Problem 3, May August 1998 - The Paper Mill

A paper mill has two papermaking machines, PM1 and PM2. The
employees of the mill work in continuous shift, so if the
machines function as they shall, paper is produced 24 hours a
day. Sometimes, however, a machine fails, and then the machine
must be repaired before production by that machine is resumed.
Therefor, at any time one repairer is present at the mill.

In this problem we will assume the following simple model of
the situation. Time since a machine is repaired before it fails
again has the exponential distribution with parameter [mu] (per
hour), and this parameter is common for both machines.
Moreover, we assume that the machines fail independently of
each other. Time it takes to repair a machine has the
exponential distribution with parameter [lambda] (per time
unit). Moreover, we shall assume that repair times are
independent of each other and independent of time passed before
the machines fail. Note that, since the mill has only one
repairer, only one machine can be repaired at one time, so if
machine PM2 fails while machine PM1 is under repair, repair of
PM2 will only start when PM1 is repaired.

Let X(t) denote the number of machines that is working at time
t.

(a) Give a short argument for the fact that X(t) is a
birth-and-death process with state space [formula]. Specify
birth and death rates.

(b) Write down the Kolmogorov equations for the present
situation.

From these, derive a system of equations for the limiting
distribution.

Show, by solving the system of equations, that

[formulas],

where [formula].

To reduce the time the machines are under repair, the mill
considers to introduce a new scheme with two repairers
available at any time. Thus repair can always be started
immediately when a machine fails. Note, however, that the
errors occurring is of such a kind that there is no time to
save by both repairers working simultaneously on the same
machine. Thus, if PM1 is working while PM2 has failed, one
repairer will work on repairing PM1, while the other will sit
down and watch.

Each machine that has failed gives the mill a loss of [alpha]
kr. per hour. On the other hand, it will cost the mill an
additional[beta] kr. per hour to introduce the scheme with two
repairers.

(c) For the current scheme with one repairer, find how much the
mill on average loses per hour because the machines
periodically fail.

Derive an equation for [formula] such that the long-run average
income equals the average income with two repairers. (Because
it is a cubic equation, you are not required to solve it.)

For [alpha]=100000 and [beta]=1000 the equation has the
following three solutions: [formulas]. Discuss for which values
of [theta] it pays for the mill to introduce the scheme with
two repairers and for which values it pays to retain the
current scheme. Further, give an explanation, based on the
situation described above, of what happens for the various
intervals for [theta].