Lecture Notes
The graph of the exponential distribution is shown in Figure 1.
Figure 1: Graph of Markov Process for Exponential Distribution. (Figure by MIT OpenCourseWare.)
The transition equations are
\[ p(0,t + \delta t) = \mu \delta tp(1,t) + p(0,t) + o(\delta t) \tag 1 \]
\[ p(1,t + \delta t) = (1 - \mu \delta t) p (1,t) + (0)p (0,t) + o(\delta t) \tag 2 \]
or,
\[\frac{dp(0,t)}{dt}= \mu p(1,t) \tag 3 \]
\[\frac{dp(1,t)}{dt}= -\mu p(1,t) \tag 4 \]
Solve differential equations (3), (4) and we get
\[p(0,t) = 1 - e^{-\mu t}\tag 5\]
\[p(1,t) = e^{-\mu t}\tag 6\]
Function \(p(0,t)\) is actually the cumulative density function of the exponential distribution
\[F(t) = p(0,t)\tag 7\]
Then the density function of the exponential distribution is
\[f(t) = \frac{dF(t)}{dt}= \mu e^{- \mu t}\tag 8\]
The memorylessness property of the exponential distribution was described in the probability lecture slides as
\[p(T>t + x | T>x) = p(T>t) \tag 1\]
where
\(p(T>t) = e^{-\lambda t}\) for \(t\geq0\); \(p(T>t) = 1\) otherwise
- To prove (1), note that
\[p(T>t+x|T>x)= \frac{p(\{T>t+x\}\bigcap\{T>x\})}{p(T>x)} \tag 2\]
But
\[\{T>t+x\}\bigcap\{T>x\} = \{T>t+x\}\]
because if \(T > t + x\), then \(T>x\). Therefore
\[p(\{T > t + x\}\bigcap\{T > x\}) = p(\{T > t + x\})\]
so, from (2), assuming \(x>0\),
\[p(T > t + x | T > x) = \frac{p(\{T > t + x\})}{p(T > x)} = \frac{e^{-\lambda(t + x)}}{e^{-\lambda x}}\]
and the last fraction is just equal to \(e^{-\lambda t}\), so (1) is proved.
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What does (1) mean in intuitive terms?
Let random variable \(T\) be the time until a certain event (like the failure of a machine) occurs after time 0. That time is exponentially distributed with parameter \(\lambda\). Then (1) means that if we look at the machine at time \(x>0\) and the failure has not yet occurred, the time until the failure does occur \((t)\) is distributed exponentially with parameter \(\lambda\).
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How can this possibly be true? Machines wear (at least a little) during each operation. The more operations a machine does since its last maintenance, the sooner it is likely to fail than it would immediately after maintenance. That is, suppose that immediately after maintenance, the expected number of operations it can perform before it fails is 100. Then if it somehow does 1000 operations without failure, the exponential distribution model says that the expected number it will do before it fails is still 100 (for a total lifetime of 1,100 operations). But the expected number of operations until the failure ought to be a lot less than 100 because of all the wear that must have occurred.
That’s correct. But remember: Models are not exact representations of reality. As engineers, we don’t judge models by how accurately they represent reality; we judge them by how accurately they make the predictions we need. The exponential distribution can be a good approximation for some (but not all) purposes. For example, suppose we have to order expensive cutting tools infrequently in large batches. (You will see why that might be true in the inventory lecture.) Suppose these tools have average lifetimes of one week and suppose we can only order tools once a year. How many tools should we have on hand at the start of a year (that is, after the tools we order have arrived)?
Suppose we require that the probability of running out of tools before we order the next batch is less than .01. (If we run out of tools, production stops.) The number we need depends on the variance of the time to fail. If the tool life were perfectly deterministic, then 52 tools would be plenty. The greater the variance of the tool life, the more tools would be needed. The small deviation from the exponential distribution is not important to the answer to this question.
Challenge: Solve the problem assuming exponentially distributed life times. (Treat a year as having exactly 52 weeks.)
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Sometimes it is important to model a given non-exponential probability distribution more accurately than an exponential distribution can. This can be done, but describing it is too complicated for this note.
