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Kishore_Kumar
Fluorite | Level 6

Hi All,

 

I am pretty new to using SAS OR module, I have a very simple problem which can easily be solved using excel solver. However I want to do this exercise using Proc OPTMODEL provided in SAS OR. In my case, I have three decision variables whose values needs to be changed by maximizing the objective function based on some constraints. I have listed down constraints, objective function & decision variables below:-

Constraint 1: All the three decision variables(Var1,Var2 & Var3) have to be binary in nature.

Constraint 2: SUM of all the three decision variables (Total column) should be equal to 1 for each row. (e.g any one of the decision variable can take value as 1 rest decision vars have to be 0)

Maximize Objective function:  SUMPRODUCT(Total, Price)/SUM(Total) <= 100

I have pasted below the sample table.

 

S.No.Var1Var2Var3TotalPrice
1010130
2100177
3100199
4001126
5010149

 

Can anybody help me with this?

 

 

Thanks!

Kishore

1 ACCEPTED SOLUTION

Accepted Solutions
RobPratt
SAS Super FREQ

Yes, that is clearer.  You do have 15 binary variables.  I think the following does what you want, although I have changed 100 to 10 because otherwise the first constraint would be redundant and the optimal solution would be trivial.

 

data indata;
   input S Var4 Var5 Var6 Price;
   datalines;
1	0.2000	0.2500	0.9900	30
2	0.1000	0.2000	0.9000	77
3	0.1500	0.2500	0.9500	99
4	0.2500	0.3500	0.9500	26
5	0.3500	0.4500	0.9900	49
;

proc optmodel;
   set SSET;
   set JSET = 1..3;
   num a {SSET, JSET};
   num price {SSET};
   read data indata into SSET=[s] {j in JSET} <a[s,j]=col('Var'||(j+3))> price;
   print a price;

   var X {SSET, JSET} binary;
   impvar NumPeople {s in SSET} = sum {j in JSET} j*X[s,j];
   max Objective = (sum {s in SSET, j in JSET} price[s]*a[s,j]*X[s,j]) / (sum {s in SSET} price[s]);
   con TotalPeople:
      sum {s in SSET} NumPeople[s] <= 10;
   con OnceChoice {s in SSET}:
      sum {j in JSET} X[s,j] = 1;

   solve;
   print X NumPeople;
quit;

SAS Output

X
  1 2 3
1 1 0 0
2 0 0 1
3 0 0 1
4 1 0 0
5 0 1 0

[1] NumPeople
1 1
2 3
3 3
4 1
5 2

View solution in original post

7 REPLIES 7
RobPratt
SAS Super FREQ

I think something is missing in your description.  How do the decision variables relate to the objective function?  What is the <= 100?

 

What do the rows in the table represent?  Do you maybe have 15 decision variables and not just 3?

 

What is the optimal solution returned by Excel?  And what is the corresponding objective value?

Kishore_Kumar
Fluorite | Level 6

Hi Rob,

 

Thanks for your quick reply, Indeed I forgot to mention lot of things. Apologies for creating confusion. Here is the updated table mentioned below:-

 

 123       

S.No.

Var1Var2Var3Var4Var5Var6SUM_PNo_of_PeopleTotalPrice
10100.20000.25000.99000.25002130
21000.10000.20000.90000.10001177
31000.15000.25000.95000.15001199
40010.25000.35000.95000.95003126
50100.35000.45000.99000.45002149

 

So column "No_of_people" is the sumproduct of var1,var2 & var3 with values 1,2 & 3. SUM_P is also the sumproduct between va1 , var2, var3 & var4,var5,var6. Now I mentioned about <=100, that is one of my constraint for column "No_of_people". So sum(No_of_people) <=100.

 

I have mentioned corrected objective function below:-

Maximize Objective function:  SUMPRODUCT(SUM_P, Price)/SUM(Price)

Constraint 1:- sum(No_of_people) <=100

Constraint 2: SUM of all the three decision variables (Total column) should be equal to 1 for each row. (e.g any one of the decision variable can take value as 1 rest decision vars have to be 0)

Constraint 3: Var1,Var2 & Var3 can take only binary values.

Var4,Var5 & Var6 are the constant values.

 

I hope this time it is clear that objective functions depends on SUM_P which is the sumproduct between var1,var2,var3 & var4,var5,var6.

 

 

Thanks!

Kishore

 

 

RobPratt
SAS Super FREQ

Yes, that is clearer.  You do have 15 binary variables.  I think the following does what you want, although I have changed 100 to 10 because otherwise the first constraint would be redundant and the optimal solution would be trivial.

 

data indata;
   input S Var4 Var5 Var6 Price;
   datalines;
1	0.2000	0.2500	0.9900	30
2	0.1000	0.2000	0.9000	77
3	0.1500	0.2500	0.9500	99
4	0.2500	0.3500	0.9500	26
5	0.3500	0.4500	0.9900	49
;

proc optmodel;
   set SSET;
   set JSET = 1..3;
   num a {SSET, JSET};
   num price {SSET};
   read data indata into SSET=[s] {j in JSET} <a[s,j]=col('Var'||(j+3))> price;
   print a price;

   var X {SSET, JSET} binary;
   impvar NumPeople {s in SSET} = sum {j in JSET} j*X[s,j];
   max Objective = (sum {s in SSET, j in JSET} price[s]*a[s,j]*X[s,j]) / (sum {s in SSET} price[s]);
   con TotalPeople:
      sum {s in SSET} NumPeople[s] <= 10;
   con OnceChoice {s in SSET}:
      sum {j in JSET} X[s,j] = 1;

   solve;
   print X NumPeople;
quit;

SAS Output

X
  1 2 3
1 1 0 0
2 0 0 1
3 0 0 1
4 1 0 0
5 0 1 0

[1] NumPeople
1 1
2 3
3 3
4 1
5 2
Kishore_Kumar
Fluorite | Level 6

Hi Rob,

Just a follow up question, If I have to create an output datsest for variable X then How do I use "Create data" statement to do the same? I was able to create dataset for variable "NumPeople", however wasn't able to do it for variable "X".

 

I used this syntax " Create dataset output_model from [s] [NumPeople]";

 

 

 

Thanks!

Kishore

 

RobPratt
SAS Super FREQ

That statement is not correct and will generate errors.  The correct syntax is:

   Create data output_model from [s] NumPeople;

To create a data set for X, you can mimic the READ DATA statement from earlier, as follows:

   create data output_x from [s]=SSET {j in JSET} <col('Var'||(j+3))=X[s,j]>;

You can find lots of variations of both READ DATA and CREATE DATA in this book of examples

Kishore_Kumar
Fluorite | Level 6
Once again thank you so much for your help, I really appreciate it.

Thanks again!
Kishore
Kishore_Kumar
Fluorite | Level 6
Thank you so much Rob, that's what I have been looking for. I really appreciate your help on this.
Once again thanks a ton!

Kishore

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