lingo模型在评价版显示超限制，但在12 API破解版，请大家帮忙分析！

hubq_pro

  LINGO/WIN32 12.0.2.21 (20 Oct 10)

  LINDO API 6.1.1.515 (Oct 20 2010 20:14:13)

  Copyright (C) 2005-2009 LINDO Systems Inc. Licensed material,
  all rights reserved. Copying except as authorized in license
  agreement is prohibited.

  License location: F:\TDDownload\LINDO Document\lingo12破解\lndlng12.lic
  License location: F:\TDDownload\LINDO Document\lingo12破解\lndlng12.lic

  Eval Use Only

  License expires: 23 Nov 2029

  Licensed for commercial use.
  Branch-and-bound solver enabled.
  Nonlinear solver enabled.
  Barrier solver enabled.
  Global solver enabled.
  Integer solver enabled.
  Stochastic solver enabled.


   Parameter        Old Value     New Value
    ECHOIN             0             1

:  take EnergyPlan.lng
: model:
? Title EnergyPlanModel By HuBingQing in 20140720;
? sets:
?    District/@FILE('cDistrict.LDT')/: GDP0, GRateMin, GRateMax, EPG0, EPGRateMin, EP
? endsets
? data:
?    T = 8;
?    TGDP = 98500;
?    TEPG = 0.75;
?    ! GDP0;
?    GDP0 = @pointer(1);
?    ! Min Increase Rate of GDP;
?    GRateMin = @pointer(2);
?    ! Max Increase Rate of GDP;
?    GRateMax = @pointer(3);
?    ! EPG0=Energy Consumption Percent GDP in Base
?    EPG0 = @pointer(4);
?    ! Min Decrease Rate of EPG;
?    EPGRateMin = @pointer(5);
?    ! Max Decrease Rate of EPG;
?    EPGRateMax = @pointer(6);
?    ! Forcast Years
?    T = @pointer(7);
?    ! Total GDP Lower Demand
?    TGDP = @pointer(8);
?    ! Total GDP Lower Demand
?    TEPG = @pointer(9);
?    !Output File
?    @Text('InputData.txt')=GDP0,GRateMin,GRateMax,EPG0,EPGRateMin,EPGRateMax,T,TGDP,
? enddata
?    !Real EPGt;
?    !@for (District(i) : EPGt(i) = EPG0(i) * ( 1-EPGRate(i) ) );
?    !Real GDPt;
?    !@for (District(i) : GDPt(i) = GDP0(i) * (1 + GRate(i))^T);
?    !Real Energy Consumption Now
?    !@for (District(i) : E0(i) = GDP0(i)*EPG0(i));
?    !Real Energy Consumption
?    !@for (District(i) : Et(i) = GDPt(i)*EPGt(i));
?    !Minimize Total Energy Consumption;
?    [rObj] min = @sum(District(i) : (GDP0(i) * (1 + GRate(i))^T)*(EPG0(i) * (1-EPGRa
?    !TE = @Sum (District(i) : Et(i));
?    ![rObj] min = TE;
?    !GDP Increase limit for Minimun or Maximun Rate of GDP;
?    !@for (District(i) : GRate(i) >= GRateMin(i) , GRate(i) <= GRateMax(i));
?    @for(District(i) : @bnd(GRateMin(i),GRate(i),GRateMax(i)));
?    !EPG Decrease limit for Minimun or Maximun Rate of EPG;
?    !@for (District(i) : EPGRate(i) >= EPGRateMin(i) , EPGRate(i) <= EPGRateMax(i));
?    @for(District(i) : @bnd(EPGRateMin(i),EPGRate(i),EPGRateMax(i)));
?    !Real GDPt;
?    @for(District(i) : GDPt(i) = GDP0(i) * (1 + GRate(i))^T);
?    !Real EPGt;
?    @for(District(i) : EPGt(i) = EPG0(i) * (1-EPGRate(i)));
?    !Real Energy Consumption
?    @for(District(i) : Et(i) = GDPt(i)*EPGt(i));
?    !Region Total GDP Lower Demand;
?    CurTGDP = @sum (District(i) : GDPt(i));
?    CurTGDP >= TGDP;
?    !Region Total EPG Higher Demand
?    TE = @sum (District(i) : Et(i));
?    CurTEPG = TE/CurTGDP ;
?    !CurTEPG >= TEPG;
? data:
?    @pointer(10) = GDPt;
?    @pointer(11) = EPGt;
?    @pointer(12) = Et;
?    @pointer(13) = GRate;
?    @pointer(14) = EPGRate;
?    @pointer(15) = rObj;
?    @pointer(16) = @status();
?    @pointer(17) = CurTGDP;
?    @pointer(18) = CurTEPG;
? enddata
? end
:  go
  Compiling model ...
  Structural analysis, pass 1 ...
  Reading API parameters from: LINDO.PAR
  Scalarizing model ...
  Generating nonzero matrix ...
  Solving ...

[Error Code:  191]

  The LINDO API returned the following error code:        2026
  LINDO API routine: LSsolve
  License is too small for the given problem.



[Error Code:  193]

  A solution is not available for this model.

  Running output operations ...

:  quit

hubq_pro

没有人遇到过2026错误？请高手帮助！

wujianjack2

    LINDO API 12破解不完全，LINGO 11 Crack对应的API也不能调用，Team zwt的LINGO 9好像也有问题，你试试LINGO 10看看。主要是我不怎么熟悉C这些编程语言，LINGO使用这块关于API的调用了解得比较少，连用个mxLINDO我都差点看吐，也没有什么时间去静下去看，你要是急的话告诉我如何调用，我有真正无限制的API，可以帮你试试，不过不能分享。