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我正在尝试使用 AMPL 来优化一些返回数据的日志实用程序功能。程序正确读取数据但吐出错误

LOQO 6.07：LOQO ERROR(50)：无法在初始解决方案中评估 obj 和/或约束

我的代码很简单：

set ASSETS; # asset names
set TIME; # dates

param R {TIME, ASSETS}; # return of asset A in month T
param w0:= 100; # initial wealth

var x{ASSETS} >=0; # amt of capital in each asset

maximize expected_utility:
    sum {i in TIME} (1/card(TIME)) * log(sum {j in ASSETS} x[j]*(1+R[i,j])); # x*(1+r) = w

subject to tot_mass:
    sum{j in ASSETS} x[j] <= w0; # amount in each asset must total to w0

data;

set ASSETS := VT SPY TLT EEM EFA GSG DBC DBV TIP;
set TIME := January-07 February-07 March-07 April-07 May-07 June-07 July-07 August-07 September-07 October-07 November-07 December-07 January-08 February-08 March-08 April-08 May-08 June-08;

param R: VT SPY TLT EEM EFA GSG DBC DBV TIP :=

January-07  0.01248713  0.015065017 -0.009943182    0.001149425 0.013974909 -0.025455453    -0.023718763    0.007237636 0.002407806
February-07 -0.013271444    -0.019606312    0.034002869 -0.032433984    -0.001252937    0.037131882 0.050759219 0.008383234 0.019721871
March-07    0.025685366 0.011552864 -0.017066741    0.053396618 0.029010507 0.024691358 0.006606111 0.023357086 0.003347384
April-07    0.040152115 0.044265911 0.009034444 0.037172627 0.037488571 0.004578313 0.007383101 0.021276596 0.006548869
May-07      0.031768575 0.033941771 -0.02308338 0.050773826 0.023501763 -0.011753418    -0.002035831    0.023863636 -0.013012521
June-07     -0.001752913    -0.01459002 -0.01016755 0.036692506 -0.00315729 0.03131068  0.007343941 0.027376989 -0.002238806
July-07     -0.021742556    -0.031314776    0.033130787 0.007228315 -0.022890872    0.047775947 0.019036047 -0.007922218    0.023061581
August-07   0.003843633 0.012839129 0.01792466  0.010393467 -0.006335642    -0.031895777    -0.019077901    -0.039564428    0.008163763
September-07    0.056778104 0.038708217 0.002338699 0.115601274 0.053232503 0.094431555 0.094003241 0.030234316 0.012811216
October-07  0.042289615 0.013511550 0.018116937 0.118770582 0.042376461 0.05999576  0.085185185 0.029713866 0.011694511
November-07 -0.043249736    -0.038704128    0.053518469 -0.076530612    -0.03619665 -0.00460000     -0.006143345    -0.034912718    0.041165369
December-07 -0.020039022    -0.011258574    -0.006269994    -0.014024649    -0.029848655    0.059674503 0.066964286 0.00036914  0.000566444
January-08  -0.07287025 -0.060478094    0.020988926 -0.089224138    -0.078434205    -0.004550626    0.029288703 -0.016605166    0.035665761
February-08 -0.011941005    -0.025844771    -0.004540295    0.019876952 -0.010188088    0.112571429 0.111319575 -0.007129456    0.01421231
March-08    -0.007344683    -0.00889841 0.021284683 -0.037587007    0.004117181 -0.009416196    -0.009285312    -0.041950113    -0.001401315
April-08    0.057279035 0.047634882 -0.024810818    0.091610415 0.054407822 0.08192188  0.060494178 0.038658777 -0.019753886
May-08      0.016184106 0.015156324 -0.026841369    0.031581272 0.011965301 0.084664537 0.069094804 0.017850361 0.0011012
June-08     -0.096123568    -0.08363949 0.026535948 -0.102975808    -0.104936447    0.103240059 0.104208417 -0.005597015    0.017929821
;

#display(ASSETS);
#display(TIME);
#display R;

solve;

display {i in ASSETS} x[i];

有任何想法吗？

再次感谢stackoverflow

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1 回答 1

2

由于您没有为变量 x[j] 指定初始值，因此默认情况下它们都被赋予初始值 0。因此，您的目标函数涉及 0 的对数，而 LOQO 无法评估。尝试设置显式初始值，例如

var x{ASSETS} >=0, := w0 / card(ASSETS);

在这里，所有变量都以相等的总资产份额进行初始化，这可能是一个合理的开始。

于 2012-10-11T12:49:02.270 回答