数学建模社区-数学中国
标题:
求助,代码第二段啥意思
[打印本页]
作者:
魏关亭侯
时间:
2012-7-25 10:30
标题:
求助,代码第二段啥意思
背包问题,12件物品质量为2,5,18,3,2,5,10,4,11,7,14,6,价值为5,10,13,4,3,11,13,10,8,16.7.4,包的最大重量是46
9 [7 n" w/ X8 T% Y, Y9 M3 ] b: e
求最优方案
7 y# @ c$ b) ~6 v; P) h6 `% z* A2 u) {2 ]
代码如下,第二段看不懂,从那个=~开始,那符号啥意思
' s4 I& S) F b3 u4 U# i
a=0.95
9 c* o8 `) S8 `( `8 U) L
k=[5;10;13;4;3;11;13;10;8;16;7;4];
" C4 b: f, c( w
k=-k;
6 _4 q- J) X( z
d=[2;5;18;3;2;5;10;4;11;7;14;6];
- f- v1 p+ _/ q x: X4 E1 @6 q
restriction=46;
/ |( t0 z+ G9 h( `( B
num=12;
# a4 R0 F& d4 r9 ?2 x
sol_new=ones(1,num);
2 E+ E5 X/ l( y2 W$ T
E_current=inf;E_best=inf;
9 r9 t# I1 {) m; t+ h. Q$ f
sol_current=sol_new; sol_best=sol_new;
5 h, A7 O# h& G, h
t0=97;tf=3;t=t0;
8 p5 u% [5 o7 l& L
p=1;
, I1 I* f5 V- b) b0 W) z4 P% X8 m
7 N9 E8 F/ b7 Y! K4 ~4 ^+ P
while t>=tf
9 S0 j7 e0 O' P; z) E1 g( X3 l
for r=1:100
9 }* |% ^1 c0 U
tmp=ceil(rand.*num);
/ p; s6 W: v6 }2 @0 |/ W* D
sol_new(1,tmp)=~sol_new(1,tmp);
* U U# w) ?' d) z
while 1
, G7 h3 l0 W0 L5 \& K# F
q=(sol_new*d<=restriction)
& p7 ?/ x6 E) c( y7 U& x$ c
if ~q
. k. _6 ]3 `+ \0 B* g: V
p=~p
: X c0 U1 R3 \0 ]3 f
tmp=find(sol_new==1);
- n5 l- h3 Y+ u9 ]5 M$ X: |
if p
0 d: X1 g0 Y6 X: L- C$ s
sol_new(1,tmp)=0;
' R, p+ O& w0 |- H7 D
else
* D! w0 l5 E4 D
sol_new(1,tmp(end))=0;
7 g* c7 P+ q+ v4 l+ c
end
& r1 Y, d$ ~7 W. c8 Q
else
# Y c; j+ M7 d. {
break
2 k+ [4 j5 v- Q
end
2 D6 N, c6 z2 G | W l
end
- q3 S1 s- j& a$ ^3 Y S* G3 Z
作者:
赵煜星
时间:
2012-7-25 21:01
好啊 我喜欢
作者:
梦天涯M
时间:
2012-8-5 11:54
model:
3 |! \9 _" L7 b; n% \5 ^
sets:
" R6 |" W! t# | P
beibao/1..12/:zhiliang,jiazhi,x;
& G9 J) O3 m c# Q
endsets
0 c! u# z$ f) L' X- F
data:
$ R2 p2 v2 i9 |! y, P
zhiliang=2 5 18 3 2 5 10 4 11 7 14 6;
1 j$ c0 @5 J; u& S
jiazhi=5 10 13 4 3 11 13 10 8 16 7 4;
: ~* d; N( q/ f5 Z$ r9 m9 k
enddata
' ]* t/ [7 I; u7 e
max=@sum(beibao(i):jiazhi(i)*x(i));
8 h8 t8 s' ^) I! X' E
@sum(beibao(i):zhiliang(i)*x(i))<46;
! T& P. z8 B/ I8 m
@for(beibao(i):
/ E- P9 A+ u3 a& g0 M7 F# C9 n
@bin(x(i)));
! h1 n& b9 |" z J
9 \% W0 d4 q5 |" U9 @, T, H6 B
结果:
9 V8 [2 R3 c, u( I: T
Global optimal solution found.
