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and consequently we have
Z
300
Z
1
1
2
f.x/dx
D
f.x/dx:
1
1
Since f is a probability density function, we must have
2
Z
1
f.x/dx
D
1;
1
and therefore
Z
300
f.x/dx
D
1
2
;
which indeed is consistent with the argument indicated above, that if b
D
300,we
would have a sufficient supply on 50% of the days.
By using this result, we can reformulate the problem stated in (1.5)as
1
C
Z
b
300
1
2
p
D
f.x/dx;
(1.6)
where
1
e
.x300/
2
f.x/
D
p
:
(1.7)
220
2
220
We want to be able to compute the value of p for b larger than 300. By the
arguments given above, we will be interested in p for b ranging from 300 to 370.
Thus, the computational problem is now reduced to that of computing a definite
integral on a bounded domain.
1.3
The Trapezoidal Method
We will derive a numerical algorithm based on dividing the interval of integra-
tion into very small intervals, where the integral can be approximated by simple
formulas. Then, we add all the contributions from the small intervals to yield an
approximation of the entire integral. We will start this process by considering only
one interval - the original interval - and then proceed by using two intervals and
finally we will divide the original interval into an arbitrarily large number of n
subintervals.
2
This property can be derived in a straightforward manner. By a change of variable, we have
1
p
220
Z
1
Z
1
.
x
300/
2
2
20
2
1
p
e
y
2
dy
e
dx
D
D
1:
1
1
The latter integral can be found in any table of definite integrals, see, e.g., [24].
