 by percentage that describes your
facility)
No
work order or inventory system, 75-100%
Manual work order system and inventory system, 50-75%
Manual work order system and inventory system, 30-50%
Computerized inventory and manual work order system,
25-40%
This increases
inventory holding costs and makes materials
unnecessarily expensive. The inability of stores to
service maintenance department’s needs results in
unnecessary storage depots for just in case spares.
This practice also increases the cost of maintenance
materials.
Good inventory
control enables companies to lower the value of the
inventory and continue to maintain a service level of at
least 95%. This enables the maintenance department to
be responsive to the operations group, while increasing
the maintenance department’s own personal productivity.
Successful computerized maintenance management system
users have averaged 19% lower material costs and an
overall 18% reduction in total inventory.
A procedure for
calculating potential savings in maintenance material
costs is shown in Table II.
Major
projects, outage and overhaul savings –
In
many companies, maintenance is involved in projects,
outage or refurbishing activities. These activities, if
not properly controlled can have a dramatic impact on a
company’s production capacity. The reason is, that
these activities are usually performed with the
equipment in a down condition. That means there is no
production during this time. For this reason, any time
that can be eliminated from the project, outage or
refurbishing activity, can be converted to production
time.
Improved planning and
coordination can be achieved with a computerized
maintenance management system. This will often assist
in shortening the downtime, even if the company is
currently using a project management system. Successful
computerized maintenance management system users have
indicated an average 5% reduction in outage time.
A procedure for
calculating potential savings is shown in Table III.
TABLE II – Procedure for
calculating savings in
maintenance material costs
1.
Total dollar value of maintenance spares
purchased per year, $ _____
2.
Percentage of time spares are already in stores
when _____
others are purchased, %
No
inventory system, 25-30%
Manual inventory system, 10-20%
Computerized inventory system, 5-15%
3.
Savings total (cost avoidance),
$ _____
4.
Additional savings (inventory overhead),
$ _____
(multiply line 3 by 30%)
5.
Estimated total inventory valuation, $
6.
Estimated inventory reduction, %
_____
No
inventory system, 15-20%
Manual system, 5-10%
(Obsolete of unnecessary spares)
7.
Estimated one-time inventory reduction,
$ _____
(multiply lines 5 and 6)
8.
Estimated additional savings,
$ _____
(multiply line 7 by 30%)
(holding cost reduction)
9.
Number of stock-outs causing
downtime _____
10.
Amount of downtime,
hr
_____
11.
Cost of downtime,
$/hr
_____
12.
Total cost of materials related downtime,
$ _____
(multiply lines 10 and 11)
13.
Percentage of savings obtainable,
% _____
Current controls poor, 75%
Current controls fair, 50%
Current controls good, 25%
14.
Savings in materials-related equipment downtime,
$ _____
(multiply lines 12 and 13)
15.
Total savings,
$
_____
(add lines 3, 4, 7, 8 and 14)
TABLE
III – Procedure for calculating major project
outage and overhaul
savings
1.
Number of major outages and overhauls per
year _____
2.
Average length of outage or overhaul,
days _____
3.
Cost of equipment downtime in lost sales, $
(hourly downtime rate multiplied by total hours of
outages)
4.
Total estimated cost per year,
$ _____
(multiply lines 1, 2 and 3)
5.
Estimated savings percentage,
% _____
No
computerized work order system, 5-10%
Project management system, 3-8%
Project management system and inventory control
system, 2-5%
6.
Total cost savings, $
_____
Equipment downtime costs –
Downtime/availability costs are major potential savings
for a company that improves maintenance policies and
practices. Downtime costs for equipment may very from
several hundred to hundreds of thousands of dollars per
hour. (One company has several production lines in its
plants, with the downtime on each being $1 million/24
hr.)
In some companies, the
levels of downtime can run as high as 30% or more. By
dedicating the company to enforcing good maintenance
policies and practices, and utilizing the computerized
maintenance management system as a tracking tool,
equipment downtime can be reduced dramatically.
Successful users have averaged a 20% reduction in
equipment downtime losses.
A procedure for
calculating potential savings in equipment downtime is
shown in Table IV.