# C5 ^" G3 a7 d) N
Objective value: 76.00000
/ O {8 |: e, L+ W
Objective bound: 76.00000
- n$ [8 G ?, S+ J
Infeasibilities: 0.000000
. ]) \" W# p* U6 I4 k4 j& @
Extended solver steps: 0
" u! g9 O* w0 w* Y& h' ]5 P
Total solver iterations: 0
9 A6 a2 P! Y$ A0 E2 Z: S& Z
- K" b0 {# @, i0 e
Model Class: PILP
; T9 D0 y" W |3 a" O. G/ k( s
) c. v1 u W; D; E
Total variables: 12
" |5 q: H9 E P% R
Nonlinear variables: 0
9 i. u" n4 j9 x& t5 h* I
Integer variables: 12
: H% a: A9 _- e& K
. T8 H* g3 S- o+ W0 O# e8 q" ~
Total constraints: 2
0 G7 V! S* E9 |) b9 w: ^
Nonlinear constraints: 0
/ b2 Y4 P, \/ V5 Y& Y- R6 p
! y, k5 W! @1 l; {1 J
Total nonzeros: 24
8 k/ `; \1 U5 U$ S
Nonlinear nonzeros: 0
3 G W5 r# x) F" c9 V* B! R3 T, z
2 S3 s/ W, N ]/ d6 b" i
2 \/ e+ d9 T: w# W) u3 ]
Variable Value Reduced Cost
# p ^# |9 j- A# @$ y, `/ P# C
ZHILIANG( 1) 2.000000 0.000000
1 R' G8 w0 Z& u4 v& a6 g
ZHILIANG( 2) 5.000000 0.000000
! n& Q# a+ |) K' |% X
ZHILIANG( 3) 18.00000 0.000000
4 ~% y- b) L1 x% C7 C8 [6 ~
ZHILIANG( 4) 3.000000 0.000000
" {0 J9 \: s0 `" l5 ^6 y3 T+ }
ZHILIANG( 5) 2.000000 0.000000
, I$ ^. y" `9 R# _) G1 b7 d* u
ZHILIANG( 6) 5.000000 0.000000
2 |# H$ Y# Z, d" I
ZHILIANG( 7) 10.00000 0.000000
1 `% G. F6 P. C. {- c/ U
ZHILIANG( 8) 4.000000 0.000000
- d. b! V: D# j5 @5 X
ZHILIANG( 9) 11.00000 0.000000
$ O( C4 j! [2 c8 F% A: P" o6 j
ZHILIANG( 10) 7.000000 0.000000
. X+ |, U* J" Y# M
ZHILIANG( 11) 14.00000 0.000000
) ^6 y" ^2 k! z( c) f
ZHILIANG( 12) 6.000000 0.000000
, n/ T$ x M3 M) X f6 i
JIAZHI( 1) 5.000000 0.000000
1 I, ^- W, v2 b# w
JIAZHI( 2) 10.00000 0.000000
4 y# U) m! s/ N6 Y w5 D
JIAZHI( 3) 13.00000 0.000000
e3 x9 b, A/ k% \+ E. y
JIAZHI( 4) 4.000000 0.000000
, K% D( v4 @% i4 G2 P
JIAZHI( 5) 3.000000 0.000000
7 S; l8 A5 n. {0 Q5 ^3 K+ e3 J8 p
JIAZHI( 6) 11.00000 0.000000
* x% @, o& ? m$ N0 I G. Z; l
JIAZHI( 7) 13.00000 0.000000
1 q3 ]9 X- \( B+ b3 V5 d! m1 w
JIAZHI( 8) 10.00000 0.000000
% c2 m, G* k1 O8 G. r3 ?
