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.
Jawaban :
§ 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.
2. 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
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.
Jawaban :
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 :
a)
the mean time
to between failure (MTBF ) and mean downtime (MDT) ?
b)
the annual
labor cost of corrective maintenance (CMal) ?
Whereas
: SOH = 3500 hour, MTTR = 30 h, LCH = $ 40.
Jawab :
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.
Jawaban
:
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
)?
Jawaban
:
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 )?
Jawaban :
§ Cem =
= $75,000
- TLC = $75,000 x 50
= $3,750,000
