082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

972 Chapter 29 Statistics and probability distributions
Solution
The mean failure rate of each controller is 1/6 breakdowns per month. Given that
thereare12controllerstheoverallfailurerateforcontrollersis12/6 = 2breakdowns
per month. Using Equation (29.1) withk = 2 andt = 1 gives the probability,P(t),
of the process line being stopped because of a controller failure within a period of
one month as
P(t) =1−e −kt =1−e −2 =0.865
So far we have ignored the ‘downtime’ associated with an item waiting to be repaired
after it has failed. This can be a significant factor with many engineering systems. The
simplestpossibilityisthatanitemisoutofactionforafixedperiodoftimeT r
whileitis
being repaired. If there areN failures during a periodT then the total downtime isNT r
.
The time the item isavailable,T a
, isgiven by
T a
=T−NT r
Ausefulquantityisthefractionaldeadtime,D,whichistheratioofthemeantimethe
item isinthe dead statetothe total time.Inthiscase
D = NT r
(29.2)
T
Anotherusefulquantityistheavailability,A,whichistheratioofthemeantimeinthe
working statetothe total time. For the present model
A = T−NT r
=1−D
T
Thefailureratemodeldevelopedpreviouslywasbasedonthetimetheitemwasworking
rather thanthe total time.When the repair time isincludedkbecomes
and so
k = N T a
N =kT a
=k(T −NT r
)
N+kNT r
=kT
and therefore
N =
kT
1+kT r
So using Equation (29.2)
Also
D =
kT
1+kT r
T r
T =
A=1−D=1−
kT r
1+kT r
kT r
1
=
(29.3)
1+kT r
1+kT r
For more complicated repair characteristics the equations forDandAare correspondingly
more complicated.