BookmarkSubscribeRSS Feed
🔒 This topic is solved and locked. Need further help from the community? Please sign in and ask a new question.
usamasm10
Calcite | Level 5
proc optmodel;

var xc1w1, xc1w2, xt1w1, xt1w2, xfw1, xfw2, xc2w1, xc2w2, xt2w1, xt2w2 >=0 integer;

var yc2 binary;
var yt2 binary;

maximize z=82*xc1w1+83*xc1w2+45*xt1w1+49*xt1w2+30*xfw1+35*xfw2+84*xc2w1+81*xc2w2+42*xt2w1+54*xt2w2+41000*yc2+21700*yt2;

con Demand1: xc1w1+xc2w1+xt1w1+xt2w1+xfw1=6000;
con Demand2: xc1w2+xc2w2+xt1w2+xt2w2+xfw2=5200;

con china1: xc1w1+xc1w2<=1500;
con china2: xc2w1+xc2w2<=2000;
con turkey1: xt1w1+xt1w2<=1800;
con turkey2: xt2w1+xt2w2<=2200;
con france: xfw1+xfw2<=6000;

con linkc2: xc2w1+xc2w2-2000*yc2<=0;
con linkt2: xt2w1+xt2w2-2200*yc2<=0;

con Fixc2: xc2w1+xc2w2<=1000000*yc2;
con Fixt2: xt2w1+xt2w2<=1000000*yt2;

con repurpose: yc2+yt2<=1;

solve;

print xc1w1 xc1w2 xt1w1 xt1w2 xfw1 xfw2 xc2w1 xc2w2 xt2w1 xt2w2 z;

quit;
1 ACCEPTED SOLUTION

Accepted Solutions
RobPratt
SAS Super FREQ

I suspect that you meant for all the x* variables to be >= 0 and integer, but your first VAR statement imposes those restrictions on only the last one.  The log provides a hint:

 

NOTE: The problem has 12 variables (9 free, 0 fixed).
NOTE: The problem has 2 binary and 1 integer variables.

 

You can see the detail by using the EXPAND statement:

 

SAS Output

Var xc1w1                                                                                      
Var xc1w2                                                                                      
Var xt1w1                                                                                      
Var xt1w2                                                                                      
Var xfw1                                                                                       
Var xfw2                                                                                       
Var xc2w1                                                                                      
Var xc2w2                                                                                      
Var xt2w1                                                                                      
Var xt2w2 INTEGER >= 0                                                                         
Var yc2 BINARY                                                                                 
Var yt2 BINARY                                                                                 
Maximize z=82*xc1w1 + 83*xc1w2 + 45*xt1w1 + 49*xt1w2 + 30*xfw1 + 35*xfw2 + 84*xc2w1 + 81*xc2w2 
+ 42*xt2w1 + 54*xt2w2 + 41000*yc2 + 21700*yt2                                                  
Constraint Demand1: xc1w1 + xc2w1 + xt1w1 + xt2w1 + xfw1 = 6000                                
Constraint Demand2: xc1w2 + xc2w2 + xt1w2 + xt2w2 + xfw2 = 5200                                
Constraint china1: xc1w1 + xc1w2 <= 1500                                                       
Constraint china2: xc2w1 + xc2w2 <= 2000                                                       
Constraint turkey1: xt1w1 + xt1w2 <= 1800                                                      
Constraint turkey2: xt2w1 + xt2w2 <= 2200                                                      
Constraint france: xfw1 + xfw2 <= 6000                                                         
Constraint linkc2: xc2w1 + xc2w2 - 2000*yc2 <= 0                                               
Constraint linkt2: xt2w1 + xt2w2 - 2200*yc2 <= 0                                               
Constraint Fixc2: xc2w1 + xc2w2 - 1000000*yc2 <= 0                                             
Constraint Fixt2: xt2w1 + xt2w2 - 1000000*yt2 <= 0                                             
Constraint repurpose: yc2 + yt2 <= 1  

 

The simplest fix is the following:

var xc1w1 >=0 integer, xc1w2 >=0 integer, xt1w1 >=0 integer, xt1w2 >=0 integer, xfw1 >=0 integer, xfw2 >=0 integer, xc2w1 >=0 integer, xc2w2 >=0 integer, xt2w1 >=0 integer, xt2w2 >=0 integer;

But your code would be more readable and less error-prone if you instead use index sets.  For example, you could declare x as follows:

set ISET = /c1 t1 f c2 t2/;
set JSET = /w1 w2/;
var x {ISET, JSET} >=0 integer;

Also, arbitrarily large values like 1000000 tend to cause unnecessary numerical difficulty.  In this case, you can just omit the Fix* constraints because they are implied by the link* constraints.

