Alyssa P. Hacker is designing a system to help people solve
engineering problems. One feature she wants to provide in her system
is the ability to manipulate inexact quantities (such as measured
parameters of physical devices) with known precision, so that when
computations are done with such approximate quantities the results
will be numbers of known precision.
Electrical engineers will be using Alyssa's system to compute
electrical quantities. It is sometimes necessary for them to compute
the value of a parallel equivalent resistance
$R_{p}$ of two resistors
$R_{1}$ and $R_{2}$
using the formula
\[
\begin{array}{lll}
R_{p} & = & \dfrac{1}{1/R_{1}+1/R_{2}}
\end{array}
\]
Resistance values are usually known only up to some
tolerance guaranteed by the manufacturer of the resistor. For example, if
you buy a resistor labeled 6.8 ohms with 10% tolerance you can
only be sure that the resistor has a resistance between
$6.8-0.68=6.12$ and
$6.8+0.68=7.48$ ohms. Thus, if you have a
6.8-ohm 10% resistor in parallel with a 4.7-ohm
5% resistor, the resistance of the combination can range from about
2.58 ohms (if the two resistors are at the lower bounds) to about 2.97 ohms
(if the two resistors are at the upper bounds).
Alyssa's idea is to implement interval arithmetic as a
set of arithmetic operations for combining intervals (objects
that represent the range of possible values of an inexact quantity). The
result of adding, subtracting, multiplying, or dividing two intervals is
itself an interval, representing the range of the result.
Alyssa postulates the existence of an abstract object called an
interval that has two endpoints: a lower bound and an upper bound.
She also presumes that, given the endpoints of an interval, she can
construct the interval using the data constructor
make-interval.make_interval.
Alyssa first writes a
procedurefunction
for adding two intervals. She reasons that the minimum value the sum could
be is the sum of the two lower bounds and the maximum value it could be is
the sum of the two upper bounds:
Alyssa also works out the product of two intervals by finding the
minimum and the maximum of the products of the bounds and using them
as the bounds of the resulting interval.
(Min(The functions math_min
and
maxmath_max
are
primitives that find the minimum or maximum of any number of arguments.)
To divide two intervals, Alyssa multiplies the first by the reciprocal of
the second. Note that the bounds of the reciprocal interval are
the reciprocal of the upper bound and the reciprocal of the lower bound, in
that order.
Original
JavaScript
(define (div-interval x y)
(mul-interval x
(make-interval (/ 1.0 (upper-bound y))
(/ 1.0 (lower-bound y)))))
Exercise 2.7
Alyssa's program is incomplete because she has not specified the
implementation of the interval abstraction. Here is a definition of
the interval constructor:
Original
JavaScript
(define (make-interval a b) (cons a b))
function make_interval(x, y) { return pair(x, y); }
Define selectors
upper-boundupper_bound
and
lower-boundlower_bound
to complete the implementation.
Original
JavaScript
function make_interval(x, y) {
return pair(x, y);
}
function lower_bound(x) {
return head(x);
}
function upper_bound(x) {
return tail(x);
}
Exercise 2.8
Using reasoning analogous to Alyssa's, describe how the difference
of two intervals may be computed. Define a corresponding subtraction
procedure,function,
called
sub-interval.sub_interval.
Exercise 2.9
The
width of an interval is half of the difference between its
upper and lower bounds. The width is a measure of the uncertainty of
the number specified by the interval. For some arithmetic operations
the width of the result of combining two intervals is a function only
of the widths of the argument intervals, whereas for others the width
of the combination is not a function of the widths of the argument
intervals. Show that the width of the sum (or difference) of two
intervals is a function only of the widths of the intervals being
added (or subtracted). Give examples to show that this is not true
for multiplication or division.
Let us denote the width of interval $i$
with $W(i)$, and its lower and upper
bound with $L(i)$ and
$U(i)$, respectively. Two
intervals $i_1$ and
$i_2$ have by definition widths of
$(U(i_1) - L(i_1))/2$ and
$(U(i_2) - L(i_2))/2$, respectively.
Adding the two intervals
leads to the interval
$[ L(i_1) + L(i_2), U(i_1) + U(i_2)]$,
whose width is
\[(U(i_1) + U(i_2) - (L(i_1) + L(i_2)))/2\]
\[= (U(i_1) - L(i_1))/2 + (U(i_2) - L(i_2))/2\]
\[= W(i_1) + W(i_2)\]
The argument for subtraction is similar.
The widths of the result of multiplying intervals does not have
such a nice property. For example, multiplying any interval
with the zero-width interval
$[ 0, 0 ]$ yields a zero-width interval
whereas multiplying any interval
$i$
with the zero-width interval
$[ 1, 1 ]$ yields an interval with width
$W(i)$.
The argument for division is similar.
Exercise 2.10
Ben Bitdiddle, an expert systems programmer, looks over Alyssa's
shoulder and comments that it is not clear what it means to
divide by an interval that spans zero. Modify Alyssa's program to
check for this condition and to signal an error if it occurs.
Exercise 2.11
In passing, Ben also cryptically comments: By testing the signs of
the endpoints of the intervals, it is possible to break
mul-intervalmul_interval
into nine cases, only one of which requires more than two
multiplications. Rewrite this
procedurefunction
using Ben's suggestion.
