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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: 538

Re: Mixed integer Linear Programing

Posted in reply to KafeelBasha

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: 538

Re: Mixed integer Linear Programing

Posted in reply to KafeelBasha

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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