The OPTLP Procedure

Example 12.1 Oil Refinery Problem

Consider an oil refinery scenario. A step in refining crude oil into finished oil products involves a distillation process that splits crude into various streams. Suppose there are three types of crude available: Arabian light (a_l), Arabian heavy (a_h), and Brega (br). These crudes are distilled into light naphtha (na_l), intermediate naphtha (na_i), and heating oil (h_o). These in turn are blended into two types of jet fuel. Jet fuel j_1 is made up of 30% intermediate naphtha and 70% heating oil, and jet fuel j_2 is made up of 20% light naphtha and 80% heating oil. What amounts of the three crudes maximize the profit from producing jet fuel (j_1, j_2)? This problem can be formulated as the following linear program:

\[ \begin{array}{rrcrcrccccc} \max & \multicolumn{9}{l}{ -\, 175\, \mr{a\_ l} - 165\, \mr{a\_ h} - 205\, \mr{br} + 350\, \mr{j\_ 1} + 350\, \mr{j\_ 2}} \\ \mr{subject\ to} & & & & & & & & & \\ (\mr{napha\_ l}) & 0.035\, \mr{a\_ l} & +& 0.03\, \mr{a\_ h} & +& 0.045 \, \mr{br} & =& \mr{na\_ l} & & \\ (\mr{napha\_ i}) & 0.1\, \mr{a\_ l} & +& 0.075\, \mr{a\_ h} & +& 0.135\, \mr{br} & =& \mr{na\_ i} & & \\ (\mr{htg\_ oil}) & 0.39\, \mr{a\_ l} & +& 0.3\, \mr{a\_ h} & +& 0.43\, \mr{br} & =& \mr{h\_ o} & & \\ (\mr{blend1}) & & & 0.3\, \mr{j\_ 1} & & & \leq & \mr{na\_ i} & & \\ (\mr{blend2}) & & & & & 0.2\, \mr{j\_ 2} & \leq & \mr{na\_ l} & & \\ (\mr{blend3}) & & & 0.7\, \mr{j\_ 1} & +& 0.8\, \mr{j\_ 2} & \leq & \mr{h\_ o} & & \\ & \mr{a\_ l} & & & & & \leq & 110 & & \\ & & & \mr{a\_ h} & & & \leq & 165 & & \\ & & & & & \mr{br} & \leq & 80 & & \\ & \multicolumn{5}{r}\mr {a\_ l, a\_ h, br, na\_ 1, na\_ i, h\_ o, j\_ 1, j\_ 2} & \geq & 0 & & \end{array} \]

The constraints "blend1" and "blend2" ensure that j_1 and j_2 are made with the specified amounts of na_i and na_l, respectively. The constraint "blend3" is actually the reduced form of the following constraints:

\[ \begin{array}{ccccc} \mr{h\_ o1} & & & \geq & 0.7\, \mr{j\_ 1} \\ & & \mr{h\_ o2} & \geq & 0.8\, \mr{j\_ 2} \\ \mr{h\_ o1} & + & \mr{h\_ o2} & \leq & \mr{h\_ o} \end{array} \]

where h_o1 and h_o2 are dummy variables.

You can use the following SAS code to create the input data set ex1: 

data ex1;
   input field1 $ field2 $ field3 $ field4 field5 $ field6;
   datalines;
NAME        .          EX1      .     .         .
ROWS        .          .        .     .         .
 N          profit     .        .     .         .
 E          napha_l    .        .     .         .
 E          napha_i    .        .     .         .
 E          htg_oil    .        .     .         .
 L          blend1     .        .     .         .
 L          blend2     .        .     .         .
 L          blend3     .        .     .         .
COLUMNS     .          .        .     .         .
.           a_l        profit   -175  napha_l   .035
.           a_l        napha_i  .100  htg_oil   .390
.           a_h        profit   -165  napha_l   .030
.           a_h        napha_i  .075  htg_oil   .300
.           br         profit   -205  napha_l   .045
.           br         napha_i  .135  htg_oil   .430
.           na_l       napha_l  -1    blend2    -1
.           na_i       napha_i  -1    blend1    -1
.           h_o        htg_oil  -1    blend3    -1
.           j_1        profit   350   blend1    .3
.           j_1        blend3   .7    .         .
.           j_2        profit   350   blend2    .2
.           j_2        blend3   .8    .         .
BOUNDS      .          .        .     .         .
UP          .          a_l      110   .         .
UP          .          a_h      165   .         .
UP          .          br       80    .         .
ENDATA      .          .        .     .         .
;

You can use the following call to PROC OPTLP to solve the LP problem: 

proc optlp data=ex1
   objsense  = max
   algorithm = primal
   primalout = ex1pout
   dualout   = ex1dout
   logfreq   = 1;
run;
%put &_OROPTLP_;

Note that the OBJSENSE=MAX option is used to indicate that the objective function is to be maximized.

The primal and dual solutions are displayed in Output 12.1.1.

