運籌學(xué)--對策論

1、 ),(jiji（1，1）（1，2） （2，1）（2，2）0430400330020320Ak,),(*121SxxxxTmkikixi, 0, 1ijjjijTnijiyayyayiEyE),)()(0 ,0 , 1 , 0(),(),(1iiijTijmjxaaxxjxExE) 0, 1 , 0)()(,(),(),(1 iiiiijijjijiijjxyExyayxayxE),()(),( jjjjiijiijiijjxEyxayxayxE),()(),(*)*,(*)*,(*),(*),(yxExyxExyEyxEiiiii*)*,(*)*,()*,()*,(yxEyyxEyxEyxE

2、jjjjj)*,(*)*,(*),(yxEyxEyxE定理定理14-4：設(shè)設(shè) ，則，則 為為 的解的充要的解的充要條件：存在數(shù)條件：存在數(shù) 使得使得 分別是不等式組（分別是不等式組（ ）和）和（）的解。）的解。*,*21SySx*)*,(yxGv*,yxmixxnjvxaiiiiiij, 2 , 1, 01, 2 , 1,)(njyymivyajjjjjij, 2 , 1, 01, 2 , 1,)(證明：由定理證明：由定理3，（ ）和（）和（）成立。）成立。jjjjijimiyvyayiEyE, 2 , 1, 1,),(),(iiiiijjmjxvxajxExE, 2 , 1, 1,),(),

3、(定理定理14-5：任意一個給定的矩陣對策在混合策略意義下一定有解。任意一個給定的矩陣對策在混合策略意義下一定有解。 矩陣對策的解可能不只一個，但對策值是唯一。矩陣對策的解可能不只一個，但對策值是唯一。證明：證明：考慮兩個線性規(guī)劃問題考慮兩個線性規(guī)劃問題mixxnjwxaPwiiiiiij, 2 , 1, 01, 2 , 1,)(maxnjyymivyaDvjjjjjij, 2 , 1, 01, 2 , 1,)(min這是兩個互為對偶的線性規(guī)劃問題（這是兩個互為對偶的線性規(guī)劃問題（P403），）， jjTawx1min,) 0 , 0 , 1 (1max,) 0 , 0 , 1 (iiTavy

4、*)*,(*),*,(vywx*vw *,*21SySxiiijjijjxavya*,;,22121211ASSGASSG)(),(21LaAaAijijL)()()2() 1 (2112GTGTLVVGG)(),(21GTGT,;,22121211ASSGASSG0)()()2() 1 (2112GTGTVVGG;,21ASSG)()()2(0) 1 (21GTGTVGTAA;,21ASSG)(,212211ijmmaASSkjijki優(yōu)優(yōu)超超于于ilijlj優(yōu)優(yōu)超超于于)(),(21GTGT5432154321388065 . 57864959379520503023A3114233880

5、65 . 5786495937優(yōu)優(yōu)于于優(yōu)優(yōu)于于，優(yōu)優(yōu)于于A2152154214354354364373065 . 56493738065 . 5764953742優(yōu)優(yōu)于于0,164370,163472121212143434343yyyyvyyvyyxxxxvxxvxx5,) 0 , 0 , 0 , 2/ 1 , 2/ 1 (*,) 0 , 3/2 , 3/ 1 , 0 , 0(*GTTVyx07max971907min22211211212122212121211121212221122211110,10,1aaaaDyyyyvyayavyayaxxxxvxaxavxaxa,25711323

6、212211SSA 1 , 0,)1 ,(*xxxxTTxx)1 ,(321,vxxvxxvxx)1 ( 211,)1 ( 53 ,)1 ( 72OAxV752) 3()1 (211)2(,)1 (53) 1 (,)1 (72vxxvxxvxx1x(1)3(2)11AB(3)OTTyyyyyyyyyyxvxvxxvxx)11/2 ,11/9 , 0 (*11/211/9,111/492511/49113, 0*,)11/8 ,11/3 (*,11/49,11/3,)1 ( 211)1 ( 53323232321，得得故故由由圖圖最最優(yōu)優(yōu)策策略略為為2116672ATyy)1 ,(.)0 , 1

