Vincent Trinquet IVP 2 Avril 2007 - EIVP
Vincent Trinquet IVP 2 Avril 2007 - EIVP
Vincent Trinquet IVP 2 Avril 2007 - EIVP
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+ désignera +1 et – désignera –1<br />
l0 l1 l2 l3 l4 l5 l12 l13 l14 l15 l23 l24 l25 l34 l35 l45<br />
( 1) + + + + + + + + + + + + + + + +<br />
( 2) + + + + – – + + – – + – – – – +<br />
( 3) + + + – + – + – + – – + – – + –<br />
( 4) + + + – – + + – – + – – + + – +<br />
( 5) + + – + + – – + + – – + + + + +<br />
( 6) + + – + – + – + – + – + – – + –<br />
( 7) + + – – + + + – – + – – + + – –<br />
( 8) + + – – – – – – – – + + + + + +<br />
( 9) + – + + + – – – – + + + – + – –<br />
( 10) + – + + – + – – + + + – + – + –<br />
( 11) + + + + + + + + + + + + + + + +<br />
( 12) + – + – – – – + + + – – + + + +<br />
( 13) + – – + + + + – – – – – – + + +<br />
( 14) + – – + – – + – + + – + + – – +<br />
( 15) + – – – + – + + – + + – + – + –<br />
( 16) + – – – – + + + + – + + – + – –<br />
L1 : ∀ i ∈[ 1,4], xi = +1<br />
L2 : x1 = +1, x2 = +1, x3 = +1, x4 = – 1, et donc x5 = – 1, car sg (x1, x2, x3, x4, x5) = 1<br />
C’est le plan factoriel 2 5-2<br />
Pour obtenir le plan factoriel 2 5-2 , cad diviser encore par deux le nombre d’expériences, on sélectionne le<br />
demi plan tel que sg (x1) = 1. C’est le demi plan supérieur.<br />
sg (x1) = 1, d’où sg ( x2, x3, x4, x5) = 1 aussi<br />
16