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不支持自定义 Discuz! 代码
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本帖最后由 cqwj 于 2010-4-18 13:47 编辑 " P9 ?+ U R `% I. }- N
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指针处怎么写??6 e; y8 I, Z( M9 D+ ^2 a7 i
我的意思就是说@pointer在模型里面怎么定义?如@pointer(1)=??或 ??=@pointer(1)- f& V- F+ i- O% y) I# g
表示什么意思。我个人理解@pointer(1)=??表示把数模变量的值??装进@pointer(1)这个储存单元,外部程序程序可调用此储存单元数据。??=@pointer(1)表示外部数据值??通过DLL传来的数据放进数模变量,这样理解对吗?( C5 G5 ], q2 ?0 s. W6 e# _3 [
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xinxi:cl190,cl160,cl270,gzsj,sl190,sl160,sl270,sj190,sj160,sj270,k,zys,x,y,z;
: u3 _! q4 ^# `7 f# J% i. rendsets
) `0 D# ~0 u; odata:7 Y& K8 |5 N# ]5 r/ M8 i, I8 D8 p
xinxi,cl190,cl160,cl270,gzsj,sl190,sl160,sl270,sj190,sj160,sj270,k= @file("li_yong.txt"). D5 X2 B+ Y( E) R, g, ?1 e
@TEXT()=@status();
* m9 T" [4 A6 j4 {enddata
3 S% g# {) C j' @' c5 `[bb] min=@sum(xinxi:sj190*x+sj160*y+sj270*z);
/ e6 k/ o" i& f$ x@for(xinxi(i):sj190(i)*x(i)+sj160(i)*y(i)+sj270(i)*z(i)<=12);& e% w( X3 u- v1 L. \1 y
@for(xinxi(i):sj190(i)*x(i)+sj160(i)*y(i)+sj270(i)*z(i)>=0);0 S. l: c, ~$ m0 X+ i
@sum(xinxi:x)=sr190;* i% n; e6 o2 j2 c4 P
@sum(xinxi:y)=sr160;5 _- t2 l+ b, K: q
@sum(xinxi:z)=sr270;/ ~- E; G! U9 r& w* @
@sum(xinxi:x)>=0;
9 v" V, a+ x. e6 v. w/ S f' b@sum(xinxi:y)>=0;( I0 n& _: _4 l, y# n% i
@sum(xinxi:z)>=0;0 Z4 L9 I/ z4 F9 B4 x |" z
@for(xinxi(i):x(i)<=cl190(i)*sl190(i));
; h1 P" z3 b4 u; U" i0 W1 X# J@for(xinxi(i):x(i)>=0);$ q( A8 J- r" ?$ u1 g; v- k
@for(xinxi(i):y(i)<=cl160(i)*sl160(i));/ e, P8 @/ z$ G0 N/ K& P
@for(xinxi(i):y(i)>=0);
( Z3 `4 Z" r/ K@for(xinxi(i):z(i)<=cl270(i)*sl270(i));0 q. ^/ X; W, Q) D. a) f
@for(xinxi(i):z(i)>=0);& m t6 l5 f: V" ^- z- `5 W
@for(xinxi(i):@gin(x(i)));; j% t. h6 b. w6 Z
@for(xinxi(i):@gin(y(i)));0 a* \3 s3 ~% O' t
@for(xinxi(i):@gin(z(i)));# t3 _5 l1 S7 P9 I( U6 ~
mm=@sum(xinxi:k);
: p; o- z: @: g# G* yxx=@sum(xinxi:sj190*x+sj160*y+sj270*z)/mm;4 L4 B9 K9 O7 {3 N3 u
1 t/ |6 s7 C$ e@for(xinxi(i):zys(i)=sj190(i)*x(i)+sj160(i)*y(i)+sj270(i)*z(i));
' e0 _" P/ c5 j( iDATA:/ \; A" \( X; @( r1 O' f3 p! }- ~
@POINTER
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' d. r7 @9 w- j" e5 v; M!此处结果想连接VB6.0,却不知道怎么写;
" T3 t2 ~; h. q2 B9 cENDDATA9 X) B; c( y+ H" T- W9 ]
- k" o, B9 a8 v4 t' Dend0 c2 c2 `# b: s; y/ m
结果为 主要是想取 X,Y,Z的结果1 c# P) R1 ?, `) Y, ~# o
Global optimal solution found.
0 V u! P6 H( P Objective value: 0.000000 ~4 j. k% w' O, r: x' k m' _5 O# ?( {
Extended solver steps: 05 q5 y- ?8 d9 Z; x7 `9 p, E7 q. W
Total solver iterations: 07 N, t4 h, ] O9 q
2 S# K/ P4 \3 d" U3 c
" }3 X7 s7 Z/ N% B
0
9 S' W" a% {6 W1 |* ]
% N( t3 ?, s5 s Variable Value Reduced Cost
, p. x7 S" V7 X! o" e1 Z/ B SR190 0.000000 0.000000
% f2 E. S) g* X L SR160 0.000000 0.000000
) w8 t6 G f# R+ B5 H, N! y SR270 0.000000 0.0000004 ~6 [4 _8 C5 y: R4 N/ E
MM 2.000000 0.0000000 W m! U) q3 k8 z, ?