View solution in original post

4 REPLIES 4
ballardw
Super User

It mean that given the constraints you have provided there is no solution.

 

I would be strongly tempted to graph the constraint areas China2 LInkC2 and Fixc2 to see if one or more of those is actually redundant,

Similar with Turkey2 LinkT2 and Fixt2

 

and consider if you actually want your demand functions to be equalities.

 

RobPratt
SAS Super FREQ

I suspect that you meant for all the x* variables to be >= 0 and integer, but your first VAR statement imposes those restrictions on only the last one.  The log provides a hint:

 

NOTE: The problem has 12 variables (9 free, 0 fixed).
NOTE: The problem has 2 binary and 1 integer variables.

 

You can see the detail by using the EXPAND statement:

 

SAS Output

Var xc1w1                                                                                      
Var xc1w2                                                                                      
Var xt1w1                                                                                      
Var xt1w2                                                                                      
Var xfw1                                                                                       
Var xfw2                                                                                       
Var xc2w1                                                                                      
Var xc2w2                                                                                      
Var xt2w1                                                                                      
Var xt2w2 INTEGER >= 0                                                                         
Var yc2 BINARY                                                                                 
Var yt2 BINARY                                                                                 
Maximize z=82*xc1w1 + 83*xc1w2 + 45*xt1w1 + 49*xt1w2 + 30*xfw1 + 35*xfw2 + 84*xc2w1 + 81*xc2w2 
+ 42*xt2w1 + 54*xt2w2 + 41000*yc2 + 21700*yt2                                                  
Constraint Demand1: xc1w1 + xc2w1 + xt1w1 + xt2w1 + xfw1 = 6000                                
Constraint Demand2: xc1w2 + xc2w2 + xt1w2 + xt2w2 + xfw2 = 5200                                
Constraint china1: xc1w1 + xc1w2 <= 1500                                                       
Constraint china2: xc2w1 + xc2w2 <= 2000                                                       
Constraint turkey1: xt1w1 + xt1w2 <= 1800                                                      
Constraint turkey2: xt2w1 + xt2w2 <= 2200                                                      
Constraint france: xfw1 + xfw2 <= 6000                                                         
Constraint linkc2: xc2w1 + xc2w2 - 2000*yc2 <= 0                                               
Constraint linkt2: xt2w1 + xt2w2 - 2200*yc2 <= 0                                               
Constraint Fixc2: xc2w1 + xc2w2 - 1000000*yc2 <= 0                                             
Constraint Fixt2: xt2w1 + xt2w2 - 1000000*yt2 <= 0                                             
Constraint repurpose: yc2 + yt2 <= 1  

 

The simplest fix is the following:

var xc1w1 >=0 integer, xc1w2 >=0 integer, xt1w1 >=0 integer, xt1w2 >=0 integer, xfw1 >=0 integer, xfw2 >=0 integer, xc2w1 >=0 integer, xc2w2 >=0 integer, xt2w1 >=0 integer, xt2w2 >=0 integer;

But your code would be more readable and less error-prone if you instead use index sets.  For example, you could declare x as follows:

set ISET = /c1 t1 f c2 t2/;
set JSET = /w1 w2/;
var x {ISET, JSET} >=0 integer;

Also, arbitrarily large values like 1000000 tend to cause unnecessary numerical difficulty.  In this case, you can just omit the Fix* constraints because they are implied by the link* constraints.

usamasm10
Calcite | Level 5

I did not notice in that in the notes. That helped.

Thanks a lot 🙂

usamasm10
Calcite | Level 5
It is a requirement to keep demand functions to be equalities.
As for the Fixc2 and Fixt2, only one of them can have a value

sas-innovate-2024.png

Don't miss out on SAS Innovate - Register now for the FREE Livestream!

Can't make it to Vegas? No problem! Watch our general sessions LIVE or on-demand starting April 17th. Hear from SAS execs, best-selling author Adam Grant, Hot Ones host Sean Evans, top tech journalist Kara Swisher, AI expert Cassie Kozyrkov, and the mind-blowing dance crew iLuminate! Plus, get access to over 20 breakout sessions.

 

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
  • 4 replies
  • 893 views
  • 0 likes
  • 3 in conversation