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Mixed integer Linear Programing

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Mixed integer Linear Programing

Hello

 

I am getting optimal feasible solution as 31250 with decision variable values 0, 7.5 and 20.

 

proc optmodel;
var x{1..3}>=0;
max z=900*x[1]+1500*x[2]+1000*x[3];
con c1: 50000*x[1]+12000*x[2]+8000*x[3]<=250000;
con c2: 1*x[1]+0*x[2]+0*x[3]<=4;
con c3: 0*x[1]+1*x[2]+0*x[3]<=15;
con c4: 0*x[1]+0*x[2]+1*x[3]<=20;
solve with milp/presolver=none;
print x[1] x[2] x[3];
quit;

 

x1,x3>=0 or integer

x2>=0.

But in excel it is given as 42,250 optimal feasible solution with decision variable values 4, 4.16,0. Please help if I am going wrong.


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Solution
‎12-04-2016 10:08 PM
SAS Employee
Posts: 414

Re: Mixed integer Linear Programing

You have not specified any integer variables.  Use the INTEGER option in the VAR statement and then the .RELAX suffix to relax x[2].  Also, you need not enter the zero coefficients explicitly.  In fact, you can avoid those last three constraints by instead using the .UB variable suffix:

 

proc optmodel;
   var x{1..3}>=0 integer;
   x[2].relax = 1;
   max z=900*x[1]+1500*x[2]+1000*x[3];
   con c1: 50000*x[1]+12000*x[2]+8000*x[3]<=250000;
   x[1].ub = 4;
   x[2].ub = 15;
   x[3].ub = 20;
   solve with milp/presolver=none;
   print x;
quit;

 

If you change the objective coefficient for x[1] from 900 to 9000, you do get an optimal objective value of 42,250.

View solution in original post


All Replies
Solution
‎12-04-2016 10:08 PM
SAS Employee
Posts: 414

Re: Mixed integer Linear Programing

You have not specified any integer variables.  Use the INTEGER option in the VAR statement and then the .RELAX suffix to relax x[2].  Also, you need not enter the zero coefficients explicitly.  In fact, you can avoid those last three constraints by instead using the .UB variable suffix:

 

proc optmodel;
   var x{1..3}>=0 integer;
   x[2].relax = 1;
   max z=900*x[1]+1500*x[2]+1000*x[3];
   con c1: 50000*x[1]+12000*x[2]+8000*x[3]<=250000;
   x[1].ub = 4;
   x[2].ub = 15;
   x[3].ub = 20;
   solve with milp/presolver=none;
   print x;
quit;

 

If you change the objective coefficient for x[1] from 900 to 9000, you do get an optimal objective value of 42,250.

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