XX 0.000000 0.000000
: l0 [/ g# l& O CL190( ZHANG) 35.00000 0.000000
5 H3 ~$ _: D8 _; W4 F! I+ e CL190( LI) 35.00000 0.000000- I2 G5 r# Y" C; ~* j
CL160( ZHANG) 30.00000 0.000000
5 ?& @/ t; d8 A6 A$ B% P: o CL160( LI) 30.00000 0.0000004 B" M; S+ i0 e7 W6 ]4 B
CL270( ZHANG) 5.000000 0.0000000 d7 o' i! u) `2 I8 C) ?2 a
CL270( LI) 5.000000 0.000000
2 N5 Z s$ ^+ P1 a; ]4 k5 h GZSJ( ZHANG) 12.00000 0.000000
- ^. {1 g( @2 z5 b2 V GZSJ( LI) 12.00000 0.000000' \% ]; U0 X( }) F: @* V6 r4 R
SL190( ZHANG) 0.9000000 0.000000
$ l& H. \4 Y3 x4 F* T/ {( ?' q# ? `( G SL190( LI) 0.8000000 0.000000
8 m# g) J# B9 |) H+ s. o; |. n# `: O SL160( ZHANG) 0.9000000 0.000000
" F3 N/ w1 {1 X SL160( LI) 0.9000000 0.000000; ^, D4 \( ?+ E% B
SL270( ZHANG) 0.8000000 0.000000
. G& V! `. C+ W3 A! J; e SL270( LI) 0.6000000 0.000000
3 P- W: {2 ~7 ^1 O4 Y SJ190( ZHANG) 0.3770000 0.000000+ }% `+ }! j, L( x# y7 n+ Q% Z4 h7 G
SJ190( LI) 0.4120000 0.000000
1 z# C- J a3 }9 F0 X SJ160( ZHANG) 0.4400000 0.000000
) r0 [: F. p/ b% g( I: K. ` SJ160( LI) 0.4400000 0.0000000 }$ D# z$ z" y3 ^" G+ F! f
SJ270( ZHANG) 2.880000 0.000000; A( t' p, ] T8 @4 ?1 W) p8 ^8 l
SJ270( LI) 3.360000 0.0000004 ?1 _7 O& b3 f/ ?
K( ZHANG) 1.000000 0.0000005 Y. [9 y. J& D5 U, ]
K( LI) 1.000000 0.000000
/ [6 H7 R8 i0 l4 D% C# W$ F ZYS( ZHANG) 0.000000 0.000000: E+ E: \/ [$ ]% \, f: a# N1 ^
ZYS( LI) 0.000000 0.000000
- T$ B) Q6 b, |0 Q X( ZHANG) 0.000000 0.37700001 G5 g" D' B" E8 O4 _5 m
X( LI) 0.000000 0.4120000
q" B+ G3 r$ V8 n" \- f Y( ZHANG) 0.000000 0.4400000* N$ V ^- B; d! `
Y( LI) 0.000000 0.4400000
% \0 a' G. n# v3 h! u4 }! b Z( ZHANG) 0.000000 2.880000
+ w# C l+ ~1 l [" U* y, u Z( LI) 0.000000 3.360000! `3 y" }( R4 W% |
" g, O! }( |9 y/ Z Row Slack or Surplus Dual Price
, \) G x" T. s: _ BB 0.000000 -1.000000 h: F; B6 E7 e; r4 P' H: u1 H
2 12.00000 0.0000001 g% {! _1 M$ n8 |' }1 {! F
3 12.00000 0.000000
k r% `: J/ T% z) H8 Q 4 0.000000 0.000000
# g+ ~ N: ?: w) ? {- i4 {; g- v 5 0.000000 0.000000. d$ @6 z- i3 p" H
6 0.000000 0.000000 S, W) D$ G; i1 z
7 0.000000 0.000000/ H6 y0 e8 K( s/ ?5 t* J. z+ `
8 0.000000 0.000000 G$ A. _6 ]6 g. d
9 0.000000 0.000000
6 U$ Z) N' P2 r3 Q* ?8 S) f& B 10 0.000000 0.000000: _) _8 I1 p$ [7 d5 d1 }+ Y) S3 K
11 0.000000 0.000000
0 c& v1 ?3 i6 i3 ^4 m9 Q 12 31.50000 0.000000: Z* t! j/ c$ R0 I8 y
13 28.00000 0.000000
' B! C- }# K8 X2 m 14 0.000000 0.000000- l7 p9 c# o( s2 ?9 R( K0 ~
15 0.000000 0.000000
' y! z1 [) v* Z0 Q: J 16 27.00000 0.0000001 p( _+ f) k2 u2 [5 [$ E# W
17 27.00000 0.0000004 z" N% K/ T5 e/ d5 C) E
18 0.000000 0.000000. W- k1 k% ]* P. u1 M$ Z/ O
19 0.000000 0.000000$ _. C" p9 L. K5 S$ I/ R8 g7 ?
20 4.000000 0.000000
Y" A& T& |) Z3 h6 ?2 D I" ~ 21 3.000000 0.000000" U1 W9 _: h1 ?- U
22 0.000000 0.000000+ A6 s9 Y& g3 k" n# P5 C$ r8 A6 {# ?
23 0.000000 0.000000
' u4 o# s! D& d% l; Z( s# S 24 0.000000 0.0000003 [% ?8 w/ E8 ?- K1 M" B& e A8 j
25 0.000000 0.000000
- v- t$ b7 g! v" W 26 0.000000 0.0000000 B }/ V9 y9 i
27 0.000000 0.000000 |
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