1、一般隱枚舉法(1)標(biāo)準(zhǔn)問(wèn)題r min z = c()+ cx+.+ cnxn再內(nèi)+絢 2 兀 2+ %叫汕假定:1)c0 (j=ln)2)e,c20Cj，-Cj(A)9%內(nèi)+%2耳+ %二4石益取0或I（2）定義：子問(wèn)題2有等式約束（2）定義：子問(wèn)題C niin z = c0+ q十c3十c4x4十.十cnxn口4兀+舛”兀”二耳_絢|_門(B) 1,2,3x4兀取0或1令（A）的決策變量（Xi定(3)判定子問(wèn)題已查明條件：(a)若( (0, 0,0)為問(wèn)題可行解，則必為 此問(wèn)題最優(yōu)解。V 方1,bm均(b)若某約束條件中，正系數(shù)和小于常數(shù)項(xiàng)，則 此問(wèn)題無(wú)解。( (C) )若已有某可行解，相應(yīng)

2、目標(biāo)函數(shù)值乙 (上界)若可斷定，問(wèn)題既使有解，z也z。則可停止，認(rèn)為該問(wèn)題已査明。例:min z = 2兀+ 3x2+ 8x34西+ x2+ 3X3 32X2+ 1兀“2 人取 0, 1步1： M常數(shù)項(xiàng)bpb2, b3= (-4, 3, 1)(a)各約束正系數(shù)和：5,&34,3,1(b)原問(wèn)題未査明步2：1 Xi=l,x29x3未定min z = 2+ 3x2+ 8x35兀 3兀3 2bib22天2+花二b3(a)正系數(shù)和(5, 4, 3) 73x? 2兀3二1兀3取01C bi, b?, b3 ) =(7, 2, 1) 32X2+ 兀3 N 1兀2，兀3取0,1(為,厶2,厶3)=(

3、43,1)(a)(5,4,3) v(43,l)0+c. =3 -93 2兀3 2 _ 1兀3取0,1(5,厶2,/?3)=(-9,2,-1)(G)(0,3,1) A：(-9,2,-1)(b)3 +C2= 11 5(c)寸步6：1,2 min z =8X3步4：1(b)-3二- 43“ 3x3 1冷取0,1(也,/?2,方3)= (-4,3,1)(a)(),3,1)弋(4,3,1)(h)0 + 5 = 8 5(c) yj子問(wèn)題k或kmin z= 5 +伽 + +c”&C%+16+1 十+呦,嚴(yán)|”二為hm(-)判定子問(wèn)題的條件(是否査明)若(a)全部bi 0(/ =1丿和)，則Z = Z。，忑+i= =兀=。，為子問(wèn) 題的最優(yōu)解。已査明。(b)若有i,使EaiJ0(C)目前的最優(yōu)解z = zo ,Zo+C“ zo則子問(wèn)題解不會(huì)好于目前解。已査明。nS+1+1 + +（二）步驟：（1）定界。Z。（可行解必）;仙（3）子問(wèn)題k未查明； 分為k,k+l,和乜訂1若k,k+l滿足（a）,得解門,目標(biāo)Z1以Zi Zo。停，轉(zhuǎn)k。若.k, k+1滿足（B）或（C）,停，轉(zhuǎn)去乩廠門;若k