On the coefficients of variation of uptime and downtime in manufacturing equipment
International Journal of
It was reported in the literature that the coefficients of variation of uptime and downtime of manufacturing equipment are often less than 1. This technical paper is intended to provide an analytical explanation of this phenomenon. Specifically, it shows that the distributions of uptime and downtime have the coefficients of variation less than 1 if the breakdown and repair rates are increasing functions of time.
1. Introduction
The coefficients of variation of uptime and downtime (CVup and CVdown) of the
equipment on the factory floor have a significant effect on the design of lean production
systems. Indeed, as has been shown in [
2
], lean buffering (i.e., the buffering necessary and
sufficient to achieve a desired production rate of a manufacturing system) is an increasing
function of CVup and CVdown, and therefore smaller CVs lead to leaner systems.
In designing lean systems, it is often assumed that the uptime and downtime (tup and
tdown) are distributed exponentially and, thus, CVup = CVdown = 1. Since in reality tup and
tdown may not be exponential, it is important to determine if lean buffering, based on the
exponential assumption, overestimates or underestimates the real lean buffering. If CVup
and CVdown are less than 1, the lean buffers, designed using the exponential assumption,
are too large; otherwise, they are too small.
The empirical evidence [
3
] indicates that CVup is very often less than 1; this is also true
for CVdown although to a lesser extent. As follows from the above, this is an important
fact since it ensures that lean buffering, designed using the exponential assumption, will
ensure the desired throughput (however, without the maximum leanness).
A question arises: what are ?natural? causes which lead to CVup and CVdown being
less than 1? The purpose of this technical paper is to discuss this question and, thereby,
characterize practical situations that lead to small CVup and CVdown.
To accomplish this, Section 2 describes the model of machines under consideration,
Section 3 discusses a special case, in Section 4 a general scenario is considered, and in
Section 5 conclusions are formulated.
2. Model
We consider machines, which could be in two states, up and down. When up, parts are
produced; when down, no production takes place and the machine is under repair.
Transition rates from up to down and from down to up are p(t) and r(t), respectively, where
t is the time the machine spent in the respective state. In other words, if the machine
went up at t = 0, it goes down during the infinitesimal interval (t, t + ?t) with probability
p(t)?t. Similarly, transition from down to up occurs during (t, t + ?t) with probability
r(t)?t.
Obviously, the uptime and downtime defined by this model are random variables with
probability density functions (pdf ?s) induced by p(t) and r(t). If both p(t) and r(t) are
constant, that is, p(t) = p0 and r(t) = r0, t ? 0, the resulting random variables are
exponential, and CVup = CVdown = 1. What happens when p(t) and r(t) are not constant is
discussed next.
3. Special case
We analyze below the pdf and the coefficient or variation of the uptime; the downtime is
treated similarly.
Assume that p(t) is given by
p(t) = p0ta,
p0 = const, a > ?1, t ? 0.
(3.1)
This implies that the breakdown rate is strictly monotonically increasing in time when
a > 0, constant when a = 0 (the exponential case), and strictly monotonically decreasing
when ?1 < a < 0. The values of a < ?1 are not included since they do not lead to valid
pdf ?s (in the sense that the integrals of these functions are not finite).
The pdf and the CV of the uptime, induced by assumption (3.1), can be calculated as
follows.
By the total probability formula, the probability that the machine is up at time t + ?t
is given by
P ?(t + ?t) = 1 = P ?(t + ?t) = 1 | ?(t) = 1 P ?(t) = 1 ,
(3.2)
(3.3)
(3.4)
where ?(t) represents the state of the machine, that is,
?1 if the machine is up at time t,
?(t) = ?
?0 if the machine is down at time t.
As follows from the machine model of Section 2 and (3.1), the conditional probability
that the machine is up at t + ?t, given that it is up at t, is
P ?(t + ?t) = 1 | ?(t) = 1 = 1 ? p0ta?t.
Substituting this into (3.2) and rearranging terms, we obtain
= ?p0taP ?(t) = 1 .
When ?t ? 0, this becomes a first-order linear differential equation
with initial condition
dP ?(t) = 1
dt
= ?p0taP ?(t) = 1 ,
Solving this initial value problem, we obtain the probability that the machine is up at time
t assuming that it was up at time 0
P ?(t) = 1 = e? 0t p0tadt = e?p0ta+1/(a+1).
To determine the pdf of tup, we calculate the joint probability that the machine is up at
time t and down at time t + ?t. Using the formula for the probability of the intersection
of two events and (3.8), we obtain
P ?(t) = 1 ? ?(t + ?t) = 0
= P ?(t + ?t) = 0 | ?(t) = 1 P ?(t) = 1
= p0ta?t e?p0ta+1/(a+1)
= p0tae?p0ta+1/(a+1)?t.
ftup (t) = p0tae?(p0/(a+1))ta (...truncated)