Tài Liệu Đề Thi Tối Ưu Hóa, Đề Thi Môn Tối Ưu Hóa (Quy Hoạch Tuyến Tính)

PDFaid.Com #1 Pdf Solutions Truong D~i h9C Kinh Te TP HeM Khoa Toan – Thong ke He;> va ten:_ -._~~ SBD: -r ~ ~._,; – ~OcHOIA ~ DE TID MON TOI lfU HOA (QUY HO~CH TUYEN TiNH) ThOi giaD lam bai: 75 ph6t (Kh6ng sO” d1Jng tai li~n) (Ch6 y: Sinh vi~n phai nQp l~i d
1 can (3d) Giiii bAi toan quy ho~ch tuye”n tlnh sau: f(x) =Xl + 6X2 + 4X3 ~ max .c om – Xl + X2 – X3 ~ – Xl + 2X2 + X3 Xl + 2X2 + 2X3 ~ 12 Xj ~ U=I,2,3) = can (3d) Giai bAi toan v~n t,Ii v(ji s6lic$u cho nhu sau Tim phuong an t6i un khae, nSu co (aj) = (110, 100,30); (bj ) = (60, 130,40,30) vA 10 13J 6 <5 47 um C= can (2d) Cho bAi tOlin QHTT sau co phuong an t6i u"U lA x*= (0, f{x) = 5Xl + 3X2 - X3 + 4X4 ~ 14,0,0): or Xl + X2 2X3 + X4 ~ 10 -Xl + ~ ~ ~ 2X2 3X3 + X4 = 28 Xj ~ U= 1,2,3,4) hf Hay vie"t bai toan d6i ng~u va tim phuong an t6i u"U eua bai toan d6i ngau ue can (2d) MQt doanh nghic$p dn mua thie"t bi mdi dua vao kinh doanh san xu!t voi s6 ti~n d~u tu d~ mua thie"t bi dl;t tinh IA 300 ngan $ Co hai lo~i tille"t bj co th~ mua vdi cae thong tin nhu sau: Thie"t bi A: don gia 12 ngan $, thong s6 k9 thu~t eho biGt mOt gia hOf!,t dOng tieu hao nhien lic$u Ia 30 don vi, nang su!t trung blnh Ia 60 san phd"mlgiC1, m3i thie"t bi ehie"m di~n tieh m2 ThiGt bi B: don gia 15 ngAn $, thong s6 kY thu~t eho bie"t mOt giC1 ho~t dOng tieu hao nhien li~u Ia 45 don vi, nang su!t trung blnh la 80 san phd"mlgiC1, m3i thie"t bi ehie"m di~n tieh m2 Theo b~n, doanh nghi~p se ehQn mua thie"t bi nhu the" nao d~ eho tieu hao nhien li~u la it nha"t, bie"t dng ke" ho~eh eua doanh nghic$p IA phiii d~t It nha"t 1000 san phd"m IgiC1, va di~n tieh phan xudng la 150 m Yeu du eml:Ơp mo hInh bai toan, khong ghIi PDFaid.Com #1 Pdf Solutions CU a) SP loi A SP loi B TNG S b) Gii Toỏn Gii ngoi ng Gii Toỏn&NN TNG S 7/24 SP loi A SP loi B Th phm TNG S X ~H( Y ~H( Z ~H( om (2 im) a) A, B xung khc P(A) = 0,1 v P(B) = 0,3 1/ ỳng, P(A|B) = P(AB)/P(B) = 2/ ỳng, P(AB) = P(A) + P(B) = 0,4 / / / 3/ Sai, P(AB) = 4/ ỳng, P(A B ) = P(AB) = P() = / / / 5/ ỳng, P(A B ) = P(AB) = - P(AB) = - (P(A) + P(B)) = 0,6 b) X, Y c lp v X ~ B( 0,2 ) v Y ~ H( 12 ) Z=X+Y P(Z 3) = P(X = 0, Y = 0) + P(X = 1, Y = 0) + P(X = 0, Y = 1) + P(X = 1, Y = 1) + P(X = 0, Y = 2) + P(X = 2, Y = 0) + P(X = 1, Y = 2) + P(X = 2, Y = 1) + P(X = 0, Y = 3) + P(X = 3, Y = 0) = P(X = 0, Y = 2) + P(X = 1, Y = 2) + P(X = 0, Y = 3) = 0,1018 (1 im) KiN 10 10 10 10 c CU hay (1 im) Xỏc sut sinh viờn c chn: ch gii mụn v khụng gii mụn no LP Gii mụn p = C(8,1)C(37,1)/C(48,2) = 0,2624 hay 37/141 48 37 (khụng gii mụn no) Y 1/45 2/9 2/9 7/15 2/15 1/3 X1 P(X|Y = 0) 1/21 E(X|Y = 0) = 10/21 1,43 X TNG or um CU (2 im) (1 im) Xỏc sut khỏch hng mua c sn phm loi A KiN X ~ H( 10 ) X PX 0,0667 0,4667 0,4667 10 Phõn phi s sn phm loi A cũn li kin Y = - X Y PY 0,0667 0,4667 0,4667 p dng cụng thc xỏc sut y P(Y =3) = 0,0417 0,1667 0,0833 0,2917 2 2 ) ) ) 7/15 TNG 1/15 2/9 5/9 2/9 1/15 10/21 hay ue hf 10/7 CU (2 im) Nc (m3) 15 25 35 45 55 S h 22 29 20 15 a) Kim nh trung bỡnh phớa (phớa phi) (1 im) m3 Trung bỡnh mu = 28,00 Phng sai mu = 179,00 = m3 Phng sai mu cú c = 180,81 13,45 Kim nh = m 24,5 Mc ý ngha = 0,03 z = 1,88 TCK z = 2,60 |z| > z bỏc b H0 Mc tiờu th nc hin tng b) c lng trung bỡnh ca nhng h tiờu th nc bỡnh thng m3 Trung bỡnh mu BT= 24,72 Phng sai mu BT= 59,08 = m3 Phng sai mu cú c = 59,92 7,74 tin cy – = 95% z = 1,96 Sai s = 1,80 àBT ( 22,92 26,52 ) m CU (2 im) Kớch thc TT KT mu T s mu 3000 400 290 tin cy – = a) T l mu f = 0,7250 97% Sai s = 0,0484 P( 0,6766 0,7734 ) ( 2030 2321 ) sn phm A (1 im) chớnh xỏc = b) 5% z = 2,24 tin cy – = 0,9749 hay 97,49% (1 im) CU TNG TH 20 23 26 29 Ly mu khụng hon li, kớch thc 3, s cú C(4,3) = mu, vi phõn phi phng sai mu c nh sau: S2 S2 9,0 9,0 21,0 21,0 hay 9,0 21,0 P 1/4 1/4 1/4 1/4 P 1/2 1/2 Trung bỡnh phng sai mu E(S2) = 15,0 (1 im) PDFaid.Com #1 Pdf Solutions HQ TrLlCJng fJ?i hQc Kinh Te TP HeM Khoa Toan – ThtJng va ke SSD: “~JIOJ J (b~ DE THI MON TOI uu HOA (QUY HO~CH TUYEN TiNH) sa ThO”i gian lam bai: 75 phut (Khong d”ng tai li~u) (Chu y: Sinh vi~n phai nQP la)i d~ kern v8i bai thi) D~2 cau (ld) Cac phat bi~u sau day dung hay sai ? Gicii thich 19 t~i sao? a) Bai toan QHTI: max f(x), chi co PACB la Xl, X2, X3 thi gia trj t6i U”U hI: xeX max {f(x ), f(x ), f(x )} cau (3,?d) Giili bai toan quy ho~ch tuyn f Hnh sau day 6x( + 4X2 + 3X3 )- max 2xJ + X2 + X3 :5 24 2xI + 2X2 + X3 = 18 XI + X2 + X3 ?: 18 X), X2, X3 2: O U”U khac, nu co um Tim phuong an t6i c om b) Blii toan QHTI (P), them bi~n gici bai toan rna rQng (PM), nu (PM) co PATH thi (P) se luon luon co PATH cau (2d) Cho bai tot”m quy ho~ch tuy~n tfnh sau day: or 4Xl + 6X2 + 2X3 )- 2Xl + X2 + X3 :5 24 18 2x( + 2X2 + X3 Xl + X2 + X3 > 18 XI X2, X3 2: O hf a) Vit hai toan d6i ng~u