Un problema de investigación de operaciones: problema de inversión, 2 millones de dólares, 4 opciones.

Clear@"`*";

vars = Flatten@{x[1, #] & /@ Range@5, x[2, #] & /@ Rango@4, x[3, 3],

x[4, 2]};

con0 = # >= 0 & /@ vars;

con1 = {x[2, #] <= 30 & /@ Rango@4, x[3, 3] <= 80, x[4, 2] <= 100};

con2 = {

x[1, 1] + x[2, 1] & lt;= 200,

x[1, 2] + x[2 , 2 ] + x[4, 2] <= 200 + 0,1 x[1, 1] + 0,25 x[2, 1],

x[1, 3] + x[2, 3] + x[3, 3] <=

200 + 0,1 (x[1, 1] + x[1, 2]) + 0,25 (x[2, 1] + x[2, 2] ),

x[1, 4] + x[2, 4] <=

200 + 0,1 (x[1, 1] + x[1, 2] + x [1 , 3]) +

0,25 (x[2, 1] + x[2, 2] + x[2, 3]),

x[1, 5 ] < =

200 + 0,1 (x[2, 1] + x[1, 2] + x[1, 2] + x[2, 3]),

x[ 1, 5] <=

200 + 0.1 (x[1, 1] + x[1, 2] + x[1, 3] + x[1, 4]) + < / p>

0.25 (x[2, 1] + x[2, 2] + x[2, 3] + x[2, 4])};

contras = Aplanar@{ con0 , con1, con2};

obj = 200 + 0.1 Total[x[1, #] & /@ Rango@5] +

0.

0,25 Total[x[2, #] & /@ Rango@4] + 0,4 x[3, #] & /@ Rango@4+ 0,4 x[3, 3] + 0,55 x[2, 4];

Maximizar[obj, cons. vars]

El resultado es:

{383.89495, {x[1, 1] -> 170., x[1, 2] -> 194,5, x[1, 3] -> 141,45,

x[1, 4] -> 243,095, x[1, 5] ->. -> 30.,

x[2, 2] -> 30., x[2, 3] -> 30., x[2, 4] -> 30., x[3, 3 ] - > 80.,

x[4, 2] -> 0.}}

.