JIAZHI( 9) 8.000000 0.000000
- v) B7 v7 Z+ P9 Z. d9 J8 H
JIAZHI( 10) 16.00000 0.000000
' Q! C, d" K# |- z% e' _) j! G( u
JIAZHI( 11) 7.000000 0.000000
: X+ H$ g2 e# b d
JIAZHI( 12) 4.000000 0.000000
. K- ~4 k q4 S% s
X( 1) 1.000000 -5.000000
; ^! n2 x6 p9 A% `
X( 2) 1.000000 -10.00000
9 i# O8 k5 l; W8 g0 ^" A
X( 3) 0.000000 -13.00000
7 F8 [9 e6 f* [5 F9 C
X( 4) 1.000000 -4.000000
( X; i% d% D- W2 K/ f5 W+ v- u! y
X( 5) 1.000000 -3.000000
: m9 { w- x+ G3 z- {/ k
X( 6) 1.000000 -11.00000
6 @1 F9 m% A2 q+ }6 ?* a) X2 e
X( 7) 1.000000 -13.00000
- L" k) V8 E" V/ M
X( 8) 1.000000 -10.00000
3 m8 d1 L/ y; f/ A f# L4 T
X( 9) 0.000000 -8.000000
) ?( }5 H p8 [" {
X( 10) 1.000000 -16.00000
# }) T n: G+ ~3 D7 _7 ]) T" ]
X( 11) 0.000000 -7.000000
" O. O& z3 D8 J( h7 |
X( 12) 1.000000 -4.000000
$ v' `/ c( a8 Y' V0 a' j5 M
/ C8 }( [& m- Z+ F1 e
Row Slack or Surplus Dual Price
8 ?+ f. ?7 g: ~' ^, I: l* P* F
1 76.00000 1.000000
& W+ P/ Y7 P* o, f; S% b( R% _
2 2.000000 0.000000
& S4 ^! n" u: v1 U t
3 w5 i* ~& ~' f8 U
作者:
梦天涯M
时间:
2012-8-5 11:56
不好意思,上边写成bin了,应该是gin
0 F3 o0 Z' x9 M; P9 H/ _
正确程序:
& ]% i! a* Z7 \3 F
model:
a# ^+ w3 v7 K9 i4 ?
sets:
* ~! N/ ~3 G) o% `. g; i
beibao/1..12/:zhiliang,jiazhi,x;
+ ]+ M+ y3 f8 B* [: ~
endsets
4 o# V" h+ g7 ^
data:
8 T& Z+ D! O4 d' m* m
zhiliang=2 5 18 3 2 5 10 4 11 7 14 6;
6 P% b r2 `- B! d/ C3 r
jiazhi=5 10 13 4 3 11 13 10 8 16 7 4;
* V6 q6 w6 ?1 i
enddata
i9 k9 L4 ~. W5 ~* s
max=@sum(beibao(i):jiazhi(i)*x(i));
2 x: V+ z! `" H ?& E% z9 {
@sum(beibao(i):zhiliang(i)*x(i))<46;
4 P: V% g( D3 W) @7 t
@for(beibao(i):
5 {1 t- _* U8 H G5 o8 f. S d0 T8 n9 _
@gin(x(i)));
作者:
梦天涯M
时间:
2012-8-5 11:58
Global optimal solution found.
+ n) T3 Y- o' j9 d/ K. G% q# I9 u7 a
Objective value: 115.0000
3 N& C) B, K- ~0 [; r& O" H9 j) z
Objective bound: 115.0000
+ X6 v$ e& f" h, E: @; c/ p/ A* p
Infeasibilities: 0.000000
, |. |( i" \: _- J# [5 a; n/ I
Extended solver steps: 0
) u4 w0 E4 K9 h% o8 o+ ?
