BookmarkSubscribeRSS Feed
🔒 This topic is solved and locked. Need further help from the community? Please sign in and ask a new question.
jesperph
Calcite | Level 5

Hello. I'm really new to SAS Studio programming and I have a question about my code.

 

I have a dataset consisting of numbers regarding different products. I have in total 31 products with different values of the numbers stated in [box] (c1, D, cap1) etc. The thing is that I want to loop the minimalization object function Z for all 31 products so that they don't affect each other's values of the variable Q and Z.

 

Which it seems like they do now since I'm not doing a loop for one product at the time. But I just can't figure out how I'm going to do it.

 

proc import out=indata 
 DATAFILE='/home/phjesper0/Master/Minresultater1.xlsx'
 DBMS=XLSX replace;
 GETNAMES=YES;
 /*USEDATE=YES;*/

title 'Vanlige tall';
proc print data=indata;
run;

Proc optmodel;
   set <str> box;
   var Q{box} >=0;
   number   c1{box};
   number    D{box};
   number cap1{box};
   number   r1{box};
   number   r2{box};
   number   o1{box};
   number    f{box};
   number   o2{box};
   number   c2{box};
   number   p1{box};
   number   p2{box};
   number cap2{box};
   number life{box};
   number    w{box};

read data indata into box=[type] c1 D cap1 r1 r2 o1 f o2 c2 p1 p2 cap2 life w;
   print c1 D cap1 r1 r2 o1 f o2 c2 p1 p2 cap2 life w;

minimize Z = sum {i in box} (((Q[i]/2)*((c1[i]*r1[i])+(c2[i]*r2[i]*f[i])))
+(D[i]/Q[i])*(((o1[i]*p1[i]) + (o2[i]*p2[i]*f[i]))*300 + (w[i]*c1[i])));

con MinShelfLife{i in box}:(life[i]-((360)/(D[i]/Q[i]))) >= 120;
con MaxShiftSize1{i in box}: Q[i]/cap1[i] <= 2;
con MaxShiftSize2{i in box}: Q[i]/(f[i]*cap2[i]) <= 2;
con MinBatch{i in box}: Q[i] >= 0;
solve;
print Z Q;
quit;

Thanks!

1 ACCEPTED SOLUTION

Accepted Solutions
RobPratt
SAS Super FREQ

If you want to solve a separate problem for each i in box, you can do the following:

Proc optmodel;
   set <str> box;
   var Q >= 0;
   number   c1{box};
   number    D{box};
   number cap1{box};
   number   r1{box};
   number   r2{box};
   number   o1{box};
   number    f{box};
   number   o2{box};
   number   c2{box};
   number   p1{box};
   number   p2{box};
   number cap2{box};
   number life{box};
   number    w{box};

   read data indata into box=[type] c1 D cap1 r1 r2 o1 f o2 c2 p1 p2 cap2 life w;
   print c1 D cap1 r1 r2 o1 f o2 c2 p1 p2 cap2 life w;

   str i_this;
   minimize Z = ((Q/2)*((c1[i_this]*r1[i_this])+(c2[i_this]*r2[i_this]*f[i_this])))
   +(D[i_this]/Q)*(((o1[i_this]*p1[i_this]) + (o2[i_this]*p2[i_this]*f[i_this]))*300 + (w[i_this]*c1[i_this]));

   con MinShelfLife:(life[i_this]-((360)/(D[i_this]/Q))) >= 120;
   con MaxShiftSize1: Q/cap1[i_this] <= 2;
   con MaxShiftSize2: Q/(f[i_this]*cap2[i_this]) <= 2;

   num Qsol {box};
   num Zsol {box};
   for {i in box} do;
      i_this = i;
      solve;
      print Q Z;
      Qsol[i] = Q;
      Zsol[i] = Z;
   end;
   print Qsol Zsol;
quit;

Note that I removed the MinBatch constraint, which is redundant because of the >= 0 in the VAR statement.

 

By the way, you can also solve these independent problems concurrently by changing FOR to COFOR.

 

For a similar example, see Efficiency Analysis: How to Use Data Envelopment Analysis to Compare Efficiencies of Garages.

View solution in original post

2 REPLIES 2
RobPratt
SAS Super FREQ

If you want to solve a separate problem for each i in box, you can do the following:

Proc optmodel;
   set <str> box;
   var Q >= 0;
   number   c1{box};
   number    D{box};
   number cap1{box};
   number   r1{box};
   number   r2{box};
   number   o1{box};
   number    f{box};
   number   o2{box};
   number   c2{box};
   number   p1{box};
   number   p2{box};
   number cap2{box};
   number life{box};
   number    w{box};

   read data indata into box=[type] c1 D cap1 r1 r2 o1 f o2 c2 p1 p2 cap2 life w;
   print c1 D cap1 r1 r2 o1 f o2 c2 p1 p2 cap2 life w;

   str i_this;
   minimize Z = ((Q/2)*((c1[i_this]*r1[i_this])+(c2[i_this]*r2[i_this]*f[i_this])))
   +(D[i_this]/Q)*(((o1[i_this]*p1[i_this]) + (o2[i_this]*p2[i_this]*f[i_this]))*300 + (w[i_this]*c1[i_this]));

   con MinShelfLife:(life[i_this]-((360)/(D[i_this]/Q))) >= 120;
   con MaxShiftSize1: Q/cap1[i_this] <= 2;
   con MaxShiftSize2: Q/(f[i_this]*cap2[i_this]) <= 2;

   num Qsol {box};
   num Zsol {box};
   for {i in box} do;
      i_this = i;
      solve;
      print Q Z;
      Qsol[i] = Q;
      Zsol[i] = Z;
   end;
   print Qsol Zsol;
quit;

Note that I removed the MinBatch constraint, which is redundant because of the >= 0 in the VAR statement.

 

By the way, you can also solve these independent problems concurrently by changing FOR to COFOR.

 

For a similar example, see Efficiency Analysis: How to Use Data Envelopment Analysis to Compare Efficiencies of Garages.

jesperph
Calcite | Level 5
Thank you so much sir!
Worked well.

That was quick and super helpful 🙂

- Jesper

SAS Innovate 2025: Register Now

Registration is now open for SAS Innovate 2025 , our biggest and most exciting global event of the year! Join us in Orlando, FL, May 6-9.
Sign up by Dec. 31 to get the 2024 rate of just $495.
Register now!

Multiple Linear Regression in SAS

Learn how to run multiple linear regression models with and without interactions, presented by SAS user Alex Chaplin.

Find more tutorials on the SAS Users YouTube channel.

Discussion stats
  • 2 replies
  • 1190 views
  • 0 likes
  • 2 in conversation