1.
MAINTENANCE
LABOR COST ESTIMATION
The cost of labor is an important component of the
maintenance cost. Labor costs are made up of payroll information that is
usually obtained from labor distribution reports prepared by the accounting
department. The information is required for four key areas:
1. Total number of hours worked annually on a
per-employee basis
2. Hourly cost of employee benefits on a
per-employee basis
3. Ratio of cost of annual benefits to yearly wages
4. Base pay rates per hour by labor grade
·
The
cost per employee is expressed by :
Cem
= LR(1 + BR)TAH……………………………………..Eq(1.1)
Where :
Cem = cost per employee,
LR = hourly labor rate,
BR = benefit ratio,
TAH = total number of annual hours.
·
The
total labor cost is given by :
TLC
= CemN……………………………………………….Eq(1.2)
Where :
TLC = total labor cost,
N = number of employees
Example
:
Assume that in a maintenance organization we have
the following data:
TAH = 2000 h
BR = 0.2
LR = $15 per hour
N = 40
Calculating the total labor cost associated with the
maintenance activity by
inserting the above values into Eqs. (1.1) and
(1.2), we get :
Cem
= 15 × (1 + 0.2) × 2,000 = $36,000
And
TLC = 36,000 × 40 = $1,440,000
Thus, the total labor cost associated with the
maintenance activity will be $1.44 million.
2.
CORRECTIVE M AINTENANCE LABOR COST ESTIMATION
In this case corrective maintenance labor cost is
estimated when the item/system mean time between failures (MTBF) and mean time
to repair (MTTR) are known. Consequently, the annual labor cost of corrective
maintenance is expressed by :
.........……..Eq(2.1)
Where :
CMal = annual
corrective maintenance labor cost,
LCH = corrective maintenance
labor cost per hour,
SOH = annual scheduled operating
hours.
Example :
A system is scheduled to operate
for 2000 hours per year. The system’s MTBF and MTTR are 400 h and 20 h,
respectively. Calculate the annual labor cost of corrective maintenance if the
maintenance labor cost is $20 per hour. By substituting the specified data into
Eq. (2.1), we get :
The annual labor cost of
corrective maintenance is $2000.
3.
EQUIPMENT
OWNERSHIP CYCLE MAINTENANCE COST
ESTIMATION
This section is concerned with
estimating the maintenance cost of the entire ownership cycle of equipment, so
that its present value can be added to acquisition cost to obtain equipment
life cycle cost. The following two formulas are often used to find present
value of a sum of money. They can equally be used to obtain present values of
equipment ownership cycle maintenance costs :
FORMULA I
This
formula is used to estimate the present value of a single amount of money after
k periods and is expressed by :
....……….Eq(3.1)
Where :
PV = present
value,
AM = single
amount,
k = number of
interest or conversion periods (normally taken as years),
i = interest rate
per period.
FORMULA
II
This formula is concerned with
obtaining the present value of equal amounts of, say, maintenance costs
occurring at the end of each of k conversion periods (usually years). The
present value is given by :
...………Eq(3.2)
Where :
MC
= maintenance cost occurring at the end of each (interest) conversion period.
Example
:
A
maintenance department considers procuring an engineering system. Two
manufacturers are bidding to provide the system and their corrective
maintenance costrelated data are given in Table 3.1. Determine which of the two
systems will be less costly with respect to present
values
of corrective maintenance and by how much :
Tabel 3.1 Maintenance Cost-Related Data
Manufacturer
A System
The annual expected cost, MC, of
a corrective maintenance action is :
MCa
= 1000 × 2.5 = $2500
The present value, PVA,
of the engineering system life cycle corrective maintenance
cost using Eq. (3.2) and the
specified and calculated values is :
Manufacturer
B System
The annual expected cost, MC, of
a corrective maintenance action is :
MCb
= 1400 × 2 = $2800
The present value, PVB,
of the engineering system life cycle corrective maintenance
cost using Eq. (3.2) and the
specified and calculated values is :
Manufacturer A’s engineering
system will be less costly with respect to present
values
of corrective maintenance by $2,260.81.
4.
ACTIVITY EXPECTED DURATION TIME ESTIMATION
The PERT scheme calls for three estimates of
activity duration time using the following
formula
to calculate the final time:
……………Eq(4.1)
Where :
Ta
= activity expected duration time,
OT = optimistic
or minimum time an activity will require for completion,
PT = pessimistic
or maximum time an activity will require for completion,
MT = most likely
time an activity will require for completion. This is the time
used for CPM activities.
Example 3.1
Assume that we have the following
time estimates to accomplish an activity:
OT = 55 days
PT = 80 days
MT = 60 days
Calculate the activity expected
duration time. Substituting the
given data into Eq. (4.1), we get :
The expected duration time for
the activity is 62.5 days.
5.
MAINTAINABILITY
MEASURES AND FUNCTIONS
Various measures
are used in maintainability analysis: for example, mean time to repair
(MTTR), mean preventive
maintenance time, and mean maintenance downtime. Maintainability
functions are used to predict the
probability that a repair, starting at time t= 0, will be completed in a
time. Some maintainability measures and functions are presented
below.
MEAN TIME TO R
EPAIR
Mean
time to repair (MTTR) is probably the most widely used maintainability measure.
It
measures the elapsed time required to perform a given maintenance activity.
MTTR
is
expressed by :
…………..Eq(5.1)
Where :
k number
of units or parts,
li failure
rate of unit/part, for i 1, 2, 3, k
CMTi
corrective maintenance/repair time required to repair unit/part i, for i
1, 2, 3.….,k
Usually, times to repair follow
exponential, lognormal, and normal probability
distributions.
