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发表于 2009-8-16 18:59
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修改一下循环的下标可以使得程序编译通过,但是不知道是不是和你的模型相符合,你仔细分析一下吧!
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- model:( }1 ~& [+ q7 J9 z% X0 D% N. t+ ^
- !n个货车8收点运输问题;
8 c) d\" j& ~3 z i - sets:
2 b1 X* F1 r( ~ - place/place1..place10/:time1,time2,timexie;
% f6 j) \7 i# }' M6 g. F# S5 l - car/car1..car3/:time;
7 d7 ~! f\" i0 w( g5 W
& s4 M/ ^/ ]( a2 S( d- jiu/1..9/;
* y- D) L) X4 Z, ] - er/1..2/;
/ R1 G$ M1 S1 P, e: E - 1 ?# w. I5 O* k2 q- {
- lc(car,jiu):lucheng,ct;* z6 K8 V, K* u6 D7 d h5 t
- plan(place,place,car):a;
, o- E\" ?& x: T* X - distance(place,place):dd,dt,which;/ x8 Y$ ^. j% y
- endsets
- r: r+ _' B B- A% ?7 P/ p- u s - x% ]+ u# I# |9 N0 M2 T
+ x6 v* q$ j\" C. C' [2 e5 g- !这里是数据;
' W; d1 ]6 ]' n0 H - data:2 ^9 ~) p7 u0 G, I
- s=1;- v3 Y' ^\" p6 \4 D! g
- time1=-10 1 4 1 4 3.5 2 5 1.5 1; 3 m- `1 z) N) h
- time2=7 4 6 2 7 5 5 8 4 10 ;+ e! }6 S5 ^5 m& x( [8 W, s% b8 `
- timexie=0 1 2 1 3 2 2.5 3 0.8 0;
3 R. _+ P5 O7 F1 \ - 3 M0 Q+ q- \8 f% X9 D5 X+ K) p
; D# Y* l. j& e- dd=
/ r& r/ d- ^4 ]0 ~& C* s - 0 40 60 75 90 90 100 135 80 0
: X: e# ]) ^# M) |( R5 e( p - 40 0 65 40 100 50 75 110 100 405 z2 z- t\" y$ s$ z8 T4 @2 J$ r
- 60 65 0 75 100 100 75 75 75 60
6 x2 g% j! [: e - 75 40 75 0 100 50 90 90 125 75
( R+ q- H\" J# u( i2 B9 o - 90 100 100 100 0 100 75 75 100 90# g @! [. B' R
- 90 50 100 50 100 0 70 90 75 90! F3 l1 T' A$ d/ k) k
- 100 75 75 90 75 70 0 70 100 100, P, P: c; \6 G1 f\" ~
- 135 110 75 90 75 90 70 0 100 135. f6 Y7 c# t3 l) U7 E0 d
- 80 100 75 125 100 75 100 100 0 80# U) t; z6 Z\" i) k) I& w0 ^/ \+ o4 a K' z: W
- 0 40 60 75 90 90 100 135 80 0 ;
9 F4 t5 R% Z% L! z, N9 r - dt= 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 06 A\" h' G7 ], M. d- G$ Z\" E- |. I
- 0.8000 0 1.3000 0.8000 2.0000 1.0000 1.5000 2.2000 2.0000 0.8000
, c* |- l: p6 s+ v' C - 1.2000 1.3000 0 1.5000 2.0000 2.0000 1.5000 1.5000 1.5000 1.20005 e# M: X A: G( X3 P: B
- 1.5000 0.8000 1.5000 0 2.0000 1.0000 1.8000 1.8000 2.5000 1.5000! z) h0 i) s5 |: U& H
- 1.8000 2.0000 2.0000 2.0000 0 2.0000 1.5000 1.5000 2.0000 1.8000
' t o, i0 c, c4 g - 1.8000 1.0000 2.0000 1.0000 2.0000 0 1.4000 1.8000 1.5000 1.80009 b. E7 ~; L$ }* o7 o
- 2.0000 1.5000 1.5000 1.8000 1.5000 1.4000 0 1.4000 2.0000 2.0000
+ F/ u9 h/ k% P& N. z - 2.7000 2.2000 1.5000 1.8000 1.5000 1.8000 1.4000 0 2.0000 2.7000; V3 a/ k+ U9 u; z
- 1.6000 2.0000 1.5000 2.5000 2.0000 1.5000 2.0000 2.0000 0 1.6000
