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I HC QUC GIA H NII HC KHOA HC T NHIN
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THI KT THC MN HCNM HC 2011-2012
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Mn thi: Ti u haM mn hc: 49ABCD S tn ch: 2 s: 1Dnh cho sinh vin kho: Lp K52A2 Ngnh hc: Ton - Tin ng dng
Thi gian lm bi 90 pht (khng k thi gian pht )
Cu 1. (2 im) Cho hai tp li X, Y v cc s thc ,. Chng minh rng cc tp sau ycng l tp li:
(i) X Y;(ii) X = {x|x X};
(iii) X + Y = {x + y|x X,y Y}.
Cu 2. (4 im) Cho bi ton quy hoch tuyn tnh
L(x) = x1 + 3x2 + x3 min
x1 + x2 + x3 5
x2 + 2x3 = 8
x1 + 2x2 + x3 10
xi 0(i = 1,2,3)
i) a v dng chnh tc;ii) Gii bi ton trn bng phng php n hnh;
Cu 3. (4 im) Trong v bo lt va qua c 4 im B1, B2, B3, B4 b ngp nng, cn tip tlng thc vi yu cu tng ng l b = (10, 15, 20, 20) theo n v tn. Nh nc b trlng thc cu tr ba kho A1, A2, A3 vi tr lng tng ng l a = (15, 20, 30). Cc phvn chuyn t Ai n Bj l cij nh trong ma trn sau y:
3 4 7 8
4 3 7 9
6 5 4 11
Hy lp k hoch vn chuyn ti u sao cho cc im cn cu tr nhn s lng lngthc v tng s tn /km l nh nht.
1. Thit lp bng vn ti vi phng n c s theo phng php gc ty bc.2. Gii bi ton bng phng php th v.
Ch : Cn b coi thi khng gii thch g thm.
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I HC QUC GIA H NII HC KHOA HC T NHIN
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P N V THANG IMMnn hc: Ti u ha
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M mn hc: 49ABCD S tn ch: 2 s: 1Dnh cho sinh vin kho: Lp K52A2 Ngnh hc: Ton - Tin ng dng
Li gii 1. [2 im]
(i) Cho x,y X Y v 0 1. Taphi chng minh x + (1 )y X Y. Tht vy,do x,y X Y nn x,y X. Do X l li nn x + (1 )y X; tng t ta cng cx + (1 )y Y. T suy ra x + (1 )y X Y.
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(ii) Gi s ta c x1, x2 X vi x1, x2 X v 0 1. V X l li nn x1 + (1 )x2 X. Ta xt (x1) + (1 )x2 = (x1 + (1 )x2) X.
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(iii) Chng minh tng t phn trn. 0.5Li gii 2. [4 im]
i) Dng chnh tc: a thm bin ph x4, x5 l bin ph
L(x) = x1 + 3x2 + x3 min
x1 + x2 + x3 x4 = 5
x2 + 2x3 = 8
x1 + 2x2 + x3 + x5 = 10
xi 0(i = 1,2,3,4,5)
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ii) Gii bng phng php n hnh: a thm x6, x7 l bin gi, bi ton tr thnh
L(x) = x1 + 3x2 + x3 + Mx6 + Mx7 min
x1 + x2 + x3 x4 + x6 = 5
x2 + 2x3 + x7 = 8
x1 + 2x2 + x3 + x5 = 10
xi 0(i = 1,2,3,4,5,6,7)
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J cJ xJ 1 2 3 4 5 6 7
1 3 1 0 0 M M6 M 5 1 1 1 -1 0 1 0
7 M 8 0 1 [2] 0 0 0 1
5 0 10 1 2 1 0 1 0 0
L(x) 0 -1 -3 -1 0 0 0 0
13 1 2 [3] -1 0 0 0
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Do cn tn ti gi tr ln hn 0 nn cha c phng n ti u ta cn tm bin a voCt c gi ln nh nht ng vi x3 vy bin a vo l : x3 Hng c gi tr nh nht ngvi ct l hng 2
J cJ xJ 1 2 3 4 5 6 71 3 1 0 0 M M
6 M 1 [1] 1/2 0 -1 0 1
3 1 4 0 1/2 1 0 0 0
5 0 6 1 3/2 0 0 1 0
L(x) 4 [-1] -5/2 0 0 0 0
1 1 1/2 0 -1 0 0
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Do cn tn ti gi tr ln hn 0 nn cha c phng n ti u ta cn tm bin a voCt c gi ln nh nht ng vi x1 vy bin a vo l : x1 Hng c gi tr nh nht ng
vi ct l hng 1
J cJ xJ 1 2 3 4 5 6 7
1 3 1 0 0 M M
1 1 1 1 1/2 0 -1 0
3 1 4 0 1/2 1 0 0
5 0 5 0 1 0 0 1
L(x) 5 0 -2 0 -1 0
0 0 0 0 0 0
Phng n ti u ca bi ton m rng l : x = (1,0,4,0,5,0,0)Gi tr hm mc tiu t c l : F(x) = 5
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Li gii 3. [4 im]
1. y l bi ton vn ti dng min. Ni cung cp hng a1 = 15, a2 = 20, a3 = 30,v ninhn hng b1 = 10, b2 = 15, b3 = 20, b4 = 20. D thy y l bi ton vn ti cn bngthu pht.2. Phng n c s theo phng php gc ty bc trong bng 1.
