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Unformatted text preview: back properly (i.e. using (0 ,,1 , 1) 1 or (0 , 1 ,1 , 1) to compare against (0 , 2 , 1 , 0) 2 and (0 , 2 ,4 , 0) 2 ). Also, some students arrived at the condition that x 6 enters if1 <2, and then multiplied both sides by and concluded that < 1 2 . However, the multiplication step implicitly assumes that > 0, which in this case you can ( and need to ) assume, because if 0, ( x 4 ,x 6 ) would have been an invalid pivot. Finally, technically, one needs to verify the case when = 1 2 and the 3rd entry in (0 , 2 ,4 , 0) 2 and (0 , 1 2 ,1 , 1 2 are tied (which a handful of you did). However, those who neglected to do this was not penalized. 1...
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 Fall '07
 S.Furino,B.Guenin

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