Unformatted text preview: largest ﬂoaJng point number? x = (1.5)*2^1 => (1.5+eps)*2^1 = x + eps*2^1 = 3 + 2*eps is next largest ﬂoaJng point number 8 2/20/14 eps • x = (1+f)2^e • For x = 9, what is the next largest ﬂoaJng point number? x = (1.125)*2^3 => (1.125+eps)*2^3 = x + eps*2^3 = 9 + 8*eps is next largest ﬂoaJng point number ﬂoatgui.m • Numbers within [2^e, 2^(e+1)] are equally spaced • One and eps and are in red • t = number bits used to store f (52 in IEEE 754) – 2^t*f is an integer, 0<=2^t*f<=2^t • emin < e < emax ( 1022 <= e < 1 23 in IEEE 754) • logscale shows distribuJon of numbers within each binary interval is the same • What is the paern? – For x=(1+f)2^e – The diﬀerence between x and the next largest ﬂoaJng point numbers is (2^e)*eps! • What does this mean in pracJce? – Big numbers are “further apart” than small numbers – We can represent small values very precisely, as values increase, the absolute precision drops – But relaJve precision remains the same (this is great) Regarding the QuesJon… The loop terminates, final value of k
is 53, i.e. 1==1+eps/2!
!
k = 0;
while 1 + 1/2^k > 1
k = k+1;
end
k
Insight Through CompuJng The Moral The 1991 Patriot Missile Disaster Elementary
misperceptions
about the
finiteness
of computer
arithmetic.
30+ died. To produce reliable numerical results you must appreciate ﬂoaJng point arithmeJc. Insight Through CompuJng Insight Through CompuJng 9 2/20/14 The Sevng External clock counts Jme in tenths of seconds. TargeJng soXware needs Jme to compute trajectories. The method: Time = (# external clock Jcks) x (1/10) The problem is here
Insight Through CompuJng One Tenth in Binary Exact: .00011001100110011001100110011… Patriot System used: .00011001100110011001100110011… Error = .000000095sec every clock Jck Insight Through CompuJng Error Time = (# external clock Jcks) x (1/10) Error = (# external clock Jcks) x (.000000095) Insight...
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