b) Biet bai toan tren co phuong an t6i U”U hI: xã (0,0, 18) Hay dung Diob It d9 I~cb bil ySu d~ tim tftt cil cac phuong an t6i U”U cua bai toan d6i ng~u cau (3,5d) Cho bai toan v~n tili vo; s6 li~u sau day: ue 63 4>7; <6 (c jj ) = (aJ=(10, 50, 90); (b) = (10, 40, 70) a) Gi:ii bili loan v~n tM tren b) Giiii bai toan v~n tai tren vdi di~u ld~n di~m ph:it thti" pMt he"t hang PDFaid.Com #1 Pdf Solutions (2 im) a) Xỏc sut sinh viờn c chn khụng gii mụn no LP Gii mụn (1 im) Gii Toỏn Gii Anh 10 Gii Toỏn&AV p = C(34,3)/C(46,3) = 0,3942 TNG S 46 34 (khụng gii mụn no) b) X ~ N( 5,2 1,96 ) P(X >= )= 0,2839 (1 im) CU (2 im) a) A, B xung khc P(A) = 0,3 v P(B) = 0,4 (1 im) 1/ ỳng, P(A|B) = P(AB)/P(B) = 2/ ỳng, P(AB) = P(A) + P(B) = 0,7 / / / 3/ Sai, P(AB) = 5/ ỳng, P(A B ) = P(AB) = P() = / / / 4/ ỳng, P(A B ) = P(AB) = – P(AB) = – (P(A) + P(B)) = 0,3 b) KIN KIN KIN SP loi I SP loi II TNG S 10 10 10 Cỏc bin c chn c kin K12, K13 v K23 l h y (1 im) A l bin c sn phm chn u l sn phm loi I P(A) = P(A|K12)P(K12) + P(A|K13)P(K13) + P(A|K23)P(K23) = 0,2267 hay 17/75 CU (1 im) Xi (i = 1, 2) s sn phm loi I mỏy th i sn xut X1 ~ B( 50 0,6 ) X2 ~ B( 50 0,7 ) S tin thu c Y = 18(X1 + X2) + 14(100 – X1 – X2) = 4X1 + 4X2 + 1400 VarY = 16VarX1 + 16VarX2 = 360 ngn ng CU (2 im) Trng lng (gr) 150 250 350 450 550 650 S trỏi 10 40 140 110 80 20 400 (1 im) a) c lng trung bỡnh ca loi trỏi cõy ny Trung bỡnh mu = 417,50 gr Phng sai mu = 12693,75 Phng sai mu cú c = 12725,56 112,81 gr = tin cy – = 95% z = 1,96 Sai s = 11,05 ( 406,45 428,55 ) gram b) Kim nh trung bỡnh phớa Trung bỡnh mu trỏi loi I = 507,14 gr Phng sai mu trỏi loi I = 4353,74 Phng sai mu cú c = 4374,57 66,14 gr = Kim nh = 550 gr Mc ý ngha = 0,03 z = 2,17 TCK z = -9,39 |z| > z bỏc b H0 Trng lng trung bỡnh ca trỏi loi I l 550gr khụng chp nhn c (1 im) CU (2 im) Kớch thc TT KT mu T s mu 3000 400 350 tin cy – = a) T l mu f = 0,8750 97% Sai s = 0,0359 P( 0,8391 0,9109 ) ( 2518 2733 ) sn phm A (1 im) b) TNG TH 21 23 27 29 Ly mu khụng hon li, kớch thc 2, s cú C(4,2) = mu, vi phõn phi phng sai mu c nh sau: 2,0 18,0 32,0 8,0 18,0 2,0 S2 P 1/6 1/6 1/6 1/6 1/6 1/6 hay (1 im) 2,0 8,0 18,0 32,0 S P 1/3 1/6 1/3 1/6 CU k (ni npi ) 2 = = 27,3 > 0,05 (3) = 7,8147 Bỏc b H0 ue hf or um c om CU i =1 npi Bỏo cỏo ca nh mỏy khụng ỳng vi lụ hng trờn (1 im) PDFaid.Com #1 Pdf Solutions e HQ va ten: Trliong D.;li hQc Kinh T TP HeM Khoa Toan – Thtfng ke … ke