z = 1.875 = (1.111)2 = (0.1111)2 x 21
fl(z) = (0.11110)2
x 21
fl(x – y + z) = 0 + fl(z) = fl(z) = (0.11110)2 x 21
neutralizarea termenilor
rearanjarea calculelor pentru evitarea neutralizării termenilor:
fl(x + z - y) = fl( fl(x + z) - y)
fl(x + z) = (0.11110)2 x 23 + (0.11110)2 x 21 ⇒ (0.11110)2 x 23 + (0.0011110)2 x 23
fl(x + z) = (0.11110)2 x 23 + (0.00111)2 x 23 ⇒ (1.00101)2 x 23 → (0.10010) x 24
fl(x + z - y) = fl(x+z) – fl(y) = (0.10010)2 x 24 - (0.11110)2 x 23 ⇒ (0.10010)2 x 24 - (0.01111)2 x 24
fl(x – y + z) = (0.00011)2 x 24 = (0.11000) x 21
normalizare
METODE NUMERICE – pregătire 23.06.2021