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All this applies equally to geometrically distributed discrete times.
\[X,Y\,independent \rightarrow cov(X,Y) = 0\]
Assume random variables \(X\) and \(Y\) are discrete. That is, assume that there is a finite or denumerable sample space which is a set of \(w_i\) and a set of quantities \(x_i\) and \(y_i\) defined.
Definition \(X\) and \(Y\) are independent if
\[prob((X = x)\,and\,(Y = y)) = prob(X = x)prob(Y = y)\]
in which \(x\) is some \(x_i\) and \(y\) is some \(y_j\).
Then if \(X\) and \(Y\) are independent,
\[E(XY) = E(X)E(Y)\]
Proof:
\[E(XY) = \sum _{i,j} \ x_iy_jprob(XY = x_iy_j)\]
\[= \sum _{i,j} \ x_iy_jprob((X = x_i)\,and\,(Y = y_j))\]
\[= \sum _{i,j} \ x_iy_jprob(X = x_i)prob(Y = y_j)\]
\[= \sum _{i} \ x_iprob(X = x_i)\sum _{j} \ y_jprob(Y = y_j) = E(X)E(Y)\]
Then if \(X\) and \(Y\) are independent,
\[cov(X,Y) = E[(X - E(X))(Y - E(Y))]\]
\[=E[XY - XE(Y) - YE(X)+E(X)E(Y)]\]
\[=E[XY] - E(X)E(Y) - E(Y)E(X) + E(X)E(Y) = 0\]
There seems to be a lot of confusion as the parameters of the production line models. I’ll try to clear it up without creating too much more.
1. Continuous Time, Continuous Material Models
There should be no confusion about this case. \(r \delta t\) is the probability of a repair during an interval of length \(\delta t\) and is therefore dimensionless. \(p\delta \)t is the probability of a failure during an interval of length \(\delta t\) and is also dimensionless. Since \(\delta t\) is in units of time, \(r\) and \(p\) are in units of \(\frac{1}{time}\). The time unit could be seconds or minutes or hours, years, or centuries, whichever is more convenient. \(\frac{1}{r}\) is the MTTR of a machine and \(\frac{1}{p}\) is the MTTF of a machine, and they have units of time.
\(\mu\) is in units of stuff / time. (“Stuff” could be parts, it could be a weight unit, it could be a volume unit, etc.)
Just be sure that the time units are consistent for all parameters and that the stuff units of \(\mu \) is the same as the stuff units for the buffer sizes \(N\).
2. Discrete Time, Discrete Material (AKA Deterministic Processing Time) Models
\(τ\) is the common operation time of all the machines in the system being studied. It is a time and is expressed in natural time units.
However, the time unit for this model is the number of operation times. Events only occur at integer multiples of \(τ\). In this model, \(r\) and \(p\) are probabilities and therefore dimensionless. \(r\) is the probability of a repair during an operation time; \(p\) is the probability of a failure during an operation time. \(\frac {1}{r}\) is the MTTR of the machine expressed in units of operation times.
That is, MTTR is the mean number of operation times until a machine is repaired. We treat MTTR as dimensionless. Similarly, \(\frac{1}{p}\) is the MTTF of the machine, also expressed in units of operation times.
The stuff unit is a part, and is treated as dimensionless. The buffer sizes are therefore also dimensionless.
The confusion arises when I present problems in natural time units (seconds, minutes, etc.). For example, suppose I tell you that the operation time \(τ\) of a machine is 15 seconds and that its MTTR is 30 minutes. In order to use this information in the deterministic processing time model, we have to express MTTR as a multiple of \(τ\). The relationship is
\[{MTTR\,in\,seconds} = [MTTR\,in\,operation\,times] \times [the\,number\,of\,seconds\,in\,an\,operation\,time]\]
or
\[{MTTR\,in\,seconds} = [MTTR\,in\,operation\,times] \times \,{τ \;in \,seconds}\]
so in this case,
\[30 \times 60 = [MTTR\,in\,operation\,times] \times 15\]
or
\[[MTTR\,in\,operation\,times] = \frac{30 \times 60}{15} = 120\]
so
\[r = \frac{1}{MTTR} = 0.0083333\]
and similarly for MTTF and \(p\).
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Lecture Notes