Output 12.1.1: Example 1: Primal and Dual Solution Output

Primal Solution

Obs Objective Function ID RHS ID Variable
Name 		Variable
Type 		Objective
Coefficient 		Lower
Bound 		Upper Bound Variable Value Variable
Status 		Reduced Cost
1 profit   a_l D -175 0 110 110.000 U 10.2083
2 profit   a_h D -165 0 165 0.000 L -22.8125
3 profit   br D -205 0 80 80.000 U 2.8125
4 profit   na_l N 0 0 1.7977E308 7.450 B 0.0000
5 profit   na_i N 0 0 1.7977E308 21.800 B 0.0000
6 profit   h_o N 0 0 1.7977E308 77.300 B 0.0000
7 profit   j_1 N 350 0 1.7977E308 72.667 B 0.0000
8 profit   j_2 N 350 0 1.7977E308 33.042 B 0.0000

Dual Solution

Obs Objective Function ID RHS ID Constraint Name Constraint
Type 		Constraint
RHS 		Constraint
Lower
Bound 		Constraint
Upper
Bound 		Dual Variable
Value 		Constraint
Status 		Constraint Activity
1 profit   napha_l E 0 . . 0.000 L 0.00000
2 profit   napha_i E 0 . . -145.833 U 0.00000
3 profit   htg_oil E 0 . . -437.500 U 0.00000
4 profit   blend1 L 0 . . 145.833 L -0.00000
5 profit   blend2 L 0 . . 0.000 B -0.84167
6 profit   blend3 L 0 . . 437.500 L 0.00000 




            

            


            
The progress of the solution is printed to the log as follows. 

            
  

            


Output 12.1.2:  Log: Solution Progress

               

                  

                     
                     

                        
                        
                        

                           
                           

                              
                              





NOTE: The OPTLP procedure is executing in single-machine mode.                  
                                       
                                    



NOTE: The problem EX1 has 8 variables (0 free, 0 fixed).                        
                                       
                                    



NOTE: The problem has 6 constraints (3 LE, 3 EQ, 0 GE, 0 range).                
                                       
                                    



NOTE: The problem has 19 constraint coefficients.                               
                                       
                                    



WARNING: The objective sense has been changed to maximization.                  
                                       
                                    



NOTE: The LP presolver value AUTOMATIC is applied.                              
                                       
                                    



NOTE: The LP presolver removed 3 variables and 3 constraints.                   
                                       
                                    



NOTE: The LP presolver removed 6 constraint coefficients.                       
                                       
                                    



NOTE: The presolved problem has 5 variables, 3 constraints, and 13 constraint   
                                       
                                    



      coefficients.                                                             
                                       
                                    



NOTE: The LP solver is called.                                                  
                                       
                                    



NOTE: The Primal Simplex algorithm is used.                                     
                                       
                                    



                           Objective                Entering      Leaving       
                                       
                                    



      Phase Iteration        Value         Time     Variable      Variable      
                                       
                                    



       P 1          1    0.000000E+00         0                                 
                                       
                                    



       P 2          2    0.000000E+00         0        j_1         blend1 (S)   
                                       
                                    



       P 2          3    1.405640E-01         0        j_2         blend3 (S)   
                                       
                                    



       P 2          4    1.454487E-01         0        a_l         blend2 (S)   
                                       
                                    



       P 2          5    2.379819E-01         0         br            a_l       
                                       
                                    



       P 2          6    1.202394E+03         0     blend2 (S)         br       
                                       
                                    



       P 2          7    1.348074E+03         0                                 
                                       
                                    



       D 2          8    1.347917E+03         0                                 
                                       
                                    



NOTE: Optimal.                                                                  
                                       
                                    



NOTE: Objective = 1347.9166667.                                                 
                                       
                                    



NOTE: The Primal Simplex solve time is 0.00 seconds.                            
                                       
                                    



NOTE: The data set WORK.EX1POUT has 8 observations and 10 variables.            
                                       
                                    



NOTE: The data set WORK.EX1DOUT has 6 observations and 10 variables.            
                                       
                                    






                           
                        

                        



                     
                     
                  





            

            


            
Note that the %put statement immediately after the OPTLP procedure prints value of the macro variable _OROPTLP_ to the log
               as follows. 
            

            
  

            


Output 12.1.3:  Log: Value of the Macro Variable _OROPTLP_

               

                  

                     
                     

                        
                        
                        

                           
                           

                              
                              





STATUS=OK ALGORITHM=PS SOLUTION_STATUS=OPTIMAL OBJECTIVE=1347.9166667           
                                       
                                    



PRIMAL_INFEASIBILITY=2.888315E-15 DUAL_INFEASIBILITY=0 BOUND_INFEASIBILITY=0    
                                       
                                    



ITERATIONS=8 PRESOLVE_TIME=0.00 SOLUTION_TIME=0.00                              
                                       
                                    






                           
                        

                        



                     
                     
                  





            

            


            
The value briefly summarizes the status of the OPTLP procedure upon termination.