7、 , 0(*14/95/1,)1 ,(*, 4/9, 5/1; 6,)1 (211,)1 (66 ,)1 (72221即即只只取取得得最最優(yōu)優(yōu)策策略略是是，局局中中人人最最優(yōu)優(yōu)策策略略是是TTxyyyyyyvvyyvyyvyyyyA12116672762) 3()1 (211)2(,)1 (66) 1 (,)1 (72vyyvyyvyyy(1)2(2)11(3)O6A1B2A2B17551243/84Avyyvyyvxxvxx)1 ( 7,)1 ( 24,)1 ( 72 ,)1 (4413132421432121712475123/847551243/8421431優(yōu)優(yōu)超超優(yōu)優(yōu)超超A4/13,

8、) 8/3 , 0 , 0 , 8/5 (*,) 4/ 1 , 4/3 (*vyxTT 1 , 0,)1 ,(*xxxxT)4()1 (72)3()1 (54)2(,)1 (53/8)1 (,)1 (4vxxvxxvxxvxx751y(1)2(2)13/14(3)O43/4(4)*,*,4/13,*), 0*,*,(*, 0*,) 4/ 1 , 4/3 (*4214213yyyvyyyyyxTT1, 4/1375, 4/1323/84421421421yyyyyyyyy 矩陣對策的解法矩陣對策的解法1.優(yōu)超原則化簡優(yōu)超原則化簡,若有鞍點若有鞍點,則為純矩陣對策則為純矩陣對策. 2.若沒有鞍點若

9、沒有鞍點,贏得矩陣為贏得矩陣為22陣陣.用公式法用公式法3.贏得矩陣為贏得矩陣為2n,或或m2陣陣.用圖解法用圖解法4.方程組法方程組法5.線性規(guī)劃法線性規(guī)劃法 3.2 線性規(guī)劃法線性規(guī)劃法定理定理14-4：設(shè)設(shè) ，則，則 為為 的解的充要的解的充要條件：存在數(shù)條件：存在數(shù) 使得使得 分別是不等式組（分別是不等式組（ ）和）和（）的解。）的解。*,*21SySx*)*,(yxGv*,yxmixxnjvxaiiiiiij, 2 , 1, 01, 2 , 1,)(njyymivyajjjjjij, 2 , 1, 01, 2 , 1,)(等等價價于于,0, 2 , 1,maxiiiijxnjvxav

10、v1V)(,0, 2 , 1, 1maxGDvyyymiyaywjjjjjijjj1109092927Avxxii )(min1miiijjxav令令,0, 2 , 1,iiiijxnjvxa)(,0, 2 , 1, 1minz)min()1min(maxPxnjxaxxvviiiijiiii) 1 (011191921927min3213121321321xxxxxxxxxxxxxw，)2(011191921927max3213121321321yyyyyyyyyyyyyw，) 3(0,11191921927max3213213312211321321uuuyyyuyyuyyuyyyyyyw

11、，1110000 1 7291001/70 1 2900101/20 19 0110011/9111000jc1y1u2y3y2u3ubXCBB321uuujjjzc ) 3(0,11191921927max3213213312211321321uuuyyyuyyuyyuyyyyyyw，1110000 2/9 024/910-7/90 7/9 09-22/9 01-2/91 1/9 1011/9001/901-2/900-1/9jc1y1u2y3y2u3ubXCBB121yuujjjzc 1110000 4/810080/811-2/9-59/811 7/8101-22/8101/9-2/811 1/91011/9001/9014/810-1/9-7/81jc1y1u2y3y2u3ubXCBB121yyujjjzc 1110001 4/8000181/80-18/80-59/801 1/1001022/804/80-18/801 1/20100-99/8022/8081/80000-4/80-8/80-4/80jc1y1u2y3y2u3ubXCB