Total solver iterations: 0
/ k$ u) m& f9 Z( N- Y' p
6 Z+ F5 }' i1 n; f+ W- ?
Model Class: PILP
2 b8 k; }; m: f+ O$ @. @& U
I: d% i; t4 P- F4 [% I' x W
Total variables: 12
: K9 d3 h% ?# i
Nonlinear variables: 0
6 v0 O3 k1 ~8 T: _/ N1 }
Integer variables: 12
. D5 Q; w2 Z0 v2 m2 k
5 F& d& r) _1 N; m1 F! _2 I" I$ M
Total constraints: 2
S [0 H2 S( g& {1 N/ n
Nonlinear constraints: 0
: E' B) t6 g& p* I9 U) k" q3 [: w
) E4 p' A# a$ u5 W# ` n
Total nonzeros: 24
9 N1 j- H6 a F2 {+ k
Nonlinear nonzeros: 0
, `5 N& [8 h e
, ~$ a* c p; L2 k5 r+ e
, F2 e, |" `& y3 [# g7 d
Variable Value Reduced Cost
6 z7 |5 N- S+ y# Z0 U
ZHILIANG( 1) 2.000000 0.000000
+ Z4 |. \! C/ a) P
ZHILIANG( 2) 5.000000 0.000000
( V o! V1 Z1 [( [' E
ZHILIANG( 3) 18.00000 0.000000
# K" c# H* h2 D: k A
ZHILIANG( 4) 3.000000 0.000000
5 o* ^( Q9 f3 [' C
ZHILIANG( 5) 2.000000 0.000000
( I: A9 I! C: C$ B7 i2 p* t8 K
ZHILIANG( 6) 5.000000 0.000000
& g+ `2 W- i# A7 [& M. D8 P: E
ZHILIANG( 7) 10.00000 0.000000
1 K s2 ]# N& R: J( j( H% m
ZHILIANG( 8) 4.000000 0.000000
: Q: W6 ^, u: C# r1 E8 Y2 X9 R0 }
ZHILIANG( 9) 11.00000 0.000000
, m$ |$ J9 O9 a! a. F
ZHILIANG( 10) 7.000000 0.000000
, B1 L/ r! }! B
ZHILIANG( 11) 14.00000 0.000000
9 J& {/ ]# k+ _$ ^- m9 w/ }3 R8 r
ZHILIANG( 12) 6.000000 0.000000
7 f9 }, p, @7 p; r5 {/ m. d
JIAZHI( 1) 5.000000 0.000000
; s; l" [8 H- ?1 ?2 K. E0 Q( F2 V
JIAZHI( 2) 10.00000 0.000000
. C0 U( Q0 D' x) A1 n& V# G% o/ ? n
JIAZHI( 3) 13.00000 0.000000
$ H! l# l/ B3 Z+ S8 p- A
JIAZHI( 4) 4.000000 0.000000
6 q1 H1 l, E6 T& c
JIAZHI( 5) 3.000000 0.000000
$ g q$ g+ I1 ?- H9 y0 O
JIAZHI( 6) 11.00000 0.000000
) ]8 { o0 P& E7 n
JIAZHI( 7) 13.00000 0.000000
3 f. X7 S& @! f4 f i: h! _4 D
JIAZHI( 8) 10.00000 0.000000
. b5 d! B* @# Z( e5 }7 `
JIAZHI( 9) 8.000000 0.000000
+ n4 ]+ b7 a6 i; |
JIAZHI( 10) 16.00000 0.000000
9 {( b* @& X* o" d$ }$ |
JIAZHI( 11) 7.000000 0.000000
' C6 H: G( z2 K9 P, a3 R
JIAZHI( 12) 4.000000 0.000000
3 R. O- m* Z% M- }0 {
X( 1) 1.000000 -5.000000
1 b' O3 g7 Y5 M; B* Y- Q) e7 W
X( 2) 0.000000 -10.00000
: f4 X, \8 G0 p: U! y: q/ b7 E
X( 3) 0.000000 -13.00000
; ]& R/ Z- T1 h
X( 4) 0.000000 -4.000000
9 W2 H/ G" F. a. L- f( t& x' u0 y- ?" P
X( 5) 0.000000 -3.000000
7 i9 o ~' o u- |+ c) P
X( 6) 0.000000 -11.00000
5 b% |! P t3 E: n a
X( 7) 0.000000 -13.00000
( s* o6 X* U, X, I! z! o
X( 8) 11.00000 -10.00000
- x, N. n2 _2 S' u$ {
X( 9) 0.000000 -8.000000
) Y/ f9 M. y# y4 \! L' Y$ ^9 [
X( 10) 0.000000 -16.00000
+ d; d- W& i1 K' T V1 ^% {" e- O
X( 11) 0.000000 -7.000000
* n q. _: E: M
X( 12) 0.000000 -4.000000
- @: f7 u5 P4 D% D7 S5 @6 i4 ]" Z
. C: }8 I! S+ A2 H4 Y' _( G2 R" p, ]
Row Slack or Surplus Dual Price
4 v! C) c8 |% y
1 115.0000 1.000000
3 R- z- l; j# F! |, c) B5 Z0 ?' C! |
2 0.000000 0.000000
& o9 j6 W7 m4 N$ Y ^& N
作者:
梦天涯M
时间:
2012-8-5 11:59
第一件物品取1件,第8件物品取11件,最大价值115
欢迎光临 数学建模社区-数学中国 (http://www.madio.net/)
Powered by Discuz! X2.5