Example
:
MAIN
TIME BETWEEN FAILURE
Mean
time between failure (MTBF) is probably
the most time between the occurring of failure in maintainability measure.
It measures
the elapsed time required to perform a given maintenance activity.
EXAMPLE :
Data historical of machine in
industry :
a. Hour
( 0 to 200) = up-time
b. Hour
( 200 to 280 ) = down-time
c. Hour
( 280 to 420 ) = up-time
d. Hour
( 420 to 490) = down-time
e. Hour
( 490 to 590 ) = up-time
f. Hour
( 590 to 650 ) = down-time
g. Hour
(650 to 780 ) = up-time
To
calculate of :
·
the mean time to between failure (MTBF )
and mean downtime (MDT) ?
MTBF ( Main time between Failure )
= 
MDT ( Main down time ) =
Tugas
2 - Perawatan Mesin
Tanggal
: 4 April 2017
Materi
: ( Maintenance Costing ) / Biaya Pemeliharaan 3IC09
NB : 1. Tugas dikirm ke Blog dan Student site + Lampirkan Materi Maintenance Costing
nya !!
2.
Dikirim ke email matkul perawatan mesin : tekper2017@gmail.com
Kerjakanlah soal – soal maintenance costing berikut
ini !!
1. To
Calculate the mean time to repair (MTTR ) ?
a. Eight
Subsytem : a, b, c, d, e, f, g, h form an electronic system.
b. The
constants failure rates of these subsystems are :
la = 0.002 , lb = 0.004 , lc = 0.006 , ld = 0.008 , le = 0.010 , lf = 0.012 , lg = 0.014 , lh = 0.016 failures per hour.
c. The
corresponding estimate corrective maintenance time are :
Ta
= 2 hour, Tb = 3 hour, Tc = 4 hour, Td =
5 hour, Te = 6 hour, Tf = 7 hour, Tg = 8 hour, Th = 9 hour.
2. Data
historical of machine in industry :
h. Hour
( 0 to 200) = up-time
i.
Hour ( 200 to 210 ) = down-time
j.
Hour ( 210 to 400 ) = up-time
k. Hour
( 400 to 415) = down-time
l.
Hour ( 415 to 590 ) = up-time
m. Hour
( 590 to 600 ) = down-time
n. Hour
(600 to 750 ) = up-time
To
calculate of :
·
the mean time to between failure (MTBF )
and mean downtime (MDT) ?
·
the annual labor cost of corrective
maintenance (CMal) ?
Whereas
: SOH = 3500 hour, MTTR = 30 h, LCH = $ 40.
3. Maintenance
Cost-Related Data :
Description
|
Manufacture
A system
|
Manufacture
B system
|
Manufacture
C System
|
Expected
Life
|
15 years
|
15 years
|
15 years
|
Expected
cost of a corrective maintenance
action
|
$2000
|
$2500
|
$3500
|
Annual
failure rate
|
3 failures per year
|
3.5 failures per year
|
2.5 failures per year
|
Annual
interest rate
|
4 %
|
4 %
|
4 %
|
Determine which of the three system
will be less costly with present value of corrective maintenance ( Find the
value of (PV)?
NB : Using the Formula II.
4. Assume
that we have the following time estimates to accomplish an activity :
OT = 85 days
PT = 105 days
MT = 90 days
To Calculate the activity expected
duration time ( Ta )?
5. Assume
that in maintenance organization we have the following data :
TAH = 2500 hour
BR = 0.2
LR = $25
N
= 50
To Calculate the cost per employee
(Cem)and The total labor cost ( TLC )?
My Answer To The Question Of
Maintenance Costing:
Carilah MTTR ( MAIN TIME BETWEEN FAILURE )
1. Diketahui
ada 8 sub system yang membentuk
elektronik system dengan tingkat
konstanta kegagalan sub system ini :
la
= 0.002 , lb = 0.004 , lc = 0.006 , ld = 0.008 , le = 0.010 , lf = 0.012 , lg = 0.014 , lh = 0.016 failures per hour.
Dengan estimasi waktunya:
Ta = 2 hour, Tb = 3 hour, Tc = 4 hour, Td = 5 hour, Te = 6
hour, Tf = 7 hour, Tg = 8 hour, Th = 9 hour
JAWAB :
2. Diketahui
:
Data historical of machine in
industry :
a. Hour
( 0 to 200) = up-time
b. Hour
( 200 to 210 ) = down-time
c. Hour
( 210 to 400 ) = up-time
d.
Hour ( 400 to 415) = down-time
e. Hour
( 415 to 590 ) = up-time
f.
Hour ( 590 to 600 ) = down-time
g. Hour
(600 to 750 ) = up-time
Ditanya :
·
the mean time to between failure (MTBF )
and mean downtime (MDT) ?
·
the annual labor cost of corrective
maintenance (CMal) ?
Whereas
: SOH = 3500 hour, MTTR = 30 h, LCH = $ 40.
Jawab :
Ditanya
: Menemukan nilai biaya perbandingan 1 perusahaan ( PV) :
Jawab
:
Langkah
pertama cari
4. Diketahui
:
OT = 85 days
PT = 105 days
MT = 90 days
Ditanya : Berapa lama membuat suatu
produk ( Ta )?
Jawab :
- Diketahui :
BR = 0.2
LR = $25
N
= 50
Ditanya: To Calculate the cost per
employee (Cem)and The total labor cost ( TLC ) atau gaji karyawan
dan total biaya karyawan?
Jawab :
a. Mencari
gaji karyawan (personal) / Cem
= $75,000
= $75,000 x 50
= $3,750,000
Thus, the total labor cost associated with the
maintenance activity will be $3,750,000million.
