1 Основни појмови егзактно посредног претраживања Релаксација и рестрикција

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1

2

3 X 1 X 2 2 f (x)

4 B B A A X 1 X 2 () () f (x) x X 1 f (x) x X 2 X 2 X 1 x 1 x 2 f (x 1) f (x 2) B f (x 1) = f (x 2) A f (x 1) > f (x 2)

5 X 1 X 3 x Z 2 B B A A X 1 X 3 () f (x) x X 3 x 3 f (x 1) f (x 3), B f (x 1) > f (x 2) A X 3 f (x 1) = f (x 2) G = (V, E)

6 () f (x) = (i,j) E c ij x ij x ij = 1, j V i V x ij = 1, i V j V i u j + nx ij n 1, (i, j) E : i, j 1 x ij {0, 1}, (i, j) E c ij (i, j) E x ij (i, j) u i n G 0 x ij 1 x 3 X 3 f (x 1) f (x 3) f (x 3 ).

7 x 3 f = f (x 1) f = f (x 3 ) P P P P P j i i i i j f (x i ) f ( x j) x d P f (x d ) f (x i ) j

8 рестрикција i f(x i * ) рестрикција j f(x j * )... f (x d ) j i

9 P : () 15x x 2 + 4x 3 + 2x 4 8x 1 + 5x 2 + 3x 3 + 2x 4 10 x k {0, 1} k = 1, 2, 3, 4 x 1 =0 x1=1 : () 15x x 2 + 4x 3 + 2x 4 8x 1 + 5x 2 + 3x 3 + 2x x k 1 k = 1, 2, 3, 4 x 2 =0 x2= x = (0, 625; 1; 0; 0) f ( x ) = 21, 375 x x 1 0 < x 1 < 1 P 1 2

10 x 1 =0 x 2 =0 x1=1 x2= : () 15x x 2 + 4x 3 + 2x 4 8x 1 + 5x 2 + 3x 3 + 2x 4 10 x 1 = 0 0 x k 1 k = 2, 3, 4 2 : () 15x x 2 + 4x 3 + 2x 4 8x 1 + 5x 2 + 3x 3 + 2x 4 10 x 1 = 1 0 x k 1 k = 2, 3, 4 1 x 1 = (0; 1; 1; 1) f(x 1 ) = 18 P P x 1 = 0 x 1 x 1 P 2 x 2 = (1; 0, 9; 0; 0) f(x 2 ) = 19, 8 P x 1 = 1 x x 1 =0 x 2 =0 x1=1 x 2 = : () 15x x 2 + 4x 3 + 2x 4 8x 1 + 5x 2 + 3x 3 + 2x 4 10 x 1 = 1 x 2 = 0 0 x k 1 k = 3, 4 22 : () 15x x 2 + 4x 3 + 2x 4 8x 1 + 5x 2 + 3x 3 + 2x 4 10 x 1 = 1 x 2 = 1 0 x k 1 k = 3, x 21 = (1; 0; 0, 667; 0) f(x 21 ) = 17, 667 P

11 P 22 P x 1 P P : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x 1, x 2 N 0 : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x 1, x 2 0 X P x = (1, 96; 2, 70) f ( x ) = 21, 30

12 21,30 (1,96; 2,70) x 20,67 (1; 3,33) 21 (2; 2,60) 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje 1 < x 1 < 2 x 1 1 x 1 2 X X 1 X 2 X 1 < x 1 < 2 P P 21,30 (1,96; 2,70) 20,67 (1; 3,33) 21 (2; 2,60) X X 1 X 2 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje x j 0

13 1 : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x ,30 (1,96; 2,70) x 1 20,67 (1; 3,33) 21 (2; 2,60) 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje x 1 = (1; 3, 33) f ( x 1) = 20, 67 x 2 X 1 X 11 X 12 X 1 P 3 < x 2 < 4

14 21,30 (1,96; 2,70) 20,67 (1; 3,33) 21 (2; 2,60) 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje X 1 X 11 X 12 1 X 1 11 : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x 1 1 x : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x 1 1 x ,30 (1,96; 2,70) x 11 20,67 (1; 3,33) 21 (2; 2,60) (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje 11

15 11 P x 11 = (1; 3) f ( x 11) = 19 [19; 21, 30] X 11 x x 12 21,30 (1,96; 2,70) 20,67 (1; 3,33) 21 (2; 2,60) (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje X x 12 = (0; 4) f ( x 12) = ,30 (1,96; 2,70) 20,67 (1; 3,33) 21 (2; 2,60) 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje x 2 = (2; 2, 60) f ( x 2) = 21

16 2 x 2 X 2 X 21 X 22 21,30 (1,96; 2,70) 20,67 (1; 3,33) 21 (2; 2,60) 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje X 2 X 21 X 22 2 X 2 21 : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x 1 2 x : () f(x) = 4x 1 + 5x 2 2x 1 + 3x x 1 + 5x 2 35 x 1 2 x 2 3

17 21,30 (1,96; 2,70) 20,67 (1; 3,33) 21 (2; 2,60) 19 (1; 3) (0; 4) ,09 (2,27; 2) Nema rešenje P 21 X x 21 = (2, 27; 2) f ( x 21) = 19, 06 P X 21 X P x = (0; 4) f (x ) = 20

18 P : () f (x) x X, P X P T x, f x, f P f L = (( 1, f 1 ),..., ( l, f l )) L L L [1] L ((, f )) \

19 P P L ((, )) f x / ( L ) > 0 ( ) T, f T L [1] L L\ T, f T T x, f f > f x X x x f f L f x {1,..., p} k {1,..., p} T k T k L L ((T k, f)) x P f T p T P T L p = 2

20 T f f

2 2 + x =

2 2 + x = Lecture 30: Power series A Power Series is a series of the form c n = c 0 + c 1 x + c x + c 3 x 3 +... where x is a variable, the c n s are constants called the coefficients of the series. n = 1 + x +

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