Page 119 - Advanced engineering mathematics

3.4 Convolution 99

Then
4 4
F(s) = + ,
s 3 s 4
which we invert to obtain
2
f (t) = 2t + t .
3
2
3
A Replacement Scheduling Problem
We will develop an integral equation that arises in the context of planning replacements for items
(such as pieces of equipment that wear out or stored drugs that lose their effectiveness over time).
Suppose a company or organization uses large numbers of a certain item. An example might
be portable computers for use by the military, copying machines in a business, or vaccine doses
in a hospital. The organization’s plan of operation includes an estimate of how many of these
items it wants to have on hand at any time. We will imagine that this number is large enough
that it can be approximated by a piecewise continuous availability function f (t) that gives the
number of items available for use at time t. Experience and familiarity with the items enables
the organization and the supplier to produce a function m(t), called a mortality function, that is
a measure of the number of items still working satisfactorily (surviving) up to time t. We will be
more explicit about m(t) shortly.
Given f (t) and m(t) (items needed and how long items remain good), planners want to
develop a replacement function r(t) that measures the total number of replacements that must be
made up to time t.
To begin the analysis, assign the time t = 0 to that time when these items of equipment were
introduced into use, so at this initial time all the items are new. We also set r(0) = 0.
Inatimeintervalfrom τ to τ + τ, there have been

r(τ + τ) −r(τ) ≈r (τ) τ
replacements. Here is where the mortality function comes in. We assume that, at any later time
t, the number of surviving items, out of these replacements in this time interval, is

r (τ)( τ)m( τ),

which we write as

r (τ)m(t − τ) τ.
The total number f (t) of items available for use at time t is the sum of the number of items
surviving from the new items introduced at time 0 plus the number of items surviving from
replacements made over every interval of length τ from τ = 0to τ = t. This means that
t

f (t) = f (0)m(t) + r (τ)m(t − τ)dτ.
0
This is an integral equation for the derivative of the replacement function r(t).Given f (t) and
m(t), we attempt to solve this integral equation to obtain r(t).
The reason this strategy works in some instances is that this integral is a convolution,
suggesting the use of the Laplace transform. Application of L to the integral equation yields
F(s) = f (0)M(s) + L[r (t)](s)L[m(t)](s)

= f (0)M(s) + (sR(s) −r(0))M(s)
= f (0)M(s) + sR(s)M(s).

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October 14, 2010 14:14 THM/NEIL Page-99 27410_03_ch03_p77-120