SICP Exercise 2.13: Tolerance is Simple

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.

Okay, so let's assume we have two intervals, a and b, with centers c_a and c_b and percentage tolerances t_a and t_b. We can then say that the widths of a and b are given by:

wa = cata / 100
wb = cbtb / 100

...and so we can express the bounds of our intervals, a and b, as:

a = [la, ua] = [ca - wa, ca + wa]
b = [lb, ub] = [cb - wb, cb + wb]

If we further assume that a and b are non-negative integers (as we defined in the exercise 2.11) then we know that the bounds of the product of the two intervals, ab, are:

  ab
= [lalb, uaub]
= [(ca - wa)(cb - wb), (ca + wa)(cb + wb)]
= [cacb - cawb - wacb + wawb, cacb + cawb + wacb + wawb]

Now as t_a and t_b are small, we know that w_aw_b will also be small. Let's assume that w_aw_b is close to zero, or at least negligible with respect to the other terms in the lower and upper bounds, and so discard it to give:

  ab
= [cacb - cawb - wacb + wawb, cacb + cawb + wacb + wawb]
≈ [cacb - cawb - wacb, cacb + cawb + wacb]
= [cacb - (cawb + wacb), cacb + (cawb + wacb)]

Note that these bounds take the same form as that of the bounds of an interval expressed in terms of its center point and widths. So we can say that:

cab ≈ cacb
wab ≈ cawb + wacb

...and so we can express the percentage tolerance as:

  tab
= 100wa/ca
≈ 100(cawb + wacb)/cacb
= 100(cawb)/cacb + 100(wacb)/cacb
= 100wb/cb + 100wa/ca
= tb + ta