Total
projected savings –
A summary calculation of total projected savings and
return on investment is shown in Table V.
Detailed cost savings
The savings suggested in
this section are more difficult to calculate for most
companies because they require prior data or accurate
estimates. Where available, industry averages or ranges
are given as guidelines for companies not possessing
complete internal data.
TABLE IV – Procedure for
calculating savings in
equipment downtime costs
1.
Percentage of equipment downtime per year,
% _____
(if
not known, use estimate-average for industry
is 5
to 25%)
2.
Total number of production hours for equipment
for year, hr _____
3.
Total of all lost production hours for year, hr
(multiply lines 1 and
2) _____
4.
Multiply total lost production hours for year
(line 3)
by
percentage that currently describes your facility,
% _____
No
work order system, 25%
Work
order system, 20%
Work
order and stores inventory system, 10%
5.
Total of downtime saved,
hr _____
6.
Cost of downtime,
$/hr
_____
7.
Total downtime cost savings,
$ _____
(multiply lines 5 and 6)
Optional savings considerations:
§
Total
direct labor wages and benefits multiplied by total of
all lost production hours, $
§
Lost
sales for year (divide total sales for year by the total
number of yearly production hours and multiply figure by
total downtime hours saved), $
§
Increased production costs to make up production lost
due to downtime. This would include the extra labor
required on week-ends or off shifts to operate the
equipment, extra energy costs to operate the equipment,
etc.
TABLE V – Procedure for calculating total projected
savings and return on investment
1.
Total from Table
I
_____
2.
Total from Table
II
_____
3.
Total from Table
III
_____
4.
Total from Table
IV
_____
5.
Total savings possible from improvement program,
$ _____
(Total of lines 1 to 4)
6.
Total projected price for improvement program,
$ _____
7.
Return on investment (divide line 6 by line 5)
_____
Warranty
costs for equipment –
In many
companies that have recently purchased equipment,
warranty costs is an area of potential savings. In many
instances, some maintenance repairs made on equipment
under warranty are reimbursable under the purchase and
service agreement with the equipment supplier. The
amount of the reimbursement can vary but companies have
found that 5 to 10% of all work performed on equipment
covered by warranties can be reimbursable.
There are some
considerations a company might make when investigating
this area of savings. These areas may make it difficult
to comply with warranty provisions:
-
To
be covered by the warranty, do the repairs have to
made or supervised by a representative of the supplier
company?
-
If the
repairs are made by internal technicians, does it void
the warranty?
-
What
level of documentation must be provided to the
supplier to collect under the terms of the warranty?
If these, or
similar provisions, impact the warranty, the company may
want to consider whether it is worth the effort. For
example, what if a critical piece of equipment would
have to remain shut down waiting for the supplier’s
representative to arrive and make or oversee the
repairs? The cost of downtime could quickly accumulate
and exceed the money that could be regained from
warranty claims.
There are opportunities
to receive reimbursements for repairs made to equipment
under warranty. However, a company would want to make a
cost/benefit evaluation before these opportunities are
actively pursued.
Energy
cost savings –
To
effectively calculate any energy costs savings, it is
necessary for a company to know its energy usage. If
this is not known, industry averages can be used as an
estimate. Studies by engineering institutes and
international companies have shown that a company can
reduce energy consumption at a plant by 5 to 10%
depending on current maintenance effort. Companies with
good maintenance programs would see savings in the 5%
range. Companies with little or no preventative
maintenance inspections and services would realize
savings in the 10% range. Some examples of energy
savings for typical systems (mechanical, electrical,
steam and fluid) follow.
With mechanical
systems, some of the energy savings can be defined by
the type of preventative maintenance performed on some
of the basic mechanical components. For example, how
accurate are couplings aligned? Misalignment by as
small as 0.003 in. can lead to energy loss through the
coupling. This loss is typically displayed as heat
energy in the flex member of the coupling and the
supporting shaft bearings – even elastomer couplings
will display energy loss. A second type of mechanical
loss is V-belt slippage. Chain and gear misalignment
will also lose energy in the transmission area and
bearings. Poor maintenance practices and preventive
maintenance will contribute a 5 to 10% energy loss for
mechanical power transmission.