+ w+ o: n# G1 _/ b! d) f. c- I# D/ r8 { - 0 0.8000 1.2000 1.5000 1.8000 1.8000 2.0000 2.7000 1.6000 0;8 \; j% P, a/ |; {) J$ w; Y4 {7 {
- enddata3 o7 n! V) y# X6 g* @# U
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/ M3 M/ v: c) W3 c& A- !目标函数;
) x/ W+ O) K! T$ y - min=@sum(lc(I,J): lucheng(I,J));
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# w- e# Q\" j7 ]8 l' {- !整数约束;
/ y6 E\" J0 F! D; ]% p2 O - @for(plan:@bin(a));
9 O- f9 r9 T4 N9 c - @for(car(I):@free(time(I)) ;
; X4 d& Y- L' [9 |
) {4 u- K: r0 m7 z- !从零点开始最终回来; |% y) h/ ^/ h- e' j* i7 h
- @for(car(I):
. F1 v& J6 R) Z3 Z0 M3 ~ - a(1,1,I)=1;a(10,10,I)=1);, m+ `7 j. X9 N
, s9 Y5 l- ^7 [) D- t/ q- !每个任务只能有一个;$ Q# |4 g5 p {4 |* m2 ]
- @sum(car(I):& ]( b5 e0 ^) ^6 v
- @sum(place(K):% e2 U( v. X, N/ y) E+ z\" g
- @sum(place(J): a(J,K,I))))=1;
\" h* R) ^9 b- }; G M - S6 l. M3 ^1 ? D
- !每个车的任务最多有四个;; M\" U0 \/ i: [- ^, M3 T
- @for(car(I):
( h; x2 u0 N, V\" N- P - @sum(plan(J,K,I):a(J,k,I))<=6);
9 z9 `4 o% t3 t
l+ B. ]( f9 u$ O8 z- !前一辆车任务大于等于后一辆车;7 u1 h0 T7 Q5 @! c$ o. M- p- W; N
- @for(car(I)|I#lt#1:5 E# x) z- C2 d\" k
- @sum(plan:a(J,K,I))>=@sum(plan:a(J,K,I+1) );* X( M5 g# i# l+ U
- . r. V8 c, r' g s
- !求which 例子 若a(1,:)=[1 0 0 1 0 1 0 0 0 1];
: e2 X, z1 w, d3 ?- W0 s - ! 则 which(1,:)=[1 1 1 4 4 6 6 6 6 10] ;
# l' {: w' V* |# B - @for(car(I):
8 I& [7 f: K6 s! I, } - @for(place(kk)|@sum(place(jj):a(kk,jj,I)) #eq#1:
7 H9 q# M3 y8 v# J# ? - @for(place(jj)|a(kk,jj,I) #eq#1:6 I3 I4 l7 e9 w* s
- which(I,kk)=jj ););
/ z6 J! G+ i' U7 }9 K - @for(place(kk)|@sum(place(kk):a(kk,jj,I)) #ne#1:8 j6 U6 {* d\" ?5 n* x4 x7 j
- which(I,kk)=which(I,kk-1));0 y3 a2 T$ n. s, X% Q: K3 b/ N+ r
4 S) b2 i# m$ h8 U- !时间限制; r x9 a* K% C( F8 V; j& |
- @for(car(jjj):' U\" m8 Q% h: n, Z, d7 s# ]
- time(jjj)>=-10;time(jjj)<=7);
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- @for(jiu(jjj):4 C; j* P, C' G0 h
- ct(I,jjj)=@if(which(I,jjj)#ne#which(I,jjj+1),dt(which(I,jjj),which(I,jjj+1))+timexie(which(I,jjj)),0));; y! U7 U) U\" q! U
! M9 {) `% M. y7 o1 c$ n- @for(jiu(jjj):' e4 k9 p3 z* T# k @5 g, D
- time(I)+ct(I,jjj)>=time1(jjj+1);time(I)+ct(I,jjj)<=time2(jjj+1));
* I1 S! C* u; p% \$ l
1 h e# B& v8 k% ?- U1 A- _: a- !求每辆车行驶的距离;
0 }, y' ~4 M7 l3 C- u/ R- `6 X - @for(jiu(jjj): k1 ~2 X1 x' K$ m3 l) q8 D
- lucheng(I,jjj)=@if(which(I,jjj)#ne#which(I,jjj+1),dd(which(I,jjj),which(I,jjj+1)),0) );
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- end\" R1 j l! l6 D$ e0 W# B) i
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