Spring COT4501 2012
Solution I
Problem 1 Provide short answers to the following questions:
1.8 False
1,10 False
1.23
1. No. Propagated data error is measured by the conditioning of an algorithm
and is not directly tied to stability.
2. Yes. Accuracy is affected by both the conditioning and stability of a problem
3. No. The conditioning and stability of an algorithm are separate qualities that
both deal with error(computational and data error, respectively)
1.29.
1. Every number has a unique representation
2. No digits are wasted by leading 0s maximizing precision
1.32 (1/2) 1p
1.38 1/10
.1 in decimal form .00011001100110011 . . . in binary form
Problem 2 (1.9.)
5.101 108 km2 .
New surface area is 5.102 108 , so change is 1.0 105
With approximate formula, change in surface area is 1.601 105
With six-digit arithmetic, original surface area is 5.00906 108 ,new surface area
is 5.10065 108 , difference between these is 1.59 105 ,and change in surface
area from approximate formula is 1.60095 105 , so approximate formula is more
nearly correct than exact formula.
Problem 3 1.14. Assuming a and b have the same sign, formula is (b) better because
the result is guaranteed to lie in the interval [a, b], and it also avoids the possibility of
overow. With 2-digit arithmetic, for example, formula (a) gives a midpoint of 0.7 for
the interval [0.67, 0.69], whereas formula (b) gives a midpoint of 0.68.
Problem 4
1
1.18.(a) 2( 1) p1 (U L + 1) + 1. (b) 2( p1 1).
1.19.(a) OFL = 2128 (1 224 ), UFL = 2126 . (b) OFL unchanged, UFL = 2149
Problem 5
1.20. (a) 0.00011001100110011001100. (b) 0.00011001100110011001101.
1.21. (a) Not necessarily (see part b). For practical oating-point systems, however, mach is a machine number.
1.21 (b) The toy system of Example 1.9 is an example, as mach = 0.25 and UFL
= 0.5.
Problem 6 1.22. NOTE: In the posted homework c = 2.88 was typed instead of c =
2.28. Following solution is with respect question in the book. For c = 2.88 just replace
this value in the following solution
(a)b2 4ac = (3.34)2 4 1.22 2.28 = 11.2 11.1 = 0.1.
(b) 0.0292.
(c) (0.10.0292)/0.0292 2.425.
Problem 7 1.23.
(a) UFL = L = 1098
(b) x y= 6.87 1097 6.81 1097 = 0.06 1097 = 6 : 00 1099 0.
(c) With gradual under ow, result is 0.60 1098
2