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Ai Bj B1 : 10 B2 : 15 B3 : 20 B4 : 20
3 4 7 8A1 : 15 10 5
4 3 7 9A2 : 20 10 10
6 5 4 11A3 : 30 10 20
Bng 1.
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2. Vng lp th nht:Bc 1. Phng n x khng thoi ha v |G| = m + n 1 = 7. Tm cc th v
u1 + v1 = 3
u1 + v2 = 4
u2 + v2 = 3
u2 + v3 = 7
u3 + v3 = 4
u3 + v4 = 11
Cho u1 = 0 suy ra v1 = 3, v2 = 4 v u2 = 1, v3 = 8, u3 = 8, v4 = 15.
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Tnh ij nh bng 2
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Ai Bj B1 : 10 B2 : 15 B3 : 20 B4 : 20
0 0 1 7 + u1 = 0A1 : 15 10 5
-2 0 + 0 5 u2 = 1A2 : 20 10 10
-7 -5 0 + 0 u3 = 4A3 : 30 10 20
v1 = 3 v2 = 4 v3 = 8 v4 = 15Bng 2
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vi phm du hiu ti u (2,3) l a vo. Ta c vng
V = {(1, 4)+
, (3, 4)
, (3, 3)+
, (2, 3)
, (2, 2)+
, (1, 2)
}
Lng iu chnh q = min{x34, x23, x12} = min{20, 10, 5} = 5. Tng ng vi (1,2) yl loi ra ta c bng mi bng cch tnh li
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Ai Bj B1 : 10 B2 : 15 B3 : 20 B4 : 20
0 -7 -6 0 + u1 = 0A1 : 15 10 5
5 + 0 0 5 u2 = 7A2 : 20 15 5
0 -5 0 + 0 u1 = 7
A3 : 30 15 15v1 = 0 v2 = -7 v3 = -7 v4 = -7
Bng 3.
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Tm cc th vu1 + v1 = 0
u2 + v2 = 0
u2 + v3 =0
u3 + v3 = 0
u1 + v4 = 7
u3 + v4 = 0
Cho u1 = 0 suy ra v4 = 7, v1 = 0, v3 = 7, v2 = 7, u2 = 7, u3 = 7.
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Tnh ij nh bng 3 vi phm du hiu ti u (2,1) l a vo. Ta c vng
V = {(2, 1)+, (1, 1), (1, 4)+, (3, 4), (3, 3)+, (2, 3)}
Lng iu chnh q = min{x11, x34, x23} = min{10, 15, 5} = 5. Tng ng vi (2,3) y
l loi ra ta c bng mi bng cch tnh liaaaaaa
Ai Bj B1 : 10 B2 : 15 B3 : 20 B4 : 20
0 -2 -6 0 u1 = 0A1 : 15 5 10
0 0 -5 0 u2 = 5A2 : 20 5 15 5
0 0 0 0 u1 = 0A3 : 30 20 10
v1 = 0 v2 = 5 v3 = 0 v4 = 0Bng 4
Vy phng n ti u phi tm l
X =
5 0 0 10
5 15 0 0
0 0 20 10
Chi ph ti u l F(x) = 350.
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H ni, ngy 15 thng 12 nm 2011NGI LM P N
(k v ghi r h tn)
PGS.TS. Nguyn Hu in
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