After debugging her program, Alyssa shows it to a potential user, who
complains that her program solves the wrong problem. He wants a program
that can deal with numbers represented as a center value and an additive
tolerance; for example, he wants to work with intervals such as
$3.5\pm 0.15$ rather than
$[3.35, 3.65]$. Alyssa returns to her desk and
fixes this problem by supplying an alternate constructor and alternate
selectors:
function make_center_width(c, w) {
return make_interval(c - w, c + w);
}
function center(i) {
return (lower_bound(i) + upper_bound(i)) / 2;
}
function width(i) {
return (upper_bound(i) - lower_bound(i)) / 2;
}
Unfortunately, most of Alyssa's users are engineers. Real engineering
situations usually involve measurements with only a small uncertainty,
measured as the ratio of the width of the interval to the midpoint of the
interval. Engineers usually specify percentage tolerances on the parameters
of devices, as in the resistor specifications given earlier.
Exercise 2.12
Define a constructor
make-center-percentmake_center_percent
that takes a center and a percentage tolerance and produces the desired
interval. You must also define a selector
percent that produces the percentage tolerance
for a given interval. The center selector is
the same as the one shown above.
function make_center_percent(center, percent) {
const width = center * (percent / 100);
return make_center_width(center, width);
}
function percent(i) {
return (width(i) / center(i)) * 100;
}
Exercise 2.13
Show that under the assumption of small percentage tolerances there is
a simple formula for the approximate percentage tolerance of the
product of two intervals in terms of the tolerances of the factors.
You may simplify the problem by assuming that all numbers are
positive.
Let us denote the maximal error of an interval with center
$i$ by
$\Delta i$
the maximal error of an interval with center
$j$ by
$\Delta j$, and
the maximal error of the multiplication result with center
$k$ by
$\Delta k$. Then:
\[ k + \Delta k = (i+\Delta i) * (j+\Delta j)
= ij + j \Delta i + i\Delta j + \Delta i \Delta j \]
Since $k = i j$
\[ \Delta k = j\Delta i + i \Delta j + \Delta i\Delta j \]
Since we assume that
$\Delta i \ll i$ and
$\Delta j \ll j$, we can neglect
the term
$\Delta i \Delta j$ and obtain
\[ \Delta k = j \Delta i + i \Delta j \]
Expressed in tolerances, we obtain:
\[ \Delta k / k
= (j \Delta i + i \Delta j) / ij = \Delta i/i + \Delta j/j \]
Thus, the tolerance of the result of an interval multiplication
is (roughly) the sum of the tolerances of its arguments.
After considerable work, Alyssa P. Hacker delivers her finished
system. Several years later, after she has forgotten all about it, she
gets a frenzied call from an irate user, Lem E. Tweakit.
It seems that Lem has
noticed that the
formula for parallel resistors can be written in two
algebraically equivalent ways:
\[
\dfrac{R_{1}R_{2}}{R_{1}+R_{2}}
\]
and
\[
\dfrac{1}{1/R_{1}+1/R_{2}}
\]
He has written the following two programs, each of which computes the
parallel-resistors formula differently:
Original
JavaScript
(define (par1 r1 r2)
(div-interval (mul-interval r1 r2)
(add-interval r1 r2)))
(define (par2 r1 r2)
(let ((one (make-interval 1 1)))
(div-interval one
(add-interval (div-interval one r1)
(div-interval one r2)))))
function par1(r1, r2) {
return div_interval(mul_interval(r1, r2),
add_interval(r1, r2));
}
function par2(r1, r2) {
const one = make_interval(1, 1);
return div_interval(one,
add_interval(div_interval(one, r1),
div_interval(one, r2)));
}
Lem complains that Alyssa's program gives different answers for
the two ways of computing. This is a serious complaint.
Exercise 2.14
Demonstrate that Lem is right. Investigate the behavior of the
system on a variety of arithmetic expressions. Make some intervals
$A$ and $B$,
and use them in computing the expressions $A/A$
and $A/B$. You will get the most insight by
using intervals whose width is a small percentage of the center value.
Examine the results of the computation in center-percent form (see
exercise 2.12).
The expression $A/A$ is interesting, because if
the interval is meant to represent a specific (albeit imprecisely known)
value, the result should be exactly 1 (width 0), whereas the interval
division will result in an interval with positive width. Multiple
occurrences of the same term are not recognized as such in the approaches
above and thus they will suffer from this problem.
Exercise 2.15
Eva Lu Ator, another user, has also noticed the different intervals
computed by different but algebraically equivalent expressions. She
says that a formula to compute with intervals using Alyssa's system
will produce tighter error bounds if it can be written in such a form
that no
variable
name
that represents an uncertain number is repeated. Thus, she says,
par2 is a better program for
parallel resistances than par1. Is she right?
Why?
She is right. The so-called dependency problem in interval
arithmetic arises when the same input values (or intermediate terms)
appear in a function on intervals. The second formulation is better
because each input occurs only once, and therefore the result of a naive
interval calculation is optimal.
Exercise 2.16
Explain, in general, why equivalent algebraic expressions may lead to
different answers. Can you devise an interval-arithmetic package that
does not have this shortcoming, or is this task impossible? (Warning:
This problem is very difficult.)
The dependency problem in interval arithmetic is solved using linear
and polynomial approximations, leading to affine arithmetic and
Taylor series methods, respectively.