Electrical systems, as
with mechanical systems, will be determined by the
condition of the system and level of maintenance service
performed. Typical energy losses occur in loose
connections, poor motor conditions, and contamination of
the insulation that increases the temperature of the
motor and, subsequently, its energy consumption.
Improper or insufficient maintenance on mechanical
drives will also increase the amount of energy required
by the motor to drive the system. This, together with
other losses, will contribute to excessive energy
requirements by electrical systems. A 5 to 10% energy
loss due to poor electrical system maintenance can be
expected.
Steam generation
systems have long been recognized as having the
potential for substantial energy savings at most
plants. Steam trap inspection programs, energy
efficient boilers and leak detection programs have been
utilized in reducing steam system losses. Depending on
the amount of maintenance performed on the steam system,
energy savings from 5 to 15% have been reported by
companies initiating good maintenance practices.
Fluid power systems
include both hydraulic and pneumatic systems. Wasted
energy in these systems in generally related to leaks.
Leaks can be internal or external. External leaks are
easier to find since air leaks will make a noise and oil
leaks leave a pool of fluid. These leaks waste energy
since the compressor or pump will have to run more
frequently for the system to operate correctly. In
addition, hydraulic systems will require cleaning up the
leaks, another form of energy waste. Internal leaks are
more difficult to detect particularly when they are
small. They are usually identified by sluggish
performance and, in hydraulic systems, by excessive
component heat. Pumps and compressors must also run
more frequently to compensate for the leaks. These, and
other energy losses, will account for energy losses of 5
to 15% in fluid power systems.
Quality
cost savings –
Since
the maintenance department is responsible for the
condition of equipment, quality costs are impacted by
poor maintenance practices. For example, what
percentage of all quality problems eventually are solved
by a maintenance activity? Even if the activity is
performed by the operator, the activity is one of
maintaining the equipment condition. In some companies,
60% or more of the quality problems are equipment
related. To calculate to possible cost savings, the
value of the annual production for the plant should be
calculated. Next, the current first-pass quality rate
should be determined and the difference between this
rate and 100% is the current reject rate.
This next step would
be to determine the reasons for the rejects. Usually, a
top-ten list will provide the majority of the rejects.
After examining the list, determine which causes have a
maintenance solution. This is the percentage amount
that could be possibly reduced. An estimate of what
proportion of all the maintenance related losses could
be eliminated by a good maintenance program must be
made. This proportion, multiplied by the dollar value
of the company’s annual product, will produce the
possible quality related savings. This number should
then be added as a line item to all of the previous
savings.
Overall equipment
effectiveness related savings
Overall equipment
effectiveness related savings are a more recent addition
to the return on investment calculation shown in Table
V. This function evaluates all causes of lost
production, not only maintenance. However, the exact
problems causing the losses tend to be specific to the
operating and organization. The following case study
illustrates the potential losses that may be found at a
plant.
Idling and minor
stoppage losses are problems that are not closely
observed. These are relatively minor problems equipment
may experience and the company tolerates. The
following example is taken from a disk-drive
manufacturer involving the assembling of hard drives.
If the robot drops a platter, it requires operator
intervention. The operator’s intervention requires 5
min to reset the equipment each time a disk is dropped.
The operators will tolerate one intervention every 2 hr
before they complain. This line operates 7 days/24-hr
per day/12-hr shifts. By extrapolation, six drops per
shift equals 30 min delay/shift. If the 30 min is
multiplied by 14 shifts, this equals 420 min/week. The
420 min equals 7 hr of downtime. If one event of 7 hr
of downtime occurs, action would be taken to determine
the cause and remedy the situation. But 5-min delays
are tolerated and eventually cost more than the longer
delays. Examining the cost of one hour of downtime
allows the appropriate amount of expenditures to be
allocated to eliminate the root cause of the problem.
Conclusion
Maintenance improvement
can be measured and a return on investment clearly
determined.
Summary
The benefits of
increased maintenance labor productivity, material cost
savings, project cost savings and reduced equipment
downtime have been highlighted, and procedures for
calculating potential savings provided.
The role played by
maintenance in obtaining energy savings is also
reviewed. |